# PROBLEMS IN PLANE AND SOLID GEOMETRY v.1 Plane Geometry Viktor Prasolov

#### Document technical information

Format pdf
Size 4.1 MB
First found Jun 9, 2017

#### Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

#### Transcript

```PROBLEMS IN PLANE AND SOLID
GEOMETRY
v.1 Plane Geometry
Viktor Prasolov
translated and edited by Dimitry Leites
Abstract. This book has no equal. The priceless treasures of elementary geometry are
nowhere else exposed in so complete and at the same time transparent form. The short
solutions take barely 1.5 − 2 times more space than the formulations, while still remaining
complete, with no gaps whatsoever, although many of the problems are quite difficult. Only
this enabled the author to squeeze about 2000 problems on plane geometry in the book of
volume of ca 600 pages thus embracing practically all the known problems and theorems of
elementary geometry.
The book contains non-standard geometric problems of a level higher than that of the
problems usually offered at high school. The collection consists of two parts. It is based on
three Russian editions of Prasolov’s books on plane geometry.
The text is considerably modified for the English edition. Many new problems are added
and detailed structuring in accordance with the methods of solution is adopted.
The book is addressed to high school students, teachers of mathematics, mathematical
clubs, and college students.
Contents
Editor’s preface
From the Author’s preface
11
12
Chapter 1. SIMILAR TRIANGLES
Background
Introductory problems
§1. Line segments intercepted by parallel lines
§2. The ratio of sides of similar triangles
§3. The ratio of the areas of similar triangles
§4. Auxiliary equal triangles
***
§5. The triangle determined by the bases of the heights
§6. Similar figures
Problems for independent study
Solutions
15
15
15
15
17
18
18
19
19
20
20
21
CHAPTER 2. INSCRIBED ANGLES
Background
Introductory problems
§1. Angles that subtend equal arcs
§2. The value of an angle between two chords
§3. The angle between a tangent and a chord
§4. Relations between the values of an angle and the lengths of the arc and chord
associated with the angle
§5. Four points on one circle
§6. The inscribed angle and similar triangles
§7. The bisector divides an arc in halves
§8. An inscribed quadrilateral with perpendicular diagonals
§9. Three circumscribed circles intersect at one point
§10. Michel’s point
§11. Miscellaneous problems
Problems for independent study
Solutions
33
33
33
34
35
35
CHAPTER 3. CIRCLES
Background
Introductory problems
§1. The tangents to circles
§2. The product of the lengths of a chord’s segments
§3. Tangent circles
§4. Three circles of the same radius
§5. Two tangents drawn from one point
57
57
58
58
59
59
60
61
3
36
36
37
38
39
39
40
40
41
41
4
CONTENTS
∗∗∗
§6. Application of the theorem on triangle’s heights
§7. Areas of curvilinear figures
§8. Circles inscribed in a disc segment
§9. Miscellaneous problems
§10. The radical axis
Problems for independent study
Solutions
CHAPTER 4. AREA
Background
Introductory problems
§1. A median divides the triangle
into triangles of equal areas
§2. Calculation of areas
§3. The areas of the triangles into which
a quadrilateral is divided
§4. The areas of the parts into which
a quadrilateral is divided
§5. Miscellaneous problems
***
§6. Lines and curves that divide figures
into parts of equal area
§7. Formulas for the area of a quadrilateral
§8. An auxiliary area
§9. Regrouping areas
Problems for independent study
Solutions
CHAPTER 5. TRIANGLES
Background
Introductory problems
1. The inscribed and the circumscribed circles
***
***
§2. Right triangles
§3. The equilateral triangles
***
§4. Triangles with angles of 60◦ and 120◦
§5. Integer triangles
§6. Miscellaneous problems
§7. Menelaus’s theorem
***
§8. Ceva’s theorem
§9. Simson’s line
§10. The pedal triangle
§11. Euler’s line and the circle of nine points
§12. Brokar’s points
§13. Lemoine’s point
61
61
62
62
63
63
65
65
79
79
79
79
80
81
81
82
82
83
83
84
85
86
86
99
99
99
100
100
100
101
101
101
102
102
103
104
105
106
107
108
109
110
111
CONTENTS
***
Problems for independent study
Solutions
5
111
112
112
Chapter 6. POLYGONS
Background
Introductory problems
§1. The inscribed and circumscribed quadrilaterals
***
***
§2. Quadrilaterals
§3. Ptolemy’s theorem
§4. Pentagons
§5. Hexagons
§6. Regular polygons
***
***
§7. The inscribed and circumscribed polygons
***
§8. Arbitrary convex polygons
§9. Pascal’s theorem
Problems for independent study
Solutions
137
137
137
137
138
138
139
140
141
141
142
142
143
144
144
144
145
145
146
Chapter 7. LOCI
Background
Introductory problems
§1. The locus is a line or a segment of a line
***
§2. The locus is a circle or an arc of a circle
***
§3. The inscribed angle
§4. Auxiliary equal triangles
§5. The homothety
§6. A method of loci
§7. The locus with a nonzero area
§8. Carnot’s theorem
§9. Fermat-Apollonius’s circle
Problems for independent study
Solutions
169
169
169
169
170
170
170
171
171
171
171
172
172
173
173
174
Chapter 8. CONSTRUCTIONS
§1. The method of loci
§2. The inscribed angle
§3. Similar triangles and a homothety
§4. Construction of triangles from various elements
§5. Construction of triangles given various points
§6. Triangles
§7. Quadrilaterals
§8. Circles
183
183
183
183
183
184
184
185
185
6
CONTENTS
§9. Apollonius’ circle
§10. Miscellaneous problems
§11. Unusual constructions
§12. Construction with a ruler only
§13. Constructions with the help of a two-sided ruler
§14. Constructions using a right angle
Problems for independent study
Solutions
186
186
186
186
187
188
188
189
Chapter 9. GEOMETRIC INEQUALITIES
Background
Introductory problems
§1. A median of a triangle
§2. Algebraic problems on the triangle inequality
§3. The sum of the lengths of quadrilateral’s diagonals
§4. Miscellaneous problems on the triangle inequality
***
§5. The area of a triangle does not exceed a half product of two sides
§6. Inequalities of areas
§7. Area. One figure lies inside another
***
§8. Broken lines inside a square
§9. The quadrilateral
§10. Polygons
***
§11. Miscellaneous problems
***
Problems for independent study
Supplement. Certain inequalities
Solutions
205
205
205
205
206
206
207
207
207
208
209
209
209
210
210
211
211
211
212
212
213
Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
§1. Medians
§2. Heights
§3. The bisectors
§4. The lengths of sides
§5. The radii of the circumscribed, inscribed and escribed circles
§6. Symmetric inequalities between the angles of a triangle
§7. Inequalities between the angles of a triangle
§8. Inequalities for the area of a triangle
***
§9. The greater angle subtends the longer side
§10. Any segment inside a triangle is shorter than the longest side
§11. Inequalities for right triangles
§12. Inequalities for acute triangles
§13. Inequalities in triangles
Problems for independent study
Solutions
235
235
235
235
236
236
236
237
237
238
238
238
238
239
239
240
240
Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM
255
CONTENTS
Background
Introductory problems
§1. The triangle
***
§2. Extremal points of a triangle
§3. The angle
§4. The quadrilateral
§5. Polygons
§6. Miscellaneous problems
§7. The extremal properties of regular polygons
Problems for independent study
Solutions
7
255
255
255
256
256
257
257
257
258
258
258
259
Chapter 12. CALCULATIONS AND METRIC RELATIONS
Introductory problems
§1. The law of sines
§2. The law of cosines
§3. The inscribed, the circumscribed and escribed circles; their radii
§4. The lengths of the sides, heights, bisectors
§5. The sines and cosines of a triangle’s angles
§6. The tangents and cotangents of a triangle’s angles
§7. Calculation of angles
***
§8. The circles
***
§9. Miscellaneous problems
§10. The method of coordinates
Problems for independent study
Solutions
271
271
271
272
272
273
273
274
274
274
275
275
275
276
277
277
Chapter 13. VECTORS
Background
Introductory problems
§1. Vectors formed by polygons’ (?) sides
§2. Inner product. Relations
§3. Inequalities
§4. Sums of vectors
§5. Auxiliary projections
§6. The method of averaging
§7. Pseudoinner product
Problems for independent study
Solutions
289
289
289
290
290
291
292
292
293
293
294
295
Chapter 14. THE CENTER OF MASS
Background
§1. Main properties of the center of mass
§2. A theorem on mass regroupping
§3. The moment of inertia
§4. Miscellaneous problems
§5. The barycentric coordinates
307
307
307
308
309
310
310
8
CONTENTS
Solutions
311
Chapter 15. PARALLEL TRANSLATIONS
Background
Introductory problems
§1. Solving problems with the aid of parallel translations
§2. Problems on construction and loci
***
Problems for independent study
Solutions
319
319
319
319
320
320
320
320
Chapter 16. CENTRAL SYMMETRY
Background
Introductory problems
§1. Solving problems with the help of a symmetry
§2. Properties of the symmetry
§3. Solving problems with the help of a symmetry. Constructions
Problems for independent study
Solutions
327
327
327
327
328
328
329
329
Chapter 17. THE SYMMETRY THROUGH A LINE
Background
Introductory problems
§1. Solving problems with the help of a symmetry
§2. Constructions
***
§3. Inequalities and extremals
§4. Compositions of symmetries
§5. Properties of symmetries and axes of symmetries
§6. Chasles’s theorem
Problems for independent study
Solutions
335
335
335
335
336
336
336
336
337
337
338
338
Chapter 18. ROTATIONS
Background
Introductory problems
§1. Rotation by 90◦
§2. Rotation by 60◦
§3. Rotations through arbitrary angles
§4. Compositions of rotations
***
***
Problems for independent study
Solutions
345
345
345
345
346
347
347
348
348
348
349
Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
Background
Introductory problems
§1. Homothetic polygons
§2. Homothetic circles
§3. Costructions and loci
359
359
359
359
360
360
CONTENTS
***
§4. Composition of homotheties
§5. Rotational homothety
***
***
§6. The center of a rotational homothety
§7. The similarity circle of three figures
Problems for independent study
Solutions
9
361
361
361
362
362
362
363
364
364
Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
Background
§1. The least and the greatest angles
§2. The least and the greatest distances
§3. The least and the greatest areas
§4. The greatest triangle
§5. The convex hull and the base lines
§6. Miscellaneous problems
Solutions
375
375
375
376
376
376
376
378
378
Chapter 21. DIRICHLET’S PRINCIPLE
Background
§1. The case when there are finitely many points, lines, etc.
§2. Angles and lengths
§3. Area
Solutions
385
385
385
386
387
387
Chapter 22. CONVEX AND NONCONVEX POLYGONS
Background
§1. Convex polygons
***
§2. Helly’s theorem
§3. Non-convex polygons
Solutions
397
397
397
397
398
398
399
Chapter 23. DIVISIBILITY, INVARIANTS, COLORINGS
Background
§1. Even and odd
§2. Divisibility
§3. Invariants
§4. Auxiliary colorings
§5. More auxiliary colorings
***
§6. Problems on colorings
***
Solutions
409
409
409
410
410
411
412
412
412
413
413
Chapter 24. INTEGER LATTICES
§1. Polygons with vertices in the nodes of a lattice
§2. Miscellaneous problems
Solutions
425
425
425
426
10
CONTENTS
Chapter 25. CUTTINGS
§1. Cuttings into parallelograms
§2. How lines cut the plane
Solutions
431
431
431
432
Chapter 26. SYSTEMS OF POINTS AND SEGMENTS.
EXAMPLES AND COUNTEREXAMPLES
§1. Systems of points
§2. Systems of segments, lines and circles
§3. Examples and counterexamples
Solutions
437
437
437
438
438
Chapter 27. INDUCTION AND COMBINATORICS
§1. Induction
§2. Combinatorics
Solutions
445
445
445
445
Chapter 28. INVERSION
Background
§1. Properties of inversions
§2. Construction of circles
§3. Constructions with the help of a compass only
§4. Let us perform an inversion
§5. Points that lie on one circle and circles passing through one point
§6. Chains of circles
Solutions
449
449
449
450
450
451
452
454
455
Chapter 29. AFFINE TRANSFORMATIONS
§1. Affine transformations
§2. How to solve problems with the help of affine transformations
Solutions
465
465
466
466
Chapter 30. PROJECTIVE TRANSFORMATIONS
473
§1. Projective transformations of the line
473
§2. Projective transformations of the plane
474
§3. Let us transform the given line into the infinite one
477
§4. Application of projective maps that preserve a circle
478
§5. Application of projective transformations of the line
479
§6. Application of projective transformations of the line in problems on construction 479
§7. Impossibility of construction with the help of a ruler only
480
Solutions
480
Index
493
EDITOR’S PREFACE
11
Editor’s preface
The enormous number of problems and theorems of elementary geometry was considered
too wide to grasp in full even in the last century. Even nowadays the stream of new problems
is still wide. (The majority of these problems, however, are either well-forgotten old ones or
those recently pirated from a neighbouring country.)
Any attempt to collect an encyclopedia of all the problems seems to be doomed to failure
for many reasons.
First of all, this is an impossible task because of the huge number of the problems, an
enormity too vast to grasp. Second, even if this might have been possible, the book would
be terribly overloaded, and therefore of no interest to anybody.
However, in the book Problems in plane geometry followed by Problems in solid geometry
this task is successfully perfomed.
In the process of writing the book the author used the books and magazines published
in the last century as well as modern ones. The reader can judge the completeness of the
book by, for instance, the fact that American Mathematical Monthly yearly1 publishes, as
“new”, 1–2 problems already published in the Russian editions of this book.
The book turned out to be of interest to a vast audience: about 400 000 copies of the
first edition of each of the Parts (Parts 1 and 2 — Plane and Part 3 — Solid) were sold;
the second edition, published 5 years later, had an even larger circulation, the total over
1 000 000 copies. The 3rd edition of Problems in Plane Geometry was issued in 1996 and
the latest one in 2001.
The readers’ interest is partly occasioned by a well-thought classification system.
The collection consists of three parts.
Part 1 covers classical subjects of plane geometry. It contains nearly 1000 problems with
complete solutions and over 100 problems to be solved on one’s own. Still more will be added
for the English version of the book.
Part 2 includes more recent topics, geometric transformations and problems more suitable
for contests and for use in mathematical clubs. The problems cover cuttings, colorings, the
pigeonhole (or Dirichlet’s) principle, induction, and so on.
Part 3 is devoted to solid geometry.
A rather detailed table of contents serves as a guide in the sea of geometric problems. It
helps the experts to easily find what they need while the uninitiated can quickly learn what
exactly is that they are interested in in geometry. Splitting the book into small sections (5
to 10 problems in each) made the book of interest to the readers of various levels.
FOR THE ENGLISH VERSION of the book about 150 new problems are already added
and several hundred more of elementary and intermideate level problems will be added to
make the number of more elementary problems sufficient to use the book in the ordinary
school: the Russian editions are best suited for coaching for a mathematical Olympiad than
for a regular class work: the level of difficulty increases rather fast.
Problems in each section are ordered difficulty-wise. The first problems of the sections
are simple; they are a match for many. Here are some examples:
1Here
are a few samples: v. 96, n. 5, 1989, p. 429–431 (here the main idea of the solution is the
right illustration — precisely the picture from the back cover of the 1st Russian edition of Problems in Solid
Geometry, Fig. to Problem 13.22); v. 96, n. 6, p. 527, Probl. E3192 corresponds to Problems 5.31 and
18.20 of Problems in Plane Geometry — with their two absolutely different solutions, the one to Problem
5.31, unknown to AMM, is even more interesting.
12
CONTENTS
Plane 1.1. The bases of a trapezoid are a and b. Find the length of the segment that
the diagonals of the trapezoid intersept on the trapezoid’s midline.
Plane 1.52. Let AA1 and BB1 be the altitudes of △ABC. Prove that △A1 B1 C is
similar to △ABC. What is the similarity coefficient?
Plane 2.1. A line segment connects vertex A of an acute △ABC with the center O of
the circumscribed circle. The altitude AH is dropped from A. Prove that ∠BAH = ∠OAC.
Plane 6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with
the intersection point of the quadrilateral’s diagonals, then the quadrilateral is a rhombus.
Solid 1. Arrange 6 match sticks to get 4 equilateral triangles with side length equal to
the length of a stick.
Solid 1.1. Consider the cube ABCDA1 B1 C1 D1 with side length a. Find the angle and
the distance between the lines A1 B and AC1 .
Solid 6.1. Is it true that in every tetrahedron the heights meet at one point?
The above problems are not difficult. The last problems in the sections are a challenge
for the specialists in geometry. It is important that the passage from simple problems to
complicated ones is not too long; there are no boring and dull long sequences of simple
similar problems. (In the Russian edition these sequences are, perhaps, too short, so more
problems are added.)
The final problems of the sections are usually borrowed from scientific journals. Here are
some examples:
√
Plane 10.20. Prove that la + lb + mc ≤ 3p, where la , lb are the lengths of the bisectors
of the angles ∠A and ∠B of the triangle △ABC, mc is the length of the median of the side
AB, and p is the semiperimeter.
Plane 19.55. Let O be the center of the circle inscribed in △ABC, K the Lemoine’s
point, P and Q Brocard’s points. Prove that P and Q belong to the circle with diameter
KO and that OP = OQ.
Plane 22.29. The numbers α1 , . . . , αn , whose sum is equal to (n−2)π, satisfy inequalities
0 < αi < 2π. Prove that there exists an n-gon A1 . . . An with the angles α1 , . . . , αn at the
vertices A1 , . . . , An , respectively.
Plane 24.12. Prove that for any n there exists a circle on which there lie precisely n
points with integer coordinates.
Solid 4.48. Consider several arcs of great circles on a sphere with the sum of their angle
measures < π. Prove that there exists a plane that passes through the center of the sphere
but does not intersect any of these arcs.
Solid 14.22. Prove that if the centers of the escribed spheres of a tetrahedron belong
to the circumscribed sphere, then the tetrahedron’s faces are equal.
Solid 15.34. In space, consider 4 points not in one plane. How many various parallelipipeds with vertices in these points are there?
From the Author’s preface
The book underwent extensive revision. The solutions to many of the problems were
rewritten and about 600 new problems were added, particularly those concerning the geometry of the triangle. I was greatly influenced in the process by the second edition of the
book by I. F. Sharygin Problems on Geometry. Plane geometry, Nauka, Moscow,1986 and a
wonderful and undeservedly forgotten book by D. Efremov New Geometry of the Triangle,
Matezis, Odessa, 1902.
The present book can be used not only as a source of optional problems for students
but also as a self-guide for those who wish (or have no other choice but) to study geometry
FROM THE AUTHOR’S PREFACE
13
independently. Detailed headings are provided for the reader’s convenience. Problems in the
two parts of Plane are spread over 29 Chapters, each Chapter comprising 6 to 14 sections.
The classification is based on the methods used to solve geometric problems. The purpose of
the division is basically to help the reader find his/her bearings in this large array of problems.
Otherwise the huge number of problems might be somewhat depressingly overwhelming.
Advice and comments given by Academician A. V. Pogorelov, and Professors A. M. Abramov,
A. Yu. Vaintrob, N. B. Vasiliev, N. P. Dolbilin, and S. Yu. Orevkov were a great help to me
in preparing the first Soviet edition. I wish to express my sincere gratitude to all of them.
To save space, sections with background only contain the material directly pertinent to
the respective chapter. It is collected just to remind the reader of notations. Therefore, the
basic elements of a triangle are only defined in chapter 5, while in chapter 1 we assume that
their definition is known. For the reader’s convenience, cross references in this translation
are facilitated by a very detailed index.
Chapter 1. SIMILAR TRIANGLES
Background
1) Triangle ABC is said to be similar to triangle A1 B1 C1 (we write △ABC ∼ △A1 B1 C1 )
if and only if one of the following equivalent conditions is satisfied:
a) AB : BC : CA = A1 B1 : B1 C1 : C1 A1 ;
b) AB : BC = A1 B1 : B1 C1 and ∠ABC = ∠A1 B1 C1 ;
c) ∠ABC = ∠A1 B1 C1 and ∠BAC = ∠B1 A1 C1 .
2) Triangles AB1 C1 and AB2 C2 cut off from an angle with vertex A by parallel lines are
similar and AB1 : AB2 = AC1 : AC2 (here points B1 and B2 lie on one leg of the angle and
C1 and C2 on the other leg).
3) A midline of a triangle is the line connecting the midpoints of two of the triangle’s
sides. The midline is parallel to the third side and its length is equal to a half length of the
third side.
The midline of a trapezoid is the line connecting the midpoints of the trapezoid’s sides.
This line is parallel to the bases of the trapezoid and its length is equal to the halfsum of
their lengths.
4) The ratio of the areas of similar triangles is equal to the square of the similarity
coefficient, i.e., to the squared ratio of the lengths of respective sides. This follows, for
example, from the formula SABC = 21 AB · AC sin ∠A.
5) Polygons A1 A2 . . . An and B1 B2 . . . Bn are called similar if A1 A2 : A2 A3 : · · · : An A1 =
B1 B2 : B2 B3 : · · · : Bn B1 and the angles at the vertices A1 , . . . , An are equal to the angles
at the vertices B1 , . . . , Bn , respectively.
The ratio of the respective diagonals of similar polygons is equal to the similarity coefficient. For the circumscribed similar polygons, the ratio of the radii of the inscribed circles
is also equal to the similarity coefficient.
Introductory problems
1. Consider heights AA1 and BB1 in acute triangle ABC. Prove that A1 C · BC =
B1 C · AC.
2. Consider height CH in right triangle ABC with right angle ∠C. Prove that AC 2 =
AB · AH and CH 2 = AH · BH.
3. Prove that the medians of a triangle meet at one point and this point divides each
median in the ratio of 2 : 1 counting from the vertex.
4. On side BC of △ABC point A1 is taken so that BA1 : A1 C = 2 : 1. What is the
ratio in which median CC1 divides segment AA1 ?
5. Square P QRS is inscribed into △ABC so that vertices P and Q lie on sides AB and
AC and vertices R and S lie on BC. Express the length of the square’s side through a and
ha .
§1. Line segments intercepted by parallel lines
1.1. Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a > b).
15
16
CHAPTER 1. SIMILAR TRIANGLES
a) Find the length of the segment that the diagonals intercept on the midline.
b) Find the length of segment M N whose endpoints divide AB and CD in the ratio of
AM : M B = DN : N C = p : q.
1.2. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of
a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a rhombus, a
square?
1.3. Points A1 and B1 divide sides BC and AC of △ABC in the ratios BA1 : A1 C = 1 : p
and AB1 : B1 C = 1 : q, respectively. In what ratio is AA1 divided by BB1 ?
1.4. Straight lines AA1 and BB1 pass through point P of median CC1 in △ABC (A1
and B1 lie on sides BC and CA, respectively). Prove that A1 B1 k AB.
1.5. The straight line which connects the intersection point P of the diagonals in quadrilateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Prove
that it also bisects BC.
1.6. A point P is taken on side AD of parallelogram ABCD so that AP : AD = 1 : n;
let Q be the intersection point of AC and BP . Prove that AQ : AC = 1 : (n + 1).
1.7. The vertices of parallelogram A1 B1 C1 D1 lie on the sides of parallelogram ABCD
(point A1 lies on AB, B1 on BC, etc.). Prove that the centers of the two parallelograms
coincide.
1.8. Point K lies on diagonal BD of parallelogram ABCD. Straight line AK intersects
lines BC and CD at points L and M , respectively. Prove that AK 2 = LK · KM .
1.9. One of the diagonals of a quadrilateral inscribed in a circle is a diameter of the
circle. Prove that (the lengths of) the projections of the opposite sides of the quadrilateral
on the other diagonal are equal.
1.10. Point E on base AD of trapezoid ABCD is such that AE = BC. Segments CA
and CE intersect diagonal BD at O and P , respectively. Prove that if BO = P D, then
AD2 = BC 2 + AD · BC.
1.11. On a circle centered at O, points A and B single out an arc of 60◦ . Point M
belongs to this arc. Prove that the straight line passing through the midpoints of M A and
OB is perpendicular to that passing through the midpoints of M B and OA.
1.12. a) Points A, B, and C lie on one straight line; points A1 , B1 , and C1 lie on another
straight line. Prove that if AB1 k BA1 and AC1 k CA1 , then BC1 k CB1 .
b) Points A, B, and C lie on one straight line and A1 , B1 , and C1 are such that AB1 k
BA1 , AC1 k CA1 , and BC1 k CB1 . Prove that A1 , B1 and C1 lie on one line.
1.13. In △ABC bisectors AA1 and BB1 are drawn. Prove that the distance from any
point M of A1 B1 to line AB is equal to the sum of distances from M to AC and BC.
1.14. Let M and N be the midpoints of sides AD and BC in rectangle ABCD. Point
P lies on the extension of DC beyond D; point Q is the intersection point of P M and AC.
Prove that ∠QN M = ∠M N P .
1.15. Points K and L are taken on the extensions of the bases AD and BC of trapezoid
ABCD beyond A and C, respectively. Line segment KL intersects sides AB and CD at M
and N , respectively; KL intersects diagonals AC and BD at O and P , respectively. Prove
that if KM = N L, then KO = P L.
1.16. Points P , Q, R, and S on sides AB, BC, CD and DA, respectively, of convex
quadrilateral ABCD are such that BP : AB = CR : CD = α and AS : AD = BQ : BC = β.
Prove that P R and QS are divided by their intersection point in the ratios β : (1 − β) and
α : (1 − α), respectively.
§2. THE RATIO OF SIDES OF SIMILAR TRIANGLES
17
§2. The ratio of sides of similar triangles
1.17. a) In △ABC bisector BD of the external or internal angle ∠B is drawn. Prove
that AD : DC = AB : BC.
b) Prove that the center O of the circle inscribed in △ABC divides the bisector AA1 in
the ratio of AO : OA1 = (b + c) : a, where a, b and c are the lengths of the triangle’s sides.
1.18. The lengths of two sides of a triangle are equal to a while the length of the third
side is equal to b. Calculate the radius of the circumscribed circle.
1.19. A straight line passing through vertex A of square ABCD intersects side CD at
1
1
1
E and line BC at F . Prove that AE
2 + AF 2 = AB 2 .
1.20. Given points B2 and C2 on heights BB1 and CC1 of △ABC such that AB2 C =
AC2 B = 90◦ , prove that AB2 = AC2 .
1.21. A circle is inscribed in trapezoid ABCD (BC k AD). The circle is tangent to sides
AB and CD at K and L, respectively, and to bases AD and BC at M and N , respectively.
a) Let Q be the intersection point of BM and AN . Prove that KQ k AD.
b) Prove that AK · KB = CL · LD.
1.22. Perpendiculars AM and AN are dropped to sides BC and CD of parallelogram
ABCD (or to their extensions). Prove that △M AN ∼ △ABC.
1.23. Straight line l intersects sides AB and AD of parallelogram ABCD at E and F ,
respectively. Let G be the intersection point of l with diagonal AC. Prove that AB
+ AD
=
AE
AF
AC
.
AG
1.24. Let AC be the longer of the diagonals in parallelogram ABCD. Perpendiculars
CE and CF are dropped from C to the extensions of sides AB and AD, respectively. Prove
that AB · AE + AD · AF = AC 2 .
1.25. Angles α and β of △ABC are related as 3α + 2β = 180◦ . Prove that a2 + bc = c2 .
1.26. The endpoints of segments AB and CD are gliding along the sides of a given angle,
so that straight lines AB and CD are moving parallelly (i.e., each line moves parallelly to
AM ·BM
is
itself) and segments AB and CD intersect at a point, M . Prove that the value of CM
·DM
a constant.
1.27. Through an arbitrary point P on side AC of △ABC straight lines are drawn
parallelly to the triangle’s medians AK and CL. The lines intersect BC and AB at E and
F , respectively. Prove that AK and CL divide EF into three equal parts.
1.28. Point P lies on the bisector of an angle with vertex C. A line passing through P
intercepts segments of lengths a and b on the angle’s legs. Prove that the value of a1 + 1b does
not depend on the choice of the line.
1.29. A semicircle is constructed outwards on side BC of an equilateral triangle ABC
as on the diameter. Given points K and L that divide the semicircle into three equal arcs,
prove that lines AK and AL divide BC into three equal parts.
1.30. Point O is the center of the circle inscribed in △ABC. On sides AC and BC
points M and K, respectively, are selected so that BK · AB = BO2 and AM · AB = AO2 .
Prove that M , O and K lie on one straight line.
1.31. Equally oriented similar triangles AM N , N BM and M N C are constructed on
segment M N (Fig. 1).
Prove that △ABC is similar to all these triangles and the center of its curcumscribed
circle is equidistant from M and N .
1.32. Line √
segment BE divides △ABC into two similar triangles, their similarity ratio
being equal to 3.
Find the angles of △ABC.
18
CHAPTER 1. SIMILAR TRIANGLES
Figure 1 (1.31)
§3. The ratio of the areas of similar triangles
1.33. A point E is taken on side AC of △ABC. Through E pass straight lines DE
and EF parallel to sides BC and AB,
√ respectively; D and E are points on AB and BC,
respectively. Prove that SBDEF = 2 SADE · SEF G .
1.34. Points M and N are taken on sides AB and CD, respectively, of trapezoid ABCD
so that segment M N is parallel to the bases and divides the area of the trapezoid in halves.
Find the length of M N if BC = a and AD = b.
1.35. Let Q be a point inside △ABC. Three straight lines are pass through Q parallelly to the sides of the triangle. The lines divide the triangle into six parts, three of
which
are triangles¢of areas S1 , S2 and S3 . Prove that the area of △ABC is equal to
¡√
√
√ 2
S1 + S2 + S3 .
1.36. Prove that the area of a triangle whose sides are equal to the medians of a triangle
of area S is equal to 34 S.
1.37. a) Prove that the area of the quadrilateral formed by the midpoints of the sides of
convex quadrilateral ABCD is half that of ABCD.
b) Prove that if the diagonals of a convex quadrilateral are equal, then its area is the
product of the lengths of the segments which connect the midpoints of its opposite sides.
1.38. Point O lying inside a convex quadrilateral of area S is reflected symmetrically
through the midpoints of its sides. Find the area of the quadrilateral with its vertices in the
images of O under the reflections.
§4. Auxiliary equal triangles
1.39. In right triangle ABC with right angle ∠C, points D and E divide leg BC of into
three equal parts. Prove that if BC = 3AC, then ∠AEC + ∠ADC + ∠ABC = 90◦ .
1.40. Let K be the midpoint of side AB of square ABCD and let point L divide diagonal
AC in the ratio of AL : LC = 3 : 1. Prove that ∠KLD is a right angle.
1.41. In square ABCD straight lines l1 and l2 pass through vertex A. The lines intersect
the square’s sides. Perpendiculars BB1 , BB2 , DD1 , and DD2 are dropped to these lines.
Prove that segments B1 B2 and D1 D2 are equal and perpendicular to each other.
1.42. Consider an isosceles right triangle ABC with CD = CE and points D and E on
sides CA and CB, respectively. Extensions of perpendiculars dropped from D and C to AE
intersect the hypotenuse AB at K and L. Prove that KL = LB.
1.43. Consider an inscribed quadrilateral ABCD. The lengths of sides AB, BC, CD,
and DA are a, b, c, and d, respectively. Rectangles are constructed outwards on the sides of
§5. THE TRIANGLE DETERMINED BY THE BASES OF THE HEIGHTS
19
the quadrilateral; the sizes of the rectangles are a × c, b × d, c × a and d × b, respectively.
Prove that the centers of the rectangles are vertices of a rectangle.
1.44. Hexagon ABCDEF is inscribed in a circle of radius R centered at O; let AB =
CD = EF = R. Prove that the intersection points, other than O, of the pairs of circles
circumscribed about △BOC, △DOE and △F OA are the vertices of an equilateral triangle
with side R.
***
1.45. Equilateral triangles BCK and DCL are constructed outwards on sides BC and
CD of parallelogram ABCD. Prove that AKL is an equilateral triangle.
1.46. Squares are constructed outwards on the sides of a parallelogram. Prove that their
centers form a square.
1.47. Isosceles triangles with angles 2α, 2β and 2γ at vertices A′ , B ′ and C ′ are constructed outwards on the sides of triangle ABC; let α + β + γ = 180◦ . Prove that the angles
of △A′ B ′ C ′ are equal to α, β and γ.
1.48. On the sides of △ABC as on bases, isosceles similar triangles AB1 C and AC1 B
are constructed outwards and an isosceles triangle BA1 C is constructed inwards. Prove that
AB1 A1 C1 is a parallelogram.
1.49. a) On sides AB and AC of △ABC equilateral triangles ABC1 and AB1 C are
constructed outwards; let ∠C1 = ∠B1 = 90◦ , ∠ABC1 = ∠ACB1 = ϕ; let M be the
midpoint of BC. Prove that M B1 = M C1 and ∠B1 M C1 = 2ϕ.
b) Equilateral triangles are constructed outwards on the sides of △ABC. Prove that the
centers of the triangles constructed form an equilateral triangle whose center coincides with
the intersection point of the medians of △ABC.
1.50. Isosceles triangles AC1 B and AB1 C with an angle ϕ at the vertex are constructed
outwards on the unequal sides AB and AC of a scalene triangle △ABC.
a) Let M be a point on median AA1 (or on its extension), let M be equidistant from B1
and C1 . Prove that ∠B1 M C1 = ϕ.
b) Let O be a point of the midperpendicular to segment BC, let O be equidistant from
B1 and C1 . Prove that ∠B1 OC = 180◦ − ϕ.
1.51. Similar rhombuses are constructed outwards on the sides of a convex rectangle
ABCD, so that their acute angles (equal to α) are adjacent to vertices A and C. Prove
that the segments which connect the centers of opposite rhombuses are equal and the angle
between them is equal to α.
§5. The triangle determined by the bases of the heights
1.52. Let AA1 and BB1 be heights of △ABC. Prove that △A1 B1 C ∼ △ABC. What
is the similarity coefficient?
1.53. Height CH is dropped from vertex C of acute triangle ABC and perpendiculars
HM and HN are dropped to sides BC and AC, respectively. Prove that △M N C ∼ △ABC.
1.54. In △ABC heights BB1 and CC1 are drawn.
a) Prove that the tangent at A to the circumscribed circle is parallel to B1 C1 .
b) Prove that B1 C1 ⊥ OA, where O is the center of the circumscribed circle.
1.55. Points A1 , B1 and C1 are taken on the sides of an acute triangle ABC so that
segments AA1 , BB1 and CC1 meet at H. Prove that AH · A1 H = BH · B1 H = CH · C1 H
if and only if H is the intersection point of the heights of △ABC.
1.56. a) Prove that heights AA1 , BB1 and CC1 of acute triangle ABC bisect the angles
of △A1 B1 C1 .
20
CHAPTER 1. SIMILAR TRIANGLES
b) Points C1 , A1 and B1 are taken on sides AB, BC and CA, respectively, of acute triangle
ABC. Prove that if ∠B1 A1 C = ∠BA1 C1 , ∠A1 B1 C = ∠AB1 C1 and ∠A1 C1 B = ∠AC1 B1 ,
then points A1 , B1 and C1 are the bases of the heights of △ABC.
1.57. Heights AA1 , BB1 and CC1 are drawn in acute triangle ABC. Prove that the
point symmetric to A1 through AC lies on B1 C1 .
1.58. In acute triangle ABC, heights AA1 , BB1 and CC1 are drawn. Prove that if
A1 B1 k AB and B1 C1 k BC, then A1 C1 k AC.
1.59. Let p be the semiperimeter of acute triangle ABC and q the semiperimeter of the
triangle formed by the bases of the heights of △ABC. Prove that p : q = R : r, where R
and r are the radii of the circumscribed and the inscribed circles, respectively, of △ABC.
§6. Similar figures
1.60. A circle of radius r is inscribed in a triangle. The straight lines tangent to the
circle and parallel to the sides of the triangle are drawn; the lines cut three small triangles
off the triangle. Let r1 , r2 and r3 be the radii of the circles inscribed in the small triangles.
Prove that r1 + r2 + r3 = r.
1.61. Given △ABC, draw two straight lines x and y such that the sum of lengths of
the segments M XM and M YM drawn parallel to x and y from a point M on AC to their
intersections with sides AB and BC is equal to 1 for any M .
1.62. In an isosceles triangle ABC perpendicular HE is dropped from the midpoint of
base BC to side AC. Let O be the midpoint of HE. Prove that lines AO and BE are
perpendicular to each other.
1.63. Prove that projections of the base of a triangle’s height to the sides between which
it lies and on the other two heights lie on the same straight line.
1.64. Point B lies on segment AC; semicircles S1 , S2 , and S3 are constructed on one side
of AC, as on diameter. Let D be a point on S3 such that BD ⊥ AC. A common tangent
line to S1 and S2 touches these semicircles at F and E, respectively.
a) Prove that EF is parallel to the tangent to S3 passing through D.
b) Prove that BF DE is a rectangle.
1.65. Perpendiculars M Q and M P are dropped from an arbitrary point M of the circle
circumscribed about rectangle ABCD to the rectangle’s two opposite sides; the perpendiculars M R and M T are dropped to the extensions of the other two sides. Prove that lines
P R ⊥ QT and the intersection point of P R and QT belongs to a diagonal of ABCD.
1.66. Two circles enclose non-intersecting areas. Common tangent lines to the two
circles, one external and one internal, are drawn. Consider two straight lines each of which
passes through the tangent points on one of the circles. Prove that the intersection point of
the lines lies on the straight line that connects the centers of the circles.
Problems for independent study
1.67. The (length of the) base of an isosceles triangle is a quarter of its perimeter. From
an arbitrary point on the base straight lines are drawn parallel to the sides of the triangle.
How many times is the perimeter of the triangle greater than that of the parallelogram?
1.68. The diagonals of a trapezoid are mutually perpendicular. The intersection point
divides the diagonals into segments. Prove that the product of the lengths of the trapezoid’s
bases is equal to the sum of the products of the lengths of the segments of one diagonal and
those of another diagonal.
1.69. A straight line is drawn through the center of a unit square. Calculate the sum of
the squared distances between the four vertices of the square and the line.
SOLUTIONS
21
1.70. Points A1 , B1 and C1 are symmetric to the center of the circumscribed circle of
△ABC through the triangle’s sides. Prove that △ABC = △A1 B1 C1 .
1.71. Prove that if ∠BAC = 2∠ABC, then BC 2 = (AC + AB)AC.
1.72. Consider points A, B, C and D on a line l. Through A, B and through C, D
parallel straight lines are drawn. Prove that the diagonals of the parallelograms thus formed
(or their extensions) intersect l at two points that do not depend on parallel lines but depend
on points A, B, C, D only.
1.73. In △ABC bisector AD and midline A1 C1 are drawn. They intersect at K. Prove
that 2A1 K = |b − c|.
1.74. Points M and N are taken on sides AD and CD of parallelogram ABCD such
that M N k AC. Prove that SABM = SCBN .
1.75. On diagonal AC of parallelogram ABCD points P and Q are taken so that
AP = CQ. Let M be such that P M k AD and QM k AB. Prove that M lies on diagonal
BD.
1.76. Consider a trapezoid with bases AD and BC. Extensions of the sides of ABCD
meet at point O. Segment EF is parallel to the bases and passes through the intersection
point of the diagonals. The endpoints of EF lie on AB and CD. Prove that AE : CF =
AO : CO.
1.77. Three straight lines parallel to the sides of the given triangle cut three triangles off
it leaving an equilateral hexagon. Find the length of the side of the hexagon if the lengths
of the triangle’s sides are a, b and c.
1.78. Three straight lines parallel to the sides of a triangle meet at one point, the sides
of the triangle cutting off the line segments of length x each. Find x if the lengths of the
triangle’s sides are a, b and c.
1.79. Point P lies inside △ABC and ∠ABP = ∠ACP . On straight lines AB and AC,
points C1 and B1 are taken so that BC1 : CB1 = CP : BP . Prove that one of the diagonals
of the parallelogram whose two sides lie on lines BP and CP and two other sides (or their
extensions) pass through B1 and C1 is parallel to BC.
Solutions
1.1. a) Let P and Q be the midpoints of AB and CD; let K and L be the intersection
points of P Q with the diagonals AC and BD, respectively. Then P L = a2 and P K = 12 b
and so KL = P L − P K = 21 (a − b).
b) Take point F on AD such that BF k CD. Let E be the intersection point of M N
with BF . Then
M N = M E + EN =
q(a − b) + (p + q)b
qa + pb
q · AF
+b=
=
.
p+q
p+q
p+q
1.2. Consider quadrilateral ABCD. Let K, L, M and N be the midpoints of sides AB,
BC, CD and DA, respectively. Then KL = M N = 12 AC and KL k M N , that is KLM N is
a parallelogram. It becomes clear now that KLM N is a rectangle if the diagonals AC and
BD are perpendicular, a rhombus if AC = BD, and a square if AC and BD are of equal
length and perpendicular to each other.
1.3. Denote the intersection point of AA1 with BB1 by O. In △B1 BC draw segment
A1 A2 so that A1 A2 k BB1 . Then BB11AC2 = 1 + p and so AO : OA1 = AB1 : B1 A2 = B1 C :
qB1 A2 = (1 + p) : q.
22
CHAPTER 1. SIMILAR TRIANGLES
1.4. Let A2 be the midpoint of A1 B. Then CA1 : A1 A2 = CP : P C1 and A1 A2 : A1 B =
1 : 2. So CA1 : A1 B = CP : 2P C1 . Similarly, CB1 : B1 A = CP : 2P C1 = CA1 : A1 B.
1.5. Point P lies on the median QM of △AQD (or on its extension). It is easy to
verify that the solution of Problem 1.4 remains correct also for the case when P lies on the
extension of the median. Consequently, BC k AD.
1.6. We have AQ : QC = AP : BC = 1 : n because △AQP ∼ △CQB. So AC =
AQ + QC = (n + 1)AQ.
1.7. The center of A1 B1 C1 D1 being the midpoint of B1 D1 belongs to the line segment
which connects the midpoints of AB and CD. Similarly, it belongs to the segment which
connects the midpoints of BC and AD. The intersection point of the segments is the center
of ABCD.
1.8. Clearly, AK : KM = BK : KD = LK : AK, that is AK 2 = LK · KM .
1.9. Let AC be the diameter of the circle circumscribed about ABCD. Drop perpendiculars AA1 and CC1 to BD (Fig. 2).
Figure 2 (Sol. 1.9)
We must prove that BA1 = DC1 . Drop perpendicular OP from the center O of the
circumscribed circle to BD. Clearly, P is the midpoint of BD. Lines AA1 , OP and CC1 are
parallel to each other and AO = OC. So A1 P = P C1 and, since P is the midpoint of BD,
it follows that BA1 = DC1 .
1.10. We see that BO : OD = DP : P B = k, because BO =P D. Let BC = 1. Then
AD = k and ED = k1 . So k = AD = AE + ED = 1 + k1 , that is k 2 = 1 + k. Finally, observe
that k 2 = AD2 and 1 + k = BC 2 + BC · AD.
1.11. Let C, D, E and F be the midpoints of sides AO, OB, BM and M A, respectively,
of quadrilateral AOM B. Since AB = M O = R, where R is the radius of the given circle,
CDEF is a rhombus by Problem 1.2. Hence, CE ⊥ DF .
1.12. a) If the lines containing the given points are parallel, then the assertion of the
problem is obviously true. We assume that the lines meet at O. Then OA : OB = OB1 : OA1
and OC : OA = OA1 : OC1 . Hence, OC : OB = OB1 : OC1 and so BC1 k CB1 (the ratios
of the segment should be assumed to be oriented).
b) Let AB1 and CA1 meet at D, let CB1 and AC1 meet at E. Then CA1 : A1 D = CB :
BA = EC1 : C1 A. Since △CB1 D ∼ △EB1 A, points A1 , B1 and C1 lie on the same line.
1.13. A point that lies on the bisector of an angle is equidistant from the angle’s legs.
Let a be the distance from point A1 to lines AC and AB, let b be the distance from point B1
to lines AB and BC. Further, let A1 M : B1 M = p : q, where p + q = 1. Then the distances
SOLUTIONS
23
from point M to lines AC and BC are equal to qa and pb, respectively. On the other hand,
by Problem 1.1 b) the distance from point M to line AB is equal to qa + pb.
1.14. Let the line that passes through the center O of the given rectangle parallel to BC
intersect line segment QN at point K (Fig. 3).
Figure 3 (Sol. 1.14)
Since M O k P C, it follows that QM : M P = QO : OC and, since KO k BC, it follows
that QO : OC = QK : KN . Therefore, QM : M P = QK : KN , i.e., KM k N P . Hence,
∠M N P = ∠KM O = ∠QN M .
1.15. Let us draw through point M line EF so that EF k CD (points E and F lie on
lines BC and AD). Then P L : P K = BL : KD and OK : OL = KA : CL = KA : KF =
BL : EL. Since KD = EL, we have P L : P K = OK : OL and, therefore, P L = OK.
1.16. Consider parallelogram ABCD1 . We may assume that points D and D1 do not
coincide (otherwise the statement of the problem is obvious). On sides AD1 and CD1 take
points S1 and R1 , respectively, so that SS1 k DD1 and RR1 k DD1 . Let segments P R1 and
QS1 meet at N ; let N1 and N2 be the intersection points of the line that passes through N
parallel to DD1 with segments P R and QS, respectively.
−−→
−−→
−−→
−−→
−−→
−−→
Then N1 N = β RR1 = αβ DD1 and N2 N = αSS1 = αβ DD1 . Hence, segments P R and
QS meet at N1 = N2 . Clearly, P N1 : P R = P N : P R1 = β and QN2 : QS = α.
Remark. If α = β, there is a simpler solution. Since BP : BA = BQ : BC = α, it
follows that P Q k AC and P Q : AC = α. Similarly, RS k AC and RS : AC = 1 − α.
Therefore, segments P R and QS are divided by their intersection point in the ratio of
α : (1 − α).
1.17. a) From vertices A and C drop perpendiculars AK and CL to line BD. Since
∠CBL = ∠ABK and ∠CDL = ∠KDA, we see that △BLC ∼ △BKA and △CLD ∼
△AKD. Therefore, AD : DC = AK : CL = AB : BC.
b) Taking into account that BA1 : A1 C = BA : AC and BA1 + A1 C = BC we get
ac
BA1 = b+c
. Since BO is the bisector of triangle ABA1 , it follows that AO : OA1 = AB :
BA1 = (b + c) : a.
1.18. Let O be the center of the circumscribed circle of isosceles triangle ABC, let B1
be the midpoint of base AC and A1 the midpoint of the lateral side BC. Since △BOA1 ∼
2
△BCB1 , it follows that BO : BA1 = BC : BB1 and, therefore, R = BO = √4aa2 −b2 .
24
CHAPTER 1. SIMILAR TRIANGLES
1.19. If ∠EAD = ϕ, then AE =
AD
cos ϕ
=
AB
cos ϕ
and AF =
AB
.
sin ϕ
Therefore,
1
1
cos2 ϕ + sin2 ϕ
1
+
=
=
.
AE 2 AF 2
AB 2
AB 2
1.20. It is easy to verify that AB22 = AB1 · AC = AC1 · AB = AC22 .
1.21. a) Since BQ : QM = BN : AM = BK : AK, we have: KQ k AM .
b) Let O be the center of the inscribed circle. Since ∠CBA + ∠BAD = 180◦ , it follows
that ∠ABO + ∠BAO = 90◦ . Therefore, △AKO ∼ △OKB, i.e., AK : KO = OK : KB.
Consequently, AK ·KB = KO2 = R2 , where R is the radius of the inscribed circle. Similarly,
CL · LD = R2 .
1.22. If angle ∠ABC is obtuse (resp. acute), then angle ∠M AN is also obtuse (resp.
acute). Moreover, the legs of these angles are mutually perpendicular. Therefore, ∠ABC =
∠M AN . Right triangles ABM and ADN have equal angles ∠ABM = ∠ADN , therefore,
AM : AN = AB : AD = AB : CB, i.e., △ABC ∼ △M AN .
1.23. On diagonal AC, take points D′ and B ′ such that BB ′ k l and DD′ k l. Then
AB : AE = AB ′ : AG and AD : AF = AD′ : AG. Since the sides of triangles ABB ′
and CDD′ are pairwise parallel and AB = CD, these triangles are equal and AB ′ = CD′ .
Therefore,
AB AD
AB ′ AD′
CD′ + AD′
AC
+
=
+
=
=
.
AE
AF
AG
AG
AG
AG
1.24. Let us drop from vertex B perpendicular BG to AC (Fig. 4).
Figure 4 (Sol. 1.24)
Since triangles ABG and ACE are similar, AC · AG = AE · AB. Lines AF and CB are
parallel, consequently, ∠GCB = ∠CAF . We also infer that right triangles CBG and ACF
are similar and, therefore, AC · CG = AF · BC. Summing the equalities obtained we get
AC · (AG + CG) = AE · AB + AF · BC.
Since AG + CG = AC, we get the equality desired.
1.25. Since α + β = 90◦ − 21 α, it follows that γ = 180◦ − α − β = 90◦ + 12 α. Therefore,
it is possible to find point D on side AB so that ∠ACD = 90◦ − 12 α, i.e., AC = AD. Then
△ABC ∼ △CBD and, therefore, BC : BD = AB : CB, i.e., a2 = c(c − b).
1.26. As segments AB and CD move, triangle AM C is being replaced by another triangle
BM
AM
remains a constant. Analogously, DM
similar to the initial one. Therefore, the quantity CM
remains a constant.
1.27. Let medians meet at O; denote the intersection points of median AK with lines
F P and F E by Q and M , respectively; denote the intersection points of median CL with
lines EP and F E by R and N , respectively (Fig. 5).
Clearly, F M : F E = F Q : F P = LO : LC = 1 : 3, i.e., F M = 13 F E. Similarly,
EN = 13 F E.
SOLUTIONS
25
Figure 5 (Sol. 1.27)
1.28. Let A and B be the intersection points of the given line with the angle’s legs.
On segments AC and BC, take points K and L, respectively, so that P K k BC and
P L k AC. Since △AKP ∼ △P LB, it follows that AK : KP = P L : LB and, therefore,
(a − p)(b − p) = p2 , where p = P K = P L. Hence, a1 + 1b = p1 .
1.29. Denote the midpoint of side BC by O and the intersection points of AK and AL
with side BC by P and Q, respectively. We may assume that BP < BQ. Triangle LCO
is an equilateral one and LC k AB. Therefore, △ABQ ∼ △LCQ, i.e., BQ : QC = AB :
LC = 2 : 1. Hence, BC = BQ + QC = 3QC. Similarly, BC = 3BP .
1.30. Since BK : BO = BO : AB and ∠KBO = ∠ABO, it follows that △KOB ∼
△OAB. Hence, ∠KOB = ∠OAB. Similarly, ∠AOM = ∠ABO. Therefore,
∠KOM = ∠KOB + ∠BOA + ∠AOM = ∠OAB + ∠BOA + ∠ABO = 180◦ ,
i.e., points K, O and M lie on one line.
1.31. Since ∠AM N = ∠M N C and ∠BM N = ∠M N A, we see that ∠AM B = ∠AN C.
Moreover, AM : AN = N B : N M = BM : CN . Hence, △AM B ∼ △AN C and, therefore,
∠M AB = ∠N AC. Consequently, ∠BAC = ∠M AN . For the other angles the proof is
similar.
Let points B1 and C1 be symmetric to B and C, respectively, through the midperpendicular to segment M N . Since AM : N B = M N : BM = M C : N C, it follows that
M A · M C1 = AM · N C = N B · M C = M B1 · M C. Therefore, point A lies on the circle
circumscribed about trapezoid BB1 CC1 .
1.32. Since ∠AEB+∠BEC = 180◦ , angles ∠AEB and ∠BEC cannot be different angles
of similar triangles ABE and BEC, i.e., the angles are equal and BE is a perpendicular.
Two cases are possible: either ∠ABE = ∠CBE or ∠ABE = ∠BCE. The first case
should be discarded because in this case △ABE = △CBE.
◦
In the second case we have ∠ABC = ∠ABE + ∠CBE = ∠ABE
√ + ∠BAE = 90 . In
right triangle ABC the ratio of the legs’ lengths is equal to 1 : 3; hence, the angles of
triangle ABC are equal to 90◦ , 60◦ , 30◦ .
q
SBDEF
SBDE
EF
DB
EF C
1.33. We have 2SADE = SADE = AD = AD = SSADE
. Hence,
p
SBDEF = 2 SADE · SEF C .
1.34. Let M N = x; let E be the intersection point of lines AB and CD. Triangles
EBC, EM N and EAD are similar, hence, SEBC : SEM N : SEAD = a2 : x2 : b2 . Since
SEM N − SEBC = SM BCN = SM ADN = SEAD − SEM N , it follows that x2 − a2 = b2 − x2 , i.e.,
x2 = 21 (a2 + b2 ).
26
CHAPTER 1. SIMILAR TRIANGLES
1.35. Through point Q inside triangle ABC draw lines DE, F G and HI parallel to BC,
CA and AB, respectively, so that points F and H would lie on side BC, points E and I on
side AC, points D and G on side AB (Fig. 6).
Figure 6 (Sol. 1.35)
Set S = SABC , S1 = SGDQ , S2 = SIEQ , S3 = SHF Q . Then
r
r
r
GQ
IE
FQ
AI + IE + EC
S1
S2
S3
+
+
=
+
+
=
= 1,
S
S
S
AC AC AC
AC
√
√
√
i.e., S = ( S1 + S2 + S3 )2 .
1.36. Let M be the intersection point of the medians of triangle ABC; let point A1 be
symmetric to M through the midpoint of segment BC. The ratio of the lengths of sides
of triangle CM A1 to the lengths of the corresponding medians of triangle ABC is to 2 : 3.
Therefore, the area to be found is equal to 94 SCM A1 . Clearly, SCM A1 = 31 S (cf. the solution
of Problem 4.1).
1.37. Let E, F , G and H be the midpoints of sides AB, BC, CD and DA, respectively.
a) Clearly, SAEH + SCF G = 41 SABD + 14 SCBD = 14 SABCD . Analogously, SBEF + SDGH =
1
S
; hence, SEF GH = SABCD − 14 SABCD − 14 SABCD = 21 SABCD .
4 ABCD
b) Since AC = BD, it follows that EF GH is a rhombus (Problem 1.2). By heading a)
we have SABCD = 2SEF GH = EG · F H.
1.38. Let E, F , G and H be the midpoints of sides of quadrilateral ABCD; let points
E1 , F1 , G1 and H1 be symmetric to point O through these points, respectively. Since EF
is the midline of triangle E1 OF1 , we see that SE1 OF1 = 4SEOF . Similarly, SF1 OG1 = 4SF OG ,
SG1 OH1 = 4SGOH , SH1 OE1 = 4SHOE . Hence, SE1 F1 G1 H1 = 4SEF GH . By Problem 1.37 a)
SABCD = 2SEF GH . Hence, SE1 F1 G1 H1 = 2SABCD = 2S.
1.39. First solution. Let us consider square BCM N and divide its side M N by points
P and Q into three equal parts (Fig. 7).
Then △ABC = △P DQ and △ACD = △P M A. Hence, triangle △P AD is an isosceles
right triangle and ∠ABC + ∠ADC = ∠P DQ +√
∠ADC = 45◦ .
√
√
Second solution. Since DE = 1, EA = 2, EB = 2, AD = 5 and BA = 10,
it follows that DE : AE = EA : EB = AD : BA and △DEA ∼ △AEB. Therefore,
∠ABC = ∠EAD. Moreover, ∠AEC = ∠CAE = 45◦ . Hence,
∠ABC + ∠ADC + ∠AEC = (∠EAD + ∠CAE) + ∠ADC
= ∠CAD + ∠ADC = 90◦ .
SOLUTIONS
27
Figure 7 (Sol. 1.39)
1.40. From point L drop perpendiculars LM and LN on AB and AD, respectively.
Then KM = M B = N D and KL = LB = DL and, therefore, right triangles KM L and
DN L are equal. Hence, ∠DLK = ∠N LM = 90◦ .
1.41. Since D1 A = B1 B, AD2 = BB2 and ∠D1 AD2 = ∠B1 BB2 , it follows that
△D1 AD2 = △B1 BB2 . Sides AD1 and BB1 (and also AD2 and BB2 ) of these triangles
are perpendicular and, therefore, B1 B2 ⊥ D1 D2 .
1.42. On the extension of segment AC beyond point C take point M so that CM = CE
(Fig. 8).
Figure 8 (Sol. 1.42)
Then under the rotation with center C through an angle of 90◦ triangle ACE turns into
triangle BCM . Therefore, line M B is perpendicular to line AE; hence, it is parallel to line
CL. Since M C = CE = DC and lines DK, CL and M B are parallel, KL = LB.
1.43. Let rectangles ABC1 D1 and A2 BCD2 be constructed on sides AB and BC; let
P , Q, R and S be the centers of rectangles constructed on sides AB, BC, CD and DA,
respectively. Since ∠ABC + ∠ADC = 180◦ , it follows that △ADC = △A2 BC1 and, therefore, △RDS = △P BQ and RS = P Q. Similarly, QR = P S. Therefore, P QRS is a
parallelogram such that one of triangles RDS and P BQ is constructed on its sides outwards
and on the other side inwards; a similar statement holds for triangles QCR and SAP as
well. Therefore, ∠P QR + ∠RSP = ∠BQC + ∠DSA = 180◦ because ∠P QB = ∠RSD and
∠RQC = ∠P SA. It follows that P QRS is a rectangle.
1.44. Let K, L and M be the intersection points of the circumscribed circles of triangles
F OA and BOC, BOC and DOE, DOE and F OA, respectively; 2α, 2β and 2γ the angles
at the vertices of isosceles triangles BOC, DOE and F OA, respectively (Fig. 9).
28
CHAPTER 1. SIMILAR TRIANGLES
Figure 9 (Sol. 1.44)
Point K lies on arc ⌣ OB of the circumscribed circle of the isosceles triangle BOC and,
therefore, ∠OKB = 90◦ + α. Similarly, ∠OKA = 90◦ + γ. Since α + β + γ = 90◦ , it follows
that ∠AKB = 90◦ + β. Inside equilateral triangle AOB there exists a unique point K that
serves as the vertex of the angles that subtend its sides and are equal to the given angles.
Similar arguments for a point L inside triangle COD show that △OKB = △CLO.
Now, let us prove that △KOL = △OKB. Indeed, ∠COL = ∠KBO; hence, ∠KOB +
∠COL = 180◦ − ∠OKB = 90◦ − α and, therefore, ∠KOL = 2α + (90◦ − α) = 90◦ + α =
∠OKB. It follows that KL = OB = R. Similarly, LM = M K = R.
1.45. Let ∠A = α. It is easy to verify that both angles ∠KCL and ∠ADL are equal to
240◦ − α (or 120◦ + α). Since KC = BC = AD and CL = DL, it follows that △KCL =
△ADL and, therefore, KL = AL. Similarly, KL = AK.
1.46. Let P , Q and R be the centers of the squares constructed on sides DA, AB and
BC, respectively, in parallelogram ABCD with an acute angle of α at vertex A. It is easy
to verify that ∠P AQ = 90◦ + α = ∠RBQ; hence, △P AQ = △RBQ. Sides AQ and BQ of
these triangles are perpendicular, hence, P Q ⊥ QR.
1.47. First, observe that the sum of the angles at vertices A, B and C of hexagon
AB ′ CA′ BC ′ is equal to 360◦ because by the hypothesis the sum of its angles at the other
vertices is equal to 360◦ . On side AC ′ , construct outwards triangle △AC ′ P equal to triangle
△BC ′ A′ (Fig. 10).
Figure 10 (Sol. 1.47)
Then △AB ′ P = △CB ′ A′ because AB ′ = CB ′ , AP = CA′ and
∠P AB ′ = 360◦ − ∠P AC ′ − ∠C ′ AB ′ = 360◦ − ∠A′ BC ′ − ∠C ′ AB ′ = ∠A′ CB ′ .
Hence, △C ′ B ′ A′ = △C ′ B ′ P and, therefore, 2∠A′ B ′ C ′ = ∠P B ′ A′ = ∠AB ′ C because
∠P B ′ A = ∠A′ B ′ C.
SOLUTIONS
29
1.48. Since BA : BC = BC1 : BA1 and ∠ABC = ∠C1 BA1 , it follows that △ABC ∼
△C1 BA1 . Similarly, △ABC ∼ △B1 A1 C. Since BA1 = A1 C, it follows that △C1 BA1 =
△B1 A1 C. Therefore, AC1 = C1 B = B1 A1 and AB1 = B1 C = C1 A1 . It is also clear that
quadrilateral AB1 A1 C1 is a convex one.
1.49. a) Let P and Q be the midpoints of sides AB and AC. Then M P = 12 AC = QB1 ,
M Q = 21 AB = P C1 and ∠C1 P M = ∠C1 P B + ∠BP M = ∠B1 QC + ∠CQM = ∠B1 QM .
Hence, △M QB1 = △C1 P M and, therefore, M C1 = M B1 . Moreover,
and
∠P M C1 + ∠QM B1 = ∠QB1 M + ∠QM B1 = 180◦ − ∠M QB1
∠M QB1 = ∠A + ∠CQB1 = ∠A + (180◦ − 2ϕ).
Therefore, ∠B1 M C1 = ∠P M Q + 2ϕ − ∠A = 2ϕ. (The case when ∠C1 P B + ∠BP M > 180◦
is analogously treated.)
b) On sides AB and AC, take points B ′ and C ′ , respectively, such that AB ′ : AB = AC ′ :
AC = 2 : 3. The midpoint M of segment B ′ C ′ coincides with the intersection point of the
medians of triangle ABC. On sides AB ′ and AC ′ , construct outwards right triangles AB ′ C1
and AB1 C ′ with angle ϕ = 60◦ as in heading a). Then B1 and C1 are the centers of right
triangles constructed on sides AB and AC; on the other hand, by heading a), M B1 = M C1
and ∠B1 M C1 = 120◦ .
Remark. Statements of headings a) and b) remain true for triangles constructed inwards, as well.
1.50. a) Let B ′ be the intersection point of line AC and the perpendicular to line AB1
erected from point B1 ; define point C ′ similarly. Since AB ′ : AC ′ = AC1 : AB1 = AB : AC,
it follows that B ′ C ′ k BC. If N is the midpoint of segment B ′ C ′ , then, as follows from
Problem 1.49, N C1 = N B1 (i.e., N = M ) and ∠B1 N C1 = 2∠AB ′ B1 = 180◦ −2∠CAB1 = ϕ.
b) On side BC construct outwards isosceles triangle BA1 C with angle 360◦ −2ϕ at vertex
A1 (if ϕ < 90◦ construct inwards a triangle with angle 2ϕ). Since the sum of the angles at
the vertices of the three constructed isosceles triangles is equal to 360◦ , it follows that the
angles of triangle A1 B1 C1 are equal to 180◦ −ϕ, 21 ϕ and 21 ϕ (cf. Problem 1.47). In particular,
this triangle is an isosceles one, hence, A1 = O.
1.51. Let O1 , O2 , O3 and O4 be the centers of rhombuses constructed on sides AB, BC,
CA and DA, respectively; let M be the midpoint of diagonal AC. Then M O1 = M O2 and
∠O1 M O2 = α (cf. Problem 1.49). Similarly, M O3 = M O4 and ∠O3 M O4 = α. Therefore,
under the rotation through an angle of α about point M triangle △O1 M O3 turns into
△O2 M O4 .
1.52. Since A1 C = AC| cos C| , B1 C = BC| cos C| and angle ∠C is the common angle
of triangles ABC and A1 B1 C, these triangles are similar; the similarity coefficient is equal
to | cos C|.
1.53. Since points M and N lie on the circle with diameter CH, it follows that ∠CM N =
∠CHN and since AC ⊥ HN , we see that ∠CHN = ∠A. Similarly, ∠CN M = ∠B.
1.54. a) Let l be the tangent to the circumscribed circle at point A. Then ∠(l, AB) =
∠(AC, CB) = ∠(C1 B1 , AC1 ) and, therefore, l k B1 C1 .
b) Since OA ⊥ l and l k B1 C1 , it follows that OA ⊥ B1 C1 .
1.55. If AA1 , BB1 and CC1 are heights, then right triangles AA1 C and BB1 C have
equal angles at vertex C and, therefore, are similar. It follows that △A1 BH ∼ △B1 AH,
consequently, AH · A1 H = BH · B1 H. Similarly, BH · B1 H = CH · C1 H.
If AH · A1 H = BH · B1 H = CH · C1 H, then △A1 BH ∼ △B1 AH; hence, ∠BA1 H =
∠AB1 H = ϕ. Thus, ∠CA1 H = ∠CB1 H = 180◦ − ϕ.
30
CHAPTER 1. SIMILAR TRIANGLES
Similarly, ∠AC1 H = ∠CA1 H = 180◦ − ϕ and ∠AC1 H = ∠AB1 H = ϕ. Hence, ϕ = 90◦ ,
i.e., AA1 , BB1 and CC1 are heights.
1.56. a) By Problem 1.52 ∠C1 A1 B = ∠CA1 B1 = ∠A. Since AA1 ⊥ BC, it follows that
∠C1 A1 A = ∠B1 A1 A. The proof of the fact that rays B1 B and C1 C are the bisectors of
angles A1 B1 C1 and A1 C1 B1 is similar.
b) Lines AB, BC and CA are the bisectors of the outer angles of triangle A1 B1 C1 , hence,
A1 A is the bisector of angle ∠B1 A1 C1 and, therefore, AA1 ⊥ BC. For lines BB1 and CC1
the proof is similar.
1.57. From the result of Problem 1.56 a) it follows that the symmetry through line AC
sends line B1 A1 into line B1 C1 .
1.58. By Problem 1.52 ∠B1 A1 C = ∠BAC. Since A1 B1 k AB, it follows that ∠B1 A1 C =
∠ABC. Hence, ∠BAC = ∠ABC. Similarly, since B1 C1 k BC, it follows that ∠ABC =
∠BCA. Therefore, triangle ABC is an equilateral one and A1 C1 k AC.
1.59. Let O be the center of the circumscribed circle of triangle ABC. Since OA ⊥ B1 C1
(cf. Problem 1.54 b), it follows that SAOC1 + SAOB1 = 21 (R · B1 C1 ). Similar arguments for
vertices B and C show that SABC = qR. On the other hand, SABC = pr.
1.60. The perimeter of the triangle cut off by the line parallel to side BC is equal to
the sum of distances from point A to the tangent points of the inscribed circle with sides
AB and AC; therefore, the sum of perimeters of small triangles is equal to the perimeter
of triangle ABC, i.e., P1 + P2 + P3 = P . The similarity of triangles implies that rri = PPi .
Summing these equalities for all the i we get the statement desired.
1.61. Let M = A. Then XA = A; hence, AYA = 1. Similarly, CXC = 1. Let us
prove that y = AYA and x = CXC are the desired lines. On side BC, take point D so that
AB k M D, see Fig. 11. Let E be the intersection point of lines CXC and M D. Then,
XM M + YM M = XC E + YM M . Since △ABC ∼ △M DC, it follows that CE = YM M .
Therefore, CE = YM M . Hence, XM M + YM M = XC E + CE = XC C = 1.
Figure 11 (Sol. 1.61)
1.62. Let D be the midpoint of segment BH. Since △BHA ∼ △HEA, it follows that
AD : AO = AB : AH and ∠DAH = ∠OAE. Hence, ∠DAO = ∠BAH and, therefore,
△DAO ∼ △BAH and ∠DOA = ∠BAH = 90◦ .
1.63. Let AA1 , BB1 and CC1 be heights of triangle ABC. Let us drop from point B1
perpendiculars B1 K and B1 N to sides AB and BC, respectively, and perpendiculars B1 L and
B1 M to heights AA1 and CC1 , respectively. Since KB1 : C1 C = AB1 : AC = LB1 : A1 C,
it follows that △KLB1 ∼ △C1 A1 C and, therefore, KL k C1 A1 . Similarly, M N k C1 A1 .
Moreover, KN k C1 A1 (cf. Problem 1.53). It follows that points K, L, M and N lie on one
line.
1.64. a) Let O be the midpoint of AC, let O1 be the midpoint of AB and O2 the midpoint
of BC. Assume that AB ≤ BC. Through point O1 draw line O1 K parallel to EF (point K
lies on segment EO2 ). Let us prove that right triangles DBO and O1 KO2 are equal. Indeed,
SOLUTIONS
31
O1 O2 = DO = 12 AC and BO = KO2 = 21 (BC − AB). Since triangles DBO and O1 KO2
are equal, we see that ∠BOD = ∠O1 O2 E, i.e., line DO is parallel to EO2 and the tangent
drawn through point D is parallel to line EF .
b) Since the angles between the diameter AC and the tangents to the circles at points F ,
D, E are equal, it follows that ∠F AB = ∠DAC = ∠EBC and ∠F BA + ∠DCA = ∠ECB,
i.e., F lies on line segment AD and E lies on line segment DC. Moreover, ∠AF B =
∠BEC = ∠ADC = 90◦ and, therefore, F DEB is a rectangle.
1.65. Let M Q and M P be perpendiculars dropped on sides AD and BC, let M R and
M T be perpendiculars dropped on the extensions of sides AB and CD (Fig. 12). Denote
by M1 and P1 the other intersection points of lines RT and QP with the circle.
Figure 12 (Sol. 1.65)
Since T M1 = RM = AQ and T M1 k AQ, it follows that AM1 k T Q. Similarly, AP1 k
RP . Since ∠M1 AP1 = 90◦ , it follows that RP ⊥ T Q.
Denote the intersection points of lines T Q and RP , M1 A and RP , P1 A and T Q by E,
F , G, respectively. To prove that point E lies on line AC, it suffices to prove that rectangles
AF EG and AM1 CP1 are similar. Since ∠ARF = ∠AM1 R = ∠M1 T G = ∠M1 CT , we
may denote the values of these angles by the same letter α. We have: AF = RA sin α =
M1 A sin2 α and AG = M1 T sin α = M1 C sin2 α. Therefore, rectangles AF EG and AM1 CP1
are similar.
1.66. Denote the centers of the circles by O1 and O2 . The outer tangent is tangent to
the first circle at point K and to the other circle at point L; the inner tangent is tangent to
the first circle at point M and to the other circle at point N (Fig. 13).
Figure 13 (Sol. 1.66)
Let lines KM and LN intersect line O1 O2 at points P1 and P2 , respectively. We have
to prove that P1 = P2 . Let us consider points A, D1 , D2 — the intersection points of KL
with M N , KM with O1 A, and LN with O2 A, respectively. Since ∠O1 AM + ∠N AO2 = 90◦ ,
32
CHAPTER 1. SIMILAR TRIANGLES
right triangles O1 M A and AN O2 are similar; we also see that AO2 k KM and AO1 k LN .
Since these lines are parallel, AD1 : D1 O1 = O2 P1 : P1 O1 and D2 O2 : AD2 = O2 P2 : P2 O1 .
The similarity of quadrilaterals AKO1 M and O2 N AL yields AD1 : D1 O1 = D2 O2 : AD2 .
Therefore, O2 P1 : P1 O1 = O2 P2 : P2 O1 , i.e., P1 = P2 .
CHAPTER 2. INSCRIBED ANGLES
Background
1. Angle ∠ABC whose vertex lies on a circle and legs intersect this circle is called
inscribed in the circle. Let O be the center of the circle. Then
½
1
∠AOC
if points B and O lie on one side of AC
2
∠ABC =
1
◦
180 − 2 ∠AOC
otherwise.
The most important and most often used corollary of this fact is that equal chords subtend
angles that either are equal or the sum of the angles is equal to 180◦ .
2. The value of the angle between chord AB and the tangent to the circle that passes
through point A is equal to half the angle value of arc ⌣ AB.
3. The angle values of arcs confined between parallel chords are equal.
4. As we have already said, if two angles subtend the same chord, either they are equal
or the sum of their values is 180◦ . In order not to consider various variants of the positions
of points on the circle let us introduce the notion of an oriented angle between lines. The
value of the oriented angle between lines AB and CD (notation: ∠(AB, CD)) is the value
of the angle by which we have to rotate line AB counterclockwise in order for it to become
parallel to line CD. The angles that differ by n · 180◦ are considered equal.
Notice that, generally, the oriented angle between lines CD and AB is not equal to the
oriented angle between lines AB and CD (the sum of ∠(AB, CD) and ∠(CD, AB) is equal
to 180◦ which, according to our convention, is the same as 0◦ ).
It is easy to verify the following properties of the oriented angles:
a) ∠(AB, BC) = −∠(BC, AB);
b) ∠(AB, CD) + ∠(CD, EF ) = ∠(AB, EF );
c) points A, B, C, D not on one line lie on one circle if and only if ∠(AB, BC) =
∠(AD, DC). (To prove this property we have to consider two cases: points B and D lie on
one side of AC; points B and D lie on different sides of AC.)
Introductory problems
1. a) From point A lying outside a circle rays AB and AC come out and intersect the
circle. Prove that the value of angle ∠BAC is equal to half the difference of the angle
measures of the arcs of the circle confined inside this angle.
b) The vertex of angle ∠BAC lies inside a circle. Prove that the value of angle ∠BAC is
equal to half the sum of angle measures of the arcs of the circle confined inside angle ∠BAC
and inside the angle symmetric to it through vertex A.
2. From point P inside acute angle ∠BAC perpendiculars P C1 and P B1 are dropped
on lines AB and AC. Prove that ∠C1 AP = ∠C1 B1 P .
3. Prove that all the angles formed by the sides and diagonals of a regular n-gon are
◦
.
integer multiples of 180
n
4. The center of an inscribed circle of triangle ABC is symmetric through side AB to
the center of the circumscribed circle. Find the angles of triangle ABC.
33
34
CHAPTER 2. INSCRIBED ANGLES
5. The bisector of the exterior angle at vertex C of triangle ABC intersects the circumscribed circle at point D. Prove that AD = BD.
§1. Angles that subtend equal arcs
2.1. Vertex A of an acute triangle ABC is connected by a segment with the center O of
the circumscribed circle. From vertex A height AH is drawn. Prove that ∠BAH = ∠OAC.
2.2. Two circles intersect at points M and K. Lines AB and CD are drawn through M
and K, respectively; they intersect the first circle at points A and C, the second circle at
points B and D, respectively. Prove that AC k BD.
2.3. From an arbitrary point M inside a given angle with vertex A perpendiculars M P
and M Q are dropped to the sides of the angle. From point A perpendicular AK is dropped
on segment P Q. Prove that ∠P AK = ∠M AQ.
2.4. a) The continuation of the bisector of angle ∠B of triangle ABC intersects the
circumscribed circle at point M ; O is the center of the inscribed circle, Ob is the center of
the escribed circle tangent to AC. Prove that points A, C, O and Ob lie on a circle centered
at M .
b) Point O inside triangle ABC is such that lines AO, BO and CO pass through the
centers of the circumscribed circles of triangles BCO, ACO and ABO, respectively. Prove
that O is the center of the inscribed circle of triangle ABC.
2.5. Vertices A and B of right triangle ABC with right angle ∠C slide along the sides
of a right angle with vertex P . Prove that in doing so point C moves along a line segment.
2.6. Diagonal AC of square ABCD coincides with the hypothenuse of right triangle
ACK, so that points B and K lie on one side of line AC. Prove that
BK =
|AK − CK|
√
2
and DK =
AK + CK
√
.
2
2.7. In triangle ABC medians AA1 and BB1 are drawn. Prove that if ∠CAA1 =
∠CBB1 , then AC = BC.
2.8. Each angle of triangle ABC is smaller than 120◦ . Prove that inside △ABC there
exists a point that serves as the vertex for three angles each of value 120◦ and subtending
the side of the triangle different from the sides subtended by the other angles.
2.9. A circle is divided into equal arcs by n diameters. Prove that the bases of the
perpendiculars dropped from an arbitrary point M inside the circle to these diameters are
vertices of a regular n-gon.
2.10. Points A, B, M and N on a circle are given. From point M chords M A1 and M B1
perpendicular to lines N B and N A, respectively, are drawn. Prove that AA1 k BB1 .
2.11. Polygon ABCDEF is an inscribed one; AB k DE and BC k EF . Prove that
CD k AF .
2.12. Polygon A1 A2 . . . A2n as an inscribed one. We know that all the pairs of its opposite
sides except one are parallel. Prove that for any odd n the remaining pair of sides is also
parallel and for any even n the lengths of the exceptional sides are equal.
2.13. Consider triangle ABC. Prove that there exist two families of equilateral triangles
whose sides (or extensions of the sides) pass through points A, B and C. Prove also that
the centers of triangles from these families lie on two concentric circles.
§3. THE ANGLE BETWEEN A TANGENT AND A CHORD
35
§2. The value of an angle between two chords
The following fact helps to solve problems from this section. Let A, B, C, D be points
on a circle situated in the order indicated. Then
⌣ AB+ ⌣ CD
| ⌣ AD− ⌣ CB|
∠(AC, BD) =
and ∠(AB, CD) =
.
2
2
To prove this, we have to draw a chord parallel to another chord through the endpoint of
one of the chords.
2.14. Points A, B, C, D in the indicated order are given on a circle. Let M be the
midpoint of arc ⌣ AB. Denote the intersection points of chords M C and M D with chord
AB by E and K. Prove that KECD is an inscribed quadrilateral.
2.15. Concider an equilateral triangle. A circle with the radius equal to the triangle’s
height rolls along a side of the triangle. Prove that the angle measure of the arc cut off the
circle by the sides of the triangle is always equal to 60◦ .
2.16. The diagonals of an isosceles trapezoid ABCD with lateral side AB intersect at
point P . Prove that the center O of the inscribed circle lies on the inscribed circle of triangle
AP B.
2.17. Points A, B, C, D in the indicated order are given on a circle; points A1 , B1 , C1
and D1 are the midpoints of arcs ⌣ AB, ⌣ BC, ⌣ CD and ⌣ DA, respectively. Prove
that A1 C1 ⊥ B1 D1 .
2.18. Point P inside triangle ABC is taken so that ∠BP C = ∠A + 60◦ , ∠AP C =
∠B + 60◦ and ∠AP B = ∠C + 60◦ . Lines AP , BP and CP intersect the circumscribed
circle of triangle ABC at points A′ , B ′ and C ′ , respectively. Prove that triangle A′ B ′ C ′ is
an equilateral one.
2.19. Points A, C1 , B, A1 , C, B1 in the indicated order are taken on a circle.
a) Prove that if lines AA1 , BB1 and CC1 are the bisectors of the angles of triangle ABC,
then they are the heights of triangle A1 B1 C1 .
b) Prove that if lines AA1 , BB1 and CC1 are the heights of triangle ABC, then they are
the bisectors of the angles of triangle A1 B1 C1 .
2.20. Triangles T1 and T2 are inscribed in a circle so that the vertices of triangle T2
are the midpoints of the arcs into which the circle is divided by the vertices of triangle T1 .
Prove that in the hexagon which is the intersection of triangles T1 and T2 the diagonals that
connect the opposite vertices are parallel to the sides of triangle T1 and meet at one point.
§3. The angle between a tangent and a chord
2.21. Two circles intersect in points P and Q. Through point A on the first circle lines
AP and AQ are drawn. The lines intersect the second circle in points B and C. Prove that
the tangent at A to the first circle is parallel to line BC.
2.22. Circles S1 and S2 intersect at points A and P . Tangent AB to circle S1 is drawn
through point A, and line CD parallel to AB is drawn through point P (points B and C lie
on S2 , point D on S1 ). Prove that ABCD is a parallelogram.
2.23. The tangent at point A to the inscribed circle of triangle ABC intersects line BC
at point E; let AD be the bisector of triangle ABC. Prove that AE = ED.
2.24. Circles S1 and S2 intersect at point A. Through point A a line that intersects S1
at point B and S2 at point C is drawn. Through points C and B tangents to the circles are
drawn; the tangents intersect at point D. Prove that angle ∠BDC does not depend on the
choice of the line that passes through A.
2.25. Two circles intersect at points A and B. Through point A tangents AM and AN ,
where M and N are points of the respective circles, are drawn. Prove that:
36
CHAPTER 2. INSCRIBED ANGLES
a) ∠ABN + ∠M AN = 180◦ ;
¡ AM ¢2
b) BM
.
=
BN
AN
2.26. Inside square ABCD a point P is taken so that triangle ABP is an equilateral
one. Prove that ∠P CD = 15◦ .
2.27. Two circles are internally tangent at point M . Let AB be the chord of the greater
circle which is tangent to the smaller circle at point T . Prove that M T is the bisector of
angle AM B.
2.28. Through point M inside circle S chord AB is drawn; perpendiculars M P and M Q
are dropped from point M to the tangents that pass through points A and B respectively.
1
Prove that the value of P1M + QM
does not depend on the choice of the chord that passes
through point M .
2.29. Circle S1 is tangent to sides of angle ABC at points A and C. Circle S2 is tangent
to line AC at point C and passes through point B, circle S2 intersects circle S1 at point M .
Prove that line AM divides segment BC in halves.
2.30. Circle S is tangent to circles S1 and S2 at points A1 and A2 ; let B be a point of
circle S, let K1 and K2 be the other intersection points of lines A1 B and A2 B with circles S1
and S2 , respectively. Prove that if line K1 K2 is tangent to circle S1 , then it is also tangent
to circle S2 .
§4. Relations between the values of an angle and the lengths of the arc and
chord associated with the angle
2.31. Isosceles trapezoids ABCD and A1 B1 C1 D1 with parallel respective sides are inscribed in a circle. Prove that AC = A1 C1 .
2.32. From point M that moves along a circle perpendiculars M P and M Q are dropped
on diameters AB and CD, respectively. Prove that the length of segment P Q does not
depend on the position of point M .
2.33. In triangle ABC, angle ∠B is equal to 60◦ ; bisectors AD and CE intersect at
point O. Prove that OD = OE.
2.34. In triangle ABC the angles at vertices B and C are equal to 40◦ ; let BD be the
bisector of angle B. Prove that BD + DA = BC.
2.35. On chord AB of circle S centered at O a point C is taken. The circumscribed
circle of triangle AOC intersects circle S at point D. Prove that BC = CD.
2.36. Vertices A and B of an equilateral triangle ABC lie on circle S, vertex C lies
inside this circle. Point D lies on circle S and BD = AB. Line CD intersects S at point E.
Prove that the length of segment EC is equal to the radius of circle S.
2.37. Along a fixed circle another circle whose radius is half that of the fixed one rolls
on the inside without gliding. What is the trajectory of a fixed point K of the rolling circle?
§5. Four points on one circle
2.38. From an arbitrary point M on leg BC of right triangle ABC perpendicular M N
is dropped on hypothenuse AP . Prove that ∠M AN = ∠M CN .
2.39. The diagonals of trapezoid ABCD with bases AD and BC intersect at point O;
points B ′ and C ′ are symmetric through the bisector of angle ∠BOC to vertices B and C,
respectively. Prove that ∠C ′ AC = ∠B ′ DB.
2.40. The extensions of sides AB and CD of the inscribed quadrilateral ABCD meet at
point P ; the extensions of sides BC and AD meet at point Q. Prove that the intersection
points of the bisectors of angles ∠AQB and ∠BP C with the sides of the quadrilateral are
vertices of a rhombus.
§6. THE INSCRIBED ANGLE AND SIMILAR TRIANGLES
37
2.41. The inscribed circle of triangle ABC is tangent to sides AB and AC at points M
and N , respectively. Let P be the intersection point of line M N with the bisector (or its
extension) of angle ∠B. Prove that:
a) ∠BP C = 90◦ ;
b) SABP : SABC = 1 : 2.
2.42. Inside quadrilateral ABCD a point M is taken so that ABM D is a parallelogram.
Prove that if ∠CBM = ∠CDM , then ∠ACD = ∠BCM .
2.43. Lines AP , BP and CP intersect the circumscribed circle of triangle ABC at
points A1 , B1 and C1 , respectively. On lines BC, CA and AB points A2 , B2 and C2 ,
respectively, are taken so that ∠(P A2 , BC) = ∠(P B2 , CA) = ∠(P C2 , AB). Prove that
△A2 B2 C2 ∼ △A1 B1 C1 .
2.44. About an equilateral triangle AP Q a rectangular ABCD is circumscribed so that
points P and Q lie on sides BC and CD, respectively; P ′ and Q′ are the midpoints of sides
AP and AQ, respectively. Prove that triangles BQ′ C and CP ′ D are equilateral ones.
2.45. Prove that if for inscribed quadrilateral ABCD the equality CD = AD + BC
holds, then the intersection point of the bisectors of angles ∠A and ∠B lies on side CD.
2.46. Diagonals AC and CE of a regular hexagon ABCDEF are divided by points M
and N , respectively, so that AM : AC = CN : CE = λ. Find λ if it is known that points
B, M and N lie on a line.
2.47. The corresponding sides of triangles ABC and A1 B1 C1 are parallel and sides AB
and A1 B1 lie on one line. Prove that the line that connects the intersection points of the
circumscribed circles of triangles A1 BC and AB1 C contains point C1 .
2.48. In triangle ABC heights AA1 , BB1 and CC1 are drawn. Line KL is parallel to
CC1 ; points K and L lie on lines BC and B1 C1 , respectively. Prove that the center of the
circumscribed circle of triangle A1 KL lies on line AC.
2.49. Through the intersection point O of the bisectors of triangle ABC line M N is
drawn perpendicularly to CO so that M and N lie on sides AC and BC, respectively.
Lines AO and BO intersect the circumscribed circle of triangle ABC at points A′ and B ′ ,
respectively. Prove that the intersection point of lines A′ N and B ′ M lies on the circumscribed
circle.
§6. The inscribed angle and similar triangles
2.50. Points A, B, C and D on a circle are given. Lines AB and CD intersect at point
M . Prove that
BC · BD
AC · AD
=
.
AM
BM
2.51. Points A, B and C on a circle are given; the distance BC is greater than the
distance from point B to line l tangent to the circle at point A. Line AC intersects the line
drawn through point B parallelly to l at point D. Prove that AB 2 = AC · AD.
2.52. Line l is tangent to the circle of diameter AB at point C; points M and N are the
projections of points A and B on line l, respectively, and D is the projection of point C on
AB. Prove that CD2 = AM · BN .
2.53. In triangle ABC, height AH is drawn and from vertices B and C perpendiculars
BB1 and CC1 are dropped on the line that passes through point A. Prove that △ABC ∼
△HB1 C1 .
2.54. On arc ⌣ BC of the circle circumscribed about equilateral triangle ABC, point
P is taken. Segments AP and BC intersect at point Q. Prove that
1
1
1
=
+
.
PQ
PB PC
38
CHAPTER 2. INSCRIBED ANGLES
2.55. On sides BC and CD of square ABCD points E and F are taken so that ∠EAF =
45◦ . Segments AE and AF intersect diagonal BD at points P and Q, respectively. Prove
AEF
that SSAP
= 2.
Q
2.56. A line that passes through vertex C of equilateral triangle ABC intersects base
AB at point M and the circumscribed circle at point N . Prove that
AM · BM
CM
=
.
CM · CN = AC 2 and
CN
AN · BN
2.57. Consider parallelogram ABCD with an acute angle at vertex A. On rays AB and
CB points H and K, respectively, are marked so that CH = BC and AK = AB. Prove
that:
a) DH = DK;
b) △DKH ∼ △ABK.
2.58. a) The legs of an angle with vertex C are tangent to a circle at points A and B.
From point P on the circle perpendiculars P A1 , P B1 and P C1 are dropped on lines BC,
CA and AB, respectively. Prove that P C12 = P A1 · P B1 .
b) From point O of the inscribed circle of triangle ABC perpendiculars OA′ , OB ′ , OC ′
are dropped on the sides of triangle ABC opposite to vertices A, B and C, respectively, and
perpendiculars OA′′ , OB ′′ , OC ′′ are dropped to the sides of the triangle with vertices at the
tangent points. Prove that
OA′ · OB ′ · OC ′ = OA′′ · OB ′′ · OC ′′ .
2.59. Pentagon ABCDE is inscribed in a circle. Distances from point E to lines AB,
BC and CD are equal to a, b and c, respectively. Find the distance from point E to line
AD.
2.60. In triangle ABC, heights AA1 , BB1 and CC1 are drawn; B2 and C2 are the
midpoints of heights BB1 and CC1 , respectively. Prove that △A1 B2 C2 ∼ △ABC.
2.61. On heights of triangle ABC points A1 , B1 and C1 that divide them in the ratio
2 : 1 counting from the vertex are taken. Prove that △A1 B1 C1 ∼ △ABC.
2.62. Circle S1 with diameter AB intersects circle S2 centered at A at points C and
D. Through point B a line is drawn; it intersects S2 at point M that lies inside S1 and it
intersects S1 at point N . Prove that M N 2 = CN · N D.
2.63. Through the midpoint C of an arbitrary chord AB on a circle chords KL and M N
are drawn so that points K and M lie on one side of AB. Segments KN and M L intersect
AB at points Q and P , respectively. Prove that P C = QC.
2.64. a) A circle that passes through point C intersects sides BC and AC of triangle
ABC at points A1 and B1 , respectively, and it intersects the circumscribed circle of triangle
ABC at point M . Prove that △AB1 M ∼ △BA1 M .
b) On rays AC and BC segments AA1 and BB1 equal to the semiperimeter of triangle
ABC are drawn. Let M be a point on the circumscribed circle such that CM k A1 B1 . Prove
that ∠CM O = 90◦ , where O is the center of the inscribed circle.
§7. The bisector divides an arc in halves
2.65. In triangle ABC, sides AC and BC are not equal. Prove that the bisector of angle
∠C divides the angle between the median and the height drawn from this vertex in halves
if and only if ∠C = 90◦ .
2.66. It is known that in a triangle the median, the bisector and the height drawn from
vertex C divide the angle ∠C into four equal parts. Find the angles of this triangle.
2.67. Prove that in triangle ABC bisector AE lies between median AM and height AH.
§9. THREE CIRCUMSCRIBED CIRCLES INTERSECT AT ONE POINT
39
2.68. Given triangle ABC; on its side AB point P is chosen; lines P M and P N parallel
to AC and BC, respectively, are drawn through P so that points M and N lie on sides BC
and AC, respectively; let Q be the intersection point of the circumscribed circles of triangles
AP N and BP M . Prove that all lines P Q pass through a fixed point.
2.69. The continuation of bisector AD of acute triangle ABC inersects the circumscribed
circle at point E. Perpendiculars DP and DQ are dropped on sides AB and AC from point
D. Prove that SABC = SAP EQ .
§8. An inscribed quadrilateral with perpendicular diagonals
In this section ABCD is an inscribed quadrilateral whose diagonals intersect at a right
angle. We will also adopt the following notations: O is the center of the circumscribed circle
of quadrilateral ABCD and P is the intersection point of its diagonals.
2.70. Prove that the broken line AOC divides ABCD into two parts whose areas are
equal.
2.71. The radius of the circumscribed circle of quadrilateral ABCD is equal to R.
a) Find AP 2 + BP 2 + CP 2 + DP 2 .
b) Find the sum of squared lengths of the sides of ABCD.
2.72. Find the sum of squared lengths of the diagonals of ABCD if the length of segment
OP and the radius of the circumscribed circle R are known.
2.73. From vertices A and B perpendiculars to CD that intersect lines BD and AC at
points K and L, respectively, are drawn. Prove that AKLB is a rhombus.
2.74. Prove that the area of quadrilateral ABCD is equal to 12 (AB · CD + BC · AD).
2.75. Prove that the distance from point O to side AB is equal to half the length of side
CD.
2.76. Prove that the line drawn through point P perpendicularly to BC divides side
AD in halves.
2.77. Prove that the midpoints of the sides of quadrilateral ABCD and the projections
of point P on the sides lie on one circle.
2.78. a) Through vertices A, B, C and D tangents to the circumscribed circle are drawn.
Prove that the quadrilateral formed by them is an inscribed one.
b) Quadrilateral KLM N is simultaneously inscribed and circumscribed; A and B are
the tangent points of the inscribed circle with sides KL and LM , respectively. Prove that
AK · BM = r2 , where r is the radius of the inscribed circle.
§9. Three circumscribed circles intersect at one point
2.79. On sides of triangle ABC triangles ABC ′ , AB ′ C and A′ BC are constructed
outwards so that the sum of the angles at vertices A′ , B ′ and C ′ is a multiple of 180◦ . Prove
that the circumscribed circles of the constructed triangles intersect at one point.
2.80. a) On sides (or their extensions) BC, CA and AB of triangle ABC points A1 , B1
and C1 distinct from the vertices of the triangle are taken (one point on one side). Prove
that the circumscribed circles of triangles AB1 C1 , A1 BC1 and A1 B1 C intersect at one point.
b) Points A1 , B1 and C1 move along lines BC, CA and AB, respectively, so that all
triangles A1 B1 C1 are similar and equally oriented. Prove that the intersection point of the
circumscribed circles of triangles AB1 C1 , A1 BC1 and A1 B1 C remains fixed in the process.
2.81. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 are taken.
Prove that if triangles A1 B1 C1 and ABC are similar and have opposite orientations, then
circumscribed circles of triangles AB1 C1 , ABC1 and A1 B1 C pass through the center of the
circumscribed circle of triangle ABC.
40
CHAPTER 2. INSCRIBED ANGLES
2.82. Points A′ , B ′ and C ′ are symmetric to a point P relative sides BC, CA and AB,
respectively, of triangle ABC.
a) The circumscribed circles of triangles AB ′ C ′ , A′ BC ′ , A′ B ′ C and ABC have a common
point;
b) the circumscribed circles of triangles A′ BC, AB ′ C, ABC ′ and A′ B ′ C ′ have a common
point Q;
c) Let I, J, K and O be the centers of the circumscribed circles of triangles A′ BC, AB ′ C,
ABC ′ and A′ B ′ C ′ , respectively. Prove that QI : OI = QJ : OJ = QK : OK.
§10. Michel’s point
2.83. Four lines form four triangles. Prove that
a) The circumscribed circles of these triangles have a common point. (Michel’s point.)
b) The centers of the circumscribed circles of these triangles lie on one circle that passes
through Michel’s point.
2.84. A line intersects sides (or their extensions) AB, BC and CA of triangle ABC at
points C1 , B1 and A1 , respectively; let O, Oa , Ob and Oc be the centers of the circumscribed
circles of triangles ABC, AB1 C1 , A1 BC1 and A1 B1 C, respectively; let H, Ha , Hb and Hc be
the respective orthocenters of these triangles. Prove that
a) △Oa Ob Oc ∼ △ABC.
b) the midperpendiculars to segments OH, Oa Ha , Ob Hb and Oc Hc meet at one point.
2.85. Quadrilateral ABCD is an inscribed one. Prove that Michel’s point of lines that
contain its sides lies on the segment that connects the intersection points of the extensions
of the sides.
2.86. Points A, B, C and D lie on a circle centered at O. Lines AB and CD intersect
at point E and the circumscribed circles of triangles AEC and BED ilntersect at points E
and P . Prove that
a) points A, D, P and O lie on one circle;
b) ∠EP O = 90◦ .
2.87. Given four lines prove that the projections of Michel’s point to these lines lie on
one line.
See also Problem 19.45.
§11. Miscellaneous problems
2.88. In triangle ABC height AH is drawn; let O be the center of the circumscribed
circle. Prove that ∠OAH = |∠B − ∠C|.
2.89. Let H be the intersection point of the heights of triangle ABC; let AA′ be a
diameter of its circumscribed circle. Prove that segment A′ H divides side BC in halves.
2.90. Through vertices A and B of triangle ABC two parallel lines are drawn and lines
m and n are symmetric to them through the bisectors of the corresponding angles. Prove
that the intersection point of lines m and n lies on the circumscribed circle of triangle ABC.
2.91. a) Lines tangent to circle S at points B and C are drawn from point A. Prove
that the center of the inscribed circle of triangle ABC and the center of its escribed circle
tangent to side BC lie on circle S.
b) Prove that the circle that passes through vertices B and C of any triangle ABC and
the center O of its inscribed circle intercepts on lines AB and AC chords of equal length.
2.92. On sides AC and BC of triangle ABC squares ACA1 A2 and BCB1 B2 are constructed outwards. Prove that lines A1 B, A2 B2 and AB1 meet at one point.
SOLUTIONS
41
2.93. Circles S1 and S2 intersect at points A and B so that the tangents to S1 at these
points are radii of S2 . On the inner arc of S1 a point C is taken; straight lines connect it
with points A and B. Prove that the second intersection points of these lines with S2 are
the endpoints of a diameter.
2.94. From the center O of a circle the perpendicular OA is dropped to line l. On l,
points B and C are taken so that AB = AC. Through points B and C two sections are
drawn one of which intersects the circle at points P and Q and the other one at points M
and N . Lines P M and QN intersect line l at points R and S, respectively. Prove that
AR = AS.
Problems for independent study
2.95. In triangle ABC heights AA1 and BB1 are drawn; let M be the midpoint of side
AB. Prove that M A1 = M B1 .
2.96. In convex quadrilateral ABCD angles ∠A and ∠C are right ones. Prove that
AC = BD sin ABC.
2.97. Diagonals AD, BE and CF of an inscribed hexagon ABCDEF meet at one point.
Prove that AB · CD · EF = BC · DE · AF .
2.98. In a convex quadrilateral AB = BC = CD, let M be the intersection point of
diagonals, K is the intersection point of bisectors of angles ∠A and ∠D. Prove that points
A, M , K and D lie on one circle.
2.99. Circles centered at O1 and O2 intersect at points A and B. Line O1 A intersects
the circle centered at O2 at point N . Prove that points O1 , O2 , B and N lie on one circle.
2.100. Circles S1 and S2 intersect at points A and B. Line M N is tangent to circle S1
at point M and to S2 at point N . Let A be the intersection point of the circles, which is
more distant from line M N . Prove that ∠O1 AO2 = 2∠M AN .
2.101. Given quadrilateral ABCD inscribed in a circle and such that AB = BC, prove
that SABCD = 21 (DA + CD) · hb , where hb is the height of triangle ABD dropped from vertex
B.
2.102. Quadrilateral ABCD is an inscribed one and AC is the bisector of angle ∠DAB.
Prove that AC · BD = AD · DC + AB · BC.
2.103. In right triangle ABC, bisector CM and height CH are drawn from the vertex
of the right angle ∠C. Let HD and HE be bisectors of triangles AHC and CHB. Prove
that points C, D, H, E and M lie on one circle.
2.104. Two circles pass through the vertex of an angle and a point on its bisector. Prove
that the segments cut by them on the sides of the angle are equal.
2.105. Triangle BHC, where H is the orthocenter of triangle ABC is complemented to
the parallelogram BHCD. Prove that ∠BAD = ∠CAH.
2.106. Outside equilateral triangle ABC but inside angle ∠BAC, point M is taken so
that ∠CM A = 30◦ and ∠BM A = α. What is the value of angle ∠ABM ?
2.107. Prove that if the inscribed quadrilateral with perpendicular diagonals is also a
circumscribed one, then it is symmetric with respect to one of its diagonals.
Solutions
2.1. Let us draw diameter AD. Then ∠CDA = ∠CBA; hence, ∠BAH = ∠DAC
because ∠BHA = ∠ACD = 90◦ .
2.2. Let us make use of the properties of oriented angles:
∠(AC, CK) = ∠(AM, M K) = ∠(BM, M K) = ∠(BD, DK) = ∠(BD, CK),
42
CHAPTER 2. INSCRIBED ANGLES
i.e., AC k BD.
2.3. Points P and Q lie on the circle with diameter AM . Therefore, ∠QM A = ∠QP A
as angles that intersect the same arc. Triangles P AK and M AQ are right ones, therefore,
∠P AK = ∠M AQ.
2.4. a) Since
∠A + ∠B
∠AOM = ∠BAO + ∠ABO =
2
and
∠A + ∠B
∠A
+ ∠CBM =
,
∠OAM = ∠OAC + ∠CAM =
2
2
we have M A = M O. Similarly, M C = M O.
Since triangle OAOb is a right one and ∠AOM = ∠M AO = ϕ, it follows that ∠M AOb =
∠M Ob A = 90◦ − ϕ and, therefore, M A = M Ob . Similarly, M C = M Ob .
b) Let P be the center of the circumscribed circle of triangle ACO. Then
180◦ − ∠CP O
∠COP =
= 90◦ − ∠OAC.
2
Hence, ∠BOC = 90◦ + ∠OAC. Similarly, ∠BOC = 90◦ + ∠OAB and, therefore, ∠OAB =
∠OAC. We similarly establish that point O lies on the bisectors of angles ∠B and ∠C.
2.5. Points P and C lie on the circle with diameter AB, and, therefore, ∠AP C = ∠ABC,
i.e., the value of angle ∠AP C is a constant.
Remark. A similar statement is true for any triangle ABC whose vertices are moving
along the legs of angle ∠M P N equal to 180◦ − ∠C.
2.6. Points B, D and K lie on the circle with diameter AC. Let, for definiteness sake,
∠KCA = ϕ ≤ 45◦ . Then
AC(cos ϕ − sin ϕ)
√
BK = AC sin(45◦ − ϕ) =
2
and
AC(cos ϕ + sin ϕ)
√
.
DK = AC sin(45◦ + ϕ) =
2
Clearly, AC cos ϕ = CK and AC sin ϕ = AK.
2.7. Since ∠B1 AA1 = ∠A1 BB1 , it follows that points A, B, A1 and B1 lie on one circle.
Parallel lines AB and A1 B1 intercept on it equal chords AB1 and BA1 . Hence, AC = BC.
2.8. On side BC of triangle ABC construct outwards an equilateral triangle A1 BC.
Let P be the intersection point of line AA1 with the circumscribed circle of triangle A1 BC.
Then point P lies inside triangle ABC and
∠AP C = 180◦ − ∠A1 P C = 180◦ − ∠A1 BC = 120◦ .
Similarly, ∠AP B = 120◦ .
2.9. The bases of perpendiculars dropped from point M on the diameters lie on the circle
S with diameter OM (where O is the center of the initial circle). The intersection points
of the given diameters with circle S distinct from O divide the circle into n arcs. Since the
◦
angles 180
intersect all the circles that do not contain point O, the angle measure of each of
n
◦
. Therefore, the angle measure of the arc on which point O lies is
these arcs is equal to 360
n
◦
◦
= 360
. Thus, the bases of the perpendiculars divide the circle
equal to 360◦ − (n − 1) · 360
n
n
S into n equal arcs.
2.10. Clearly,
∠(AA1 , BB1 ) = ∠(AA1 , AB1 ) + ∠(AB1 , BB1 ) = ∠(M A1 , M B1 ) + ∠(AN, BN ).
SOLUTIONS
43
Since M A1 ⊥ BN and M B1 ⊥ AN , it follows that
∠(M A1 , M B1 ) = ∠(BN, AN ) = −∠(AN, BN ).
Therefore, ∠(AA1 , BB1 ) = 0◦ , i.e., AA1 k BB1 .
2.11. Since AB k DE, it follows that ∠ACE = ∠BF D and since BC k EF , it follows
that ∠CAE = ∠BDF . Triangles ACE and BDF have two pairs of equal angles and,
therefore, their third angles are also equal. The equality of these angles implies the equality
of arcs ⌣ AC and ⌣ DF , i.e., chords CD and AF are parallel.
2.12. Let us carry out the proof by induction on n. For the quadrilateral the statement
is obvious; for the hexagon it had been proved in the preceding problem. Assume that the
statement is proved for the 2(n − 1)-gon; let us prove the statement for the 2n-gon. Let
A1 . . . A2n be a 2n-gon in which A1 A2 k An+1 An+2 , . . . , An−1 An k A2n−1 A2n . Let us consider
2(n − 1)-gon A1 A2 . . . An−1 An+1 . . . A2n−1 . By the inductive hypothesis for n odd we have
An−1 An+1 = A2n−1 A1 , and for n even we have An−1 An+1 k A2n−1 A1 .
Let us consider triangles An−1 An An+1 and A2n−1 A2n A1 . Let n be even. Then vectors
{An−1 An } and {A2n−1 A2n }, as well as {An−1 An+1 } and {A2n−1 A1 } are parallel and directed
towards each other; hence, ∠An An−1 An+1 = ∠A1 A2n−1 A2n and An An+1 = A2n A1 as chords
that cut equal arcs, as required.
Let n be odd. Then An−1 An+1 = A2n−1 A1 , i.e., A1 An−1 k An+1 A2n−1 . In hexagon
An−1 An An+1 A2n−1 A2n A1 we have A1 An−1 k An+1 A2n−1 and An−1 An k A2n−1 A2n ; hence,
thanks to the preceding problem An An+1 k A2n A1 , as required.
2.13. Let lines F G, GE and EF pass through points A, B and C, respectively, so that
triangle EF G is an equilateral one, i.e.,
∠(GE, EF ) = ∠(EF, F G) = ∠F G, GE) = ±60◦ .
Then
∠(BE, EC) = ∠(CF, F A) = ∠(AG, GB) = ±60◦ .
Selecting one of the signs we get three circles SE , SF and SG on which points E, F and G
should lie. Any point E of circle SE uniquely determines triangle EF G.
Let O be the center of triangle EF G; let P , R and Q be the intersection points of lines
OE, OF and OG with the corresponding circles SE , SF and SG . Let us prove that P , Q and
R are the centers of equilateral triangles constructed on sides of triangle ABC (outwards for
one family and inwards for the other one), and point O lies on the circumscribed circle of
triangle P QR.
Clearly,
∠(CB, BP ) = ∠(CE, EP ) = ∠(EF, EO) = ∓30◦
and
Hence,
∠(BP, CP ) = ∠(BE, EC) = ∠(GE, EF ) = ±60◦ .
∠(CB, CP ) = ∠(CB, BP ) + ∠(BP, CP ) = ±30◦ .
Therefore, P is the center of an equilateral triangle with side AB.
For points Q and R the proof is similar. Triangle P QR is an equilateral one and its
center coincides with the intersection point of medians of triangle ABC (cf. Problem 1.49
b)). As is not difficult to verify,
∠(P R, RQ) = ∓60◦ = ∠(OE, OG) = ∠(OP, OQ),
i.e., point O lies on the circumscribed circle of triangle P QR.
44
CHAPTER 2. INSCRIBED ANGLES
2.14. Clearly,
2(∠KEC + ∠KDC) = (⌣ M B+ ⌣ AC) + (⌣ M B+ ⌣ BC) = 360◦ ,
since ⌣ M B =⌣ AM .
2.15. Denote the angle measure of the arc intercepted on the circle by the sides of
triangle ABC by α. Denote the angle measure of the arc intercepted by the extensions of
the sides of the triangle on the circle by α′ . Then 21 (α + α′ ) = ∠BAC = 60◦ . But α = α′
because these arcs are symmetric through the line that passes through the center of the circle
parallel to side BC. Hence, α = α′ = 60◦ .
2.16. Since ∠AP B = 12 (⌣ AB+ ⌣ CD) = ∠AOB, point O lies on the circumscribed
circle of triangle AP B.
2.17. Let O be the point where lines A1 C1 and B1 D1 meet; let α, β, γ and δ be angle
measures of arcs AB, BC, CD and DA. Then
α+β+γ+δ
⌣ A1 B+ ⌣ BB1 + ⌣ C1 D+ ⌣ DD1
=
= 90◦ .
2
4
2.18. By summing up the equalities we get
∠A1 OB1 =
⌣ C ′ A+ ⌣ CA′ = 2(180◦ − ∠AP C) = 240◦ − 2∠B
and ⌣ AB ′ + ⌣ BA′ = 240◦ − 2∠C.
Then by subtracting from their sum the equality ⌣ BA′ + ⌣ CA′ = 2∠A we get
⌣ C ′ B ′ =⌣ C ′ A+ ⌣ AB ′ = 480◦ − 2(∠A + ∠B + ∠C) = 120◦ .
Similarly, ⌣ B ′ A′ =⌣ C ′ A′ = 120◦ .
2.19. a) Let us prove, for example, that AA1 ⊥ C1 B1 . Let M be the intersection point
of these segments. Then
∠AM B1 =
⌣ AB1 + ⌣ A1 B+ ⌣ BC1
= ∠ABB1 + ∠A1 AB + ∠BCC1 =
2
∠B + ∠A + ∠C
= 90◦ .
2
b) Let M1 and M2 be the intersection points of segments AA1 with BC and BB1 with AC.
Right triangles AM1 C and BM2 C have a common angle ∠C; hence, ∠B1 BC = ∠A1 AC.
Consequently, ⌣ B1 C =⌣ A1 C and ∠B1 C1 C = ∠A1 C1 C, i.e., CC1 is the bisector of angle
∠A1 C1 B1 .
2.20. Denote the vertices of triangle T1 by A, B and C; denote the midpoints of arcs
⌣ BC, ⌣ CA, ⌣ AB by A1 , B1 , C1 , respectively. Then T2 = A1 B1 C1 . Lines AA1 , BB1 ,
CC1 are the bisectors of triangle T1 ; hence, they meet at one point, O. Let lines AB and
C1 B1 intersect at point K. It suffices to verify that KO k AC. In triangle AB1 O, line B1 C1
is a bisector and height, hence, this triangle is an isosceles one. Therefore, triangle AKO is
also an isosceles one. Lines KO and AC are parallel, since ∠KOA = ∠KAO = ∠OAC.
2.21. Let l be tangent to the first circle at point A. Then ∠(l, AP ) = ∠(AQ, P Q) =
∠(BC, P B), hence, l k BC.
2.22. Since
∠(AB, AD) = ∠(AP, P D) = ∠(AB, BC),
we have BC k AD.
2.23. Let, for definiteness, point E lie on ray BC. Then ∠ABC = ∠EAC and
∠ADE = ∠ABC + ∠BAD = ∠EAC + ∠CAD = ∠DAE.
SOLUTIONS
45
2.24. Let P be the other intersection point of the circles. Then ∠(AB, DB) = ∠(P A, P B)
and ∠(DC, AC) = ∠(P C, P A). By summing these equalities we get
∠(DC, DB) = ∠(P C, P B) = ∠(P C, CA) + ∠(BA, P B).
The latter two angles subtend constant arcs.
2.25. a) Since ∠M AB = ∠BN A, the sum of angles ∠ABN and ∠M AN is equal to the
sum of the angles of triangle ABN .
b) Since ∠BAM = ∠BN A and ∠BAN = ∠BM A, it follows that △AM B ∼ △N AB
and, therefore, AM : N A = M B : AB and AM : N A = AB : N B. By multiplying these
equalities we get the desired statement.
2.26. Point P lies on the circle of radius BC with center B and line DC is tangent to
this circle at point C. Hence, ∠P CD = 21 ∠P BC = 15◦ .
2.27. Let A1 and B1 be intersection points of lines M A and M B, respectively, with
the smaller circle. Since M is the center of homothety of the circles, A1 B1 k AB. Hence,
∠A1 M T = ∠A1 T A = ∠B1 A1 T = ∠B1 M T .
2.28. Let ϕ be the angle between chord AB and the tangent that passes through one of
the chord’s endpoints. Then AB = 2R sin ϕ, where R is the radius of circle S. Moreover,
P M = AM sin ϕ and QM = BM sin ϕ. Hence,
µ
¶
1
2R
AM + BM
1
+
=
.
AM · BM =
PM
QM
sin ϕ
AM · BM
The value AM · BM does not depend on the choice of chord AB.
2.29. Let line AM intersect circle S2 at point D. Then ∠M DC = ∠M CA = ∠M AB;
hence, CD k AB. Further, ∠CAM = ∠M CB = ∠M DB; hence, AC k BD. Therefore,
ABCD is a parallelogram and its diagonal AD divides diagonal BC in halves.
2.30. Let us draw line l1 tangent to S1 at point A1 . Line K1 K2 is tangent to S1 if and
only if ∠(K1 K2 , K1 A1 ) = ∠(K1 A1 , l1 ). It is also clear that
∠(K1 A1 , l1 ) = ∠(A1 B, l1 ) = ∠(A2 B, A1 A2 ).
Similarly, line K1 K2 is tangent to S2 if and only if ∠(K1 K2 , K2 A2 ) = ∠(A1 B, A1 A2 ). It
remains to observe that if ∠(K1 K2 , K1 A1 ) = ∠(A2 B, A1 A2 ), then
∠(K1 K2 , K2 A2 ) = ∠(K1 K2 , A2 B) =
∠(K1 K2 , A1 B) + ∠(A1 B, A1 A2 ) + ∠(A1 A2 , A2 B) = ∠(A1 B, A1 A2 ).
2.31. Equal angles ABC and A1 B1 C1 intersect chords AC and A1 C1 , hence, AC = A1 C1 .
2.32. Let us denote the center of the circle by O. Points P and Q lie on the circle with
diameter OM , i.e., points O, P , Q and M lie on a circle of radius 21 R. Moreover, either
∠P OQ = ∠AOD or ∠P OQ = ∠BOD = 180◦ − ∠AOD, i.e., the length of chord P Q is a
constant.
2.33. Since ∠AOC = 90◦ + 21 ∠B (cf. Problem 5.3), it follows that
3
∠EBD + ∠EOD = 90◦ + ∠B = 180◦
2
and, therefore, quadrilateral BEOD is an inscribed one. Equal angles ∠EBO and ∠OBD
subtend chords EO and OD, hence, EO = OD.
2.34. On the extension of segment BD beyond point D take a point Q such that
∠ACQ = 40◦ . Let P be the intersection point of lines AB and QC. Then ∠BP C = 60◦
and D is the intersection point of the bisectors of angles of triangle BCP . By Problem 2.33
AD = DQ. Moreover, ∠BQC = ∠BCQ = 80◦ . Therefore, BC = BD + DQ = BD + DA.
46
CHAPTER 2. INSCRIBED ANGLES
2.35. It suffices to verify that the exterior angle ACD of triangle BCD is twice greater
than the angle at vertex B. Clearly, ∠ACD = ∠AOD = 2∠ABD.
2.36. Let O be the center of circle S. Point B is the center of the circumscribed circle
of triangle ACD, hence, ∠CDA = 12 ∠ABC = 30◦ and, therefore, ∠EOA = 2∠EDA = 60◦ ,
i.e., triangle EOA is an equilateral one. Moreover, ∠AEC = ∠AED = ∠AOB = 2∠AOC;
hence, point E is the center of the circumscribed circle of triangle AOC. Therefore, EC =
EO.
2.37. Let us consider two positions of the moving circle: at the first moment, when
point K just gets to the fixed circle (the tangent point of the circles at this moment will be
denoted by K1 ) and at some other (second) moment.
Let O be the center of the fixed circle, O1 and O2 be the positions of the center of
the moving circle at the first and the second moments, respectively, K2 be the position of
point K at the second moment. Let A be the tangent point of the circles at the second
moment. Since the moving circle rolls without gliding, the length of arc ⌣ K1 A is equal to
the length of arc ⌣ K2 A. Since the radius of the moving circle is one half of the radius of
the fixed circle, ∠K2 O2 A = 2∠K1 OA. Point O lies on the moving circle, hence, ∠K2 OA =
1
∠K2 O2 A = ∠K1 OA, i.e., points K2 , K1 and O lie on one line.
2
The trajectory of point K is the diameter of the fixed circle.
2.38. Points N and C lie on the circle with diameter AM . Angles ∠M AN and ∠M CN
subtend the same arc and therefore, are equal.
2.39. The symmetry through the bisector of angle ∠BOC sends lines AC and DB into
each other and, therefore, we have to prove that ∠C ′ AB ′ = ∠B ′ DC ′ . Since BO = B ′ O,
CO = C ′ O and AO : DO = CO : BO, it follows that AO · B ′ O = DO · C ′ O, i.e., the
quadrilateral AC ′ B ′ D is an inscribed one and ∠C ′ AB ′ = ∠B ′ DC ′ .
2.40. Denote the intersection points and angles as indicated on Fig. 14.
Figure 14 (Sol. 2.40)
It suffices to verify that x = 90◦ . The angles of quadrilateral BM RN are equal to
180◦ − ϕ, α + ϕ, β + ϕ and x, hence, the equality x = 90◦ is equivalent to the equality
(2α + ϕ) + (2β + ϕ) = 180◦ . It remains to notice that 2α + ϕ = ∠BAD and 2β + ϕ = ∠BCD.
2.41. a) It suffices to prove that if P1 is the point on the bisector (or its extension)
of angle ∠B that serves as the vertex of an angle of 90◦ that subtends segment BC, then
P1 lies on line M N . Points P1 and N lie on the circle with diameter CO, where O is the
intersection point of bisectors, hence,
1
∠(P1 N, N C) = ∠(P1 O, OC) = (180◦ − ∠A) = ∠(M N, N C).
2
SOLUTIONS
47
; hence,
b) Since ∠BP C = 90◦ , it follows that BP = BC · cos ∠B
2
µ
¶
∠B
SABP : SABC = BP · sin
: (BC sin B) = 1 : 2.
2
2.42. Take point N so that BN k M C and N C k BM . Then N A k CD, ∠N CB =
∠CBM = ∠CDM = ∠N AB, i.e., points A, B, N and C lie on one circle. Hence, ∠ACD =
∠N AC = ∠N BC = ∠BCM .
2.43. Points A2 , B2 , C and P lie on one circle, hence,
∠(A2 B2 , B2 P ) = ∠(A2 C, CP ) = ∠(BC, CP ).
Similarly, ∠(B2 P, B2 C2 ) = ∠(AP, AB). Therefore,
∠(A2 B2 , B2 C2 ) = ∠(BC, CP ) + ∠(AP, AB) = ∠(B1 B, B1 C1 ) + ∠(A1 B1 , B1 B)
= ∠(A1 B1 , B1 C1 ).
We similarly verify that all the other angles of triangles A1 B1 C1 and A2 B2 C2 are either equal
or their sum is equal to 180◦ ; therefore, these triangles are similar (cf. Problem 5.42).
2.44. Points Q′ and C lie on the circle with diameter P Q, hence, ∠Q′ CQ = ∠Q′ P Q =
◦
30 . Therefore, ∠BCQ′ = 60◦ . Similarly, ∠CBQ′ = 60◦ and, therefore, triangle BQ′ C is
equilateral one. By similar reasons triangle CP ′ D is an equilateral one.
2.45. Let ∠BAD = 2α and ∠CBA = 2β; for definiteness we will assume that α ≥ β.
On side CD take point E so that DE = DA. Then CE = CD − AD = CB. The angle at
vertex C of an isosceles triangle BCE is equal to 180◦ − 2α; hence, ∠CBE = α. Similarly,
∠DAE = β. The bisector of angle B intersects CD at a point F . Since ∠F BA = β =
∠AED, quadrilateral ABF E is an inscribed one and, therefore, ∠F AE = ∠F BE = α − β.
It follows that ∠F AD = β + (α − β) = α, i.e., AF is the bisector of angle ∠A.
2.46. Since ED = CB, EN = CM and ∠DEC = ∠BCA = 30◦ (Fig. 15), it follows
that △EDN = △CBM . Let ∠M BC = ∠N DE = α, ∠BM C = ∠EN D = β.
Figure 15 (Sol. 2.46)
It is clear that ∠DN C = 180◦ − β. Considering triangle BN C we get ∠BN C = 90◦ − α.
Since α + β = 180◦ − 30◦ = 150◦ , it follows that
∠DN B = ∠DN C + ∠CN B = (180◦ − β) + (90◦ − α) = 270◦ − (α + β) = 120◦ .
Therefore, points B, O, N and D, where O is the center of the hexagon, lie on one circle.
Moreover,√CO = CB = CD, i.e., C is the center of this circle, hence, λ = CN : CE = CB :
CA = 1 : 3.
2.47. Let D be the other intersection point of the circumscribed circles of triangles
A1 BC and AB1 C. Then ∠(AC, CD) = ∠(AB1 , B1 D) and ∠(DC, CB) = ∠(DA1 , A1 B).
48
CHAPTER 2. INSCRIBED ANGLES
Hence,
∠(A1 C1 , C1 B1 ) = ∠(AC, CB) = ∠(AC, CD) + ∠(DC, CB)
= ∠(AB1 , B1 D) + ∠(DA1 , A1 B) = ∠(A1 D, DB1 ),
i.e., points A1 , B1 , C1 and D lie on one circle. Therefore, ∠(A1 C1 , C1 B) = ∠(A1 B1 , B1 D) =
∠(AC, CD). Taking into account that A1 C1 k AC, we get the desired statement.
2.48. Let point M be symmetric to point A1 through line AC. By Problem 1.57 point
M lies on line B1 C1 . Therefore,
∠(LM, M A1 ) = ∠(C1 B1 ) = ∠(C1 C, CB) = ∠(LK, KA1 ),
i.e., point M lies on the circumscribed circle of triangle A1 KL. It follows that the center of
this circle lies on line AC — the midperpendicular to segment A1 M .
2.49. Let P Q be the diameter perpendicular to AB and such that Q and C lie on one
side of AB; let L be the intersection point of line QO with the circumscribed circle; let M ′
and N ′ be the intersection points of lines LB ′ and LA′ with sides AC and BC, respectively.
It suffices to verify that M ′ = M and N ′ = N .
Since ⌣ P A+ ⌣ AB ′ + ⌣ B ′ Q = 180◦ , it follows that ⌣ B ′ Q = ∠A and, therefore, ∠B ′ LQ = ∠M ′ AO. Hence, quadrilateral AM ′ OL is an inscribed one and ∠M ′ OA =
∠M ′ LA = 21 ∠B. Therefore, ∠CM O = 12 (∠A + ∠B), i.e., M ′ = M . Similarly, N ′ = N .
2.50. Since △ADM ∼ △CBM and △ACM ∼ △DBM , it follows that AD : CB =
DM : BM and AC : DB = AM : DM . It remains to multiply these equalities.
2.51. Let D1 be the intersection point of line BD with the circle distinct from point B.
Then ⌣ AB =⌣ AD1 ; hence, ∠ACB = ∠AD1 B = ∠ABD1 . Triangles ACB and ABD
have a common angle, ∠A, and, moreover, ∠ACB = ∠ABD; hence, △ACB ∼ △ABD.
Therefore, AB : AC = AD : AB.
2.52. Let O be the center of the circle. Since ∠M AC = ∠ACO = ∠CAO, it follows
that △AM C = △ADC. Similarly, △CDB = △CN B. Since △ACD ∼ △CDB, it follows
that CD2 = AD · DB = AM · N B.
2.53. Points B1 and H lie on the circle with diameter AB, hence,
∠(AB, BC) = ∠(AB, BH) = ∠(AB1 , B1 H) = ∠(B1 C1 , B1 H).
Similarly, ∠(AC, BC) = ∠(B1 C1 , C1 H).
2.54. On an extension of segment BP beyond point P take point D such that P D = CP .
Then triangle CDP is an equilateral one and CD k QP . Therefore, BP : P Q = BD : DC =
1
1
(BP + CP ) : CP , i.e., P1Q = CP
+ BP
.
2.55. Segment QE subtends angles of 45◦ with vertices at points A and B, hence,
quadrilateral ABEQ is an inscribed one. Since ∠ABE = 90◦ , it √
follows that ∠AQE =
90◦ .
√
AF
AE
= 2. Similarly, AP
= 2.
Therefore, triangle AQE is an isosceles right triangle and AQ
2.56. Since ∠AN C = ∠ABC = ∠CAB, it follows that △CAM ∼ △CN A and, therefore, CA : CM = CN : CA, i.e., CM · CN = AC 2 and AM : N A = CM : CA. Similarly,
BM : N B = CM : CB. Therefore,
AM · BM
CM 2
CM 2
CM
=
=
=
.
2
AN · BN
CA
CM · CN
CN
2.57. Since AK = AB = CD, AD = BC = CH and ∠KAD = ∠DCH, it follows that
△ADK = △CHD and DK = DH. Let us show that points A, K, H, C and D lie on one
circle. Let us circumscribe the circle about triangle ADC. Draw chord CK1 in this circle
parallel to AD and chord AH1 parallel to DC. Then K1 A = DC and H1 C = AD. Hence,
K1 = K and H1 = H, i.e., the constructed circle passes through points K and H and angles
SOLUTIONS
49
∠KAH and ∠KDH are equal because they subtend the same arc. Moreover, as we have
already proved, KDH is an isosceles triangle.
2.58. a) ∠P BA1 = ∠P AC1 and ∠P BC1 = ∠P AB1 and, therefore, right triangles P BA1
and P AC1 , P AB1 and P BC1 are similar, i.e., P A1 : P B = P C1 : P A and P B1 : P A = P C1 :
P B. By multiplying these equalities we get P A1 · P B1 = P C12 .
b) According to heading a)
√
√
√
OA′′ = OB ′ · OC ′ , OB ′′ = OA′ · OC ′ , OC ′′ = OA′ · OB ′ .
By multiplying these equalities we get the desired statement.
2.59. Let K, L, M and N be the bases of perpendiculars dropped from point E to lines
AB, BC, CD and DA, respectively. Points K and N lie on the circle with diameter AE,
hence, ∠(EK, KN ) = ∠(EA, AN ). Similarly, ∠(EL, LM ) = ∠(EC, CM ) = ∠(EA, AN )
and, therefore, ∠(EK, KN ) = ∠(EL, LM ). Similarly, ∠(EN, N K) = ∠(EM, M L) and
∠(KE, EN ) = ∠(LE, EM ). It follows that △EKN ∼ △ELM and, therefore, EK : EN =
EL : EM , i.e., EN = EK·EM
= acb .
EL
2.60. Let H be the intersection point of heights, M the midpoint of side BC. Points
A1 , B2 and C2 lie on the circle with diameter M H, hence, ∠(B2 A1 , A1 C2 ) = ∠(B2 M, M C2 ) =
∠(AC, AB). Moreover, ∠(A1 B2 , B2 C2 ) = ∠(A1 H, HC2 ) = ∠(BC, AB) and ∠(A1 C2 , C2 B2 ) =
∠(BC, AC).
2.61. Let M be the intersection point of medians, H the intersection point of heights
of triangle ABC. Points A1 , B1 and C1 are the projections of point M on the heights
and, therefore, these points lie on the circle with diameter M H. Hence, ∠(A1 B1 , B1 C1 ) =
∠(AH, HC) = ∠(BC, AB). By writing similar equalities for the other angles we get the
desired statement.
2.62. Let lines BM and DN meet S2 at points L and C1 , respectively. Let us prove that
lines DC1 and CN are symmetric through line AN . Since BN ⊥ N A, it suffices to verify
that ∠CN B = ∠BN D. But arcs ⌣ CB and ⌣ BD are equal. Arcs ⌣ C1 M and ⌣ CL
are symmetric through line AN , hence, they are equal and, therefore, ∠M DC1 = ∠CM L.
Besides, ∠CN M = ∠M N D. Thus, △M CN ∼ △DM N , i.e., CN : M N = M N : DN .
2.63. Let us drop from point Q perpendiculars QK1 and QN1 to KL and N M , respectively, and from point P perpendiculars P M1 and P L1 to N M and KL, respectively. Clearly,
QC
QK1
QN1
QC 2
1 ·QN1
=
=
,
i.e.,
= QK
. Since ∠KN C = ∠M LC and ∠N KC = ∠LM C, it
PC
P L1
P M1
P C2
P L1 ·P M1
follows that QN1 : P L1 = QN : P L and QK1 : P M1 = QK : P M . Therefore,
QK · QN
AQ · QB
(AC − QC) · (AC + QC)
AC 2 − QC 2
QC 2
=
=
=
=
.
P C2
PL · PM
P B · AP
(AC − P C) · (AC + P C)
AC 2 − P C 2
This implies that QC = P C.
2.64. a) Since ∠CAM = ∠CBM and ∠CB1 M = ∠CA1 M , it follows that ∠B1 AM =
∠A1 BM and ∠AB1 M = ∠BA1 M .
b) Let M1 be a point of the circle S with diameter CO such that CM1 k A1 B1 ; let M2
be an intersection point of circle S with the circumscribed circle of triangle ABC; let A2
and B2 be the tangent points of of the inscribed circle with sides BC and AC, respectively.
It suffices to verify that M1 = M2 . By Problem a) △AB2 M2 ∼ △BA2 M2 , hence, B2 M2 :
A2 M2 = AB2 : BA2 . Since CA1 = p − b − BA2 and CB1 = AB2 , it follows that
sin B2 CM1
sin CA1 B1
CB1
AB2
B2 M1
=
=
=
=
.
A2 M1
sin A2 CM1
sin CB1 A1
CA1
BA2
On arc ⌣ A2 CB2 of circle S, there exists a unique point X for which B2 X : A2 X = k
(Problem 7.14), hence, M1 = M2 .
50
CHAPTER 2. INSCRIBED ANGLES
2.65. Let O be the center of the circumscribed circle of the triangle, M the midpoint
of side AB, H the base of height CH, D the midpoint of the arc on which point C does
not lie and with endpoints A and B. Since OD k CH, it follows that ∠DCH = ∠M DC.
The bisector divides the angle between the median and the height in halves if and only if
∠M CD = ∠DCH = ∠M DC = ∠ODC = ∠OCD, i.e., M = O and AB is the diameter of
the circle.
2.66. Let α = ∠A < ∠B. By the preceding problem ∠C = 90◦ . Median CM divides
triangle ABC into two isosceles triangles. Since ∠ACM = ∠A = α, ∠M CB = 3α, it follows
that α + 3α = 90◦ , i.e., α = 22.5◦ . Therefore, ∠A = 22.5◦ , ∠B = 67.5◦ , ∠C = 90◦ .
2.67. Let D be a point at which line AE intersects the circumscribed circle. Point D is
the midpoint of arc ⌣ BC. Therefore, M D k AH, moreover, points A and D lie on different
sides of line M H. It follows that point E lies on segment M H.
2.68. Clearly,
∠(AQ, QP ) = ∠(AN, N P ) = ∠(P M, M B) = ∠(QP, QB).
Therefore, point Q lies on the circle such that segment AB subtends an angle of 2∠(AC, CB)
with vertex at Q and line QP divides arc ⌣ AB of this circle in halves.
2.69. Points P and Q lie on the circle with diameter AD; this circle intersects side BC
at point F . (Observe that F does not coincide with D if AB 6= AC.) Clearly,
∠(F C, CE) = ∠(BA, AE) = ∠(DA, AQ) = ∠(DF, F Q), i.e., EC k F Q.
Similarly, BE k F P . To complete the proof it suffices to notice that the areas of triangles
adjacent to the lateral sides of the trapezoid are equal.
2.70. Let ∠AOB = α and ∠COD = β. Then α2 + β2 = ∠ADP + ∠P AD = 90◦ . Since
2SAOB = R2 sin α and 2SCOD = R2 sin β, where R is the radius of the circumscribed circle,
it follows that SAOB = SCOD . Similarly, SBOC = SAOD .
2.71. Let ∠AOB = 2α and ∠COD = 2β. Then α + β = ∠ADP + ∠P AD = 90◦ . Hence,
(AP 2 + BP 2 ) + (CP 2 + DP 2 ) = AB 2 + CD2 = 4R2 (sin2 α + cos2 α) = 4R2 .
Similarly, BC 2 + AD2 = 4R2 .
2.72. Let M be the midpoint of AC, N the midpoint of BD. We have AM 2 = AO2 −
OM 2 and BN 2 = BO2 − ON 2 ; hence,
AC 2 + BD2 = 4(R2 − OM 2 ) + 4(R2 − ON 2 ) = 8R2 − 4(OM 2 + ON 2 ) = 8R2 − 4OP 2
since OM 2 + ON 2 = OP 2 .
2.73. The correspondiong legs of acute angles ∠BLP and ∠BDC are perpendicular,
hence, the angles are equal.
Therefore, ∠BLP = ∠BDC = ∠BAP . Moreover, AK k BL and AL ⊥ BK. It follows
that AKLB is a rhombus.
2.74. In the circumscribed circle take a point D′ so that DD′ k AC. Since DD′ ⊥ BD,
it follows that BD′ is a diameter and, therefore, ∠D′ AB = ∠D′ CB = 90◦ . Hence,
1
1
SABCD = SABCD = (AD′ · AB + BC · CD′ ) = (AB · CD + BC · AD).
2
2
2.75. Let us draw diameter AE. Since ∠BEA = ∠BCP and ∠ABE = ∠BP C = 90◦ ,
it follows that ∠EAB = ∠CBP . The angles that intersect chords EB and CD are equal,
hence, EB = CD. Since ∠EBA = 90◦ , the distance from point O to AB is equal to 12 EB.
2.76. Let the perpendicular dropped from point P to BC intersect BC at point H and
AD at point M (Fig. 16).
SOLUTIONS
51
Figure 16 (Sol. 2.76)
Therefore, ∠BDA = ∠BCA = ∠BP H = ∠M P D. Since angles M DP and M P D are
equal, M P is a median of right triangle AP D. Indeed,
∠AP M = 90◦ − ∠M P D = 90◦ − ∠M DP = ∠P AM,
i.e., AM = P M = M D.
2.77. The midpoints of the sides of quadrilateral ABCD are vertices of a rectangle (cf.
Problem 1.2), hence, they lie on one circle. Let K and L be the midpoints of sides AB and
CD, let M be the intersection point of lines KP and CD. By Problem 2.76 P M ⊥ CD;
hence, M is the projection of point P on side CD and point M lies on the circle with
diameter KL.
For the other projections the proof is similar.
2.78. a) It is worth to observe that since points A, B, C and D divide the circle into arcs
smaller than 180◦ each, then the quadrilateral constructed contains this circle. The angle
ϕ between the tangents drawn through points A and B is equal to 180◦ − ∠AOB and the
angle ψ between the tangents drawn through points C and D is equal to 180◦ − ∠COD.
Since ∠AOB + ∠COD = 180◦ , it follows that ϕ + ψ = 180◦ .
Remark. Conversely, the equality ϕ + ψ = 180◦ implies that ∠AOB + ∠COD = 180◦ ,
i.e., AC ⊥ BD.
b) Let O be the center of the inscribed circle. Since ∠AKO + ∠BM O = 90◦ , it follows
that ∠AKO = ∠BOM and △AKO ∼ △BOM . Therefore, AK · BM = BO · AO = r2 .
2.79. First, let us suppose that the circumscribed circles of triangles A′ BC and AB ′ C
are not tangent to each other and P is their common point distinct from C. Then
∠(P A, P B) = ∠(P A, P C) + ∠(P C, P B)
= ∠(B ′ A, B ′ C) + ∠(A′ C, A′ B) = ∠(C ′ A, C ′ B),
i.e., point P lies on the circumscribed circle of triangle ABC ′ .
If the the circumscribed circles of triangles A′ BC and AB ′ C are tangent to each other,
i.e., P = C, then our arguments require an insignificant modifications: instead of line P C
we have to take the common tangent.
2.80. a) By applying the statement of Problem 2.79 to triangles AB1 C1 , A1 BC1 and
A1 B1 C constructed on the sides of triangle A1 B1 C1 we get the desired statement.
b) Let P be the intersection point of the indicated circles. Let us prove that the value
of the angle ∠(AP, P C) is a constant. Since
∠(AP, P C) = ∠(AP, AB) + ∠(AB, BC) + ∠(BC, P C)
52
CHAPTER 2. INSCRIBED ANGLES
and angle ∠(AB, BC) is a constant, it remains to verify that the sum ∠(AP, AB)+∠(BC, P C)
is a constant. Clearly,
∠(AP, AB) + ∠(BC, CP )
= ∠(AP, AC1 ) + ∠(CA1 , CP ) =
∠(B1 P, B1 C1 ) + ∠(B1 A1 , B1 P ) =
∠(B1 A1 , B1 C1 )
and the value of the latter angle is constant by hypothesis.
We similarly prove that the values of angles ∠(AP, P B) and ∠(BP, P C) are constants.
Hence, point P remains fixed.
2.81. As follows from Problem 2.80 b) it suffices to carry out the proof for one such
triangle A1 B1 C1 only; for instance, for the triangle with vertices in the midpoints of sides
of triangle ABC. Let H be the intersection point of heights of triangle A1 B1 C1 , i.e., the
center of the circumscribed circle of triangle ABC. Since A1 H ⊥ B1 C1 and B1 H ⊥ A1 C1 ,
it follows that ∠(A1 H, HB1 ) = ∠(B1 C1 , A1 C1 ) = ∠(A1 C, CB1 ), i.e., point H lies on the
circumscribed circle of triangle A1 B1 C.
A similar argument shows that point H lies on the circumscribed circles of triangles
A1 BC1 and AB1 C1 .
2.82. a) Let X be the intersection point of the circumscribed circles of triangles ABC
and AB ′ C ′ . Then
∠(XB ′ , XC) = ∠(XB ′ , XA) + ∠(XA, XC) = ∠(C ′ B ′ , C ′ A) + ∠(BA, BC).
Since AC ′ = AP = AB ′ , triangle C ′ AB ′ is an isosceles one and ∠C ′ AB ′ = 2∠A; hence,
∠(C ′ B ′ , C ′ A) = ∠A − 90◦ . Therefore,
∠(XB ′ , XC) = ∠A − 90◦ + ∠B = 90◦ − ∠C = ∠(A′ B ′ , A′ C),
i.e., point X lies on the circumscribed circle of triangle A′ B ′ C. For the circumscribed circle
of triangle A′ BC ′ the proof is similar.
b) Let X be the intersection point of the circumscribed circles of triangles A′ B ′ C ′ and
′
A BC. Let us prove that X lies on the circumscribed circle of triangle ABC ′ . Clearly,
∠(XB, XC ′ ) = ∠(XB, XA′ ) + ∠(XA′ , XC ′ ) = ∠(CB, CA′ ) + ∠(B ′ A′ , B ′ C ′ ).
Let A1 , B1 and C1 be the midpoints of segments P A′ , P B ′ and P C ′ . Then
∠(CB, CA′ ) = ∠(CP, CA1 ) = ∠(B1 P, B1 A1 ), ∠(B ′ A′ , B ′ C ′ ) = ∠(B1 A1 , B1 C1 )
and
∠(AB, AC ′ ) = ∠(AP, AC1 ) = ∠(B1 P, B1 C1 ).
It follows that ∠(XB, XC ′ ) = ∠(AB, AC ′ ).
We similarly prove that point X lies on the circumscribed circle of triangle AB ′ C.
c) Since QA′ is the common chord of circles centered at O and I, it follows that QA′ ⊥ OI.
Similarly, QB ′ ⊥ OJ and QC ⊥ IJ. Therefore, sides of angles OJI and B ′ QC, as well as
sides of angles OIJ and A′ QC, are mutually perpendicular, hence, sin OJI = sin B ′ QC and
sin OIJ = sin A′ QC. Therefore, OI : OJ = sin OJI : sin OIJ = sin B ′ QC : sin A′ QC. It is
also clear that
sin( 21 QJC)
QI
sin QB ′ C
sin QJI
=
=
=
.
QJ
sin QIJ
sin QA′ C
sin( 21 QIC)
Taking into account that sin B ′ QC : sin QB ′ C = B ′ C : QC and sin A′ QC : sin QA′ C =
A′ C : QC we get
B ′ C A′ C
OI QI
:
=
:
= 1.
OJ QJ
QC QC
SOLUTIONS
53
2.83. a) The conditions of the problem imply that no three lines meet at one point. Let
lines AB, AC and BC intersect the fourth line at points D, E, and F , respectively (Fig.
17).
Figure 17 (Sol. 2.83)
Denote by P the intersection point of circumscribed circles of triangles ABC and CEF
distinct from point C. Let us prove that point P belongs to the circumscribed circle of
triangle BDF . For this it suffices to verify that ∠(BP, P F ) = ∠(BD, DF ). Clearly,
∠(BP, P F ) = ∠(BP, P C) + ∠(P C, P F ) = ∠(BA, AC) + ∠(EC, EF )
= ∠(BD, AC) + ∠(AC, DF ) = ∠(BD, DF ).
We similarly prove that point P belongs to the circumscribed circle of triangle ADE.
b) Let us make use of notations of Fig. 17. Thanks to heading a), the circumscribed
circles of triangles ABC, ADE and BDF pass through point P and, therefore, we can
consider them as the circumscribed circles of triangles ABP , ADP and BDP respectively.
Therefore, their centers lie on a circle that passes through point P (cf. Problem 5.86).
We similarly prove that the centers of any of the three of given circles lie on a circle that
passes through point P . It follows that all the four centers lie on a circle that passes through
point P .
2.84. a) Let P be Michel’s point for lines AB, BC, CA and A1 B1 . The angles between
rays P A, P B, P C and the tangents to circles Sa , Sb , Sc are equal to ∠(P B1 , B1 A) =
∠(P C1 , C1 A), ∠(P C1 , C1 B) = ∠(P A1 , A1 B), ∠(P A1 , A1 C) = ∠(P B1 , B1 C), respectively.
Since ∠(P C1 , C1 A) = ∠(P C1 , C1 B) = ∠(P A1 , A1 C) = ϕ, it follows that after a rotation
through an angle of ϕ about point P lines P A, P B and P C turn into the tangents to circles
Sa , Sb and Sc , respectively, and, therefore, after a rotation through an angle of 90◦ − ϕ these
lines turn into lines P Oa , P Ob and P Oc respectively. Moreover,
P Oa
P Ob
P Oc
1
=
=
= sin ϕ.
PA
PB
PC
2
Therefore, the composition of the rotation through an angle of 90◦ − ϕ and the homothety
(see ???) with center P and coefficient 12 sin ϕ sends triangle ABC to Oa Ob Oc .
b) The transformation considered in the solution of heading a) sends the center O of the
circumscribed circle of triangle ABC into the center O′ of the circumscribed circle of triangle
Oa Ob Oc and the orthocenter H of triangle ABC to orthocenter H ′ of triangle Oa Ob Oc . Let
OH
us complement triangle OO′ H ′ to parallelogram OO′ H ′ M . Since OM
= OOH
′ H ′ = 2 sin ϕ and
′ ′
◦
∠HOM = ∠(HO, O H ) = 90 − ϕ, it follows that M H = M O, i.e., point M lies on the
midperpendicular of segment OH. It remains to notice that for the inscribed quadrilateral
OOa Ob Oc point M is uniquely determined: taking instead of point O any of the points Oa ,
Ob or Oc we get the same point M (cf. Problem 13.33).
2.85. We may assume that rays AB and DC meet at point E and rays BC and AD
meet at point F . Let P be the intersection point of circumscribed circles of triangles BCE
54
CHAPTER 2. INSCRIBED ANGLES
and CDF . Then ∠CP E = ∠ABC and ∠CP F = ∠ADC. Hence, ∠CP E + ∠CP F = 180◦ ,
i.e., point P lies on segment EF .
2.86. a) Since
∠(AP, P D) = ∠(AP, P E) + ∠(P E, P D) =
∠(AC, CD) + ∠(AB, BD) + ∠(AO, OD),
points A, P , D and O lie on one circle.
b) Clearly,
∠(EP, P O) = ∠(EP, P A) + ∠(P A, P O) = ∠(DC, CA) + ∠(DA, DO) = 90◦ ,
because the arcs intersected by these angles constitute a half of the circle.
2.87. Let us make use of notations of Fig. 17. The projections of point P on lines CA
and CB coincide with its projection to CE and CF , respectively. Therefore, Simson’s lines
of point P relative triangles ABC and CEF coincide (cf. Problem 5.85 a).
2.88. Let point A′ be symmetric to point A through the midperpendicular to segment
BC. Then ∠OAH = 21 ∠AOA′ = ∠ABA′ = |∠B − ∠C|.
2.89. Since AA′ is a diameter, A′ C ⊥ AC; hence, BH k A′ C. Similarly, CH k A′ B.
Therefore, BA′ CH is a parallelogram.
2.90. Let l be a line parallel to the two given lines, D the intersection point of lines m
and n. Then
∠(AD, DB) = ∠(m, AB) + ∠(AB, n) = ∠(AC, l) + ∠(l, CB) = ∠(AC, CB)
and, therefore, point D lies on the circumscribed circle of triangle ABC.
2.91. a) Let O be the midpoint of the arc of circle S that lies inside triangle ABC.
Then ∠CBO = ∠BCO and due to a property of the angle between a tangent and a chord,
∠BCO = ∠ABO. Therefore, BO is the bisector of angle ABC, i.e., O is the center of the
inscribed circle of triangle ABC. We similarly prove that the midpoint of the arc of circle S
that lies outside triangle ABC is the center of its escribed circle.
b) We have to prove that the center of the considered circle S lies on the bisector of angle
BAC. Let D be the intersection point of the bisector of the angle with the circumscribed
circle of triangle ABC. Then DB = DO = DC (cf. Problem 2.4 a), i.e., D is the center of
circle S.
2.92. If angle ∠C is a right one, then the solution of the problem is obvious: C is the
intersection point of lines A1 B, A2 B2 , AB1 . If ∠C 6= 90◦ , then the circumscribed circles of
squares ACA1 A2 and BCB1 B2 have in addition to C one more common point, C1 . Then
∠(AC1 , A2 C1 )
= ∠(A2 C1 , A1 C1 ) = ∠(A1 C1 , C1 C) = ∠(C1 C, C1 B1 )
= ∠(C1 B1 , C1 B2 ) =
∠(C1 B2 , C1 B) = 45◦
(or −45◦ ; it is only important that all the angles are of the same sign). Hence, ∠(AC, C1 B1 ) =
4 · 45◦ = 180◦ , i.e., line AB1 passes through point C1 .
Similarly, A2 B2 and A1 B pass through point C1 .
2.93. Let P and O be the centers of circles S1 and S2 , respectively; let α = ∠AP C,
β = ∠BP C; lines AC and BC intersect S2 at points K and L, respectively. Since ∠OAP =
∠OBP = 90◦ , it follows that ∠AOB = 180◦ − α − β. Furthermore,
∠LOB = 180◦ − 2∠LBO = 2∠CBP = 180◦ − β.
Similarly, ∠KOA = 180◦ − α. Therefore,
∠LOK = ∠LOB + ∠KOA − ∠AOB = 180◦ ,
SOLUTIONS
55
i.e., KL is a diameter.
2.94. Let us consider points M ′ , P ′ , Q′ and R′ symmetric to points M , P , Q and R,
respectively, through line OA. Since point C is symmetric to point B through OA, it follows
that line P ′ Q′ passes through point C. The following equalities are easy to verify:
∠(CS, N S) = ∠(Q′ Q, N Q) = ∠(Q′ P, N P ′ ) = ∠(CP ′ , N P ′ );
∠(CR′ , P ′ R′ ) = ∠(M M ′ , P ′ M ′ ) = ∠(M N, P ′ N ) = ∠(CN, P ′ N ).
From these equalities we deduce that points C, N , P ′ , S and R′ lie on one circle. But points
S, R′ and C lie on one line, therefore, S = R′ .
CHAPTER 3. CIRCLES
Background
1. A line that has exactly one common point with a circle is called a line tangent to
the circle. Through any point A outside the circle exactly two tangents to the circle can be
drawn.
Let B and C be the tangent points and O the center of the circle. Then:
a) AB = AC;
b) ∠BAO = ∠CAO;
c) OB ⊥ AB.
(Sometimes the word “tangent” is applied not to the whole line AB but to the segment
AB. Then property a), for example, is formulated as: the tangents to one circle drawn from
one point are equal.)
2. Let lines l1 and l2 that pass through point A intersect a circle at points B1 , C1 and
B2 , C2 , respectively. Then AB1 · AC1 = AB2 · AC2 . Indeed, △AB1 C2 ∼ △AB2 C1 in three
angles. (We advise the reader to prove this making use of the properties of the inscribed
angles and considering two cases: A lies outside the circle and A lies inside the circle.)
If line l2 is tangent to the circle, i.e., B2 = C2 , then AB1 · AC1 = AB22 . The proof runs
along the same lines as in the preceding case except that now we have to make use of the
properties of the angle between a tangent and a chord.
3. The line that connects the centers of tangent circles passes through their tangent
point.
4. The value of the angle between two intersecting circles is the value of the angle between
the tangents to these circles drawn through the intersection point. It does not matter which
of the two of intersection points we choose: the corresponding angles are equal.
The angle between tangent circles is equal to 0◦ .
5. In solutions of problems from §6 a property that has no direct relation to circles is
used: the heights of a triangle meet at one point. The reader can find the proof of this fact
in solutions of Problems 5.45 and 7.41 or can take it for granted for the time being.
6. It was already in the middle of the V century A.D. that Hyppocratus from island
Chios (do not confuse him with the famous doctor Hyppocratus from island Kos who lived
somewhat later) and Pythagoreans began to solve the quadrature of the circle problem. It is
formulated as follows: with the help of a ruler and compass construct a square of the same
area as the given circle.
In 1882 the German mathematician Lindemann proved that number π is transcendental,
i.e., is not a root of a polynomial with integer coefficients. This implies, in particular, that
the problem on the quadrature of the circle is impossible to solve as stated (using other tools
one can certainly solve it).
It seems that it was the problem on Hyppocratus’ crescents (Problem 3.38) that induced
in many a person great expectations to the possibility of squaring the circle: the area of the
figure formed by arcs of circles is equal to the area of a triangle. Prove this statement and
try to understand why such expectations were not grounded in this case.
57
58
CHAPTER 3. CIRCLES
Introductory problems
1. Prove that from a point A outside a circle it is possible to draw exactly two tangents
to the circle and the lengths of these tangents (more exactly, the lengths from A to the
tangent points) are equal.
2. Two circles intersect at points A and B. Point X lies on line AB but not on segment
AB. Prove that the lengths of all the tangents drawn from point X to the circles are equal.
3. Two circles whose radii are R and r are tangent from the outside (i.e., none of them
lies inside the other one). Find the length of the common tangent to these circles.
4. Let a and b be the lengths of the legs of a right triangle, c the length of its hypothenuse.
Prove that:
a) the radius of the inscribed circle of this triangle is equal to 12 (a + b − c);
b) the radius of the circle tangent to the hypothenuse and the extensions of the legs is
equal to 12 (a + b + c).
§1. The tangents to circles
3.1. Lines P A and P B are tangent to a circle centered at O; let A and B be the tangent
points. A third tangent to the circle is drawn; it intersects with segments P A and P B at
points X and Y , respectively. Prove that the value of angle XOY does not depend on the
choice of the third tangent.
3.2. The inscribed circle of triangle ABC is tangent to side BC at point K and an
escribed circle is tangent at point L. Prove that CK = BL = 21 (a + b − c), where a, b, c are
the lengths of the triangle’s sides.
3.3. On the base AB of an isosceles triangle ABC a point E is taken and circles tangent
to segment CE at points M and N are inscribed into triangles ACE and ECB, respectively.
Find the length of segment M N if the lengths of segments AE and BE are known.
3.4. Quadrilateral ABCD is such that there exists a circle inscribed into angle ∠BAD
and tangent to the extensions of sides BC and CD. Prove that AB + BC = AD + DC.
3.5. The common inner tangent to circles whose radii are R and r intersects their
common outer tangents at points A and B and is tangent to one of the circles at point C.
Prove that AC · CB = Rr.
3.6. Common outer tangents AB and CD are drawn to two circles of distinct radii.
Prove that quadrilateral ABCD is a circumscribed one if and only if the circles are tangent
to each other.
3.7. Consider parallelogram ABCD such that the escribed circle of triangle ABD is
tangent to the extensions of sides AD and AB at points M and N , respectively. Prove
that the intersection points of segment M N with BC and CD lie on the inscribed circle of
triangle BCD.
Figure 18 (3.7)
§3. TANGENT CIRCLES
59
3.8. On each side of quadrilateral ABCD two points are taken; these points are connected as shown on Fig. 18. Prove that if all the five dashed quadrilaterals are circumscribed
ones, then the quadrilateral ABCD is also a circumscribed one.
§2. The product of the lengths of a chord’s segments
3.9. Through a point P lying on the common chord AB of two intersecting circles chord
KM of the first circle and chord LN of the second circle are drawn. Prove that quadrilateral
KLM N is an inscribed one.
3.10. Two circles intersect at points A and B; let M N be their common tangent. Prove
that line AB divides M N in halves.
3.11. Line OA is tangent to a circle at point A and chord BC is parallel to OA. Lines
OB and OC intersect the circle for the second time at points K and L, respectively. Prove
that line KL divides segment OA in halves.
3.12. In parallelogram ABCD, diagonal AC is longer than diagonal BD; let M be a
point on diagonal AC such that quadrilateral BCDM is an inscribed one. Prove that line
BD is a common tangent to the circumscribed circles of triangles ABM and ADM .
3.13. Given circle S and points A and B outside it. For each line l that passes through
point A and intersects circle S at points M and N consider the circumscribed circle of
triangle BM N . Prove that all these circles have a common point distinct from point B.
3.14. Given circle S, points A and B on it and point C on chord AB. For every circle
′
S tangent to chord AB at point C and intersecting circle S at points P and Q consider
the intersection point M of lines AB and P Q. Prove that the position of point M does not
depend on the choice of circle S ′ .
§3. Tangent circles
3.15. Two circles are tangent at point A. A common (outer) tangent line is drawn to
them; it is tangent to the circles at points C and D, respectively. Prove that ∠CAD = 90◦ .
3.16. Two circles S1 and S2 centered at O1 and O2 are tangent to each other at point A.
A line that intersects S1 at point A1 and S2 at point A2 is drawn through point A. Prove
that O1 A1 k O2 A2 .
3.17. Three circles S1 , S2 and S3 are pairwise tangent to each other at three distinct
points. Prove that the lines that connect the tangent point of circles S1 and S2 with the
other two tangent points intersect circle S3 at points that are the endpoints of its diameter.
3.18. Two tangent circles centered at O1 and O2 , respectively, are tangent from the
inside to the circle of radius R centered at O. Find the perimeter of triangle OO1 O2 .
3.19. Circles S1 and S2 are tangent to circle S from the inside at points A and B so
that one of the intersection points of circles S1 and S2 lies on segment AB. Prove that the
sum of the radii of circles S1 and S2 is equal to the radius of circle S.
3.20. The radii of circles S1 and S2 tangent at point A are equal to R and r (R > r).
Find the length of the tangent drawn to circle S2 from point B on circle S1 if AB = a
(consider the cases of the inner and outer tangent).
3.21. A point C is taken on segment AB. A line that passes through point C intersects
circles with diameters AC and BC at points K and L and the circle with diameter AB at
points M and N , respectively. Prove that KM = LN .
3.22. Given four circles S1 , S2 , S3 and S4 such that Si and Si+1 are tangent from the
outside for i = 1, 2, 3, 4 (S5 = S1 ). Prove that the tangent points are the vertices of an
inscribed quadrilateral.
60
CHAPTER 3. CIRCLES
3.23. a) Three circles centered at A, B and C are tangent to each other and line l; they
are placed as shown on Fig. 19. Let a, b and c be radii of circles centered at A, B and C,
respectively. Prove that √1c = √1a + √1b .
Figure 19 (3.23)
b) Four circles are pairwise tangent from the outside (at 6 distinct points). Let a, b, c
and d be their radii; α = a1 , β = 1b , γ = 1c and δ = d1 . Prove that
2(α2 + β 2 + γ 2 + δ 2 ) = (α + β + γ + δ)2 .
§4. Three circles of the same radius
3.24. Three circles of radius R pass through point H; let A, B and C be points of their
pairwise intersection distinct from H. Prove that
a) H is the intersection point of heights of triangle ABC;
b) the radius of the circumscribed circle of the triangle ABC is also equal to R.
Figure 20 (3.24)
3.25. Three equal circles intersect as shown on Fig. 20 a) or b). Prove that ⌣ AB1 + ⌣
BC1 ± ⌣ CA1 = 180◦ , where the minus sign is taken in case b) and plus in case a).
Figure 21 (3.26)
3.26. Three circles of the same radius pass through point P ; let A, B and Q be points
of their pairwise intersections. A fourth circle of the same radius passes through point Q
§6. APPLICATION OF THE THEOREM ON TRIANGLE’S HEIGHTS
61
and intersects the other two circles at points C and D. The triangles ABQ and CDP thus
obtained are acute ones and quadrilateral ABCD is a convex one (Fig. 21). Prove that
ABCD is a parallelogram.
§5. Two tangents drawn from one point
3.27. Tangents AB and AC are drawn from point A to a circle centered at O. Prove
that if segment AO subtends a right angle with vertex at point M , then segments OB and
OC subtend equal angles with vertices at M .
3.28. Tangents AB and AC are drawn from point A to a circle centered at O. Through
point X on segment BC line KL perpendicular to XO is drawn so that points K and L lie
on lines AB and AC, respectively. Prove that KX = XL.
3.29. On the extension of chord KL of a circle centered at O a point A is taken and
tangents AP and AQ to the circle are drawn from it; let M be the midpoint of segment P Q.
Prove that ∠M KO = ∠M LO.
3.30. From point A tangents AB and AC to a circle and a line that intersects the circle at
points D and E are drawn; let M be the midpoint of segment BC. Prove that BM 2 = DM ·
M E and either ∠DM E = 2∠DBE or ∠DM E = 2∠DCE; moreover, ∠BEM = ∠DEC.
3.31. Quadrilateral ABCD is inscribed in a circle so that tangents to this circle at points
B and D intersect at a point K that lies on line AC.
a) Prove that AB · CD = BC · AD.
b) A line parallel to KB intersects lines BA, BD and BC at points P , Q and R,
respectively. Prove that P Q = QR.
∗∗∗
3.32. A circle S and a line l that has no common points with S are given. From point
P that moves along line l tangents P A and P B to circle S are drawn. Prove that all chords
AB have a common point.
Let point P lie outside circle S; let P A and P B be tangents to the circle. Then line AB
is called the polar line of point P relative circle S.
3.33. Circles S1 and S2 intersect at points A and B so that the center O of circle S1 lies
on S2 . A line that passes through point O intersects segment AB at point P and circle S2
at point C. Prove that point P lies on the polar line of point C relative circle S1 .
§6. Application of the theorem on triangle’s heights
3.34. Points C and D lie on the circle with diameter AB. Lines AC and BD, AD and
BC meet at points P and Q, respectively. Prove that AB ⊥ P Q.
3.35. Lines P C and P D are tangent to the circle with diameter AB so that C and D
are tangent points. Prove that the line that connects P with the intersection point of lines
AC and BD is perpendicular to AB.
3.36. Given diameter AB of a circle and point C outside AB. With the help of the ruler
alone (no compasses) drop the perpendicular from C to AB if:
a) point C does not lie on the circle;
b) point C lies on the circle.
3.37. Let Oa , Ob and Oc be the centers of circumscribed circles of triangles P BC, P CA
and P AB. Prove that if points Oa and Ob lie on lines P A and P B, then point Oc lies on
line P C.
62
CHAPTER 3. CIRCLES
§7. Areas of curvilinear figures
3.38. On the hypothenuse and legs of a rectangular triangle semicircles are constructed
as shown on Fig. 22. Prove that the sum of the areas of the crescents obtained (shaded) is
equal to the area of the given triangle.
Figure 22 (3.38)
3.39. In a disc two perpendicular diameters, i.e., four radii, are constructed. Then there
are constructed four disks whose diameters are these radii. Prove that the total area of the
pairwise common parts of these four disks is equal to the area of the initial (larger) disk that
lies outside the considered four disks (Fig. 23).
Figure 23 (3.39)
3.40. On three segments OA, OB and OC of the same length (point B lies outside angle
AOC) circles are constructed as on diameters. Prove that the area of the curvilinear triangle
bounded by the arcs of these circles and not containing point O is equal to a half area of the
(common) triangle ABC.
3.41. On sides of an arbitrary acute triangle ABC as on diameters circles are constructed.
They form three “outer” curvilinear triangles and one “inner” triangle (Fig. 24). Prove that
if we subtract the area of the “inner” triangle from the sum of the areas of “outer” triangles
we get the doubled area of triangle ABC.
§8. Circles inscribed in a disc segment
In this section a segment is always a disc segment.
3.42. Chord AB divides circle S into two arcs. Circle S1 is tangent to chord AB at
point M and one of the arcs at point N . Prove that:
a) line M N passes through the midpoint P of the second arc;
b) the length of tangent P Q to circle S1 is equal to that of P A.
3.43. From point D of circle S the perpendicular DC is dropped to diameter AB. Circle
S1 is tangent to segment CA at point E and also to segment CD and to circle S. Prove that
DE is a bisector of triangle ADC.
§10. THE RADICAL AXIS
63
Figure 24 (3.41)
3.44. Two circles inscribed in segment AB of the given circle intersect at points M and
N . Prove that line M N passes through the midpoint C of arc AB complementary for the
given segment.
3.45. A circle tangent to sides AC and BC of triangle ABC at points M and N ,
respectively is also tangent to its circumscribed circle (from the inside). Prove that the
midpoint of segment M N coincides with the center of the inscribed circle of triangle ABC.
3.46. Triangles ABC1 and ABC2 are inscribed in circle S so that chords AC2 and BC1
intersect. Circle S1 is tangent to chord AC2 at point M2 , to chord BC1 at point N1 and to
circle S (???where?). Prove that the centers of the inscribed circles of triangles ABC1 and
ABC2 lie on segment M2 N1 .
§9. Miscellaneous problems
3.47. The radii of two circles are equal to R1 and R2 and the distance between the
centers of the circles is equal to d. Prove that these circles are orthogonal if and only if
d2 = R12 + R22 .
3.48. Three circles are pairwise tangent from the outside at points A, B and C. Prove
that the circumscribed circle of triangle ABC is perpendicular to all the three circles.
3.49. Two circles centered at O1 and O2 intersect at points A and B. A line is drawn
through point A; the line intersects the first circle at point M1 and the second circle at point
M2 . Prove that ∠BO1 M1 = ∠BO2 M2 .
§10. The radical axis
3.50. Circle S and point P are given on the plane. A line drawn through point P
intersects the circle at points A and B. Prove that the product P A · P B does not depend
on the choice of a line.
This product taken with the plus sign if point P is outside the circle and with minus sign
if P is inside of the circle is called the degree of point P with respect to circle S.
3.51. Prove that for a point P outside circle S its degree with respect to S is equal to
the square of the length of the tangent drawn to the circle from point P .
3.52. Prove that the degree of point P with respect to circle S is equal to d2 − R2 , where
R is the radius of S and d is the distance from P to the center of S.
3.53. Two nonconcentric circles S1 and S2 are given in plane. Prove that the locus of
points whose degree with respect to S1 is equal to the degree with respect to S2 is a line.
This line is called the radical axis of circles S1 and S2 .
64
CHAPTER 3. CIRCLES
3.54. Prove that the radical axis of two intersecting circles passes through the intersection
points.
3.55. Given three circles in plane whose centers do not lie on one line. Let us draw
radical axes for each pair of these circles. Prove that all the three radical axes meet at one
point.
This point is called the radical center of the three circles.
3.56. Consider three pairwise intersecting circles in plane. Through the intersection
points of any two of them a line is drawn. Prove that either these three lines meet at one
point or are parallel.
3.57. Two nonconcentric circles S1 and S2 are given. Prove that the set of centers of
circles that intersect both these circles at a right angle is their radical axis (without their
common chord if the given circles intersect).
3.58. a) Prove that the midpoints of the four common tangents to two nonintersecting
circles lie on one line.
b) Through two of the tangent points of common exterior tangents with two circles a line
is drawn, see Fig. . Prove that the circles cut on this line equal chords.
3.59. On sides BC and AC of triangle ABC, points A1 and B1 , respectively, are taken;
let l be the line that passes through the common points of circles with diameters AA1 and
BB1 . Prove that:
a) Line l passes through the intersection point H of heights of triangle ABC;
b) line l passes through point C if and only if AB1 : AC = BA1 : BC.
3.60. The extensions of sides AB and CD of quadrilateral ABCD meet at point F and
the extensions of sides BC and AD meet at point E. Prove that the circles with diameters
AC, BD and EF have a common radical axis and the orthocenters of triangles ABE, CDE,
ADF and BCF lie on it.
3.6l. Three circles intersect pairwise at points A1 and A2 , B1 and B2 , C1 and C2 . Prove
that A1 B2 · B1 C2 · C1 A2 = A2 B1 · B2 C1 · C2 A1 .
3.62. On side BC of triangle ABC point A′ is taken. The midperpendicular to segment
A′ B intersects side AB at point M and the midperpendicular to segment A′ C intersects
side AC at point N . Prove that point symmetric to point A′ through line M N lies on the
circumscribed circle of triangle ABC.
3.63. Solve Problem 1.66 making use of the properties of the radical axis.
3.64. Inside a convex polygon several pairwise nonintersecting disks of distinct radii are
placed. Prove that it is possible to cut the polygon into smaller polygons so that all these
small polygons are convex and each of them contains exactly one of the given disks.
3.65. a) In triangle ABC, heights AA1 , BB1 and CC1 are drawn. Lines AB and A1 B1 ,
BC and B1 C1 , CA and C1 A1 intersect at points C ′ , A′ and B ′ , respectively. Prove that
points A′ , B ′ and C ′ lie on the radical axis of the circle of nine points (cf. Problem 5.106)
and on that of the circumscribed circle.
b) The bisectors of the outer angles of triangle ABC intersect the extensions of the
opposite sides at points A′ , B ′ and C ′ . Prove that points A′ , B ′ and C ′ lie on one line
and this line is perpendicular to the line that connects the centers of the inscribed and
circumscribed circles of triangle ABC.
3.66. Prove that diagonals AD, BE and CF of the circumscribed hexagon ABCDEF
meet at one point. (Brianchon’s theorem.)
3.67. Given four circles S1 , S2 , S3 and S4 such that the circles Si and Si+1 are tangent
from the outside for i = 1, 2, 3, 4, where S5 = S1 . Prove that the radical axis of circles S1
and S3 passes through the intersection point of common outer tangents to S2 and S4 .
SOLUTIONS
65
3.68. a) Circles S1 and S2 intersect at points A and B. The degree of point P of circle
S1 with respect to circle S2 is equal to p, the distance from point P to line AB is equal to
h and the distance between the centers of circles is equal to d. Prove that |p| = 2dh.
b) The degrees of points A and B with respect to the circumscribed circles of triangles
BCD and ACD are equal to pa and pb , respectively. Prove that |pa |SBCD = |pb |SACD .
Problems for independent study
3.69. An easy chair of the form of a disc sector of radius R is swinging on a horizontal
table. What is the trajectory of the vertex of the sector?
3.70. From a point A outside a circle of radius R two tangents AB and AC are drawn,
B and C are tangent points. Let BC = a. Prove that 4R2 = r2 + ra2 + 12 a2 , where r and ra
are the radii of the inscribed and escribed circles of triangle ABC.
3.71. Two circles have an inner tangent. The line that passes through the center of a
smaller circle intersects the greater one at points A and D and the smaller one at points B
and C. Find the ratio of the radii of the circles if AB : BC : CD = 2 : 3 : 4.
3.72. The centers of three circles each of radius R, where 1 < R < 2, form an equilateral
triangle with side 2. What is the distance between the intersection points of these circles
that lie outside the triangle?
3.73. A point C is taken on segment AB and semicircles with diameters AB, AC and
BC are constructed (on one side of line AB). Find the ratio of the area of the curvilinear
triangle bounded by these semicircles to the area of the triangle formed by the midpoints of
the arcs of these semicircles.
3.74. A circle intersects side BC of triangle ABC at points A1 and A2 , side AC at
points B1 and B2 , side AB at points C1 and C2 . Prove that
¶−1
µ
AC1 BA1 CB1
AC2 BA2 CB2
.
·
·
=
·
·
C1 B A1 C B1 A
C2 B A2 C B2 A
3.75. From point A tangents AB and AC to a circle are drawn (B and C are tangent
points); P Q is a diameter of the circle; line l is tangent to the circle at point Q. Lines P A,
P B and P C intersect line l at points A1 , B1 and C1 . Prove that A1 B1 = A1 C1 .
Solutions
3.1. Let line XY be tangent to the given circle at point Z. The corresponding sides of
triangles XOA and XOZ are equal and, therefore, ∠XOA = ∠XOZ. Similarly, ∠ZOY =
∠BOY . Therefore,
∠AOB
∠AOZ + ∠ZOB
=
.
∠XOY = ∠XOZ + ∠ZOY =
2
2
3.2. Let M and N be the tangent points of the inscribed circle with sides AB and BC.
Then BK + AN = BM + AM = AB, hence, CK + CN = a + b − c.
Let P and Q be the tangent points of the escribed circle with the extensions of sides AB
and BC. Then AP = AB + BP = AB + BL and AQ = AC + CQ = AC + CL. Hence,
AP + AQ = a + b + c. Therefore, BL = BP = AP − AB = 12 (a + b − c).
3.3. By Problem 3.2 CM = 12 (AC + CE − AE) and CN = 12 (BC + CE − BE). Taking
into account that AC = BC we get M N = |CM − CN | = 12 |AE − BE|.
3.4. Let lines AB, BC, CD and DA be tangent to the circle at points P , Q, R and S,
respectively. Then CQ = CR = x, hence, BP = BC +CQ = BC +x and DS = DC +CR =
DC + x. Therefore, AP = AB + BP = AB + BC + x and AS = AD + DS = AD + DC + x.
Taking into account that AP = AS, we get the statement desired.
66
CHAPTER 3. CIRCLES
3.5. Let line AB be tangent to the circles centered at O1 and O2 at points C and
D, respectively. Since ∠O1 AO2 = 90◦ , the right triangles AO1 C and O2 AD are similar.
Therefore, O1 C : AC = AD : DO2 . Moreover, AD = CB (cf. Problem 3.2). Therefore,
AC · CB = Rr.
3.6. Let lines AB and CD intersect at point O. Let us assume for definiteness that
points A and D lie on the first circle while points B and C lie on the second one. Suppose
also that OB < OA (Fig. 25).
Figure 25 (Sol. 3.6)
The intersection point M of bisectors of angles ∠A and ∠D of quadrilateral ABCD is
the midpoint of the arc of the first circle that lies inside triangle AOD and the intersection
point N of bisectors of angles ∠B and ∠C is the midpoint of the arc of the second circle that
lies outside triangle BOC, cf. Problem 2.91 a). Quadrilateral ABCD is a circumscribed one
if and only if points M and N coincide.
3.7. Let R be the tangent point of the escribed circle with side BD, let P and Q be the
intersection points of segment M N with BC and CD, respectively (Fig. 26).
Figure 26 (Sol. 3.7)
Since ∠DM Q = ∠BP N , ∠DQM = ∠BN P and ∠DM Q = ∠BN P , it follows that
triangles M DQ, P BN and P CQ are isosceles ones. Therefore, CP = CQ, DQ = DM = DR
and BP = BN = BR. Therefore, P , Q and R are the tangent points of the inscribed circle
of triangle BCD with its sides (cf. Problem 5.1).
3.8. Denote some of the tangent points as shown on Fig. 27. The sum of the lengths of
one pair of the opposite sides of the inner quadrilateral is equal to the sum of the lengths
SOLUTIONS
67
of the pair of its other sides. Let us extend the sides of this quadrilateral to tangent points
with inscribed circles of the other quadrilaterals (ST is one of the obtained segments).
Figure 27 (Sol. 3.8)
Then both sums of lengths of pairs of opposite segments increase by the same number.
Each of the obtained segments is the common tangent to a pair of “corner” circles; each
segment can be replaced with another common outer tangent of equal length (i.e., replace
ST with QR). To prove the equality AB + CD = BC + AD, it remains to make use of
equalities of the form AP = AQ.
3.9. Let P be the intersection point of diagonals of convex quadrilateral ABCD. Quadrilateral ABCD is an inscribed one if and only if △AP B ∼ △DP C, i.e., P A · P C = P B · P D.
Since quadrilaterals ALBN and AM BK are inscribed ones, P L·P N = P A·P B = P M ·P K.
Hence, quadrilateral KLM N is an inscribed one.
3.10. Let O be the intersection point of line AB and segment M N . Then OM 2 =
OA · OB = ON 2 , i.e., OM = ON .
3.11. Let, for definiteness, rays OA and BC be codirected, M the intersection point of
lines KL and OA. Then ∠LOM = ∠LCB = ∠OKM and, therefore, △KOM ∼ △OLM .
Hence, OM : KM = LM : OM , i.e., OM 2 = KM · LM . Moreover, M A2 = M K · M L.
Therefore, M A = OM .
3.12. Let O be the intersection point of diagonals AC and BD. Then M O · OC =
BO · OD. Since OC = OA and BO = OD, we have M O · OA = BO2 and M O · OA = DO2 .
These equalities mean that OB is tangent to the circumscribed circle of triangle ABM and
OD is tangent to the circumscribed circle of triangle ADM .
3.13. Let C be the intersection point of line AB with the circumscribed circle of triangle
BM N distinct from point B; let AP be the tangent to circle S. Then AB ·AC = AM ·AN =
2
AP 2 and, therefore, AC = AP
, i.e., point C is the same for all lines l.
AB
Remark. We have to exclude the case when the length of the tangent drawn to S from
A is equal to AB.
3.14. Clearly, M C 2 = M P · M Q = M A · M B and point M lies on ray AB if AC > BC
and on ray BA if AC < BC. Let, for definiteness sake, point M lie on ray AB. Then
BC 2
(M B + BC)2 = (M B + BA) · M B. Therefore, M B = AB−2BC
and we deduce that the
position of point M does not depend on the choice of circle S ′ .
3.15. Let M be the intersection point of line CD and the tangent to circles at point A.
Then M C = M A = M D. Therefore, point A lies on the circle with diameter CD.
68
CHAPTER 3. CIRCLES
3.16. Points O1 , A and O2 lie on one line, hence, ∠A2 AO2 = ∠A1 AO1 . Triangles
AO2 A2 and AO1 A1 are isosceles ones, hence, ∠A2 AO2 = ∠AA2 O2 and ∠A1 AO1 = ∠AA1 O1 .
Therefore, ∠AA2 O2 = ∠AA1 O1 , i.e., O1 A1 k O2 A2 .
3.17. Let O1 , O2 and O3 be the centers of circles S1 , S2 and S3 ; let A, B, C be
the tangent points of circles S2 and S3 , S3 and S1 , S1 and S2 , respectively; A1 and B1 the
intersection points of lines CA and CB, respectively, with circle S3 . By the previous problem
B1 O3 k CO1 and A1 O3 k CO2 . Points O1 , C and O2 lie on one line and, therefore, points
A1 , O3 and B1 also lie on one line, i.e., A1 B1 is a diameter of cicle S3 .
3.18. Let A1 , A2 and B be the tangent points of the circles centered at O and O1 , O
and O2 , O1 and O2 , respectively. Then O1 O2 = O1 B + BO2 = O1 A1 + O2 A2 . Therefore,
OO1 + OO2 + O1 O2 = (OO1 + O1 A1 ) + (OO2 + O2 A2 ) = OA1 + OA2 = 2R.
3.19. Let O, O1 and O2 be centers of circles S, S1 and S2 ; let C be the common
point of circles S1 and S2 that lies on segment AB. Triangles AOB, AO1 C and CO2 B are
isosceles ones; consequently, OO1 CO2 is a parallelogram and OO1 = O2 C = O2 B; hence,
AO = AO1 + O1 O = AO1 + O2 B.
3.20. Let O1 and O2 be the centers of circles S1 and S2 ; let X be the other intersection
point of line AB with circle S2 . The square of the length of the tangent in question is equal
2
2 ·O O
1 2
= a (R±r)
,
to BA · BX. Since AB : BX = O1 A : O1 O2 , it follows that AB · BX = AB R
R
where the minus sign is taken for the inner tangent and the plus sign for the outer tangent.
3.21. Let O, O1 and O2 be the centers of the circles with diameters AB, AC and BC,
respectively. It suffices to verify that KO = OL. Let us prove that △O1 KO = △O2 OL.
Indeed, O1 K = 21 AC = O2 O, O1 O = 21 BC = O2 L and ∠KO1 O = ∠OO2 L = 180◦ − 2α,
where α is the value of the angle between lines KL and AB.
3.22. Let Oi be the center of circle Si and Ai the tangent point of circles Si and Si+1 .
Quadrilateral O1 O2 O3 O4 is a convex one; let α1 , α2 , α3 and α4 be the values of its angles.
It is easy to verify that ∠Ai−1 Ai Ai+1 = 12 (αi + αi+1 ) and, therefore,
1
∠A1 + ∠A3 = (α1 + α2 + α3 + α4 ) = ∠A2 + ∠A4 .
2
3.23. a) Let A1 , B1 and C1 be the projections of points A, B and C, respectively, to line l;
let C2 be the projection of point C to line AA1 . By Pythagorus theorem CC22 = AC 2 − AC22 ,
2
i.e., A1 C12 = (a + c)2 − (a − c)2 = 4ac. Similarly,
A1 B12 = 4ab. Since
√
√B1 C1 = 14bc and
√
1
A1 C1 + C1 B1 = A1 B1 , it follows that ac + bc = ab, i.e., √b + √a = √1c .
Figure 28 (Sol. 3.23 b))
SOLUTIONS
69
b) Let A, B, C be the centers of “outer” circles, D the center of the “inner” circle (Fig.
28). The semiperimeter of triangle BDC is equal to b + c + d, and, therefore,
µ
¶
¶
µ
d(b + c + d)
∠BDC
bc
∠BDC
2
2
, sin
=
=
cos
2
(b + d)(c + d)
2
(b + d)(c + d)
(cf. Problem 12.13). As is easy to see the law of cosines is equivalent to the statement:
α′ + β ′ + γ ′ = 180◦ =⇒ sin2 α′ − sin2 β ′ − sin2 γ ′ + 2 sin β ′ sin γ ′ cos α′ = 0.
(∗)
Substituting the values α′ = 21 ∠BDC, β ′ = 12 ∠ADC and γ ′ = 12 ∠ADB into formula (∗), we
get
p
a bcd(b + c + d)
ac
ab
bc
−
−
+2
= 0,
(b + d)(c + d) (a + d)(c + d) (a + d)(b + d)
(a + d)(b + d)(c + d)
i.e.,
r
a+d b+d c+d
d(b + c + d)
−
−
+2
= 0.
a
b
c
bc
Dividing this by d we get
p
α − β − γ − δ + 2 βγ + γδ + δβ = 0.
Therefore,
i.e.,
(α + β + γ + δ)2 = (α − β − γ − δ)2 + 4(αβ + αγ + αδ)+
4(βγ + γδ + δβ) + 4(αβ + αγ + αδ) =
2(α + β + γ + δ)2 − 2(α2 + β 2 + γ 2 + δ 2 ),
2(α2 + β 2 + γ 2 + δ 2 ) = (α + β + γ + δ)2 .
3.24. Let A1 , B1 and C1 be the centers of the given circles (Fig. 29). Then A1 BC1 H
is a rhombus and, therefore, BA1 k HC1 . Similarly, B1 A k HC1 ; hence, B1 A k BA1 and
B1 ABA1 is a parallelogram.
Figure 29 (Sol. 3.24)
a) Since A1 B1 ⊥ CH and A1 B1 k AB, it follows that AB ⊥ CH. We similarly prove
that BC ⊥ AH and CA ⊥ BH.
b) In the same way as we have proved that B1 A k BA1 , we can prove that B1 C k BC1
and A1 C k AC1 ; moreover, the lengths of all these six segments are equal to R. Let us
complement the triangle BA1 C to a rhombus BA1 CO. Then AB1 CO is also a rhombus.
Therefore, AO = BO = CO = R, i.e., O is the center of the circumscribed circle of triangle
ABC and its radius is equal to R.
70
CHAPTER 3. CIRCLES
3.25. It is easy to verify that
⌣ AB1 ± ⌣ B1 A1 = ⌣ AC1 + ⌣ C1 A1 ,
⌣ BC1 + ⌣ C1 B1 = ⌣ BA1 ± ⌣ B1 A1
⌣ C1 A1 ± ⌣ CA1 = ⌣ C1 B1 ± ⌣ B1 C,
where the minus sign is only taken in case b). Adding up these equalities we get
⌣ AB1 + ⌣ BC1 ± ⌣ CA1 =⌣ AC1 + ⌣ BA± ⌣ CB1 .
On the other hand, the doubled values of the angles of triangle ABC are equal to ⌣ BA1 ± ⌣
CA1 , ⌣ AB1 ± ⌣ CB1 and ⌣ BC1 + ⌣ AC1 , and their sum is equal to 360◦ .
3.26. Since ⌣ AP + ⌣ BP + ⌣ P Q = 180◦ (cf. Problem 3.25), it follows that
⌣ AB = 180◦ − ⌣ P Q. Similarly, ⌣ CD = 180◦ − ⌣ P Q, i.e., ⌣ AB =⌣ CD and,
therefore, AB = CD. Moreover, P Q ⊥ AB and P Q ⊥ CD (cf. Problem 3.24) and,
therefore, AB k CD.
3.27. Points M , B and C lie on the circle with diameter AO. Moreover, chords OB and
OC of the circle are equal.
3.28. Points B and X lie on the circle with diameter KO, and, therefore, ∠XKO =
∠XBO. Similarly, ∠XLO = ∠XCO. Since ∠XBO = ∠XCO, triangle KOL is an isosceles
one and OX is its height.
3.29. It suffices to verify that AK ·AL = AM ·AO. Indeed, if such is the case, then points
K, L, M and O lie on one circle and, therefore, ∠M KO = ∠M LO. Since △AOP ∼ △AP M ,
it follows that AM · AO = AP 2 ; it is also clear that AK · AL = AP 2 .
3.30. Let O be the center of the circle; let points D′ and E ′ be symmetric to points
D and E through line AO. By Problem 28.7 the lines ED′ and E ′ D meet at point M .
Hence, ∠BDM = ∠EBM and ∠BEM = ∠DBM and, therefore, △BDM ∼ △EBM . It
follows that BM : DM = EM : BM . Moreover, if line ED separates points B and M , then
∠DM E =⌣ DE = 2∠DCE.
The equality ∠BEM = ∠DBM implies that ∠BEM = ∠DBC = ∠DEC.
3.31. a) Since △KAB ∼ △KBC, we have AB : BC = KB : KC. Similarly, AD :
DC = KD : KC. Taking into account that KB = KD we get the desired statement.
b) The problem of this heading reduces to that of the previous one, since
PQ
sin ∠P BQ
sin ∠ABD
sin ∠ABD
AD
QR
CD
=
=
=
=
,
=
.
BQ
sin ∠BP Q
sin ∠KBA
sin ∠ADB
AB
BQ
CB
3.32. Let us drop perpendicular OM to line l from center O of circle S. Let us prove that
point X at which AB and OM intersect remains fixed. Points A, B and M lie on the circle
with diameter P O. Hence, ∠AM O = ∠ABO = ∠BAO and, therefore, △AM O ∼ △XAO,
because these triangles have a common angle at vertex O. It follows that AO : M O = XO :
2
is a constant.
AO, i.e., OX = OA
MO
3.33. Since ∠OBP = ∠OAB = ∠OCB, we deduce that △OBP ∼ △OCB and,
therefore, OB 2 = OP · OC. Let us draw tangent CD to circle S1 from point C. Then
OD2 = OB 2 = OP · OC. Therefore, △ODC ∼ △OP D and ∠OP D = ∠ODC = 90◦ .
3.34. Lines BC and AD are heights of triangle AP B and, therefore, line P Q that passes
through their intersection point Q is perpendicular to line AB.
3.35. Denote the intersection points of lines AC and BD, BC and AD by K and K1 ,
respectively. Thanks to the above problem, KK1 ⊥ AB and, therefore, it suffices to show
that the intersection point of tangents at points C and D lies on line KK1 .
Let us prove that the tangent at point C passes through the midpoint of segment KK1 .
Let M be the intersection point of the tangent at point C and segment KK1 . The respective
sides of acute angles ∠ABC and ∠CKK1 are perpendicular and, therefore, the angles are
SOLUTIONS
71
equal. Similarly, ∠CAB = ∠CK1 K. It is also clear that ∠KCM = ∠ABC and, therefore,
triangle CM K is an isosceles one. Similarly, triangle CM K1 is an isosceles one and KM =
CM = K1 M , i.e., M is the midpoint of segment KK1 .
We similarly prove that the tangent at point D passes through the midpoint of segment
KK1 .
3.36. a) Line AC intersects the circle at points A and A1 , line BC does same at points
B and B1 . If A = A1 (or B = B1 ), then line AC (or BC) is the perpendicular to be
constructed. If this is not the case, then AB1 and BA1 are heights of triangle ABC and the
line to be constructed is the line that passes through point C and the intersection point of
lines AB1 and BA1 .
b) Let us take point C1 that does not lie on the circle and drop from it perpendicular
to AB. Let the perpendicular intersect the circle at points D and E. Let us construct the
intersection point P of lines DC and AB and then the intersection point F of line P E with
the circle. The symmetry through AB sends point C to point F . Therefore, CF is the
perpendicular to be constructed.
3.37. Since P A ⊥ Ob Oc , line P A passes through point Oa if and only if line P Oa passes
through the intersection point of heights of triangle Oa Ob Oc . Similar statements are true for
points B and C as well.
The hypothesis of the problem implies that P is the intersection point of heights of
triangle Oa Ob Oc and, therefore, P Oc ⊥ Oa Ob .
3.38. Let 2a and 2b be the lengths of the legs, 2c the length of the hypothenuse. The
sum of the areas of the “crescents” is equal to πa2 +πb2 +SABC −πc2 . But π(a2 +b2 −c2 ) = 0.
3.39. It suffices to carry out the proof for each of the four parts into which the diameters
divide the initial disc (Fig. 30).
Figure 30 (Sol. 3.39)
In the disc, consider the segment cut off by the chord intercepted by the central angle
of 90◦ ; let S and s be the areas of such segments for the initial disc and any of the four
constructed disks, respectively. Clearly, S = 4s. It remains to observe that the area of the
part shaded once is equal to S − 2s = 2s and the area of the part shaded twice is equal to
2s.
3.40. Denote the intersection points of circles constructed on segments OB and OC,
OA and OC, OA and OB as on diameters by A1 , B1 , C1 , respectively (Fig. 31). Since
∠OA1 B = ∠OA1 C = 90◦ , it follows that points B, A1 and C lie on one line and since all
the circles have equal radii, BA1 = A1 C.
Points A1 , B1 , C1 are the midpoints of sides of triangle ABC, therefore, BA1 = C1 B1
and BC = A1 B1 . Since the disks are of the same radius, the equal chords BA1 and C1 B1
cut off the disks parts of equal area and equal chords C1 B and B1 A1 also cut off the disc’s
parts of equal area. Therefore, the area of curvilinear triangle A1 B1 C1 is equal to the area
of parallelogram A1 B1 C1 B, i.e., is equal to half the area of triangle ABC.
72
CHAPTER 3. CIRCLES
Figure 31 (Sol. 3.40)
Figure 32 (Sol. 3.41)
3.41. The considered circles pass through the bases of the triangle’s heights and, therefore, their intersection points lie on the triangle’s sides. Let x, y, z and u be the areas of the
considered curvilinear triangles; let a, b, c, d, e and f be the areas of the segments cut off
the circles by the sides of the triangle; let p, q and r be the areas of the parts of the triangle
that lie outside the inner curvilinear triangle (see Fig. 32). Then
x + (a + b) = u + p + q + (c + f ),
y + (c + d) = u + q + r + (e + b),
z + (e + f ) = u + r + p + (a + d)
By adding up these equalities we get
x + y + z = 2(p + q + r + u) + u.
3.42. a) Let O and O1 be the centers of circles S and S1 . The triangles M O1 N and
P ON are isosceles ones and ∠M O1 N = ∠P ON . Therefore, points P , M and N lie on one
line.
b) It is clear that P Q2 = P M · P N = P M · (P M + M N ). Let K be the midpoint of
chord AB. Then
P M 2 = P K2 + M K2
and P M · M N = AM · M B = AK 2 − M K 2 .
Therefore, P Q2 = P K 2 + AK 2 = P A2 .
3.43. By Problem 3.42 b) BE = BD. Hence,
∠DAE + ∠ADE = ∠DEB = ∠BDE = ∠BDC + ∠CDE.
Since ∠DAB = ∠BDC, it follows that ∠ADE = ∠CDE.
3.44. Let O1 and O2 be the centers of the inscribed circles, CP and CQ the tangents to
them. Then CO12 = CP 2 +P O12 = CP 2 +O1 M 2 and since CQ = CA = CP (by Problem 3.42
SOLUTIONS
73
b), we have CO22 = CQ2 + QO22 = CP 2 + O2 M 2 . It follows that CO12 − CO22 = M O12 − M O22
and, therefore, line CM is perpendicular to O1 O2 (see Problem 7.6). Therefore, line M N
passes through point C.
Remark. If the circles do not intersect but are tangent to each other the statement is
still true; in this case, however, one should replace line M N with the tangent to the circles
at their common point.
3.45. Let A1 and B1 be the midpoints of arcs ⌣ BC and ⌣ AC;let O the center of
the inscribed circle. Then A1 B1 ⊥ CO (cf. Problem 2.19 a) and M N ⊥ CO, consequently,
M N k A1 B1 . Let us move points M ′ and N ′ along rays CA and CB, respectively, so that
M ′ N ′ k A1 B1 . Only for one position of points M ′ and N ′ does point L at which lines B1 M ′
and A1 N ′ intersect lie on the circumscribed circle of triangle ABC.
On the other hand, if segment M N passes through point O, then point L lies on this
circle (cf. Problem 2.49).
3.46. The solution of this problem generalizes the solution of the preceding problem. It
suffices to prove that the center O1 of the inscribed circle of triangle ABC1 lies on segment
M2 N1 . Let A1 and A2 be the midpoints of arcs ⌣ BC1 and ⌣ BC2 ; let B1 and B2 be the
midpoints of arcs ⌣ AC1 and ⌣ AC2 ; let P Q be the diameter of circle S perpendicular to
chord AB and let points Q and C1 lie on one side of line AB. Point O1 is the intersection
point of chords AA1 and BB1 and point L of tangent of circles S and S1 is the intersection
point of lines A1 N1 and B2 M2 (Fig. 33).
Figure 33 (Sol. 3.46)
Let ∠C1 AB = 2α, ∠C1 BA = 2β, ∠C1 AC2 = 2ϕ. Then ⌣ A1 A2 = 2ϕ =⌣ B1 B2 , i.e.,
A1 B2 k B1 A2 . For the angles between chords we have:
∠(A1 B2 , BC1 ) = 21 (⌣ B2 C1 + ⌣ A1 B) = β − ϕ + α,
∠(BC1 , AC2 ) = 12 (⌣ C1 C2 + ⌣ AB) = 2ϕ + 180◦ − 2α − 2β.
Consequently, chord M2 N1 constitutes equal angles with tangents BC1 and AC2 , each angle
equal to α + β − ϕ. Therefore, M2 N1 k A1 B2 .
Now, suppose that points M2′ and N1′ are moved along chords AC2 and BC1 so that
k A1 B2 . Let lines A1 N1′ and B2 M2′ meet at point L′ . Point L′ lies on circle S for one
position of points M2′ and N1′ only. Therefore, it suffices to indicate on arc ⌣ AB a point
L1 such that if M2′′ , N1′′ are the intersection points of chords AC2 and L1 B2 , BC1 and L1 A1 ,
M2′ N1′
74
CHAPTER 3. CIRCLES
respectively, then M2′′ N1′′ k A1 B2 and point O1 lies on segment M2′′ N1′′ . Let Q1 be a point on
circle S such that 2∠(P Q, P Q1 ) = ∠(P C2 , P C1 ) and L1 the intersection point of line Q1 O1
with S.
Let us prove that L1 is the desired point. Since ⌣ B1 Q = 2α, it follows that ⌣ B2 Q1 =
2(α − 2ϕ) =⌣ C2 A1 . Hence, quadrilateral AM2′′ O1 L1 is an inscribed one and, therefore,
∠M2′′ O1 A = ∠M2′′ L1 A = ∠B2 A1 A, i.e., M2′′ O1 k B2 A1 .
Similarly, N1′′ O1 k B2 A1 .
3.47. Let circles centered at O1 and O2 pass through point A. The radii O1 A and
O2 A are perpendicular to the tangents to circles at point A and, therefore, the circles are
orthogonal if and only if ∠O1 AO2 = 90◦ , i.e., ∠O1 O22 = O1 A2 + O2 A2 .
3.48. Let A1 , B1 and C1 be the centers of the given circles so that points A, B and C
lie on segments B1 C1 , C1 A1 and A1 B1 , respectively. Since A1 B = A1 C, B1 A = B1 C and
CA = C1 B, it follows that A, B and C are the tangent points of the inscribed circle of
triangle A1 B1 C1 with its sides (cf. Problem 5.1). Therefore, the radii A1 B, BC and C1 A of
the given circles are tangent to the circumscribed circle of triangle ABC.
−−→
−−−→
3.49. It is easy to verify that the angle of rotation from vector Oi B to vector Oi Mi (counterclockwise) is equal to 2∠(AB, AMi ). It is also clear that ∠(AB, AM1 ) = ∠(AB, AM2 ).
3.50. Let us draw through point P another line that intersects the circle at points A1
and B1 . Then △P AA1 ∼ △P B1 B and, therefore, P A : P A1 = P B1 : P B.
3.51. Let us draw through point P tangent P C. Since △P AC ∼ △P CB, it follows that
P A : P C = P C : P B.
3.52. Let the line that passes through point P and the center of the circle intersect the
circle at points A and B. Then P A = d+R and P B = |d−R|. Therefore, P A·P B = |d2 −R2 |.
It is also clear that the signs of the expression d2 − R2 and of the degree of point P with
respect to to S are the same.
3.53. Let R1 and R2 be the radii of the circles. Let us consider the coordinate system in
which the coordinates of the centers of the circles are (−a, 0) and (a, 0). By Problem 3.52
the degrees of the point with coordinates (x, y) with respect to the given circles are equal to
(x + a)2 + y 2 − R12 and (x − a)2 + y 2 − R22 , respectively. By equating these expressions we
R2 −R2
get x = 14a 2 . This equation determines the perpendicular to the segment that connects
the centers of the circles.
3.54. The degrees of the intersection point of the circles with respect to each one of the
cicles are equal to zero and, therefore, the point belongs to the radical axis. If there are two
intersection points, then they uniquely determine the radical axis.
3.55. Since the centers of the circles do not lie on one line, the radical axis of the first
and the second circles intersects with the radical axis of the second and third circles. The
degrees of the intersection point with respect to all three circles are equal and, therefore,
this intersection point lies on the radical axis of the first and third circles.
3.56. By Problem 3.54 the lines that contain chords are radical axes. By Problem 3.55
the radical axes meet at one point if the centers of the circles do not lie on one line. Otherwise
they are perpendicular to this line.
3.57. Let O1 and O2 be the centers of given circles, r1 and r2 their radii. The circle S
of radius r centered at O is orthogonal to circle Si if and only if r2 = OOi2 − ri2 , i.e., the
squared radius of S is equal to the degree of point O with respect to circle Si . Therefore,
the locus of the centers of the circles to be found is the set of the points of the radical axis
whose degrees with respect to the given circles are positive.
3.58. a) The indicated points lie on the radical axis.
SOLUTIONS
75
b) The tangent points of the outer tangents with the circles are vertices of trapezoid
ABCD with base AB. The midpoints of lateral sides AD and BC belong to the radical
axis and, therefore, the midpoint O of diagonal AC also belongs to the radical axis. If line
AC intersects the circles at points A1 and C1 , then OA1 · OA = OC1 · OC; consequently,
OA1 = OC1 and AA1 = CC1 .
3.59. a) Let SA and SB be circles with diameters AA1 and BB1 ; let S be the circle with
diameter AB. The common chords of circles S and SA , S and SB are heights AHa and BHb
and, therefore, these heights (or their extensions) intersect at point H. By Problem 3.56 the
common chord of circles SA and SB passes through the intersection point of chords AHa and
BHb .
b) The common chord of circles SA and SB passes through the intersection point of lines
A1 Ha and B1 Hb (i.e., through point C) if and only if CB1 · CHb = CA1 · CHa (here we
should consider the lengths of segments as oriented). Since
CHb =
a2 + b 2 − c 2
2b
and CHa =
a2 + b 2 − c 2
,
2a
1
1
= CA
.
we deduce that CB
b
a
3.60. In triangle CDE, draw heights CC1 and DD1 ; let H be their intersection point.
The circles with diameters AC and BD pass through points C1 and D1 , respectively, therefore, the degree of point H with respect to each of these circles is equal to its degree with
respect to the circle with diameter CD (this circle passes through points C1 and D1 ). We
similarly prove that the degrees of point H with respect to to circles with diameters AC,
BD and EF are equal, i.e., the radical axes of these circles pass through point H.
For the intersection points of heights of the other three triangles the proof is carried out
in a similar way.
Remark. The centers of the considered circles lie on the Gauss’ line (cf. Problem 4.55)
and, therefore, their common radical axis is perpendicular to the Gauss line.
3.61. Lines A1 A2 , B1 B2 and C1 C2 meet at a point O (cf. Problem 3.56). Since
△A1 OB2 ∼ △B1 OA2 , it follows that A1 B2 : A2 B1 = OA1 : OB1 . Similarly, B1 C2 : B2 C1 =
OB1 : OC1 and C1 A2 : C2 A1 = OC1 : OA1 . By multiplying these equalities we get the
statement desired.
3.62. Denote by B ′ and C ′ the intersection points of lines A′ M and A′ N , respectively,
with the line drawn through point A parallel to BC (Fig. 34).
Figure 34 (Sol. 3.62)
Since triangles A′ BM and A′ N C are isosceles ones, △ABC = △A′ B ′ C ′ . Since AM ·
BM = A′ M · B ′ M , the degrees of point M with respect to circles S and S ′ circumscribed
about triangles ABC and A′ B ′ C ′ , respectively, are equal. This is true for point N as well
76
CHAPTER 3. CIRCLES
and, therefore, line M N is the radical axis of circles S and S ′ . Circles S and S ′ have equal
radii and, therefore, their radical axis is their axis of symmetry. The symmetry through line
M N sends a point A′ that lies on circle S ′ into a point that lies on circle S.
3.63. Let AC and BD be the tangents; E and K the intersection points of lines AC
and BD, AB and CD, respectively; O1 and O2 the centers of the circles (Fig. 35).
Figure 35 (Sol. 3.63)
Since AB ⊥ O1 E, O1 E ⊥ O2 E and O2 E ⊥ CD, it follows that AB ⊥ CD and, therefore,
K is the intersection point of circles S1 and S2 with diameters AC and BD, respectively.
Point K lies on the radical axis of circles S1 and S2 ; it remains to verify that line O1 O2 is this
radical axis. The radii O1 A and O1 B are tangent to S1 and S2 , respectively, and, therefore,
point O1 lies on the radical axis. Similarly, point O2 also lies on the radical axis.
3.64. Denote the given circles by S1 , . . . , Sn . For each circle Si consider the set Mi that
consists of all the points X whose degree with respect to Si does not exceed their degrees
with respect to the other circles S1 , . . . , Si−1 , Si+1 , . . . , Sn .
The set Mi is a convex one. Indeed, let Mij be the set of points X whose degree with
respect to Si does not exceed the degree with respect to Sj . The set Mij is a half plane that
consists of the points that lie on the same side of the radical axis of circles Si and Sj as Si
does. The set Mi is the intersection of the convex sets Mij for all j and, therefore, is a convex
set itself. Moreover, since each of the sets Mij contains circle Si , then Mi also contains Si .
Since for each point of the plane at least one of the degrees with respect to S1 , . . . , Sn is the
least one, the sets Mi cover the whole plane.
Now, by considering the parts of the sets Mi that lie inside the initial polygon we get the
partition statement desired.
3.65. a) Points B1 and C1 lie on the circle with diameter BC and, therefore, the degrees
of point A′ with respect to the circumscribed circles of triangles A1 B1 C1 and ABC are equal
to the degrees of point A′ with respect to this circle. This means that point A′ lies on the
radical axis of the Euler circle and the circumscribed circle of triangle ABC. For points B ′
and C ′ the proof is similar.
b) Let us consider triangle A1 B1 C1 formed by the outer bisectors of triangle ABC (triangle A1 B1 C1 is an acute one). Thanks to heading a) points A′ , B ′ and C ′ lie on the radical
axis of the circumscribed circles of triangles ABC and A1 B1 C1 . The radical axis of these
circles is perpendicular to the line that connects their centers, i.e., the Euler line of triangle
A1 B1 C1 . It remains to notice that the intersection point of the heights of triangle A1 B1 C1
is the intersection point of the bisectors of triangle ABC, cf. Problem 1.56 a).
3.66. Let a convex hexagon ABCDEF be tangent to the circle at points R, Q, T , S,
P , U (point R lies on AB, point Q lies on BC, etc.).
SOLUTIONS
77
Take a number a > 0 and construct points Q′ and P ′ on lines BC and EF so that
−−→
−−→
−−→
−→
QQ′ = P P ′ = a and vectors QQ′ and P P ′ are codirected with vectors CB and EF .
Let us similarly construct points R′ , S ′ , T ′ , U ′ (see Fig. 36, where RR′ = SS ′ = T T ′ =
U U ′ = a). Let us construct circle S1 tangent to lines BC and EF at points Q′ and P ′ . Let
us similarly construct circles S2 and S3 .
Figure 36 (Sol. 3.66)
Let us prove that points B and E lie on the radical axis of circles S1 and S2 . We have
BQ′ = QQ′ − BQ = RR′ − BR = BR′
(if QQ′ < BQ, then BQ′ = BQ − QQ′ = BR − RR′ = BR′ ) and
EP ′ = EP + P P ′ = ES + SS ′ = ES ′ .
We similarly prove that lines F C and AD are the radical axes of circles S1 and S3 , S2 and
S3 , respectively. Since the radical axes of three circles meet at one point, lines AD, BE and
CF meet at one point.
3.67. Let Ai be the tangent point of circles Si and Si+1 and X be the intersection point
of lines A1 A4 and A2 A3 . Then X is the intersection point of the common outer tangents to
circles S2 and S4 (cf. Problem 5.60). Since quadrilateral A1 A2 A3 A4 is an inscribed one (by
Problem 3.22), XA1 · XA4 = XA2 · XA3 ; consequently, point X lies on the radical axis of
circles S1 and S3 .
3.68. a) Let us consider the coordinate system whose origin O is at the center of
the segment that connects the centers of the circles and the Ox-axis is directed along this
segment. Let (x, y) be the coordinates of point P ; let R and r be the radii of circles S1 and
S2 , respectively; a = 12 d. Then (x + a)2 + y 2 = R2 and
p = (x − a)2 + y 2 − r2 = ((x + a)2 + y 2 − R2 ) − 4ax − r2 + R2 = R2 − r2 − 4ax.
Let (x0 , y0 ) be the coordinates of point A. Then
(x0 + a)2 + y02 − R2 = (x0 − a)2 + y02 − r2 ,
i.e., x0 =
R2 − r 2
.
4a
Therefore,
2dh = 4a|x0 − x| = |R2 − r2 − 4ax| = |p|.
b) Let d be the distance between the centers of the circumscribed circles of triangles ACD
and BCD; let ha and hb be the distances from points A and B to line CD. By heading a)
|pa | = 2dha and |pb | = 2dhb . Taking into account that SBCD = 12 hb CD and SACD = 21 ha CD
we get the statement desired.
CHAPTER 4. AREA
Background
1. One can calculate the area S of triangle ABC with the help of the following formulas:
a) S = 12 aha , where a = BC and ha is the length of the height dropped to BC;
b) S = 21 bc sin ∠A, where b, c are sides of the triangle, ∠A the angle between these sides;
c) S = pr, where p is a semiperimeter, r the radius of the inscribed circle. Indeed, if O
is the center of the inscribed circle, then
1
S = SABO + SAOC + SOBC = (c + b + a)r = pr.
2
2. If a polygon is cut into several polygons, then the sum of their areas is equal to the
area of the initial polygon.
Introductory problems
1. Prove that the area of a convex quadrilateral is equal to 12 d1 d2 sin ϕ, where d1 and d2
are the lengths of the diagonals and ϕ is the angle between them.
2. Let E and F be the midpoints of sides BC and AD of parallelogram ABCD. Find
the area of the quadrilateral formed by lines AE, ED, BF and F C if it is known that the
area of ABCD is equal to S.
3. A polygon is circumscribed about a circle of radius r. Prove that the area of the
polygon is equal to pr, where p is the semiperimeter of the polygon.
4. Point X is inside parallelogram ABCD. Prove that SABX + SCDX = SBCX + SADX .
5. Let A1 , B1 , C1 and D1 be the midpoints of sides CD, DA, AB, BC, respectively, of
square ABCD whose area is equal to S. Find the area of the quadrilateral formed by lines
AA1 , BB1 , CC1 and DD1 .
§1. A median divides the triangle
into triangles of equal areas
4.1. Prove that the medians divide any triangle into six triangles of equal area.
4.2. Given triangle ABC, find all points P such that the areas of triangles ABP , BCP
and ACP are equal.
4.3. Inside given triangle ABC find a point O such that the areas of triangles BOL,
COM and AON are equal (points L, M and N lie on sides AB, BC and CA so that
OL k BC, OM k AC and ON k AB; see Fig. 37).
4.4. On the extensions of the sides of triangle ABC points A1 , B1 and C1 are taken so
−−→
−→ −−→
−−→
−−→
−→
that AB1 = 2AB, BC1 = 2BC and CA1 = 2AC. Find the area of triangle A1 B1 C1 if it is
known that the area of triangle ABC is equal to S.
4.5. On the extensions of sides DA, AB, BC, CD of convex quadrilateral ABCD
−−→
−−→ −−→
−→ −−→
−−→
points A1 , B1 , C1 , D1 are taken so that DA1 = 2DA, AB1 = 2AB, BC1 = 2BC and
−−→
−−→
CD1 = 2CD. Find the area of the obtained quadrilateral A1 B1 C1 D1 if it is known that the
area of quadrilateral ABCD is equal to S.
79
80
CHAPTER 4. AREA
Figure 37 (4.3)
4.6. Hexagon ABCDEF is inscribed in a circle. Diagonals AD, BE and CF are
diameters of this circle. Prove that SABCDEF = 2SACE .
4.7. Inside a convex quadrilateral ABCD there exists a point O such that the areas of
triangles OAB, OBC, OCD and ODA are equal. Prove that one of the diagonals of the
quadrilateral divides the other diagonal in halves.
§2. Calculation of areas
4.8. The height of a trapezoid whose diagonals are mutually perpendicular is equal to 4.
Find the area of the trapezoid if it is known that the length of one of its diagonals is equal
to 5.
4.9. Each diagonal of convex pentagon ABCDE cuts off it a triangle of unit area.
Calculate the area of pentagon ABCDE.
4.10. In a rectangle ABCD there are inscribed two distinct rectangles with a common
vertex K lying on side AB. Prove that the sum of their areas is equal to the area of rectangle
ABCD.
4.11. In triangle ABC, point E is the midpoint of side BC, point D lies on side AC;
let AC = 1, ∠BAC = 60◦ , ∠ABC = 100◦ , ∠ACB = 20◦ and ∠DEC = 80◦ (Fig. 38). Find
S△ABC + 2S△CDE .
Figure 38 (4.11)
4.12. Triangle Ta = △A1 A2 A3 is inscribed in triangle Tb = △B1 B2 B3 and triangle Tb
is inscribed in triangle Tc = △C1 C2 C3 so that the sides of triangles Ta and Tc are pairwise
parallel. Express the area of triangle Tb in terms of the areas of triangles Ta and Tc .
Figure 39 (4.12)
4.13. On sides of triangle ABC, points A1 , B1 and C1 that divide its sides in ratios
BA1 : A1 C = p, CB1 : B1 A = q and AC1 : C1 B = r, respectively, are taken. The
§4. THE AREAS OF THE PARTS INTO WHICH A QUADRILATERAL IS DIVIDED
81
intersection points of segments AA1 , BB1 and CC1 are situated as depicted on Fig. 39.
Find the ratio of areas of triangles P QR and ABC.
§3. The areas of the triangles into which
a quadrilateral is divided
4.14. The diagonals of quadrilateral ABCD meet at point O. Prove that SAOB = SCOD
if and only if BC k AD.
4.15. a) The diagonals of convex quadrilateral ABCD meet at point P . The areas of
triangles ABP , BCP , CDP are known. Find the area of triangle ADP .
b) A convex quadrilateral is divided by its diagonals into four triangles whose areas are
expressed in integers. Prove that the product of these integers is a perfect square.
2
2
4.16. The diagonals of quadrilateral ABCD meet at point P and SABP
+ SCDP
=
2
2
SBCP + SADP . Prove that P is the midpoint of one of the diagonals.
4.17. In a convex quadrilateral ABCD there are three inner points P1 , P2 , P3 not on
one line and with the property that
SABPi + SCDPi = SBCPi + SADPi
for i = 1, 2, 3. Prove that ABCD is a parallelogram.
§4. The areas of the parts into which
a quadrilateral is divided
4.18. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively,
of convex quadrilateral ABCD; segments KM and LN intersect at point O. Prove that
SAKON + SCLOM = SBKOL + SDN OM .
4.19. Points K, L, M and N lie on sides AB, BC, CD and DA, respectively, of parallelogram ABCD so that segments KM and LN are parallel to the sides of the parallelogram.
These segments meet at point O. Prove that the areas of parallelograms KBLO and M DN O
are equal if and only if point O lies on diagonal AC.
4.20. On sides AB and CD of quadrilateral ABCD, points M and N are taken so that
AM : M B = CN : N D. Segments AN and DM meet at point K, and segments BN and
CM meet at point L. Prove that SKM LN = SADK + SBCL .
4.21. On side AB of quadrilateral ABCD, points A1 and B1 are taken, on side CD
points C1 and D1 are taken so that AA1 = BB1 = pAB and CC1 = DD1 = pCD, where
S B1 C1 D1
p < 0.5. Prove that AS1ABCD
= 1 − 2p.
4.22. Each of the sides of a convex quadrilateral is divided into five equal parts and
the corresponding points of the opposite sides are connected as on Fig. 40. Prove that the
area of the middle (shaded) quadrilateral is 25 times smaller than the area of the initial
quadrilateral.
Figure 40 (4.22)
82
CHAPTER 4. AREA
4.23. On each side of a parallelogram a point is taken. The area of the quadrilateral
with vertices at these points is equal to half the area of the parallelogram. Prove that at
least one of the diagonals of the quadrilateral is parallel to a side of the parallelogram.
4.24. Points K and M are the midpoints of sides AB and CD, respectively, of convex
quadrilateral ABCD, points L and N lie on sides BC and AD so that KLM N is a rectangle.
Prove that SABCD = SKLM N .
4.25. A square is divided into four parts by two perpendicular lines whose intersection
point lies inside the square. Prove that if the areas of three of these parts are equal, then
the area of all four parts are equal.
§5. Miscellaneous problems
4.26. Given parallelogram ABCD and a point M , prove that
SACM = |SABM ± SADM |.
4.27. On sides AB and BC of triangle ABC, parallelograms are constructed outwards;
let P be the intersection point of the extensions of the sides of these parallelograms parallel
to AB and BC. On side AC, a parallelogram is constructed whose other side is equal and
parallel to BP . Prove that the area of this parallelogram is equal to the sum of areas of the
first two parallelograms.
4.28. Point O inside a regular hexagon is connected with the vertices. The six triangles
obtained in this way are alternately painted red and blue. Prove that the sum of areas of
red triangles is equal to the sum of areas of blue ones.
4.29. The extensions of sides AD and BC of convex quadrilateral ABCD meet at point
O; let M and N be the midpoints of sides AB and CD; let P and Q be the midpoints of
diagonals AC and BD. Prove that:
a) SP M QN = 12 |SABD − SACD |;
b) SOP Q = 12 SABCD .
4.30. On sides AB and CD of a convex quadrilateral ABCD points E and F are taken.
Let K, L, M and N be the midpoints of segments DE, BF , CE and AF , respectively.
Prove that quadrilateral KLM N is a convex one and its area does not depend on the choice
of points E and F .
4.31. The midpoints of diagonals AC, BD, CE, . . . of convex hexagon ABCDEF are
vertices of a convex hexagon. Prove that the area of the new hexagon is 41 of that of the
initial one.
4.32. The diameter P Q and the chord RS perpendicular to it intersect in point A.
Point C lies on the circle, point B lies inside the circle and we know that BC k P Q and
BC = RA. From points A and B perpendiculars AK and BL are dropped to line CQ.
Prove that SACK = SBCL .
***
4.33. Through point O inside triangle ABC segments are drawn parallel to its sides
(Fig. 41). Segments AA1 , BB1 and CC1 divide triangle ABC into four triangles and three
quadrilaterals. Prove that the sum of areas of the triangles adjacent to vertices A, B and C
is equal to the area of the fourth triangle.
4.34. On the bisector of angle ∠A of triangle ABC a point A1 is taken so that AA1 =
p − a = 12 (b + c − a) and through point A1 line la perpendicular to the bisector is drawn. If
we similarly construct lines lb and lc , then triangle ABC will be divided into parts among
§7. FORMULAS FOR THE AREA OF A QUADRILATERAL
83
Figure 41 (4.33)
which there are four triangles. Prove that the area of one of these triangles is equal to the
sum of areas of the three other triangles.
See also problems 3.38–3.41, 13.52–13.56, 16.5, 24.5.
§6. Lines and curves that divide figures
into parts of equal area
4.35. Segment M N parallel to side CD of quadrilateral ABCD divides its area in halves
(points M and N lie on sides BC and AD). The lengths of segments drawn from points
A and B parallel to CD till they intersect with lines BC and AD are equal to a and b,
respectively. Prove that M N 2 = 21 (ab + c2 ), where c = CD.
4.36. Each of certain three lines divides the area of a figure in halves. Prove that the
area of the part of the figure confined inside the triangle formed by these lines does not
exceed 14 of the area of the whole figure.
4.37. Line l divides the area of a convex polygon in halves. Prove that this line divides
the projection
√ of the given polygon onto a line perpendicular to l in the ratio that does not
exceed 1 + 2.
4.38. Prove that any convex polygon can be cut by two mutually perpendicular lines in
four figures of equal area.
4.39. a) Prove that any line that divides the area and the perimeter of the triangle in
halves passes through the center of the inscribed circle.
b) Prove a similar statement for any circumscribed polygon.
4.40. Points A and B of circle S1 are connected by an arc of circle S2 that divides the
area of the disk bounded by S1 into equal parts. Prove that the length of the arc of S2 that
connects A and B is greater than that of the diameter of S1 .
4.41. Curve Γ divides a square into two parts of equal area. Prove that on Γ we can
select two points A and B so that line AB passes through the center O of the square.
See also problems 6.51, 6.52, 16.8, 18.29.
§7. Formulas for the area of a quadrilateral
4.42. The diagonals of quadrilateral ABCD meet at point P . The distances from
points A, B and P to line CD are equal to a, b and p, respectively. Prove that the area of
.
quadrilateral ABCD is equal to ab·CD
2p
4.43. Quadrilateral ABCD is inscribed into a circle of radius R; let ϕ be the angle
between the diagonals of ABCD. Prove that the area S of ABCD is equal to 2R2 · sin ∠A ·
sin ∠B · sin ϕ.
84
CHAPTER 4. AREA
4.44. Prove that the area of a quadrilateral whose diagonals are not perpendicular is
equal to 41 tan ϕ · |a2 + c2 − b2 − d2 |, where a, b, c and d are the lengths of the consecutive
sides and ϕ is the angle between the diagonals.
4.45. a) Prove that the area of a convex quadrilatral ABCD can be computed with the
help of the formula
¶
µ
∠B + ∠D
2
2
,
S = (p − a)(p − b)(p − c)(p − d) − abcd cos
2
where p is the semiperimeter, a, b, c, d are the lengths of the quadrilateral’s sides.
b) Prove that if quadrilateral ABCD is an inscribed one, then
S 2 = (p − a)(p − b)(p − c)(p − d).
c) Prove that if quadrilateral ABCD is a circumscribed one, then
µ
¶
∠B + ∠D
2
2
.
S = abcd sin
2
See also Problem 11.34.
§8. An auxiliary area
4.46. Prove that the sum of distances from an arbitrary point within an equilateral
triangle to the triangle’s sides is constant (equal to the length of the triangle’s height).
2bc
4.47. Prove that the length of the bisector AD of triangle ABC is equal to b+c
cos 21 α.
4.48. Inside triangle ABC, point O is taken; lines AO, BO and CO meet the sides of
the triangle at points A1 , B1 and C1 , respectively. Prove that:
OB1
1
1
+ BB
+ OC
= 1;
a) OA
AA1
CC1
1
AC BA1 CB1
b) CB · A1 C · B1 A = 1.
4.49. A (2n − 1)-gon A1 . . . A2n−1 and a point O are given. Lines Ak O and An+k−1 An+k
Bk
for k = 1, . . . , n is equal to 1.
meet at point Bk . Prove that the product of ratios AAn+k−1
n+k Bk
4.50. A convex polygon A1 A2 . . . An is given. On side A1 A2 points B1 and D2 are taken,
on side A2 A3 points B2 and D3 , etc. so that if we construct parallelograms A1 B1 C1 D1 , . . . , An Bn Cn Dn ,
then lines A1 C1 , . . . , An Cn would meet at one point O. Prove that
A1 B1 · A2 B2 · · · · · An Bn = A1 D1 · A2 D2 · · · · · An Dn .
4.51. The lengths of the sides of a triangle form an arithmetic progression. Prove that
the (length of the) radius of the inscribed circle is equal to one third of the length of one of
the triangle’s heights.
4.52. The distances from point X on side BC of triangle ABC to lines AB and AC are
BX·AC
.
equal to db and dc , respectively. Prove that ddcb = CX·AB
4.53. A polygon circumscribed about a circle of radius r is divided into triangles (in
an arbitrary way). Prove that the sum of radii of the inscribed circles of these triangles is
greater than r.
4.54. Through point M inside parallelogram ABCD lines P R and QS parallel to sides
BC and AB are drawn (points P , Q, R and S lie on sides AB, BC, CD and DA, respectively). Prove that lines BS, P D and M C meet at one point.
4.55. Prove that if no side of a quadrilateral is parallel to any other side, then the
midpoint of the segment that connects the intersection points of the opposite sides lies on
the line that connects the midpoints of the diagonals. (The Gauss line.)
4.56. In an acute triangle ABC heights BB1 and CC1 are drawn and points K and L
are taken on sides AB and AC so that AK = BC1 and AL = CB1 . Prove that line AO,
§9. REGROUPING AREAS
85
where O is the center of the circumscribed circle of triangle ABC, divides segment KL in
halves.
4.57. Medians AA1 and CC1 of triangle ABC meet at point M . Prove that if quadrilateral A1 BC1 M is a circumscribed one, then AB = BC.
4.58. Inside triangle ABC a point O is taken. Denote the distances from O to sides
BC, CA, AB of the triangle by da , db , dc , respectively, and the distances from point O to
vertices A, B, C by Ra , Rb , Rc , respectively. Prove that:
a) aRa ≥ cdc + bdb ;
b) da Ra + db Rb + dc Rc ≥ 2(da db + db dc + dc da );
c) Ra + Rb + Rc ≥ 2(da + db + dc );
R
d) Ra Rb Rc ≥ 2r
(da + db )(db + dc )(dc + da ).
See also problems 5.5, 10.6.
§9. Regrouping areas
4.59. Prove that the area of a regular octagon is equal to the product of the lengths of
its greatest and smallest diagonals.
4.60. From the midpoint of each side of an acute triangle perpendiculars are dropped
to two other sides. Prove that the area of the hexagon bounded by these perpendiculars is
equal to a half area of the initial triangle.
4.61. Sides AB and CD of parallelogram ABCD of unit area are divided into n equal
parts; sides AD and BC are divided into m equal parts. The division points are connected
as indicated on a) Fig. 42 a); b) Fig. 42 b).
Figure 42 (4.61)
What are the areas of small parallelograms obtained in this way?
4.62. a) Four vertices of a regular 12-gon lie in the midpoints of a square (Fig. 43).
1
that of the 12-gon.
Prove that the area of the shaded part is equal to 12
Figure 43 (4.62)
b) Prove that the area of a 12-gon inscribed in the unit circle is equal to 3.
86
CHAPTER 4. AREA
Problems for independent study
4.63. The sides of an inscribed quadrilateral ABCD satisfy the relation AB · BC =
AD · DC. Prove that the areas of triangles ABC and ADC are equal.
4.64. Is it possible to use two straight cuts passing through two vertices of a triangle to
divide the triangle into four parts so that three triangles (of these parts) were of equal area?
4.65. Prove that all the convex quadrilaterals with common midpoints of sides are of
equal area.
4.66. Prove that if two triangles obtained by extention of sides of a convex quadrilateral
to their intersection are of equal area, then one of the diagonals divides the other one in
halves.
4.67. The area of a triangle is equal to S, its perimeter is equal to P . Each of the lines
on which the sides of the triangle lie are moved (outwards) by a distance of h. Find the area
and the perimeter of the triangle formed by the three obtained lines.
4.68. On side AB of triangle ABC, points D and E are taken so that ∠ACD = ∠DCE =
∠ECB = ϕ. Find the ratio CD : CE if the lengths of sides AC and BC and angle ϕ are
known.
4.69. Let AA1 , BB1 and CC1 be the bisectors of triangle ABC. Prove that
S A1 B 1 C 1
2abc
.
=
SABC
(a + b) · (b + c) · (c + a)
4.70. Points M and N are the midpoints of lateral sides AB and CD of trapezoid
ABCD. Prove that if the doubled area of the trapezoid is equal to AN · N B + CM · M D,
then AB = CD = BC + AD.
4.71. If a quadrilateral with sides od distinct lengths is inscribed into a circle of radius
R, then there exist two more quadrilaterals not equal to it with the same lengths of sides
inscribed in the same circle. These quadrilaterals have not more than three distinct lengths
d2 d3
.
of diagonals: d1 , d2 and d3 . Prove that the area of the quadrilateral is equal to d14R
4.72. On sides AB, BC and CA of triangle ABC points C1 , A1 and B1 are taken; points
C2 , A2 and B2 are symmetric to these points through the midpoints of the corresponding
sides. Prove that SA1 B1 C1 = SA2 B2 C2 .
4.73. Inside triangle ABC, point P is taken. The lines that pass through P and vertices
of the triangle intersect the sides at points A1 , B1 and C1 . Prove that the area of the triangle
determined by the midpoints of segments AA1 , BB1 and CC1 is equal to 41 of the area of
triangle A1 B1 C1 .
Solutions
4.1. The triangles adjacent to one side have equal bases and a common height and,
therefore, are of equal area. Let M be the intersection point of the medians of triangle
ABC. Line BM divides each of the triangles ABC and AM C into two triangles of equal
area; consequently, SABM = SBCM . Similarly, SBCM = SCAM .
4.2. The equality of areas of triangles ABP and BCP implies that the distances from
points A and C to line BP are equal. Therefore, either line BP passes through the midpoint
of segment AC or it is parallel to it. The points to be found are depicted on Fig. 44.
4.3. Denote the intersection point of line LO with side AC by L1 . Since SLOB = SM OC
and △M OC = △L1 OC, it follows that SLOB = SL1 CO . The heights of triangles LOB and
L1 OC are equal and, therefore, LO = L1 O, i.e., point O lies on the median drawn from
vertex A.
SOLUTIONS
87
Figure 44 (Sol. 4.2)
We similarly prove that point O lies on the medians drawn from vertices B and C, i.e., O
is the intersection point of the medians of the triangle. These arguments also demonstrate
that the intersection point of the medians of the triangle possesses the necessary property.
4.4. Since SA1 BB1 = SA1 AB = SABC , it follows that SAA1 B1 = 2S. Similarly, SBB1 C1 =
SCC1 A1 = 2S. Therefore, SABC = 7S.
4.5. Since AB = BB1 , it follows that SBB1 C = SBAC . Since BC = CC1 , we have
SB1 C1 C = SBB1 C = SBAC and SBB1 C1 = 2SBAC . Similarly, SDD1 A1 = 2SACD and, consequently,
SBB1 C1 + SDD1 A1 = 2SABC + 2SACD = 2SABCD .
Similarly, SAA1 B1 + SCC1 D1 = 2SABCD , consequently,
SA1 B1 C1 D1 = SABCD + SAA1 B1 + SBB1 C1 + SCC1 D1 + SDD1 A1 = 5SABCD .
4.6. Let O be the center of the circumscribed circle. Since AD, BE and CF are
diameters,
SABO = SDEO = SAEO , SBCO = SEF O = SCEO , SCDO = SAF O = SACO .
It is also clear that SABCDEF = 2(SABO + SBCO + SCDO ) and SACE = SAEO + SCEO + SACO .
Therefore, SABCDEF = 2SACE .
4.7. Let E and F be the midpoints of diagonals AC and BD, respectively. Since
SAOB = SAOD , point O lies on line AF . Similarly, point O lies on line CF . Suppose that the
intersection point of the diagonals is not the midpoint of either of them. Then the lines AF
and CF have a unique common point, F ; hence, O = F . We similarly prove that O = E.
Contradiction.
4.8. Let the length of diagonal AC of trapezoid ABCD with base AD be equal to 5.
Let us complement triangle ACB to parallelogram ACBE. The area of trapezoid ABCD
is equal to the area of the right triangle DBE. Let BH be a height of triangle DBE. Then
2
= 25
. Therefore, SDBE = 21 ED·BH = 50
.
EH 2 = BE 2 −BH 2 = 52 −42 = 32 and ED = BE
EH
3
3
4.9. Since SABE = SABC , it follows that EC k AB. The remaining diagonals are also
parallel to the corresponding sides. Let P be the intersection point of BD and EC. If
SBP C = x, then
SABCDE = SABE + SEP B + SEDC + SBP C = 3 + x.
(we have SEP B = SABE = 1 because ABP E is a parallelogram). Since SBP C : √SDP C =
BP : DP =√ SEP B : SEP D , it follows that x : (1 − x) = 1 : x and, therefore, x = 5−1
and
2
5+5
SABCDE = 2 .
4.10. The centers of all the three rectangles coincide (see Problem 1.7) and, therefore,
two smaller rectangles have a common diagonal, KL. Let M and N be the vertices of these
rectangles that lie on side BC. Points M and N lie on the circle with diameter KL. Let O be
88
CHAPTER 4. AREA
the center of the circle, O1 the projection of O to BC. Then BO1 = CO1 and M O1 = N O1
and, therefore, BM = N C. To prove that SKLM + SKLN = SKBCL it suffices to verify that
1
1
(SKBM + SLCM ) + (SKBN + SLCN ) = SKBCL = BC(KB + CL) = BC · AB.
2
2
It remains to observe that
KB · BM + KB · BN = KB · BC,
LC · CM + LC · CN = LC · BC,
KB · BC + LC · BC = AB · BC.
4.11. Let us drop perpendicular l from point C to line AB. Let points A′ , B ′ and E ′
be symmetric to points A, B and E, respectively, through line l. Then triangle AA′ C is
an equilateral one and ∠ACB = ∠BCB ′ = ∠B ′ CA′ = 20◦ . Triangles EE ′ C and DEC are
isosceles ones with the angle of 20◦ at the vertex and a common lateral side EC. Therefore,
SABC + 2SEDC = SABC + 2SEE ′ C . Since E is the midpoint of BC, it follows that 2SEE ′ C =
SBE ′ C = 21 SBB ′ C . Hence,
√
3
SAA′ C
=
.
SABC + 2SEDC =
2
8
4.12. Let the areas of triangles Ta , Tb and Tc be equal to a, b and c, respectively.
Triangles Ta and Tc are homothetic and, therefore, the lines that connect their respective
p
vertices meett at one point, O. The similarity coefficient k of these triangles is equal to ac .
Clearly, SA1 B3 O : SC1 B3 O = A1 O : C1 O = k.
√ Writing similar equations for \$??? and adding
them, we get a : b = k and, therefore, b = ac.
4.13. Making use of the result of Problem 1.3 it is easy to verify that
(1)
It is also clear that
BQ
p + pq
,
=
BB1
1 + p + pq
q + qr
CR
=
,
CC1
1 + q + qr
SP QR
SRB1 C
=
QR
RB1
·
PR
RC
and
SRB1 C
SABC
B1 R
= qr1 + q + qr,
BB1
CP
pr
=
.
CC1
1 + r + pr
=
B1 C
AC
·
B1 R
.
BB1
Hence,
SP QR
QR P R CC1 B1 C
QR P R B1 C
·
=
·
.
=
·
·
·
SABC
BB1 RC AC
BB1 CC1 CR AC
Taking into account that
QR
BB1
and
=1−
p+pq
1+p+pq
=
qr
1
= 1+p+pq
1+q+rq
(1+q)(1−pqr)
(1+p+pq)(1+q+qr)
−
−
rq
1+q+rq
(1 + r)(1 − pqr)
PR
=
CC1
(1 + q + qr)(1 + r + pr)
we get
SP QR
(1 − pqr)2
.
=
SABC
(1 + p + pq)(1 + q + qr)(1 + r + pr)
4.14. If SAOB = SCOD , then AO · BO = CO · DO. Hence, △AOD ∼ △COB and
AD k BC. These arguments are invertible.
4.15. a) Since SADP : SABP = DP : BP = SCDP : SBCP , we have
SABP · SCDP
.
SADP =
SBCP
SOLUTIONS
89
b) Thanks to heading a) SADP · SCBP = SABP · SCDP . Therefore,
SABP · SCBP · SCDP · SADP = (SADP · SCBP )2 .
4.16. After division by 14 sin2 ϕ, where ϕ is the angle between the diagonals, we rewrite
the given equality of the areas in the form
i.e.,
(AP · BP )2 + (CP · DP )2 = (BP · CP )2 + (AP · DP )2 ,
(AP 2 − CP 2 )(BP 2 − DP 2 ) = 0.
4.17. Suppose that quadrilateral ABCD is not a parallelogram; for instance, let lines
AB and CD intersect. By Problem 7.2 the set of points P that lie inside quadrilateral
ABCD for which
1
SABP + SCDP = SBCP + SADP = SABCD
2
is a segment. Therefore, points P1 , P2 and P3 lie on one line. Contradiction.
4.18. Clearly,
1
SAKON = SAKO + SAN O = (SAOB + SAOD ).
2
1
Similarly, SCLOM = 2 (SBCO + SCOD ). Hence, SAKON + SCLOM = 12 SABCD .
4.19. If the areas of the parallelograms KBLO and M DN O are equal, then OK · OL =
OM · ON . Taking into account that ON = KA and OM = LC, we get KO : KA = LC :
LO. Therefore, △KOA ∼ △LCO which means that point O lies on diagonal AC. These
arguments are invertible.
4.20. Let h1 , h and h2 be the distances from points A, M and B to line CD, respectively.
By Problem 1.1 b) we have h = ph2 + (1 − p)h1 , where p = AM
. Therefore,
AB
h · DC
h2 p · DC + h1 (1 − p) · DC
=
= SBCN + SADN .
2
2
Subtracting SDKN + SCLN from both sides of this equality we get the desired statement.
4.21. Thanks to Problem 4.20,
SDM C =
SABD1 + SCDB1 = SABCD .
Hence,
SA1 B1 C1 D1 = SA1 B1 D1 + SC1 D1 B1
= (1 − 2p)SABD1 + (1 − 2p)SCDB1 = (1 − 2p)SABCD .
4.22. By Problem 4.21 the area of the middle quadrilateral of those deteremined by
segments that connect points of sides AB and CD is 15 of the area of the initial quadrilateral.
Since each of the considered segments is divided by segments that connect the corresponding
points of the other pair of opposite sides into 5 equal parts (see Problem 1.16). By making
use once again of the result of Problem 4.21, we get the desired statement.
4.23. On sides AB, BC, CD and AD points K, L, M and N , respectively, are taken.
Suppose that diagonal KM is not parallel to side AD. Fix points K, M and N and let us
move point L along side BC. In accordance with this movement the area of triangle KLM
varies strictly monotonously. Moreover, if LN k AB, then the equality SAKN + SBKL +
SCLM + SDM N = 21 SABCD holds, i.e., SKLM N = 21 SABCD .
4.24. Let L1 and N1 be the midpoints of sides BC and AD, respectively. Then KL1 M N1
is a parallelogram and its area is equal to a half area of quadrilateral ABCD, cf. Problem
1.37 a). Therefore, it suffices to prove that the areas of parallelograms KLM N and KL1 M N1
are equal. If these parallelograms coincide, then there is nothing more to prove and if they
do not coincide, then LL1 k N N1 and BC k AD because the midpoint of segment KM is
90
CHAPTER 4. AREA
their center of symmetry. In this case the midline KM of trapezoid ABCD is parallel to
bases BC and AD and therefore, heights of triangles KLM and KL1 M dropped to side
KM are equal, i.e., the areas of parallelograms KLM N and KL1 M N1 are equal.
4.25. Let the given lines l1 and l2 divide the square into four parts whose areas are equal
to S1 , S2 , S3 and S4 so that for the first line the areas of the parts into which it divides the
square are equal to S1 + S2 and S3 + S4 and for the second line they are equal to S2 + S3
and S1 + S4 . Since by assumption S1 = S2 = S3 , it follows that S1 + S2 = S2 + S3 . This
means that the image of line l1 under the rotation about the center of the square through
an angle of +90◦ or −90◦ is not just parallel to line l2 but coincides with it.
It remains to prove that line l1 (hence, line l2 ) passes through the center of the square.
Suppose that this is not true. Let us consider the images of lines l1 and l2 under rotations
through an angle of ±90◦ and denote the areas of the parts into which they divide the square
as plotted on Fig. 45 (on this figure both distinct variants of the disposition of the lines are
plotted).
Figure 45 (Sol. 4.25)
Lines l1 and l2 divide the square into four parts whose areas are equal to a, a+b, a+2b+c
and a + b, where numbers a, b and c are nonzero. It is clear that three of the four numbers
indicated cannot be equal. Contradiction.
4.26. All the three triangles considered have a common base AM . Let hb , hc and hd be
−→ −→ −−→
the distances from points B, C and D, respectively, to line AM . Since AC = AB + AD, it
follows that hc = |hb ± hd |.
4.27. We may assume that P is a common point of parallelograms constructed on sides
AB and BC, i.e., these parallelograms are of the form ABP Q and CBP R. It is clear that
SACRQ = SABP Q + SCBP R .
4.28. Let the length of the hexagon’s side be equal to a. The extensions of red sides of
the hexagon form an equilateral triangle with side 3a and the sum of areas of red triangles
is equal to a half product of a √by the sum of distances from point O to a side of this triangle
and, therefore, it is equal to 3 4 3 a2 , cf. Problem 4.46.
The sum of areas of blue triangles is similarly calculated.
4.29. a) The area of parallelogram P M QN is equal to 41 BC · AD sin α, where α is the
angle between lines AD and BC. The heights of triangles ABD and ACD dropped from
vertices B and C are equal to OB sin α and OC sin α, respectively; hence,
|SABD − SACD | =
|OB − OC| · AD sin α
BC · AD sin α
=
.
2
2
b) Let, for definiteness, rays AD and BC intersect. Since P N k AO and QN k CO,
point N lies inside triangle OP Q. Therefore,
SOP Q = SP QN + SP ON + SQON = 12 SP M QN + 41 SACD + 41 SBCD
= 41 (SABD − SACD + SACD + SBCD = 14 SABCD .
SOLUTIONS
91
4.30. Segments KM and LN are the midlines of triangles CED and AF B and, therefore,
they have a common point — the midpoint of segment EF . Moreover, KM = 12 CD,
LN = 12 AB and the angle between lines KM and LN is equal to the angle α between lines
AB and CD. Therefore, the area of quadrilateral KLM N is equal to 18 AB · CD sin α.
Figure 46 (Sol. 4.31)
4.31. Denote the midpoints of the diagonals of hexagon ABCDEF as shown on Fig.
46. Let us prove that the area of quadrilateral A1 B1 C1 D1 is 41 of the area of quadrilateral
ABCD. To this end let us make use of the fact that the area of the quadrilateral is equal to
a half product of the lengths of the diagonals by the sine of the angle between them. Since
A1 C1 and B1 D1 are the midlines of triangles BDF and ACE, we get the desired statement.
We similarly prove that the area of quadrilateral D1 E1 F1 A1 is 14 of the area of quadrilateral DEF A.
4.32. Let α = ∠P QC. Then
2SACK = CK · AK = (AP cos α) · (AQ sin α) = AR2 sin α · cos α
= BC 2 sin α · cos α = BL · CL = 2SBCL .
4.33. Let Sa , Sb and Sc be the areas of the triangles adjacent to vertices A, B and C;
let S be the area of the fourth of the triangles considered. Clearly,
SACC1 + SBAA1 + SCBB1 = SABC − S + Sa + Sb + Sc .
Moreover,
SABC = SAOC + SAOB + SBOC = SACC1 + SBAA1 + SCBB1 .
4.34. Let O be the center of the inscribed circle of triangle ABC, let B1 be the tangent
point of the inscribed circle and side AC. Let us cut off triangle ABC triangle AOB1 and
reflect AOB1 symmetrically through the bisector of angle OAB1 . Under this reflection line
OB1 turns into line la . Let us perform a similar operation for the remaining triangles. The
common parts of the triangles obtained in this way are three triangles of the considered
partition and the uncovered part of triangle ABC is the fourth triangle. It is also clear that
the area of the uncovered part is equal to the sum of areas of the parts covered twice.
4.35. Let, for definiteness, rays AD and BC meet at point O. Then SCDO : SM N O =
2
c : x2 , where x = M N and SABO : SM N O = ab : x2 because OA : ON = a : x and
OB : OM = b : x. It follows that x2 − c2 = ab − x2 , i.e., 2x2 = ab + c2 .
4.36. Denote the areas of the parts of the figure into which it is divided by lines as
shown on Fig. 47. Let us denote by S the area of the whole figure. Since
1
S3 + (S2 + S7 ) = S = S1 + S6 + (S2 + S7 ),
2
92
CHAPTER 4. AREA
Figure 47 (Sol. 4.36)
it follows that S3 = S1 + S6 . Adding this equality to the equality 21 S = S1 + S2 + S3 + S4 we
get
1
1
S = 2S1 + S2 + S4 + S6 ≥ 2S1 , i.e., S1 ≤ S.
2
4
4.37. Let us denote the projection of line l by B and the endpoints of the projection of
the polygon by A and C. Let C1 be a point of the polygon whose projection is C. Then line
l intersects the polygon at points K and L; let points K1 and L1 be points on lines C1 K and
C1 L that have point A as their projection (Fig. 48).
Figure 48 (Sol. 4.37)
One of the parts into which line l divides the polygon is contained in trapezoid K1 KLL1 ,
the other part contains triangle C1 KL. Therefore, SK1 KLL1 ¡ ≥ SC1 KL
¢ , i.e., AB · (KL +
AB
≥ BC. Solving this
,
we
have
AB
·
2
+
K1 L1 ) ≥ BC · KL. Since K1 L1 = KL · AB+BC
BC
BC
√
BC
inequality we get AB ≤ 1 + 2.
√
AB
Similarly, BC
≤ 1 + 2. (We have to perform the same arguments but interchange A
and C.)
4.38. Let S denote the area of the polygon, l an arbitrary line. Let us introduce a
coordinate system in which line l is Ox-axis. Let S(a) be the area of the part of the polygon
below the line y = a. Clearly, S(a) varies continuously from 0 to S as a varies from −∞ to
+∞ and, therefore (by Calculus, see, e.g., ??), S(a) = 21 S for some a, i.e., the line y = a
divides the area of the polygon in halves.
Similarly, there exists a line perpendicular to l and this perpendicular also divides the
area of the polygon in halves. These two lines divide the polygon into parts whose areas are
equal to S1 , S2 , S3 and S4 (see Fig. 49). Since S1 + S2 = S3 + S4 and S1 + S4 = S2 + S3 ,
we have S1 = S3 = A and S2 = S4 = B. The rotation of line l by 90◦ interchanges points A
SOLUTIONS
93
Figure 49 (Sol. 4.38)
and B. Since A and B vary continuously under the rotation of l, it follows that A = B for
a certain position of l, i.e., the areas of all the four figures are equal at this moment.
4.39. a) Let the line that divides the area and the perimeter of triangle ABC in halves
intersect sides AC and BC at points P and Q, respectively. Denote the center of the
inscribed circle of triangle ABC by O and the radius of the inscribed circle by r. Then
SABQOP = 21 r(AP + AB + BQ) and SOQCP = 12 r(QC + CP ). Since line P Q divides the
perimeter in halves, AP +AB+BQ = QC +CP and, therefore, SABQOP = SOQCP . Moreover,
SABQP = SQCP by the hypothesis. Therefore, SOQP = 0, i.e., line QP passes through point
O.
b) Proof is carried out similarly to that of heading a).
4.40. By considering the image of circle S2 under the symmetry through the center of
circle S1 and taking into account the equality of areas, it is possible to prove that diameter
AA1 of circle S1 intersects S2 at a point K distinct from A and so that AK > A1 K. The
circle of radius KA1 centered at K is tangent to S1 at point A1 and, therefore, BK > A1 K,
i.e., BK + KA > A1 A. It is also clear that the sum of the lengths of segments BK and KA
is smaller than the length of the arc of S2 that connects points A and B.
4.41. The case when point O belongs to Γ is obvious; therefore, let us assume that O
does not belong to Γ. Let Γ′ be the image of the curve Γ under the symmetry through point
O. If curves Γ and Γ′ do not intersect, then the parts into which Γ divides the square cannot
be of equal area. Let X be the intersection point of Γ and Γ′ ; let X ′ be symmetric to X
through point O. Since under the symmetry through point O curve Γ′ turns into Γ, it follows
that X ′ belongs to Γ. Hence, line XX ′ is the desired one.
4.42. Let the areas of triangles AP B, BP C, CP D and DP A be equal to S1 , S2 , S3 and
4
S4 , respectively. Then ap = S3S+S
and b·CD
= S3 + S2 ; consequently,
2
3
(S3 + S4 )(S3 + S2 )
ab · CD
=
.
2p
S3
Taking into account that S2 S4 = S1 S3 we get the desired statement.
4.43. By applying the law of sines to triangles ABC and ABD we get AC = 2R sin ∠B
and BD = 2R · sin ∠A. Therefore,
1
S = AC · BD sin ϕ = 2R2 sin ∠A · sin ∠B · sin ϕ.
2
4.44. Since the area of the quadrilateral is equal to 21 d1 d2 sin ϕ, where d1 and d2 are the
lengths of the diagonals, it remains to verify that 2d1 d2 cos ϕ = |a2 + c2 − b2 − d2 |. Let O be
the intersection point of the diagonals of quadrilateral ABCD and ϕ = ∠AOB. Then
AB 2 = AO2 + BO2 − 2AO · OB cos ϕ;
BC 2 = BO2 + CO2 + 2BO · CO cos ϕ.
94
CHAPTER 4. AREA
Hence,
AB 2 − BC 2 = AO2 − CO2 − 2BO · AC cos ϕ.
Similarly,
CD2 − AD2 = CO2 − AO2 − 2DO · AC cos ϕ.
By adding these equalities we get the desired statement.
Remark. Since
16S 2 = 4d21 d22 sin2 ϕ = 4d21 d22 − (2d1 d2 cos ϕ)2 ,
it follows that 16S 2 = 4d21 d22 − (a2 + c2 − b2 − d2 )2 .
4.45. a) Let AB = a, BC = b, CD = c and AD = d. Clearly,
Therefore,
S = SABC + SADC = 12 ab sin ∠B + cd sin ∠D;
a2 + b2 − 2ab cos ∠B = AC 2 = c2 + d2 − 2cd cos ∠D.
16S 2 = 4a2 b2 − 4a2 b2 cos2 ∠B + 8abcd sin ∠B sin ∠D + 4c2 d2 − 4c2 d2 cos2 ∠D,
(a2 + b2 − c2 − d2 )2 + 8abcd cos ∠B cos ∠D = 4a2 b2 · cos2 ∠B + 4c2 d2 cos2 ∠D.
By inserting the second equality into the first one we get
16S 2 = 4(ab + cd)2 − (a2 + b2 − c2 − d2 )2 − 8abcd(1 + cos ∠B cos ∠D − sin ∠B sin ∠D).
Clearly,
4(ab + cd)2 − (a2 + b2 − c2 − d2 )2 = 16(p − a)(p − b)(p − c)(p − d);
1 + cos ∠B cos ∠D − sin ∠B sin ∠D = 2 cos2 ∠B+∠D
.
2
b) If ABCD is an inscribed quadrilateral, then ∠B + ∠D = 180◦ and, therefore,
= 0.
cos2 ∠B+∠D
2
c) If ABCD is a circumscribed quadrilateral, then a + c = b + d and, therefore, p =
a + c = b + d and p − a = c, p − b = d, p − c = a, p − d = b. Hence,
¶
µ
∠B + ∠D
2
2 ∠B + ∠D
= abcd sin2
S = abcd 1 − cos
.
2
2
If ABCD is simultaneously an inscribed and circumscribed quadrilateral, then S 2 = abcd.
4.46. Let us drop perpendiculars OA1 , OB1 and OC1 to sides BC, AC and AB, respectively, of an equilateral triangle ABC from a point O inside it. In triangle ABC, let a be
the length of the side, h the length of the height. Clearly, SABC = SBCO + SACO + SABO .
Therefore, ah = a · OA1 + a · OB1 + a · OC1 , i.e., h = OA1 + OB1 + OC1 .
4.47. Let AD = l. Then
α
α
2SABD = cl sin , 2SACD = bl sin , 2SABD = bc sin α.
2
2
Hence,
α
α
α
α
cl sin + bl sin = bc sin α = 2bc sin cos .
2
2
2
2
4.48. a) Let the distances from points A and O to line BC be equal to h and h1 ,
respectively. Then SOBC : SABC = h1 : H = OA1 : AA1 . Similarly, SOAC : SABC = OB1 :
BB1 and SOAB : SABC = OC1 : CC1 . By adding these equalities and taking into account
that SOBC + SOAC + SOAB = SABC we get the desired statement.
b) Let the distances from points B and C to line AA1 be equal to db and dc , respectively.
Then SABO : SACO = db : dc = BA1 : A1 C. Similarly, SACO : SBCO = AC1 : C1 B and
SBCO : SABO = CB1 : B1 A. It remains to multiply these equalities.
SOLUTIONS
95
4.49. It is easy to verify that the ratio of the lengths of segments An+k−1 Bk and An+k Bk
is equal to the ratio of areas of triangles An+k−1 OAk and Ak OAn+k . By multiplying these
equalities we get the desired statement.
4.50. Since Ai Bi Ci Di is a parallelogram and point O lies on the extension of its diagonal
Ai Ci , it follows that SAi Bi O = SAi Di O and, therefore, Ai Bi : Ai Di = hi : hi−1 , where hi is the
distance from point O to side Ai Ai+1 . It remains to multiply these equalities for i = 1, . . . , n.
4.51. Let the lengths of sides of triangle ABC be equal to a, b and c, where a ≤ b ≤ c.
Then 2b = a + c and 2SABC = r(a + b + c) = 3rb, where r is the radius of the inscribed
circle. On the other hand, 2SABC = hb b. Therefore, r = 13 hb .
4.52. It suffices to observe that
db · AB = 2SAXB = BX · AX sin ϕ,
where ϕ = ∠AXB and dc · AC = 2SAXC = CX · AX sin ϕ.
4.53. Let r1 , . . . , rn be the radii of the inscribed circles of the obtained triangles, let
P1 , . . . , Pn their perimeters and S1 , . . . , Sn their areas. Let us denote the area and the
perimeter of the initial polygon by S and P , respectively.
It is clear that Pi < P (cf. Problem 9.27, b). Hence,
r1 + · · · + r n = 2
Sn
S
Sn
S1
S1
= 2 = r.
+ ··· + 2
> 2 + ··· + 2
P1
Pn
P
P
P
4.54. Let us draw lines Q1 S1 and P1 R1 parallel to lines QS and P R through the intersection point N of lines BS and CM (points P1 , Q1 , R1 and S1 lie on sides AB, BC, CD and
DA, respectively). Let F and G be the intersection points of lines P R and Q1 S1 , P1 R1 and
QS, respectively. Since point M lies on diagonal N C of parallelogram N Q1 CR1 , it follows
that SF Q1 QM = SM RR1 G (by Problem 4.19) and, therefore, SN Q1 QG = SN F RR1 . Point N lies
on diagonal BS of parallelogram ABQS and, therefore, SAP1 N S1 = SN Q1 QG = SN F RR1 . It
follows that point N lies on diagonal P D of parallelogram AP RD.
4.55. Let E and F be the intersection points of the extensions of sides of the given
quadrilateral. Denote the vertices of the quadrilateral so that E is the intersection point of
the extensions of sides AB and CD beyond points B and C and F is the intersection point
of rays BC and AD. Let us complement triangles AEF and ABD to parallelograms AERF
and ABLD, respectively.
The homothety with center A and coefficient 2 sends the midpoint of the diagonal BD,
the midpoint of the diagonal AC and the midpoint of segment EF to points L, C and R,
respectively. Therefore, it suffices to prove that points L, C and R lie on one line. This is
precisely the fact proved in the preceding problem.
4.56. It suffices to verify that SAKO = SALO , i.e., AO·AL sin ∠OAL = AO·AK sin ∠OAK.
Clearly,
AL = CB1 = BC cos ∠C, sin ∠OAL = cos ∠B,
AK = BC1 = BC cos ∠B, sin ∠OAK = cos ∠C.
4.57. Since quadrilateral A1 BC1 M is a circumscribed one, then, first, the sums of the
lengths of its opposite sides are equal:
a mc
c ma
+
= +
2
3
2
3
and, second, its inscribed circle is simultaneously the inscribed circle of triangles AA1 B and
CC1 B. Since these triangles have equal areas, their perimeters are equal:
c
a
c + ma + = a + mc + .
2
2
96
CHAPTER 4. AREA
By multiplying the first equality by 3 and adding to the second one we get the desired
statement.
4.58. First, let us prove a general inequality that will be used in the proof of headings
a)–d):
(∗)
BC1 · Ra ≥ B1 K · Ra + C1 L · Ra = 2SAOB1 + 2SAOC1 = AB1 · dc + AC1 · db .
On rays AB and AC take arbitrary points B1 and C1 and drop from them perpendiculars
B1 K and C1 L to line AO. Since B1 C1 ≥ B1 K + C1 L, inequality (∗) follows.
a) Setting B1 = B and C1 = C we get the desired statement.
b) By multiplying both sides of the inequality aRa ≥ cdc + bdb by daa we get
c
b
da Ra ≥ da dc + da db .
a
a
Taking the sum of this inequality with the similar inequalities for db Rb and dc Rc and taking
into account that xy + xy ≥ 2 we get the desired statement.
c) Take points B1 and C1 such that AB1 = AC and AC1 = AB. Then aRa ≥ bdc + cdb ,
i.e., Ra ≥ ab dc + ac db . Taking the sum of this inequality with similar inequalities for Rb and
Rc and taking into account that xy + xy ≥ 2 we get the desired statement.
d) Take points B1 and C1 such that AB1 = AC1 = 1; then B1 C1 = 2 sin 21 ∠A and,
therefore, 2 sin 12 Ra ≥ dc + db . By multiplying this inequality by similar inequalities for Rb
r
(by Problem 12.36 a))
and Rc and taking into account that sin 21 ∠A sin 12 ∠B sin 21 ∠C = 4R
we get the desired statement.
4.59. Let us cut triangles off a regular octagon and replace the triangles as shown on
Fig. 50. As a result we get a rectangle whose sides are equal to the longest and shortest
diagonals of the octagon.
Figure 50 (Sol. 4.59)
4.60. Let A1 , B1 and C1 be the midpoints of sides BC, CA and AB, respectively, of
triangle ABC. The drawn segments are heights of triangles AB1 C1 , A1 BC1 and A1 B1 C,
respectively. Let P , Q and R be the respective intersection points of the heights of these
triangles and O the intersection point of the heights of triangle A1 B1 C1 (Fig. 51).
The considered hexagon consists of triangle A1 B1 C1 and triangles B1 C1 P , C1 A1 Q and
A1 BR. Clearly, △B1 C1 P = △C1 B1 O, △C1 A1 Q = △A1 C1 O and △A1 B1 R = △B1 A1 O.
Therefore, the area of the considered hexagon is equal to the doubled area of triangle A1 B1 C1 .
It remains to observe that SABC = 4SA1 B1 C1 .
4.61. a) Let us cut two parts off the parallelogram (Fig. 52 a)) and replace these parts as
shown on Fig. 52 b). We get a figure composed of mn + 1 small parallelograms. Therefore,
1
the area of a small parallelogram is equal to mn+1
.
b) Let us cut off the parallelogram three parts (Fig. 53 a)) and replace these parts as
indicated on Fig. 53 b). We get a figure that consists of mn − 1 small parallelograms.
1
.
Therefore, the area of a small parallelogram is equal to mn−1
SOLUTIONS
97
Figure 51 (Sol. 4.60)
Figure 52 (Sol. 4.61 a))
Figure 53 (Sol. 4.61 b))
4.62. a) Let us cut the initial square into four squares and consider one of them (Fig. 54).
Let point B ′ be symmetric to point B through line P Q. Let us prove that △AP B = △OB ′ P .
Indeed, triangle AP B is an isosceles one and angle at its base is equal to 15◦ (Problem 2.26),
hence, triangle BP Q is an isosceles one. Therefore,
∠OP B ′ = ∠OP Q − ∠B ′ P Q = 75◦ − 60◦ = 15◦
and ∠P OB ′ = 12 ∠OP Q = 15◦ . Moreover, AB = OP . We similarly prove that △BQC =
△OB ′ Q. It follows that the area of the shaded part on Fig. 43 is equal to the area of triangle
OP Q.
Figure 54 (Sol. 4.62)
b) Let the area of the regular 12-gon inscribed in a circle of radius 1 be equal to 12x.
Thanks to heading a) the area of the square circumscribed around this circle is equal to
98
CHAPTER 4. AREA
12x + 4x = 16x; on the other hand, the area of the square is equal to 4; hence, x =
12x = 3.
1
4
and
CHAPTER 5. TRIANGLES
Background
1) The inscribed circle of a triangle is the circle tangent to all its sides. The center of an
inscribed circle is the intersection point of the bisectors of the triangle’s angles.
An escribed circle of triangle ABC is the circle tangent to one side of the triangle and
extensions of the other two sides. For each triangle there are exactly three escribed circles.
The center of an escribed circle tangent to side AB is the intersection point of the bisector
of angle C and the bisectors of the outer angles A and B.
The circumscribed circle of a triangle is the circle that passes through the vertices of the
triangle. The center of the circumscribed circle of a triangle is the intersection point of the
midperpendiculars to the triangle’s sides.
2) For elements of a triangle ABC the following notations are often used:
a, b and c are the lengths of sides BC, CA and AB, respectively;
α, β and γ are the values of angles at vertices A, B, C;
R is the radius of the circumscribed cirlce;
r is the radius of the inscribed circle;
ra , rb and rc are the radii of the escribed circles tangent to sides BC, CA and AB,
respectively;
ha , hb and hc the lengths of the heights dropped from vertices A, B and C, respectively.
3) If AD is the bisector of angle A of triangle ABC (or the bisector of the outer angle
A), then BD : CD = AB : AC (cf. Problem 1.17).
4) In a right triangle, the median drawn from the vertex of the right angle is equal to a
half the hypothenuse (cf. Problem 5.16).
5) To prove that the intersection points of certain lines lie on one line Menelaus’s theorem
(Problem 5.58) is often used.
6) To prove that certain lines intersect at one point Ceva’s theorem (Problem 5.70) is
often used.
Introductory problems
1. Prove that the triangle is an isosceles one if a) one of its medians coincides with a
height;
b) if one of its bisectors coincides with a height.
2. Prove that the bisectors of a triangle meet at one point.
3. On height AH of triangle ABC a point M is taken. Prove that AB 2 − AC 2 =
M B2 − M C 2.
4. On sides AB, BC, CA of an equilateral triangle ABC points P , Q and R, respectively,
are taken so that
AP : P B = BQ : QC = CR : RA = 2 : 1.
Prove that the sides of triangle P QR are perpendicular to the respective sides of triangle
ABC.
99
100
CHAPTER 5. TRIANGLES
1. The inscribed and the circumscribed circles
5.1. On sides BC, CA and AB of triangle ABC, points A1 , B1 and C1 , respectively, are
taken so that AC1 = AB1 , BA1 = BC1 and CA1 = CB1 . Prove that A1 , B and C1 are the
points at which the inscribed circle is tangent to the sides of the triangle.
5.2. Let Oa , Ob and Oc be the centers of the escribed circles of triangle ABC. Prove
that points A, B and C are the bases of heights of triangle Oa Ob Oc .
5.3. Prove that side BC of triangle ABC subtends (1) an angle with the vertex at the
center O of the inscribed circle; the value of the angle is equal to 90◦ + 12 ∠A and (2) an
angle with the vertex at the center Oa of the escribed circle; the value of the angle is equal
to 90◦ − 12 ∠A.
5.4. Inside triangle ABC, point P is taken such that
∠P AB : ∠P AC = ∠P CA : ∠P CB = ∠P BC : ∠P BA = x.
Prove that x = 1.
5.5. Let A1 , B1 and C1 be the projections of an inner point O of triangle ABC to the
heights. Prove that if the lengths of segments AA1 , BB1 and CC1 are equal, then they are
equal to 2r.
5.6. An angle of value α = ∠BAC is rotated about its vertex O, the midpoint of the
basis AC of an isosceles triangle ABC. The legs of this angle meet the segments AB and
BC at points P and Q, respectively. Prove that the perimeter of triangle P BQ remains
constant under the rotation.
5.7. In a scalene triangle ABC, line M O is drawn through the midpoint M of side BC
and the center O of the inscribed circle. Line M O intersects height AH at point E. Prove
that AE = r.
5.8. A circle is tangent to the sides of an angle with vertex A at points P and Q.
The distances from points P , Q and A to a tangent to this circle are equal to u, v and w,
respectively. Prove that wuv2 = sin2 12 ∠A.
***
5.9. Prove that the points symmetric to the intersection point of the heights of triangle
ABC through its sides lie on the circumscribed circle.
5.10. From point P of arc BC of the circumscribed circle of triangle ABC perpendiculars
P X, P Y and P Z are dropped to BC, CA and AB, respectively. Prove that PBC
= PAC
+ AB
.
X
Y
PZ
***
5.11. Let O be the center of the circumscribed circle of triangle ABC, let I be the center
of the inscribed circle, Ia the center of the escribed circle tangent to side BC. Prove that
a) d2 = R2 − 2Rr, where d = OI;
b) d2a = R2 + 2Rra , where da = OIa .
5.12. The extensions of the bisectors of the angles of triangle ABC intersect the circumscribed circle at points A1 , B1 and C1 ; let M be the intersection point of bisectors. Prove
C1
A·M C
that a) MM
= 2r; b) M AM1 ·M
= R.
B1
B
5.13. The lengths of the sides of triangle ABC form an arithmetic progression: a, b, c,
where a < b < c. The bisector of angle ∠B intersects the circumscribed circle at point B1 .
Prove that the center O of the inscribed circle divides segment BB1 in halves.
5.14. In triangle ABC side BC is the shortest one. On rays BA and CA, segments
BD and CE, respectively, each equal to BC, are marked. Prove that the radius of the
* * *
101
circumscribed circle of triangle ADE is equal to the distance between the centers of the
inscribed and circumscribed circles of triangle ABC.
§2. Right triangles
5.15. In triangle ABC, angle ∠C is a right one. Prove that r = a+b−c
and rc = a+b+c
.
2
2
1
5.16. In triangle ABC, let M be the midpoint of side AB. Prove that CM = 2 AB if
and only if ∠ACB = 90◦ .
5.17. Consider trapezoid ABCD with base AD. The bisectors of the outer angles at
vertices A and B meet at point P and the bisectors of the angles at vertices C and D meet at
point Q. Prove that the length of segment P Q is equal to a half perimeter of the trapezoid.
5.18. In an isosceles triangle ABC with base AC bisector CD is drawn. The line
that passes through point D perpendicularly to DC intersects AC at point E. Prove that
EC = 2AD.
5.19. The sum of angles at the base of a trapezoid is equal to 90◦ . Prove that the
segment that connects the midpoints of the bases is equal to a half difference of the bases.
5.20. In a right triangle ABC, height CK from the vertex C of the right angle is drawn
and in triangle ACK bisector CE is drawn. Prove that CB = BE.
5.21. In a right triangle ABC with right angle ∠C, height CD and bisector CF are
drawn; let DK and DL be bisectors in triangles BDC and ADC. Prove that CLF K is a
square.
5.22. On hypothenuse AB of right triangle ABC, square ABP Q is constructed outwards.
Let α = ∠ACQ, β = ∠QCP and γ = ∠P CB. Prove that cos β = cos α · cos γ.
See also Problems 2.65, 5.62.
§3. The equilateral triangles
5.23. From a point M inside an equilateral triangle ABC perpendiculars M P , M Q and
M R are dropped to sides AB, BC and CA, respectively. Prove that
AP 2 + BQ2 + CR2 = P B 2 + QC 2 + RA2 ,
AP + BQ + CR = P B + QC + RA.
5.24. Points D and E divide sides AC and AB of an equilateral triangle ABC in the
ratio of AD : DC = BE : EA = 1 : 2. Lines BD and CE meet at point O. Prove that
∠AOC = 90◦ .
***
5.25. A circle divides each of the sides of a triangle into three equal parts. Prove that
this triangle is an equilateral one.
5.26. Prove that if the intersection point of the heights of an acute triangle divides the
heights in the same ratio, then the triangle is an equilateral one.
5.27. a) Prove that if a + ha = b + hb = c + hc , then triangle ABC is a equilateral one.
b) Three squares are inscribed in triangle ABC: two vertices of one of the squares lie on
side AC, those of another one lie on side BC, and those of the third lie one on AB. Prove
that if all the three squares are equal, then triangle ABC is an equilateral one.
5.28. The circle inscribed in triangle ABC is tangent to the sides of the triangle at
points A1 , B1 , C1 . Prove that if triangles ABC and A1 B1 C1 are similar, then triangle ABC
is an equilateral one.
5.29. The radius of the inscribed circle of a triangle is equal to 1, the lengths of the
heights of the triangle are integers. Prove that the triangle is an equilateral one.
102
CHAPTER 5. TRIANGLES
See also Problems 2.18, 2.26, 2.36, 2.44, 2.54, 4.46, 5.56, 7.45, 10.3, 10.77, 11.3, 11.5,
16.7, 18.9, 18.12, 18.15, 18.17-18.20, 18.22, 18.38, 24.1.
§4. Triangles with angles of 60◦ and 120◦
5.30. In triangle ABC with angle A equal to 120◦ bisectors AA1 , BB1 and CC1 are
drawn. Prove that triangle A1 B1 C1 is a right one.
5.31. In triangle ABC with angle A equal to 120◦ bisectors AA1 , BB1 and CC1 meet
at point O. Prove that ∠A1 C1 O = 30◦ .
5.32. a) Prove that if angle ∠A of triangle ABC is equal to 120◦ then the center of the
circumscribed circle and the orthocenter are symmetric through the bisector of the outer
angle ∠A.
b) In triangle ABC, the angle ∠A is equal to 60◦ ; O is the center of the circumscribed
circle, H is the orthocenter, I is the center of the inscribed circle and Ia is the center of the
escribed circle tangent to side BC. Prove that IO = IH and Ia O = Ia H.
5.33. In triangle ABC angle ∠A is equal to 120◦ . Prove that from segments of lengths
a, b and b + c a triangle can be formed.
5.34. In an acute triangle ABC with angle ∠A equal to 60◦ the heights meet at point
H.
a) Let M and N be the intersection points of the midperpendiculars to segments BH
and CH with sides AB and AC, respectively. Prove that points M , N and H lie on one
line.
b) Prove that the center O of the circumscribed circle lies on the same line.
5.35. In triangle ABC, bisectors BB1 and CC1 are drawn. Prove that if ∠CC1 B1 = 30◦ ,
then either ∠A = 60◦ or ∠B = 120◦ .
See also Problem 2.33.
§5. Integer triangles
5.36. The lengths of the sides of a triangle are consecutive integers. Find these integers
if it is known that one of the medians is perpendicular to one of the bisectors.
5.37. The lengths of all the sides of a right triangle are integers and the greatest common
divisor of these integers is equal to 1. Prove that the legs of the triangle are equal to 2mn
and m2 − n2 and the hypothenuse is equal to m2 + n2 , where m and n are integers.
A right triangle the lengths of whose sides are integers is called a Pythagorean triangle.
5.38. The radius of the inscribed circle of a triangle is equal to 1 and the lengths of its
sides are integers. Prove that these integers are equal to 3, 4, 5.
5.39. Give an example of an inscribed quadrilateral with pairwise distinct integer lengths
of sides and the lengths of whose diagonals, the area and the radius of the circumscribed
circle are all integers. (Brakhmagupta.)
5.40. a) Indicate two right triangles from which one can compose a triangle so that the
lengths of the sides and the area of the composed triangle would be integers.
b) Prove that if the area of a triangle is an integer and the lengths of the sides are
consecutive integers then this triangle can be composed of two right triangles the lengths of
whose sides are integers.
5.41. a) In triangle ABC, the lengths of whose sides are rational numbers, height BB1
is drawn.
Prove that the lengths of segments AB1 and CB1 are rational numbers.
§6. MISCELLANEOUS PROBLEMS
103
b) The lengths of the sides and diagonals of a convex quadrilateral are rational numbers.
Prove that the diagonals cut it into four triangles the lengths of whose sides are rational
numbers.
See also Problem 26.7.
§6. Miscellaneous problems
5.42. Triangles ABC and A1 B1 C1 are such that either their corresponding angles are
equal or their sum is equal to 180◦ . Prove that the corresponding angles are equal, actually.
5.43. Inside triangle ABC an arbitrary point O is taken. Let points A1 , B1 and C1 be
symmetric to O through the midpoints of sides BC, CA and AB, respectively. Prove that
△ABC = △A1 B1 C1 and, moreover, lines AA1 , BB1 and CC1 meet at one point.
5.44. Through the intersection point O of the bisectors of triangle ABC lines parallel
to the sides of the triangle are drawn. The line parallel to AB meets AC and BC at points
M and N , respectively, and lines parallel to AC and BC meet AB at points P and Q,
respectively. Prove that M N = AM + BN and the perimeter of triangle OP Q is equal to
the length of segment AB.
5.45. a) Prove that the heigths of a triangle meet at one point.
b) Let H be the intersection point of heights of triangle ABC and R the radius of the
circumscribed circle. Prove that
AH 2 + BC 2 = 4R2
and AH = BC| cot α|.
5.46. Let x = sin 18◦ . Prove that 4x2 + 2x = 1.
5.47. Prove that the projections of vertex A of triangle ABC on the bisectors of the
outer and inner angles at vertices B and C lie on one line.
5.48. Prove that if two bisectors in a triangle are equal, then the triangle is an isosceles
one.
5.49. a) In triangles ABC and A′ B ′ C ′ , sides AC and A′ C ′ are equal, the angles at
vertices B and B ′ are equal, and the bisectors of angles ∠B and ∠B ′ are equal. Prove that
these triangles are equal. (More precisely, either △ABC = △A′ B ′ C ′ or △ABC = △C ′ B ′ A′ .)
b) Through point D on the bisector BB1 of angle ABC lines AA1 and CC1 are drawn
(points A1 and C1 lie on sides of triangle ABC). Prove that if AA1 = CC1 , then AB = BC.
5.50. Prove that a line divides the perimeter and the area of a triangle in equal ratios if
and only if it passes through the center of the inscribed circle.
5.51. Point E is the midpoint of arc ⌣ AB of the circumscribed circle of triangle ABC
on which point C lies; let C1 be the midpoint of side AB. Perpendicular EF is dropped
from point E to AC. Prove that:
a) line C1 F divides the perimeter of triangle ABC in halves;
b) three such lines constructed for each side of the triangle meet at one point.
5.52. On sides AB and BC of an acute triangle ABC, squares ABC1 D1 and A2 BCD2
are constructed outwards. Prove that the intersection point of lines AD2 and CD1 lies on
height BH.
5.53. On sides of triangle ABC squares centered at A1 , B1 and C1 are constructed
outwards. Let a1 , b1 and c1 be the lengths of the sides of triangle A1 B1 C1 ; let S and S1 be
the areas of triangles ABC and A1 B1 C1 , respectively. Prove that:
a) a21 + b21 + c21 = a2 + b2 + c2 + 6S.
b) S1 − S = 81 (a2 + b2 + c2 ).
5.54. On sides AB, BC and CA of triangle ABC (or on their extensions), points C1 ,
A1 and B1 , respectively, are taken so that ∠(CC1 , AB) = ∠(AA1 , BC) = ∠(BB1 , CA) =
104
CHAPTER 5. TRIANGLES
α. Lines AA1 and BB1 , BB1 and CC1 , CC1 and AA1 intersect at points C ′ , A′ and B ′ ,
respectively. Prove that:
a) the intersection point of heights of triangle ABC coincides with the center of the
circumscribed circle of triangle A′ B ′ C ′ ;
b) △A′ B ′ C ′ ∼ △ABC and the similarity coefficient is equal to 2 cos α.
5.55. On sides of triangle ABC points A1 , B1 and C1 are taken so that AB1 : B1 C =
n
c : an , BC1 : CA = an : bn and CA1 : A1 B = bn : cn (here a, b and c are the lengths of
the triangle’s sides). The circumscribed circle of triangle A1 B1 C1 singles out on the sides of
triangle ABC segments of length ±x, ±y and ±z, where the signs are chosen in accordance
with the orientation of the triangle. Prove that
x
y
z
+ n−1 + n−1 = 0.
n−1
a
b
c
5.56. In triangle ABC trisectors (the rays that divide the angles into three equal parts)
are drawn. The nearest to side BC trisectors of angles B and C intersect at point A1 ; let us
define points B1 and C1 similarly, (Fig. 55). Prove that triangle A1 B1 C1 is an equilateral
one. (Morlie’s theorem.)
Figure 55 (5.56)
5.57. On the sides of an equilateral triangle ABC as on bases, isosceles triangles A1 BC,
AB1 C and ABC1 with angles α, β and γ at the bases such that α + β + γ = 60◦ are
constructed inwards. Lines BC1 and B1 C meet at point A2 , lines AC1 and A1 C meet at
point B2 , and lines AB1 and A1 B meet at point C2 . Prove that the angles of triangle A2 B2 C2
are equal to 3α, 3β and 3γ.
§7. Menelaus’s theorem
−→
−−→
AB
AB
Let AB and CD be colinear vectors. Denote by CD
the quantity ± CD
, where the plus
−→
−−→
sign is taken if the vectors AB and CD are codirected and the minus sign if the vectors are
directed opposite to each other.
5.58. On sides BC, CA and AB of triangle ABC (or on their extensions) points A1 , B1
and C1 , respectively, are taken. Prove that points A1 , B1 and C1 lie on one line if and only
if
BA1 CB1 AC1
·
·
= 1.
(Menelaus’s theorem)
CA1 AB1 BC1
5.59. Prove Problem 5.85 a) with the help of Menelaus’s theorem.
* * *
105
5.60. A circle S is tangent to circles S1 and S2 at points A1 and A2 , respectively. Prove
that line A1 A2 passes through the intersection point of either common outer or common
inner tangents to circles S1 and S2 .
5.61. a) The midperpendicular to the bisector AD of triangle ABC intersects line BC
at point E. Prove that BE : CE = c2 : b2 .
b) Prove that the intersection point of the midperpendiculars to the bisectors of a triangle
and the extensions of the corresponding sides lie on one line.
5.62. From vertex C of the right angle of triangle ABC height CK is dropped and in
triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets
CK at point F . Prove that line EF divides segment AC in halves.
5.63. On lines BC, CA and AB points A1 , B1 and C1 , respectively, are taken so that
points A1 , B1 and C1 lie on one line. The lines symmetric to lines AA1 , BB1 and CC1
through the corresponding bisectors of triangle ABC meet lines BC, CA and AB at points
A2 , B2 and C2 , respectively. Prove that points A2 , B2 and C2 lie on one line.
***
5.64. Lines AA1 , BB1 and CC1 meet at one point, O. Prove that the intersection points
of lines AB and A1 B1 , BC and B1 C1 , AC and A1 C1 lie on one line. (Desargues’s theorem.)
5.65. Points A1 , B1 and C1 are taken on one line and points A2 , B2 and C2 are taken on
another line. The intersection pointa of lines A1 B2 with A2 B1 , B1 C2 with B2 C1 and C1 A2
with C2 A1 are C, A and B, respectively. Prove that points A, B and C lie on one line.
(Pappus’ theorem.)
5.66. On sides AB, BC and CD of quadrilateral ABCD (or on their extensions) points
K, L and M are taken. Lines KL and AC meet at point P , lines LM and BD meet at
point Q. Prove that the intersection point of lines KQ and M P lies on line AD.
5.67. The extensions of sides AB and CD of quadrilateral ABCD meet at point P and
the extensions of sides BC and AD meet at point Q. Through point P a line is drawn that
intersects sides BC and AD at points E and F . Prove that the intersection points of the
diagonals of quadrilaterals ABCD, ABEF and CDF E lie on the line that passes through
point Q.
5.68. a) Through points P and Q triples of lines are drawn. Let us denote their
intersection points as shown on Fig. 56. Prove that lines KL, AC and M N either meet at
one point or are parallel.
Figure 56 (5.68)
b) Prove further that if point O lies on line BD, then the intersection point of lines KL,
AC and M N lies on line P Q.
5.69. On lines BC, CA and AB points A1 , B1 and C1 are taken. Let P1 be an arbitrary
point of line BC, let P2 be the intersection point of lines P1 B1 and AB, let P3 be the
106
CHAPTER 5. TRIANGLES
intersection point of lines P2 A1 and CA, let P4 be the intersection point of P3 C1 and BC,
etc. Prove that points P7 and P1 coincide.
See also Problem 6.98.
§8. Ceva’s theorem
5.70. Triangle ABC is given and on lines AB, BC and CA points C1 , A1 and B1 ,
respectively, are taken so that k of them lie on sides of the triangle and 3 − k on the
extensions of the sides. Let
BA1 CB1 AC1
R=
·
·
.
CA1 AB1 BC1
Prove that
a) points A1 , B1 and C1 lie on one line if and only if R = 1 and k is even. (Menelaus’s
theorem.)
b) lines AA1 , BB1 and CC1 either meet at one point or are parallel if and only if R = 1
and k is odd. (Ceva’s theorem.)
5.71. The inscribed (or an escribed) circle of triangle ABC is tangent to lines BC, CA
and AB at points A1 , B1 and C1 , respectively. Prove that lines AA1 , BB1 and CC1 meet at
one point.
5.72. Prove that the heights of an acute triangle intersect at one point.
5.73. Lines AP, BP and CP meet the sides of triangle ABC (or their extensions) at
points A1 , B1 and C1 , respectively. Prove that:
a) lines that pass through the midpoints of sides BC, CA and AB parallel to lines AP ,
BP and CP , respectively, meet at one point;
b) lines that connect the midpoints of sides BC, CA and AB with the midpoints of
segments AA1 , BB1 , CC1 , respectively, meet at one point.
5.74. On sides BC, CA, and AB of triangle ABC, points A1 , B1 and C1 are taken so
that segments AA1 , BB1 and CC1 meet at one point. Lines A1 B1 and A1 C1 meet the line
that passes through vertex A parallel to side BC at points C2 and B2 , respectively. Prove
that AB2 = AC2 .
5.75. a) Let α, β and γ be arbitrary angles such that the sum of any two of them is not
less than 180◦ . On sides of triangle ABC, triangles A1 BC, AB1 C and ABC1 with angles
at vertices A, B, and C equal to α, β and γ, respectively, are constructed outwards. Prove
that lines AA1 , BB1 and CC1 meet at one point.
b) Prove a similar statement for triangles constructed on sides of triangle ABC inwards.
5.76. Sides BC, CA and AB of triangle ABC are tangent to a circle centered at O at
points A1 , B1 and C1 . On rays OA1 , OB1 and OC1 equal segments OA2 , OB2 and OC2 are
marked. Prove that lines AA2 , BB2 and CC2 meet at one point.
5.77. Lines AB, BP and CP meet lines BC, CA and AB at points A1 , B1 and C1 ,
respectively. Points A2 , B2 and C2 are selected on lines BC, CA and AB so that
BA2 : A2 C = A1 C : BA1 ,
CB2 : B2 A = B1 A : CB1 ,
AC2 : C2 B = C1 B : AC1 .
Prove that lines AA2 , BB2 and CC2 also meet at one point, Q (or are parallel).
Such points P and Q are called isotomically conjugate with respect to triangle ABC.
5.78. On sides BC, CA, AB of triangle ABC points A1 , B1 and C1 are taken so that
lines AA1 , BB1 and CC1 intersect at one point, P . Prove that lines AA2 , BB2 and CC2
symmetric to these lines through the corresponding bisectors also intersect at one point, Q.
§9. SIMSON’S LINE
107
Such points P and Q are called isogonally conjugate with respect to triangle ABC.
5.80. The opposite sides of a convex hexagon are pairwise parallel. Prove that the lines
that connect the midpoints of opposite sides intersect at one point.
5.81. From a point P perpendiculars P A1 and P A2 are dropped to side BC of triangle
ABC and to height AA3 . Points B1 , B2 and C1 , C2 are similarly defined. Prove that lines
A1 A2 , B1 B2 and C1 C2 either meet at one point or are parallel.
5.82. Through points A and D lying on a circle tangents that intersect at point S are
drawn. On arc ⌣ AD points B and C are taken. Lines AC and BD meet at point P , lines
AB and CD meet at point Q. Prove that line P Q passes through point S.
5.83. a) On sides BC, CA and AB of an isosceles triangle ABC with base AB, points
A1 , B1 and C1 , respectively, are taken so that lines AA1 , BB1 and CC1 meet at one point.
Prove that
sin ∠ABB1 · sin ∠CAA1
AC1
=
.
C1 B
sin ∠BAA1 · sin ∠CBB1
b) Inside an isosceles triangle ABC with base AB points M and N are taken so that
∠CAM = ∠ABN and ∠CBM = ∠BAN . Prove that points C, M and N lie on one line.
5.84. In triangle ABC bisectors AA1 , BB1 and CC1 are drawn. Bisectors AA1 and CC1
intersect segments C1 B1 and B1 A1 at points M and N , respectively. Prove that ∠M BB1 =
∠N BB1 .
See also Problems 10.56, 14.7, 14.38.
§9. Simson’s line
5.85. a) Prove that the bases of the perpendiculars dropped from a point P of the
circumscribed circle of a triangle to the sides of the triangle or to their extensions lie on one
line.
This line is called Simson’s line of point P with respect to the triangle.
b) The bases of perpendiculars dropped from a point P to the sides (or their extensions)
of a triangle lie on one line. Prove that point P lies on the circumscribed circle of the
triangle.
5.86. Points A, B and C lie on one line, point P lies outside this line. Prove that the
centers of the circumscribed circles of triangles ABP , BCP , ACP and point P lie on one
circle.
5.87. In triangle ABC the bisector AD is drawn and from point D perpendiculars DB ′
and DC ′ are dropped to lines AC and AB, respectively; point M lies on line B ′ C ′ and
DM ⊥ BC. Prove that point M lies on median AA1 .
5.88. a) From point P of the circumscribed circle of triangle ABC lines P A1 , P B1 and
P C1 are drawn at a given (oriented) angle α to lines BC, CA and AB, respectively, so that
points A1 , B1 and C1 lie on lines BC, CA and AB, respectively. Prove that points A1 , B1
and C1 lie on one line.
b) Prove that if in the definition of Simson’s line we replace the angle 90◦ by an angle α,
i.e., replace the perpendiculars with the lines that form angles of α, their intersection points
with the sides lie on the line and the angle between this line and Simson’s line becomes equal
to 90◦ − α.
5.89. a) From a point P of the circumscribed circle of triangle ABC perpendiculars P A1
and P B1 are dropped to lines BC and AC, respectively. Prove that P A · P A1 = 2Rd, where
R is the radius of the circumscribed circle, d the distance from point P to line A1 B1 .
b) Let α be the angle between lines A1 B1 and BC. Prove that cos α = P2RA .
108
CHAPTER 5. TRIANGLES
5.90. Let A1 and B1 be the projections of point P of the circumscribed circle of triangle
ABC to lines BC and AC, respectively. Prove that the length of segment A1 B1 is equal to
the length of the projection of segment AB to line A1 B1 .
5.91. Points P and C on a circle are fixed; points A and B move along the circle so that
angle ∠ACB remains fixed. Prove that Simson’s lines of point P with respect to triangle
ABC are tangent to a fixed circle.
5.92. Point P moves along the circumscribed circle of triangle ABC. Prove that Simson’s
line of point P with respect to triangle ABC rotates accordingly through the angle equal to
a half the angle value of the arc circumvent by P .
5.93. Prove that Simson’s lines of two diametrically opposite points of the circumscribed
circle of triangle ABC are perpendicular and their intersection point lies on the circle of 9
points, cf. Problem 5.106.
5.94. Points A, B, C, P and Q lie on a circle centered at O and the angles between vector
−→
−→ −−→ −→
−→
OP and vectors OA, OB, OC and OQ are equal to α, β, γ and 21 (α + β + γ), respectively.
Prove that Simson’s line of point P with respect to triangle ABC is parallel to OQ.
5.95. Chord P Q of the circumscribed circle of triangle ABC is perpendicular to side
BC. Prove that Simson’s line of point P with respect to triangle ABC is parallel to line
AQ.
5.96. The heights of triangle ABC intersect at point H; let P be a point of its circumscribed circle. Prove that Simson’s line of point P with respect to triangle ABC divides
segment P H in halves.
5.97. Quadrilateral ABCD is inscribed in a circle; la is Simson’s line of point A with
respect to triangle BCD; let lines lb , lc and ld be similarly defined. Prove that these lines
intersect at one point.
5.98. a) Prove that the projection of point P of the circumscribed circle of quadrilateral
ABCD onto Simson’s lines of this point with respect to triangles BCD, CDA, DAB and
BAC lie on one line. (Simson’s line of the inscribed quadrilateral.)
b) Prove that by induction we can similarly define Simson’s line of an inscribed n-gon
as the line that contains the projections of a point P on Simson’s lines of all (n − 1)-gons
obtained by deleting one of the vertices of the n-gon.
See also Problems 5.10, 5.59.
§10. The pedal triangle
Let A1 , B1 and C1 be the bases of the perpendiculars dropped from point P to lines
BC, CA and AB, respectively. Triangle A1 B1 C1 is called the pedal triangle of point P with
respect to triangle ABC.
5.99. Let A1 B1 C1 be the pedal triangle of point P with respect to triangle ABC. Prove
that B1 C1 = BC·AP
, where R is the radius of the circumscribed circle of triangle ABC.
2R
5.100. Lines AP, BP and CP intersect the circumscribed circle of triangle ABC at
points A2 , B2 and C2 ; let A1 B1 C1 be the pedal triangle of point P with respect to triangle
ABC. Prove that △A1 B1 C1 ∼ △A2 B2 C2 .
5.101. Inside an acute triangle ABC a point P is given. If we drop from it perpendiculars
P A1 , P B1 and P C1 to the sides, we get △A1 B1 C1 . Performing for △A1 B1 C1 the same
operation we get △A2 B2 C2 and then we similarly get △A3 B3 C3 . Prove that △A3 B3 C3 ∼
△ABC.
§11. EULER’S LINE AND THE CIRCLE OF NINE POINTS
109
5.102. A triangle ABC is inscribed in the circle of radius R centered at O. Prove
that
¯ the ¯area of the pedal triangle of point P with respect to triangle ABC is equal to
2 ¯
1 ¯
1 − Rd 2 ¯ SABC , where d = |P O|.
4 ¯
5.103. From point P perpendiculars P A1 , P B1 and P C1 are dropped on sides of triangle
ABC. Line la connects the midpoints of segments P A and B1 C1 . Lines lb and lc are similarly
defined. Prove that la , lb and lc meet at one point.
5.104. a) Points P1 and P2 are isogonally conjugate with respect to triangle ABC, cf.
Problem 5.79. Prove that their pedal triangles have a common circumscribed circle whose
center is the midpoint of segment P1 P2 .
b) Prove that the above statement remains true if instead of perpendiculars we draw
from points P1 and P2 lines forming a given (oriented) angle to the sides.
See also Problems 5.132, 5.133, 14.19 b).
§11. Euler’s line and the circle of nine points
5.105. Let H be the point of intersection of heights of triangle ABC, O the center of
the circumscribed circle and M the point of intersection of medians. Prove that point M
lies on segment OH and OM : M H = 1 : 2.
The line that contains points O, M and H is called Euler’s line.
5.106. Prove that the midpoints of sides of a triangle, the bases of heights and the
midpoints of segments that connect the intersection point of heights with the vertices lie on
one circle and the center of this circle is the midpoint of segment OH.
The circle defined above is called the circle of nine points.
5.107. The heights of triangle ABC meet at point H.
a) Prove that triangles ABC, HBC, AHC and ABH have a common circle of 9 points.
b) Prove that Euler’s lines of triangles ABC, HBC, AHC and ABH intersect at one
point.
c) Prove that the centers of the circumscribed circles of triangles ABC, HBC, AHC and
ABH constitute a quadrilateral symmetric to quadrilateral HABC.
5.108. What are the sides the Euler line intersects in an acute and an obtuse triangles?
5.109. a) Prove that the circumscribed circle of triangle ABC is the circle of 9 points
for the triangle whose vertices are the centers of escribed circles of triangle ABC.
b) Prove that the circumscribed circle divides the segment that connects the centers of
the inscribed and an escribed circles in halves.
5.110. Prove that Euler’s line of triangle ABC is parallel to side BC if and only if
tan B tan C = 3.
5.111. On side AB of acute triangle ABC the circle of 9 points singles out a segment.
Prove that the segment subtends an angle of 2|∠A − ∠B| with the vertex at the center.
5.112. Prove that if Euler’s line passes through the center of the inscribed circle of a
triangle, then the triangle is an isosceles one.
5.113. The inscribed circle is tangent to the sides of triangle ABC at points A1 , B1 and
C1 . Prove that Euler’s line of triangle A1 B1 C1 passes through the center of the circumscribed
circle of triangle ABC.
5.114. In triangle ABC, heights AA1 , BB1 and CC1 are drawn. Let A1 A2 , B1 B2 and
C1 C2 be diameters of the circle of nine points of triangle ABC. Prove that lines AA2 , BB2
and CC2 either meet at one point or are parallel.
110
CHAPTER 5. TRIANGLES
See also Problems 3.65 a), 13.34 b).
§12. Brokar’s points
5.115. a) Prove that inside triangle ABC there exists a point P such that ∠ABP =
∠CAP = ∠BCP .
b) On sides of triangle ABC, triangles CA1 B, CAB1 and C1 AB similar to ABC are
constructed outwards (the angles at the first vertices of all the four triangles are equal, etc.).
Prove that lines AA1 , BB1 and CC1 meet at one point and this point coincides with the
point found in heading a).
This point P is called Brokar’s point of triangle ABC. The proof of the fact that there
exists another Brokar’s point Q for which ∠BAQ = ∠ACQ = ∠CBQ is similar to the proof
of existence of P given in what follows. We will refer to P and Q as the first and the second
Brokar’s points.
5.116. a) Through Brokar’s point P of triangle ABC lines AB, BP and CP are drawn.
They intersect the circumscribed circle at points A1 , B1 and C1 , respectively. Prove that
△ABC = △B1 C1 A1 .
b) Triangle ABC is inscribed into circle S. Prove that the triangle formed by the intersection points of lines P A, P B and P C with circle S can be equal to triangle ABC for no
more than 8 distinct points P . (We suppose that the intersection points of lines P A, P B
and P C with the circle are distinct from points A, B and C.)
5.117. a) Let P be Brokar’s point of triangle ABC. Let ϕ = ∠ABP = ∠BCP = ∠CAP .
Prove that cot ϕ = cot α + cot β + cot γ.
The angle ϕ from Problem 5.117 is called Brokar’s angle of triangle ABC.
b) Prove that Brokar’s points of triangle ABC are isogonally conjugate to each other (cf.
Problem 5.79).
c) The tangent to the circumscribed circle of triangle ABC at point C and the line
passing through point B parallel to AC intersect at point A1 . Prove that Brokar’s angle of
triangle ABC is equal to angle ∠A1 AC.
5.118. a) Prove that Brokar’s angle of any triangle does not exceed 30◦ .
b) Inside triangle ABC, point M is taken. Prove that one of the angles ∠ABM , ∠BCM
and ∠CAM does not exceed 30◦ .
5.119. Let Q be the second Brokar’s point of triangle ABC, let O be the center of its
circumscribed circle; A1 , B1 and C1 the centers of the circumscribed circles of triangles CAQ,
ABQ and BCQ, respectively. Prove that △A1 B1 C1 ∼ △ABC and O is the first Brokar’s
point of triangle A1 B1 C1 .
5.120. Let P be Brokar’s point of triangle ABC; let R1 , R2 and R3 be the radii of the
circumscribed circles of triangles ABP , BCP and CAP , respectively. Prove that R1 R2 R3 =
R3 , where R is the radius of the circumscribed circle of triangle ABC.
5.121. Let P and Q be the first and the second Brokar’s points of triangle ABC. Lines
CP and BQ, AP and CQ, BP and AQ meet at points A1 , B1 and C1 , respectively. Prove
that the circumscribed circle of triangle A1 B1 C1 passes through points P and Q.
5.122. On sides CA, AB and BC of an acute triangle ABC points A1 , B1 and C1 ,
respectively, are taken so that ∠AB1 A1 = ∠BC1 B1 = ∠CA1 C1 . Prove that △A1 B1 C1 ∼
△ABC and the center of the rotational homothety that sends one triangle into another
coincides with the first Brokar’s point of both triangles.
See also Problem 19.55.
* * *
111
§13. Lemoine’s point
Let AM be a median of triangle ABC and line AS be symmetric to line AM through
the bisector of angle A (point S lies on segment BC). Then segment AS is called a simedian
of triangle ABC; sometimes the whole ray AS is referred to as a simedian.
Simedians of a triangle meet at the point isogonally conjugate to the intersection point
of medians (cf. Problem 5.79). The intersection point of simedians of a triangle is called
Lemoine’s point.
5.123. Let lines AM and AN be symmetric through the bisector of angle ∠A of triangle
2
·BN
= cb2 . In particular, if AS is a
ABC (points M and N lie on line BC). Prove that BM
CM ·CN
2
simedian, then BS
= cb2 .
CS
5.124. Express the length of simedian AS in terms of the lengths of sides of triangle
ABC.
Segment B1 C1 , where points B1 and C1 lie on rays AC and AB, respectively, is said to
be antiparallel to side BC if ∠AB1 C1 = ∠ABC and ∠AC1 B1 = ∠ACB.
5.125. Prove that simedian AS divides any segment B1 C1 antiparallel to side BC in
halves.
5.126. The tangent at point B to the circumscribed circle S of triangle ABC intersects
line AC at point K. From point K another tangent KD to circle S is drawn. Prove that
BD is a simedian of triangle ABC.
5.127. Tangents to the circumscribed circle of triangle ABC at points B and C meet at
point P . Prove that line AP contains simedian AS.
5.128. Circle S1 passes through points A and B and is tangent to line AC, circle S2
passes through points A and C and is tangent to line AB. Prove that the common chord of
these circles is a simedian of triangle ABC.
5.129. Bisectors of the outer and inner angles at vertex A of triangle ABC intersect
line BC at points D and E, respectively. The circle with diameter DE intersects the
circumscribed circle of triangle ABC at points A and X. Prove that AX is a simedian of
triangle ABC.
***
5.130. Prove that Lemoine’s point of right triangle ABC with right angle ∠C is the
midpoint of height CH.
5.131. Through a point X inside triangle ABC three segments antiparallel to its sides
are drawn, cf. Problem 5.125?. Prove that these segments are equal if and only if X is
Lemoine’s point.
5.132. Let A1 , B1 and C1 be the projections of Lemoine’s point K to the sides of triangle
ABC. Prove that K is the intersection point of medians of triangle A1 B1 C1 .
5.133. Let A1 , B1 and C1 be the projections of Lemoine’s point K of triangle ABC on
sides BC, CA and AB, respectively. Prove that median AM of triangle ABC is perpendicular to line B1 C1 .
5.134. Lines AK, BK and CK, where K is Lemoine’s point of triangle ABC, intersect
the circumscribed circle at points A1 , B1 and C1 , respectively. Prove that K is Lemoine’s
point of triangle A1 B1 C1 .
5.135. Prove that lines that connect the midpoints of the sides of a triangle with the
midpoints of the corresponding heights intersect at Lemoine’s point.
See also Problems 11.22, 19.54, 19.55.
112
CHAPTER 5. TRIANGLES
Problems for independent study
5.136. Prove that the projection of the diameter of a circumscribed circle perpendicular
to a side of the triangle to the line that contains the second side is equal to the third side.
5.137. Prove that the area of the triangle with vertices in the centers of the escribed
circles of triangle ABC is equal to 2pR.
5.138. An isosceles triangle with base a and the lateral side b, and an isosceles triangle
with base b√and the lateral side a are inscribed in a circle of radius R. Prove that if a 6= b,
then ab = 5R2 .
5.139. The inscribed circle of right triangle ABC is tangent to the hypothenuse AB at
point P ; let CH be a height of triangle ABC. Prove that the center of the inscribed circle
of triangle ACH lies on the perpendicular dropped from point P to AC.
5.140. The inscribed circle of triangle ABC is tangent to sides CA and AB at points
B1 and C1 , respectively, and an escribed circle is tangent to the extension of sides at points
B2 and C2 . Prove that the midpoint of side BC is equidistant from lines B1 C1 and B2 C2 .
5.141. In triangle ABC, bisector AD is drawn. Let O, O1 and O2 be the centers
of the circumscribed circles of triangles ABC, ABD and ACD, respectively. Prove that
OO1 = OO2 .
5.142. The triangle constructed from a) medians, b) heights of triangle ABC is similar
to triangle ABC. What is the ratio of the lengths of the sides of triangle ABC?
5.143. Through the center O of an equilateral triangle ABC a line is drawn. It intersects
lines BC, CA and AB at points A1 , B1 and C1 , respectively. Prove that one of the numbers
1
1
, 1 and OC
is equal to the sum of the other two numbers.
OA1 OB1
1
5.144. In triangle ABC heights BB1 and CC1 are drawn. Prove that if ∠A = 45◦ , then
B1 C1 is a diameter of the circle of nine points of triangle ABC.
5.145. The angles of triangle ABC satisfy the relation sin2 ∠A + sin2 ∠B + sin2 ∠C = 1.
Prove that the circumscribed circle and the circle of nine points of triangle ABC intersect
at a right angle.
Solutions
5.1. Let AC1 = AB1 = x, BA1 = BC1 = y and CA1 = CB1 = z. Then
a = y + z,
b = z + x and c = x + y.
. Hence,
Subtracting the third equality from the sum of the first two ones we get z = a+b−c
2
if triangle ABC is given, then the position of points A1 and B1 is uniquely determined.
Similarly, the position of point C1 is also uniquely determined. It remains to notice that
the tangency points of the inscribed circle with the sides of the triangle satisfy the relations
indicated in the hypothesis of the problem.
5.2. Rays COa and COb are the bisectors of the outer angles at vertex C, hence, C lies
on line Oa Ob and ∠Oa CB = ∠Ob CA. Since COc is the bisector of angle ∠BCA, it follows
that ∠BCOc = ∠ACOc . Adding these equalities we get: ∠Oa COc = ∠Oc COb , i.e., Oc C
is a height of triangle Oa Ob Oc . We similarly prove that Oa A and Ob B are heights of this
triangle.
5.3. Clearly,
∠B ∠C
∠A
∠BOC = 180◦ − ∠CBO − ∠BCO = 180◦ −
−
= 90◦ +
2
2
2
and ∠BOa C = 180◦ − ∠BOC, because ∠OBOa = ∠OCOa = 90◦ .
5.4. Let AA1 , BB1 and CC1 be the bisectors of triangle ABC and O the intersection
point of these bisectors. Suppose that x > 1. Then ∠P AB > ∠P AC, i.e., point P lies
SOLUTIONS
113
inside triangle AA1 C. Similarly, point P lies inside triangles CC1 B and BB1 A. But the
only common point of these three triangles is point O. Contradiction. The case x < 1 is
similarly treated.
5.5. Let da , db and dc be the distances from point O to sides BC, CA and AB. Then
ada + bdb + cdc = 2S and aha = bhb = chc = 2S. If ha − da = hb − db = hc − dc = x, then
(a + b + c)x = a(ha − da ) = b(hb − db ) + c(hc − dc ) = 6S − 2S = 4S.
Hence, x = 4S
= 2r.
2p
5.6. Let us prove that point O is the center of the escribed circle of triangle P BQ tangent
to side P Q. Indeed, ∠P OQ = ∠A = 90◦ − 21 ∠B. The angle of the same value with the
vertex at the center of the escribed circle subtends segment P Q (Problem 5.3). Moreover,
point O lies on the bisector of angle B. Hence, the semiperimeter of triangle P BQ is equal
to the length of the projection of segment OB to line CB.
5.7. Let P be the tangent point of the inscribed circle with side BC, let P Q be a diameter
of the inscribed circle, R the intersection point of lines AQ and BC. Since CR = BP (cf.
Problem 19.11 a)) and M is the midpoint of side BC, we have: RM = P M . Moreover, O is
the midpoint of diameter P Q, hence, M O k QR and since AH k P Q, we have AE = OQ.
5.8. The given circle can be the inscribed as well as the escribed circle of triangle ABC
cut off by the tangent from the angle. Making use of the result of Problem 3.2 we can verify
that in either case
(p − b)(p − c) sin ∠B sin ∠C
uv
.
=
w2
h2a
It remains to notice that ha = b sin ∠C = c sin ∠B and (p−b)(p−c)
= sin2 12 ∠A (Problem
bc
12.13).
5.9. Let A1 , B1 and C1 be points symmetric to point H through sides BC, CA and AB,
respectively. Since AB ⊥ CH and BC ⊥ AH, it follows that ∠(AB, BC) = ∠(CH, HA) and
since triangle AC1 H is an isosceles one, ∠(CH, HA) = ∠(AC1 , C1 C). Hence, ∠(AB, BC) =
∠(AC1 , C1 C), i.e., point C1 lies on the circumscribed circle of triangle ABC. We similarly
prove that points A1 and B1 lie on this same circle.
5.10. Let R be the radius of the circumscribed circle of triangle ABC. This circle
is also the circumscribed circle of triangles ABP , AP C and P BC. Clearly, ∠ABP =
180◦ − ∠ACP = α, ∠BAP = ∠BCP = β and ∠CAP = ∠CBP = γ. Hence,
P X = P B sin γ = 2R sin β sin γ, P Y = 2R sin α sin γ and P = 2R sin α sin β.
It is also clear that
BC = 2R sin ∠BAC = 2R sin(β + γ), AC = 2R sin(α − γ), AB = 2R sin(α + β).
It remains to verify the equality
sin(α − γ) sin(α + β)
sin(β + γ)
=
+
sin β sin γ
sin α sin γ
sin α sin β
which is subject to a direct calculation.
5.11. a) Let M be the intersection point of line AI with the circumscribed circle.
Drawing the diameter through point I we get
AI · IM = (R + d)(R − d) = R2 − d2 .
Since IM = CM (by Problem 2.4 a)), it follows that R2 − d2 = AI · CM . It remains to
observe that AI = sin 1r ∠A and CM = 2R sin 12 ∠A.
2
114
CHAPTER 5. TRIANGLES
b) Let M be the intersection point of line AIa with the circumscribed circle. Then
AIa ·Ia M = d2a −R2 . Since Ia M = CM (by Problem 2.4 a)), it follows that d2a −R2 = AIa ·CM .
It remains to notice that AIa = sin r1a∠A and CM = 2R sin 12 ∠A.
2
5.12. a) Since B1 is the center of the circumscribed circle of triangle AM C (cf. Problem
A·M C
r
. Hence, MM
= 2r.
2.4 a)), AM = 2M B1 sin ∠ACM . It is also clear that M C = sin ∠ACM
B1
b) Since
∠M BC1 = ∠BM C1 = 180◦ − ∠BM C
and ∠BC1 M = ∠A,
it follows that
BM M C1
sin ∠BCM sin ∠M BC1
sin ∠BCM
M C1
=
·
=
·
=
.
BC
BC BM
sin ∠BM C sin ∠BC1 M
sin ∠A
A1
BC
= 2 sin
= R.
Moreover, M B = 2M A1 sin ∠BCM . Therefore, M CM1 ·M
B
∠A
5.13. Let M be the midpoint of side AC, and N the tangent point of the inscribed circle
with side BC. Then BN = p − b (see Problem 3.2), hence, BN = AM because p = 23 b
by assumption. Moreover, ∠OBN = ∠B1 AM and, therefore, △OBN = △B1 AM , i.e.,
OB = B1 A. But B1 A = B1 O (see Problem 2.4 a)).
5.14. Let O and O1 be the centers of the inscribed and circumscribed circles of triangle
ABC. Let us consider the circle of radius d = OO1 centered at O. In this circle, let us draw
chords O1 M and O1 N parallel to sides AB and AC, respectively. Let K be the tangent
point of the inscribed circle with side AB and L the midpoint of side AB. Since OK ⊥ AB,
O1 L ⊥ AB and O1 M k AB, it follows that
O1 M = 2KL = 2BL − 2BK = c − (a + c − b) = b − a = AE.
Similarly, O1 N = AD and, therefore, △M O1 N = △EAD. Consequently, the radius of the
circumscribed circle of triangle EAD is equal to d.
5.15. Let the inscribed circle be tangent to side AC at point K and the escribed circle
be tangent to the extension of side AC at point L. Then r = CK and rc = CL. It remains
to make use of the result of Problem 3.2.
5.16. Since 21 AB = AM = BM , it follows that CM = 12 AB if and only if point C lies
on the circle with diameter AB.
5.17. Let M and N be the midpoints of sides AB and CD. Triangle AP B is a right one;
hence, P M = 12 AB and ∠M P A = ∠P AM and, therefore, P M k AD. Similar arguments
show that points P , M and Q lie on one line and
AB + (BC + AD) + CD
.
2
5.18. Let F be the intersection point of lines DE and BC; let K be the midpoint of
segment EC. Segment CD is simultaneosly a bisector and a height of triangle ECF , hence,
ED = DF and, therefore, DK k F C. Median DK of right triangle EDC is twice shorter
its hypothenuse EC (Problem 5.16), hence, AD = DK = 12 EC.
5.19. Let the sum of the angles at the base AD of trapezoid ABCD be equal to 90◦ .
Denote the intersection point of lines AB and CD by O. Point O lies on the line that passes
through the midpoints of the bases. Let us draw through point C line CK parallel to this
line and line CE parallel to line AB (points K and E lie on base AD). Then CK is a median
= AD−BC
(cf. Problem 5.16).
of right triangle ECD, hence, CK = ED
2
2
5.20. It is clear that ∠CEB = ∠A + ∠ACE = ∠BCK + ∠KCE = ∠BCE.
5.21. Segments CF and DK are bisectors in similar triangles ACB and CDB and,
therefore, AB : F B = CB : KB. Hence, F K k AC. We similarly prove that LF k CB.
P Q = P M + MN + NQ =
SOLUTIONS
115
Therefore, CLF K is a rectangle whose diagonal CF is the bisector of angle LCK, i.e., the
rectangle is a square.
5.22. Since sin ∠ACQ
= sin ∠AQC
, it follows that
AQ
AC
sin α
sin(180◦ − α − 90◦ − ϕ)
cos(α + ϕ)
=
=
,
a
a cos ϕ
a cos ϕ
where a is the (length of the) side of square ABP Q and ϕ = ∠CAB. Hence, cot α = 1+tan ϕ.
Similarly,
cot γ = 1 + tan(90◦ − ϕ) = 1 + cot ϕ.
It follows that
1
1
+
=1
tan α + tan γ =
1 + tan ϕ 1 + cot ϕ
and, therefore,
cos α cos γ = cos α sin γ + cos γ sin α = sin(α + γ) = cos β.
5.23. By Pythagoras theorem
and
AP 2 + BQ2 + CR2 + (AM 2 − P M 2 ) + (BM 2 − QM 2 ) + (CM 2 − RM 2 )
P B 2 + QC 2 + RA2 = (BM 2 − P M 2 ) + (CM 2 − QM 2 ) + (AM 2 − RM 2 ).
These equations are equal.
Since
AP 2 + BQ2 + CR2 = (a − P B)2 + (a − QC)2 + (a − RA)2 =
3a2 − 2a(P B + QC + RA) + P B 2 + QC 2 + RA2 ,
where a = AB, it follows that P B + QC + RA = 23 a.
5.24. Let point F divide segment BC in the ratio of CF : F B = 1 : 2; let P and Q be the
intersection points of segment AF with BD and CE, respectively. It is clear that triangle
OP Q is an equilateral one. Making use of the result of Problem 1.3 it is easy to verify that
AP : P F = 3 : 4 and AQ : QF = 6 : 1. Hence, AP : P Q : QF = 3 : 3 : 1 and, therefore,
◦
O
AP = P Q = OP . Hence, ∠AOP = 180 −∠AP
= 30◦ and ∠AOC = ∠AOP + ∠P OQ = 90◦ .
2
5.25. Let A and B, C and D, E and F be the intersection points of the circle with sides
P Q, QR, RP , respectively, of triangle P QR. Let us consider median P S. It connects the
midpoints of parallel chords F A and DC and, therefore, is perpendicular to them. Hence,
P S is a height of triangle P QR and, therefore, P Q = P R. Similarly, P Q = QR.
5.26. Let H be the intersection point of heights AA1 , BB1 and CC1 of triangle ABC.
By hypothesis, A1 H · BH = B1 H · AH. On the other hand, since points A1 and B1 lie on
the circle with diameter AB, then AH · A1 H = BH · B1 H. It follows that AH = BH and
A1 H = B1 H and, therefore, AC = BC. Similarly, BC = AC.
5.27. a) Suppose that triangle ABC is not an equilateral one; for instance, a 6= b.
Since a + ha = a + b sin γ and b + hb = b + a sin γ, it follows that (a − b)(1 − sin γ) = 0;
hence, sin γ = 0, i.e., γ = 90◦ . But then a 6= c and similar arguments show that β = 90◦ .
Contradiction.
b) Let us denote the (length of the) side of the square two vertices of which lie on side
BC by x. The similarity of triangles ABC and AP Q, where P and Q are the vertices of the
aha
2S
square that lie on AB and AC, respectively, yields xa = hah−x
, i.e., x = a+h
= a+h
.
a
a
a
Similar arguments for the other squares show that a + ha = b + hb = c + hc .
5.28. If α, β and γ are the angles of triangle ABC, then the angles of triangle A1 B1 C1
, γ+α
and α+β
. Let, for definiteness, α ≥ β ≥ γ. Then α+β
≥ α+γ
≥ β+γ
.
are equal to β+γ
2
2
2
2
2
2
β+γ
α+β
Hence, α = 2 and γ = 2 , i.e., α = β and β = γ.
116
CHAPTER 5. TRIANGLES
5.29. In any triangle a height is longer than the diameter of the inscribed circle. Therefore, the lengths of heights are integers greater than 2, i.e., all of them are not less than 3.
Let S be the area of the triangle, a the length of its longest side and h the corresponding
height.
Suppose that the triangle is not an equilateral one. Then its perimeter P is shorter than
3a. Therefore, 3a > P = Pr = 2S = ha, i.e., h < 3. Contradiction.
5.30. Since the outer angle at vertex A of triangle ABA1 is equal to 120◦ and ∠A1 AB1 =
60◦ , it follows that AB1 is the bisector of this outer angle. Moreover, BB1 is the bisector of
the outer angle at vertex B, hence, A1 B1 is the bisector of angle ∠AA1 C. Similarly, A1 C1
is the bisector of angle ∠AA1 B. Hence,
∠B1 A1 C1 =
∠AA1 C + ∠AA1 B
= 90◦ .
2
5.31. Thanks to the solution of the preceding problem ray A1 C1 is the bisector of angle
∠AA1 B. Let K be the intersection point of the bisectors of triangle A1 AB. Then
∠C1 KO = ∠A1 KB = 90◦ +
∠A
= 120◦ .
2
Hence, ∠C1 KO + ∠C1 AO = 180◦ , i.e., quadrilateral AOKC1 is an inscribed one. Hence,
∠A1 C1 O = ∠KC1 O = ∠KAO = 30◦ .
5.32. a) Let S be the circumscribed circle of triangle ABC, let S1 be the circle symmetric
to S through line BC. The orthocenter H of triangle ABC lies on circle S1 (Problem 5.9)
and, therefore, it suffices to verify that the center O of circle S also belongs to S1 and
the bisector of the outer angle A passes through the center of circle S1 . Then P OAH is a
rhombus, because P O k HA.
Let P Q be the diameter of circle S perpendicular to line BC; let points P and A lie on
one side of line BC. Then AQ is the bisector of angle A and AP is the bisector of the outer
angle ∠A. Since ∠BP C = 120◦ = ∠BOC, point P is the center of circle S1 and point O
belongs to circle S1 .
b) Let S be the circumscribed circle of triangle ABC and Q the intersection point of
the bisector of angle ∠BAC with circle S. It is easy to verify that Q is the center of circle
S1 symmetric to circle S through line BC. Moreover, points O and H lie on circle S1 and
since ∠BIC = 120◦ and ∠BIa C = 60◦ (cf. Problem 5.3), it follows that IIa is a diameter
of circle S1 . It is also clear that ∠OQI = ∠QAH = ∠AQH, because OQ k AH and
HA = QO = QH. Hence, points O and H are symmetric through line IIa .
5.33. On side AC of triangle ABC, construct outwards an equilateral triangle AB1 C.
Since ∠A = 120◦ , point A lies on segment BB1 . Therefore, BB1 = b + c and, moreover,
BC = a and B1 C = b, i.e., triangle BB1 C is the desired one.
5.34. a) Let M1 and N1 be the midpoints of segments BH and CH, respectively; let
BB1 and CC1 be heights. Right triangles ABB1 and BHC1 have a common acute angle —
the one at vertex B; hence, ∠C1 HB = ∠A = 60◦ . Since triangle BM H is an isosceles one,
∠BHM = ∠HBM = 30◦ . Therefore, ∠C1 HM = 60◦ − 30◦ = 30◦ = ∠BHM , i.e., point
M lies on the bisector of angle ∠C1 HB. Similarly, point N lies on the bisector of angle
∠B1 HC.
b) Let us make use of the notations of the preceding problem and, moreover, let B ′ and
′
C be the midpoints of sides AC and AB. Since AC1 = AC cos ∠A = 12 AC, it follows that
C1 C ′ = 12 |AB − AC|. Similarly, B1 B ′ = 21 |AB − AC|, i.e., B1 B ′ = C1 C ′ . It follows that
the parallel lines BB1 and B ′ O, CC1 and C ′ O form not just a parallelogram but a rhombus.
Hence, its diagonal HO is the bisector of the angle at vertex H.
SOLUTIONS
117
5.35. Since
∠BB1 C = ∠B1 BA + ∠B1 AB > ∠B1 BA = ∠B1 BC,
it follows that BC > B1 C. Hence, point K symmetric to B1 through bisector CC1 lies
on side BC and not on its extension. Since ∠CC1 B = 30◦ , we have ∠B1 C1 K = 60◦ and,
therefore, triangle B1 C1 K is an equilateral one. In triangles BC1 B1 and BKB1 side BB1 is
a common one and sides C1 B1 and KB1 are equal; the angles C1 BB1 and KBB1 are also
equal but these angles are not the ones between equal sides. Therefore, the following two
cases are possible:
◦
1) ∠BC1 B1 = ∠BKB1 . Then ∠BB1 C1 = ∠BB1 K = 602 = 30◦ . Therefore, if O is the
intersection point of bisectors BB1 and CC1 , then
∠BOC = ∠B1 OC1 = 180◦ − ∠OC1 B1 − ∠OB1 C1 = 120◦ .
On the other hand, ∠BOC = 90◦ + ∠A
(cf. Problem 5.3), i.e., ∠A = 60◦ .
2
◦
2) ∠BC1 B1 + ∠BKB1 = 180 . Then quadrilateral BC1 B1 K is an inscribed one and
since triangle B1 C1 K is an equilateral one, ∠B = 180◦ − ∠C1 B1 K = 120◦ .
5.36. Let BM be a median, AK a bisector of triangle ABC and BM ⊥ AK. Line AK
is a bisector and a height of triangle ABM , hence, AM = AB, i.e., AC = 2AM = 2AB.
Therefore, AB = 2, BC = 3 and AC = 4.
5.37. Let a and b be legs and c the hypothenuse of the given triangle. If numbers a and
b are odd, then the remainder after division of a2 + b2 by 4 is equal to 2 and a2 + b2 cannot
be a perfect square. Hence, one of the numbers a and b is even and another one is odd; let,
for definiteness, a = 2p. The numbers b and c are odd, hence, c + b = 2q and c − b = 2r for
some q and r. Therefore, 4p2 = a2 = c2 − b2 = 4qr. If d is a common divisor of q and r,
√
then a = 2 qr, b = q − r and c = q + r are divisible by d. Therefore, q and r are relatively
prime, ??? since p2 = qr, it follows that q = m2 and r = n2 . As a result we get a = 2mn,
b = m2 − n2 and c = m2 + n2 .
It is also easy to verify that if a = 2mn, b = m2 − n2 and c = m2 + n2 , then a2 + b2 = c2 .
5.38. Let p be the semiperimeter of the triangle and a, b, c the lengths of the triangle’s
sides. By Heron’s formula S 2 = p(p − a)(p − b)(p − c). On the other hand, S 2 = p2 r2 = p2
since r = 1. Hence, p = (p − a)(p − b)(p − c). Setting x = p − a, y = p − b, z = p − c we
rewrite our equation in the form
x + y + z = xyz.
, where n is an integer)
Notice that p is either integer or half integer (i.e., of the form 2n+1
2
and, therefore, all the numbers x, y, z are simultaneously either integers or half integers. But
if they are half integers, then x + y + z is a half integer and xyz is of the form m8 , where m
is an odd number. Therefore, numbers x, y, z are integers. Let, for definiteness, x ≤ y ≤ z.
Then xyz = x + y + z ≤ 3z, i.e., xy ≤ 3. The following three cases are possible:
1) x = 1, y = 1. Then 2 + z = z which is impossible.
2) x = 1, y = 2. Then 3 + z = 2z, i.e., z = 3.
3) x = 1, y = 3. Then 4 + z = 3z, i.e., z = 2 < y which is impossible.
Thus, x = 1, y = 2, z = 3. Therefore, p = x + y + z = 6 and a = p − x = 5, b = 4, c = 3.
5.39. Let a1 and b1 , a2 and b2 be the legs of two distinct Pythagorean triangles, c1 and
c2 their hypothenuses. Let us take two perpendicular lines and mark on them segments
OA = a1 a2 , OB = a1 b2 , OC = b1 b2 and OD = a2 b1 (Fig. 57). Since OA · OC = OB · OD,
quadrilateral ABCD is an inscribed one. By Problem 2.71
4R2 = OA2 + OB 2 + OC 2 + OD2 = (c1 c2 )2 ,
118
CHAPTER 5. TRIANGLES
i.e., R = c12c2 . Magnifying, if necessary, quadrilateral ABCD twice, we get the quadrilateral
to be found.
Figure 57 (Sol. 5.39)
5.40. a) The lengths of hypothenuses of right triangles with legs 5 and 12, 9 and 12
are equal to 13 and 15, respectively. Identifying the equal legs of these triangles we get a
= 84.
triangle whose area is equal to 12(5+9)
2
b) First, suppose that the length of the shortest side of the given triangle is an even
number, i.e., the lengths of the sides of the triangle are equal to 2n, 2n + 1, 2n + 2. Then
by Heron’s formula
16S 2 = (6n + 3)(2n + 3)(2n + 1)(2n − 1) = 4(3n2 + 6n + 2)(4n2 − 1) + 4n2 − 1.
We have obtained a contradiction since the number in the right-hand side is not divisible by
4. Consecutively, the lengths of the sides of the triangle are equal to 2n − 1, 2n and 2n + 1,
where S 2 = 3n2 (n2 − 1). Hence, S = nk, where k is an integer and k 2 = 3(n2 − 1). It is also
clear that k is the length of the height dropped to the side of length 2n. This height divides
the initial triangle into two right triangles with a common leg of length k and hypothenuses
of length 2n + 1 and 2n − 1 the squares of the lengths of the other legs of these triangles are
equal to
(2n ± 1)2 − k 2 = 4n2 ± 4n + 1 − 3n2 + 3 = (n ± 2)2 .
5.41. a) Since AB 2 − AB12 = BB12 = BC 2 − (AC ± AB1 )2 , we see that AB1 =
2
2 −BC 2
± AB +AC
.
2AC
b) Let diagonals AC and BD meet at point O. Let us prove, for example, that the
BO
BD
number q = OD
is a rational one (then the number OD = q+1
is also a rational one). In
triangles ABC and ADC draw heights BB1 and DD1 . By heading a) the numbers AB1 and
CD1 — the lengths of the corresponding sides — are rational and, therefore, the number
B1 D1 is also rational.
Let E be the intersection point of line BB1 and the line that passes through point D
parallel to AC. In right triangle BDE, we have ED = B1 D1 and the lengths of leg ED
and hypothenuse BD are rational numbers; hence, BE 2 is also a rational number. From
triangles ABB1 and CDD1 we derive that numbers BB12 and DD12 are rational. Since
BE 2 = (BB1 + DD1 )2 = BB12 + DD12 + 2BB1 · DD1 ,
number BB1 · DD1 is rational. It follows that the number
BB1
BB1 · DD1
BO
=
=
OD
DD1
DD12
is a rational one.
SOLUTIONS
119
5.42. Triangles ABC and A1 B1 C1 cannot have two pairs of corresponding angles whose
sum is equal to 180◦ since otherwise their sum would be equal to 360◦ and the third angles
of these triangles should be equal to zero. Now, suppose that the angles of the first triangle
are equal to α, β and γ and the angles of the second one are equal to 180◦ − α, β and γ.
The sum of the angles of the two triangles is equal to 360◦ , hence, 180◦ + 2β + 2γ = 360◦ ,
i.e., β + γ = 90◦ . It follows that α = 90◦ = 180◦ − α.
−−→
−−→
−−→
−→
−−−→
−→
−−−→
5.43. Clearly, A1 C = BO and CB1 = OA, hence, A1 B1 = BA. Similarly, B1 C1 =
−−→
−−−→
−→
CB and C1 A1 = AC, i.e., △ABC = △A1 B1 C1 . Moreover, ABA1 B1 and ACA1 C1 are
parallelograms. It follows that segments BB1 and CC1 pass through the midpoint of segment
AA1 .
5.44. Since ∠M AO = ∠P AO = ∠AOM , it follows that AM OP is a rhombus. Similarly,
BN OQ is a rhombus. It follows that
M N = M O + ON = AM + BN
and OP + P Q + QO = AP + P Q + QB = AB.
5.45. a) Through vertices of triangle ABC let us draw lines parallel to the triangle’s
opposite sides. As a result we get triangle A1 B1 C1 ; the midpoints of the sides of the new
triangle are points A, B and C. The heights of triangle ABC are the midperpendiculars to
the sides of triangle A1 B1 C1 and, therefore, the center of the circumscribed circle of triangle
A1 B1 C1 is the intersection point of heights of triangle ABC.
b) Point H is the center of the circumscribed circle of triangle A1 B1 C1 , hence,
4R2 = B1 H 2 = B1 A2 + AH 2 = BC 2 + AH 2 .
Therefore,
¶
1
− 1 BC 2 = (BC cot α)2 .
AH = 4R − BC =
sin2 α
5.46. Let AD be the bisector of an equilateral triangle ABC with base AB and angle
36◦ at vertex C. Then triangle ACD is an isosceles one and △ABC ∼ △BDA. Therefore,
CD = AD = AB = 2xBC and DB = 2xAB = 4x2 BC; hence,
2
2
2
µ
BC = CD + DB = (2x + 4x2 )BC.
5.47. Let B1 and B2 be the projections of point A to bisectors of the inner and outer
angles at vertex B; let M the midpoint of side AB. Since the bisectors of the inner and
outer angles are perpendicular, it follows that AB1 BB2 is a rectangular and its diagonal
B1 B2 passes through point M . Moreover,
∠B1 M B = 180◦ − 2∠M BB1 = 180◦ − ∠B.
Hence, B1 B2 k BC and, therefore, line B1 B2 coincides with line l that connects the midpoints
of sides AB and AC.
We similarly prove that the projections of point A to the bisectors of angles at vertex C
lie on line l.
5.48. Suppose that the bisectors of angles A and B are equal but a > b. Then cos 12 ∠A <
bc
ac
cos 21 ∠B and 1c + 1b > 1c + a1 , i.e., b+c
< a+c
. By multiplying these inequalities we get a
2bc cos
∠A
2ac cos
∠B
contradiction, since la = b+c 2 and lb = a+c 2 (cf. Problem 4.47).
5.49. a) By Problem 4.47 the length of the bisector of angle ∠B of triangle ABC is
equal to
2ac cos ∠B
2
a+c
and, therefore, it suffices to verify that the system of equations
ac
= p,
a+c
a2 + c2 − 2ac cos ∠B = q
120
CHAPTER 5. TRIANGLES
has (up to a transposition of a with c) a unique positive solution. Let a + c = u. Then
ac = pu and q = u2 − 2pu(1 + cos β). The product of the roots of this quadratic equation
for u is equal to −q and, therefore, it has one positive root. Clearly, the system of equations
a + c = u,
ac = pu
has a unique solution.
b) In triangles AA1 B and CC1 B, sides AA1 and CC1 are equal; the angles at vertex
B are equal, and the bisectors of the angles at vertex B are also equal. Therefore, these
triangles are equal and either AB = BC or AB = BC1 . The second equality cannot take
place.
5.50. Let points M and N lie on sides AB and AC. If r1 is the radius of the circle whose
center lies on segment M N and which is tangent to sides AB and AC, then SAM N = qr1 ,
where q = AM +AN
. Line M N passes through the center of the inscribed circle if and only if
2
SAM N
M
= SBCN
.
r1 = r, i.e., q = SABC
p
p−q
5.51. a) On the extension of segment AC beyond point C take a point B ′ such that
CB ′ = CB. Triangle BCB ′ is an isosceles one; hence, ∠AEB = ∠ACB = 2∠CBB ′ and,
therefore, E is the center of the circumscribed circle of triangle ABB ′ . It follows that point
F divides segment AB ′ in halves; hence, line C1 F divides the perimeter of triangle ABC in
halves.
b) It is easy to verify that the line drawn through point C parallel to BB ′ is the bisector
of angle ACB. Since C1 F k BB ′ , line C1 F is the bisector of the angle of the triangle with
vertices at the midpoints of triangle ABC. The bisectors of this new triangle meet at one
point.
5.52. Let X be the intersection point of lines AD2 and CD1 ; let M , E1 and E2 be the
projections of points X, D1 and D2 , respectively, to line AC. Then CE2 = CD2 sin γ =
a sin γ and AE1 = c sin α. Since a sin γ = c sin α, it follows that CE2 = AE1 = q. Hence,
XM
D2 E2
a cos γ
=
=
AM
AE2
b+q
and
XM
c cos α
=
.
CM
b+q
Therefore, AM : CM = c cos α : a cos γ. Height BH divides side AC in the same ratio.
5.53. a) By the law of cosines
B1 C12 = AC12 + AB12 − 2AC1 · AB1 · cos(90◦ + α),
i.e.,
b2 + c 2
c 2 b2
+ + bc sin α =
+ 2S.
2
2
2
Writing similar equalities for b21 and c21 and taking their sum we get the statement desired.
b) For an acute triangle ABC, add to S the areas of triangles ABC1 , AB1 C and A1 BC;
add to S1 the areas of triangles AB1 C1 , A1 BC1 and A1 B1 C. We get equal quantities (for a
triangle with an obtuse angle ∠A the area of triangle AB1 C1 should be taken with a minus
sign). Hence,
a2 + b2 + c2 ab cos γ + ac cos β + bc cos α
S1 = S +
−
.
4
4
It remains to notice that
a21 =
ab cos γ + bc cos α + ac cos β = 2S(cot γ + cot α + cot β) =
cf. Problem 12.44 a).
a2 + b 2 + c 2
;
2
SOLUTIONS
121
5.54. First, let us prove that point B ′ lies on the circumscribed circle of triangle AHC,
where H is the intersection point of heights of triangle ABC. We have
∠(AB ′ , B ′ C) = ∠(AA1 , CC1 ) =
∠(AA1 , BC) + ∠(BC, AB) + ∠(AB, CC1 ) = ∠(BC, AB).
But as follows from the solution of Problem 5.9 ∠(BC, AB) = ∠(AH, HC) and, therefore,
points A, B ′ , H and C lie on one circle and this circle is symmetric to the circumscribed
circle of triangle ABC through line AC. Hence, both these circles have the same radius, R,
consequently,
B ′ H = 2R sin B ′ AH = 2R cos α.
Similarly, A′ H = 2R cos α = C ′ H. This completes solution of heading a); to solve heading b)
it remains to notice that △A′ B ′ C ′ ∼ △ABC since after triangle A′ B ′ C ′ is rotated through
an angle of α its sides become parallel to the sides of triangle ABC.
5.55. Let a1 = BA1 , a2 = A1 C, b1 = CB1 , b2 = B1 A, c1 = AC1 and c2 = C1 B. The
products of the lengths of segments of intersecting lines that pass through one point are
equal and, therefore, a1 (a1 + x) = c2 (c2 − z), i.e.,
a1 x + c2 z = c22 − a21 .
We similarly get two more equations for x, y and z:
b1 y + a2 x = a22 − b21
and c1 z + b2 y = b22 − c21 .
Let us multiply the first equation by b2n ; multiply the second and the third ones by c2n and
a2n , respectively, and add the equations obtained. Since, for instance, c2 bn − c1 an = 0 by
the hypothesis, we get zero in the right-hand side. The coefficient of, say, x in the left-hand
side is equal to
acn b2n + abn c2n
= abn cn .
a1 b2n + a2 c2n =
bn + c n
Hence,
abn cn x + ban cn y + can bn z = 0.
Dividing both sides of this equation by (abc)n we get the statement desired.
5.56. Let in the initial triangle ∠A = 3α, ∠B = 3β and ∠C = 3γ. Let us take an
equilateral triangle A2 B2 C2 and construct on its sides as on bases isosceles triangles A2 B2 R,
B2 C2 P and C2 A2 Q with angles at the bases equal to 60◦ − γ, 60◦ − α, 60◦ − β, respectively
(Fig. 58).
Let us extend the lateral sides of these triangles beyond points A2 , B2 and C2 ; denote the
intersection point of the extensions of sides RB2 and QC2 by A3 , that of P C2 and RA2 by
B3 , that of QA2 and P B2 by C3 .Through point B2 draw the line parallel to A2 C2 and denote
by M and N the its intersection points with lines QA3 and QC3 , respectively. Clearly, B2 is
the midpoint of segment M N . Let us compute the angles of triangles B2 C3 N and B2 A3 M :
∠C3 B2 N = ∠P B2 M = ∠C2 B2 M = ∠C2 B2 P = α;
∠B2 N C3 = 180◦ − ∠C2 A2 Q = 120◦ + β;
hence, ∠B2 C3 N = 180◦ − α − (120◦ + β) = γ. Similarly, ∠A3 B2 M = γ and ∠B2 A3 M = α.
Hence, △B2 C3 N ∼ △A3 B2 M . It follows that C3 B2 : B2 A3 = C3 N : B2 M and since
B2 M = B2 N and ∠C3 B2 A3 = ∠C3 N B2 , it follows that C3 B2 : B2 A3 = C3 N : N B2 and
△C3 B2 A3 ∼ △C3 N B2 ; hence, ∠B2 C3 A3 = γ.
122
CHAPTER 5. TRIANGLES
Figure 58 (Sol. 5.56)
Similarly, ∠A2 C3 B3 = γ and, therefore, ∠A3 C3 B3 = 3γ = ∠C and C3 B3 , C3 A2 are the
trisectors of angle C3 of triangle A3 B3 C3 . Similar arguments for vertices A3 and B3 show
that △ABC ∼ △A3 B3 C3 and the intersection points of the trisectors of triangle A3 B3 C3
are vertices of an equilateral triangle A2 B2 C2 .
5.57. Point A1 lies on the bisector of angle ∠BAC, hence, point A lies on the extension
◦
2 A1 C 2
of the bisector of angle ∠B2 A1 C2 . Moreover, ∠B2 AC2 = α = 180 −∠B
. Hence, A is the
2
center of an escribed circle of triangle B2 A1 C2 (cf. Problem 5.3). Let D be the intersection
point of lines AB and CB2 . Then
∠AB2 C2 = ∠AB2 D = 180◦ − ∠B2 AD − ∠ADB2 = 180◦ − γ − (60◦ + α) = 60◦ + β.
Since
it follows that
∠AB2 C = 180◦ − (α + β) − (β + γ) = 120◦ − β,
∠CB2 C2 = ∠AB2 C − ∠AB2 C2 = 60◦ − 2β.
Similarly, ∠AB2 A2 = 60◦ − 2β. Hence,
∠A2 B2 C2 = ∠AB2 C − ∠AB2 A2 − ∠CB2 C2 = 3β.
Similarly, ∠B2 A2 C2 = 3α and ∠A2 C2 B2 = 3γ.
5.58. Let the projection to a line perpendicular to line A1 B1 send points A, B and C to
′
A , B ′ and C ′ , respectively; point C1 to Q and points A1 and B1 into one point, P . Since
A1 B
P B′
=
,
A1 C
P C′
B1 C
P C′
=
B1 A
P A′
and
C1 A
QA′
=
,
C1 B
QB ′
it follows that
A1 B B1 C C1 A
P B ′ P C ′ QA′
P B ′ QA′
b ′ a′ + x
·
·
=
·
·
=
·
=
·
,
a′ b ′ + x
A1 C B1 A C1 B
P C ′ P A′ QB ′
P A′ QB ′
′
′
+x
= 1 is equivalent to the fact that x = 0. (We have
where |x| = P Q. The equality ab ′ · ab′ +x
′
′
to take into account that a 6= b since A′ 6= B ′ .) But the equality x = 0 means that P = Q,
i.e., point C1 lies on line A1 B1 .
5.59. Let point P lie on arc ⌣ BC of the circumscribed circle of triangle ABC. Then
BA1
BP cos ∠P BC
=−
,
CP cos ∠P CB
CA1
CB1
CP cos ∠P CA
=−
,
AP cos P AC
AB1
AC1
AP cos ∠P AB
=−
.
P B cos ∠P BA
BC1
SOLUTIONS
123
By multiplying these equalities and taking into account that
∠P AC = ∠P BC,
∠P AB = ∠P CB
and ∠P AC + ∠P BA = 180◦
we get
BA1 CB1 AC1
·
·
= 1.
CA1 AB1 BC1
5.60. Let O, O1 and O2 be the centers of circles S, S1 and S2 ; let X be the intersection
point of lines O1 O2 and A1 A2 . By applying Menelaus’s theorem to triangle OO1 O2 and
points A1 , A2 and X we get
O1 X O2 A2 OA1
·
·
=1
O2 X OA2 O1 A1
and, therefore, O1 X : O2 X = R1 : R2 , where R1 and R2 are the radii of circles S1 and S2 ,
respectively. It follows that X is the intersection point of the common outer or common
inner tangents to circles S1 and S2 .
5.61. a) Let, for definiteness, ∠B < ∠C. Then ∠DAE = ∠ADE = ∠B + ∠A
; hence,
2
∠CAE = ∠B. Since
sin ∠BAE
AC
sin ∠AEC
BE
=
and
=
,
AB
sin ∠AEB
CE
sin ∠CAE
it follows that
BE
c sin ∠BAE
c sin(∠A + ∠B)
c sin ∠C
c2
=
=
=
= 2.
CE
b sin ∠CAE
b sin ∠B
b sin ∠B
b
b) In heading a) point E lies on the extension of side BC since ∠ADC = ∠BAD + ∠B >
∠CAD. Therefore, making use of the result of heading a) and Menelaus’s theorem we get
the statement desired.
5.62. Since ∠BCE = 90◦ − ∠B
, we have: ∠BCE = ∠BEC and, therefore, BE = BC.
2
Hence,
CF : KF = BE : BK = BC : BK
and AE : KE = CA : CK = BC : BK.
AD
CF
Let line EF intersect AC at point D. By Menelaus’s theorem CD
· KF
· KE
= 1. Taking
AE
into account that CF : KF = AE : KE we get the statement desired.
5.63. Proof is similar to that of Problem 5.79; we only have to consider the ratio of
oriented segments and angles.
5.64. Let A2 , B2 and C2 be the intersection points of lines BC with B1 C1 , AC with A1 C1 ,
AB with A1 B1 , respectively. Let us apply Menelaus’s theorem to the following triangles and
points on their sides: OAB and (A1 , B1 , C2 ), OBC and (B1 , C1 , A2 ), OAC and (A1 , C1 , B2 ).
Then
AA1 OB1 BC2
OC1 BB1 CA2
OA1 CC1 AB2
·
·
= 1,
·
·
= 1,
·
·
= 1.
OA1 BB1 AC2
CC1 OB1 BA2
AA1 OC1 CB2
By multiplying these equalities we get
BC2 AB2 CA2
·
·
= 1.
AC2 CB2 BA2
Menelaus’s theorem implies that points A2 , B2 , C2 lie on one line.
5.65. Let us consider triangle A0 B0 C0 formed by lines A1 B2 , B1 C2 and C1 A2 (here A0
is the intersection point of lines A1 B2 and A2 C1 , etc), and apply Menelaus’s theorem to this
triangle and the following five triples of points:
(A, B2 , C1 ), (B, C2 , A1 ), (C, A2 , B1 ), (A1 , B1 , C1 ) and (A2 , B2 , C2 ).
124
CHAPTER 5. TRIANGLES
As a result we get
(2)
(3)
B0 A A0 B2 C0 C1
·
= 1,
·
C0 A B0 B2 A0 C1
A0 C C0 A2 B0 B1
·
= 1,
·
B0 C A0 A2 C0 B1
A0 A2 B0 B2
·
C0 A2 A0 B2
C0 B B0 C2 A0 A1
·
= 1,
·
A0 B C0 C2 B0 A1
B0 A1 C0 B1 A0 C1
·
·
= 1,
A0 A1 B0 B1 C0 C1
C0 C2
·
= 1.
B0 C2
0A
By multiplying these equalities we get B
· C0 B · A0 C = 1 and, therefore, points A, B and
C 0 A A0 B B 0 C
C lie on one line.
5.66. Let N be the intersection point of lines AD and KQ, P ′ the intersection point of
lines KL and M N . By Desargue’s theorem applied to triangles KBL and N DM we derive
that P ′ , A and C lie on one line. Hence, P ′ = P .
5.67. It suffices to apply Desargues’s theorem to triangles AED and BF C and Pappus’
theorem to triples of points (B, E, C) and (A, F, D).
5.68. a) Let R be the intersection point of lines KL and M N . By applying Pappus’
theorem to triples of points (P, L, N ) and (Q, M, K), we deduce that points A, C and R lie
on one line.
b) By applying Desargues’s theorem to triangles N DM and LBK we see that the intersection points of lines N D with LB, DM with BK, and N M with LK lie on one line.
5.69. Let us make use of the result of Problem 5.68 a). For points P and Q take points
P2 and P4 , for points A and C take points C1 and P1 and for K, L, M and N take points
P5 , A1 , B1 and P3 , respectively. As a result we see that line P6 C1 passes through point P1 .
5.70. a) This problem is a reformulation of Problem 5.58 since the number BA1 : CA1
is negative if point A1 lies on segment BC and positive otherwise.
b) First, suppose that lines AA1 , BB1 and CC1 meet at point M . Any three (nonzero)
vectors in plane are linearly dependent, i.e., there exist numbers λ, µ and ν (not all equal
−−→
−−→
−−→
to zero) such that λAM + µBM + ν CM = 0. Let us consider the projection to line BC
parallel to line AM . This projection sends points A and M to A1 and points B and C into
−−→
−−→
themselves. Therefore, µBA1 + ν CA1 = 0, i.e.,
ν
BA1
=− .
µ
CA1
Similarly,
λ
µ
CB1
AC1
=−
=− .
and
ν
λ
AB1
BC1
By multiplying these three equalities we get the statement desired.
If lines AA1 , BB1 and CC1 are parallel, in order to get the proof it suffices to notice that
BA1
BA
=
CA1
C1 A
and
CB1
C1 B
=
.
AB1
AB
Now, suppose that the indicated relation holds and prove that then lines AA1 , BB1 and
CC1 intersect at one point. Let C1∗ be the intersection point of line AB with the line that
passes through point C and the intersection point of lines AA1 and BB1 . For point C1∗ the
same relation as for point C1 holds. Therefore, C1∗ A : C1∗ B = C1 A : C1 B. Hence, C1∗ = C1 ,
i.e., lines AA1 , BB1 and CC1 meet at one point.
SOLUTIONS
125
It is also possible to verify that if the indicated relation holds and two of the lines AA1 ,
BB1 and CC1 are parallel, then the third line is also parallel to them.
5.71. Clearly, AB1 = AC1 , BA1 = BC1 and CA1 = CB1 , and, in the case of the
inscribed circle, on sides of triangle ABC, there are three points and in the case of an
escribed circle there is just one point on sides of triangle ABC. It remains to make use of
Ceva’s theorem.
5.72. Let AA1 , BB1 and CC1 be heights of triangle ABC. Then
b cos ∠A c cos ∠B a cos ∠C
AC1 BA1 CB1
·
·
=
·
·
= 1.
C1 B A1 C B1 A
a cos ∠B b cos ∠C c cos ∠A
5.73. Let A2 , B2 and C2 be the midpoints of sides BC, CA and AB. The considered
lines pass through the vertices of triangle A2 B2 C2 and in heading a) they divide its sides
in the same ratios in which lines AP , BP and CP divide sides of triangle ABC whereas in
heading b) they divide them in the inverse ratios. It remains to make use of Ceva’s theorem.
5.74. Since △AC1 B2 ∼ △BC1 A1 and △AB1 C2 ∼ △CB1 A1 , it follows that AB2 · C1 B =
AC1 · BA1 and AC2 · CB1 = A1 C · B1 A. Hence,
AB2
AC1 BA1 CB1
·
·
= 1.
=
AC2
C1 B A1 C B1 A
5.75. Let lines AA1 , BB1 and CC1 intersect lines BC, CA and AB at points A1 , B2
and C2 .
a) If ∠B + β < 180◦ and ∠C + γ < 180◦ , then
BA2
AB · BA1 sin(∠B + β)
SABA1
AB sin γ sin(∠B + β)
=
=
=
·
·
.
A2 C
SACA1
AC · CA1 sin(∠C + γ)
AC sin β sin(∠C + γ)
The latter expression is equal to BA2 : A2 C in all the cases. Let us write similar expressions
for CB2 : B2 A and AC2 : C2 B and multiply them. Now it remains to make use of Ceva’s
theorem.
b) Point A2 lies outside segment BC only if precisely one of the angles β and γ is greater
than the corresponding angle ∠B or ∠C. Hence,
BA2
AB sin γ sin(∠B − β)
=
·
·
.
AC sin β sin(∠C − γ)
A2 C
5.76. It is easy to verify that this problem is a particular case of Problem 5.75.
Remark. A similar statement is also true for an escribed circle.
5.77. The solution of the problem obviously follows from Ceva’s theorem.
5.78. By applying the sine theorem to triangles ACC1 and BCC1 we get
sin ∠ACC1
CC1
sin ∠B
AC1
=
and
=
,
C1 C
sin ∠A
C1 B
sin ∠C1 CB
i.e.,
AC1
sin ∠ACC1 sin ∠B
=
·
.
C1 B
sin ∠C1 CB sin ∠A
Similarly,
sin ∠BAA1 sin ∠C
CB1
sin ∠CBB1 sin ∠A
BA1
=
·
and
=
·
.
A1 C
sin ∠A1 AC sin ∠B
B1 A
sin ∠B1 BA sin ∠C
To complete the proof it remains to multiply these equalities.
Remark. A similar statement is true for the ratios of oriented segments and angles in
the case when the points are taken on the extensions of sides.
126
CHAPTER 5. TRIANGLES
5.79. We may assume that points A2 , B2 and C2 lie on the sides of triangle ABC. By
Problem 5.78
AC2 BA2 CB2
sin ∠ACC2 sin ∠BAA2 sin ∠CBB2
·
·
=
·
·
.
C2 B A2 C B2 A
sin ∠C2 CB sin ∠A2 AC sin ∠B2 BA
Since lines AA2 , BB2 and CC2 are symmetric to lines AA1 , BB1 and CC1 , respectively,
through the bisectors, it follows that ∠ACC2 = ∠C1 CB, ∠C2 CB = ∠ACC1 etc., hence,
sin ∠C1 CB sin ∠A1 AC sin ∠B1 BA
sin ∠ACC2 sin ∠BAA2 sin ∠CBB2
·
·
=
·
·
=
sin ∠C2 CB sin ∠A2 AC sin ∠B2 BA
sin ∠ACC1 sin ∠BAA1 sin ∠CBB1
C1 B A1 C B1 A
·
·
= 1.
AC1 BA1 CB1
Therefore,
AC2 BA2 CB2
·
·
= 1,
C2 B A2 C B2 A
i.e., lines AA2 , BB2 and CC2 meet at one point.
Remark. The statement holds also in the case when points A1 , B1 and C1 are taken on
the extensions of sides if only point P does not lie on the circumscribed circle S of triangle
ABC; if P does lie on S, then lines AA2 , BB2 and CC2 are parallel (cf. Problem 2.90).
5.80. Let diagonals AD and BE of the given hexagon ABCDEF meet at point P ; let
K and L be the midpoints of sides AB and ED, respectively. Since ABDE is a trapezoid,
segment KL passes through point P (by Problem 19.2). By the law of sines
sin ∠AP K : sin ∠AKP = AK : AP
and
sin ∠BP K : sin ∠BKP = BK : BP.
Since sin ∠AKP = sin ∠BKP and AK = BK, we have
sin ∠AP K : sin ∠BP K = BP : AP = BE : AD.
Similar relations can be also written for the segments that connect the midpoints of the
other two pairs of the opposite sides. By multiplying these relations and applying the result
of Problem 5.78 to the triangle formed by lines AD, BE and CF , we get the statement
desired.
5.81. Let us consider the homothety with center P and coefficient 2. Since P A1 A3 A2 is
a rectangle, this homothety sends line A1 A2 into line la that passes through point A3 ; lines
la and A3 P are symmetric through line A3 A. Line A3 A divides the angle B3 A3 C3 in halves
(Problem 1.56 a)).
We similarly prove that lines lb and lc are symmetric to lines B3 P and C3 P , respectively,
through bisectors of triangle A3 B3 C3 . Therefore, lines la , lb and lc either meet at one point
or are parallel (Problem 1.79) and, therefore, lines A1 A2 , B1 B2 and C1 C2 meet at one point.
5.82. By Problems 5.78 and 5.70 b)) we have
sin ∠ASQ sin ∠DAQ sin ∠SDQ
sin ∠ASP sin ∠DAP sin ∠SDP
·
·
=1=
·
·
.
sin ∠P SD sin ∠P AS sin ∠P DA
sin ∠QSD sin ∠QAS sin ∠QDA
But
∠DAP = ∠SDQ, ∠SDP = ∠DAQ, ∠P AS = ∠QDA and ∠P DA = ∠QAS.
Hence,
sin ∠ASQ
sin ∠ASP
=
.
sin ∠P SD
sin ∠QSD
SOLUTIONS
127
This implies that points S, P and Q lie on one line, since the function
with respect to x: indeed,
µ
¶
sin α
d sin(α − x)
=− 2 .
dx
sin x
sin x
5.83. a) By Ceva’s theorem
sin(α−x)
sin x
is monotonous
CA1 AB1
AC1
=
·
C1 B
A1 B B1 C
and by the law of sines
CA1 =
AB1 =
CA sin ∠CAA1
,
sin ∠AA1 B
AB sin ∠ABB1
,
sin ∠AB1 B
A1 B =
B1 C =
AB sin ∠BAA1
,
sin ∠AA1 B
BC sin ∠CBB1
.
sin ∠AB1 B
Substituting the last four identities in the first identity and taking into account that AC =
BC, we get the statement desired.
b) Let us denote the intersection points of lines CM and CN with base AB by M1
and N1 , respectively. We have to prove that M1 = N1 . From heading a) it follows that
AM1 : M1 B = AN1 : N1 B, i.e., M1 = N1 .
5.84. Let segments BM and BN meet side AC at points P and Q, respectively. Then
sin ∠P BB1 sin ∠AP B
P B AB
sin ∠P BB1
=
=
.
·
·
sin P BA
sin ∠BP B1 sin ∠P BA
BB1 P A
1O
1
If O is the intersection point of bisectors of triangle ABC, then PAP
· BOB
· BC
= 1 and,
B1
C1 A
therefore,
AB B1 O BC1
sin ∠P BB1
=
·
.
·
sin ∠P BA
BB1 OB C1 A
Observe that BC1 : C1 A = BC : CA and perform similar calculations for sin ∠QBB1 :
sin ∠QBC; we deduce that
sin ∠QBB1
sin ∠P BB1
=
.
sin ∠P BA
sin ∠QBC
Since ∠ABB1 = ∠CBB1 , we have: ∠P BB1 = ∠QBB1 .
5.85. a) Let point P lie on arc ⌣ AC of the circumscribed circle of triangle ABC; let
A1 , B1 and C1 be the bases of perpendiculars dropped from point P to lines BC, CA and
AB. The sum of angles at vertices A1 and C1 of quadrilateral A1 BC1 P is equal to 180◦ ,
hence, ∠A1 P C1 = 180◦ − ∠B = ∠AP C. Therefore, ∠AP C1 = ∠A1 P C, where one of points
A1 and C1 (say, A1 ) lies on a side of the triangle and the other point lies on the extension
of a side. Quadrilaterals AB1 P C1 and A1 B1 P C are inscribed ones, hence,
∠AB1 C1 = ∠AP C1 = ∠A1 P C = ∠A1 B1 C
and, therefore, point B1 lies on segment A1 C1 .
b) By the same arguments as in heading a) we get
∠(AP, P C1 ) = ∠(AB1 , B1 C) = ∠(CB1 , B1 A1 ) = ∠(CP, P A1 ).
Add ∠(P C1 , P C) to ∠(AP, P C1 ); we get
∠(AP, P C) = ∠(P C1 , P A1 ) = ∠(BC1 , BA1 ) = ∠(AB, BC),
i.e., point P lies on the circumscribed circle of triangle ABC.
5.86. Let A1 , B1 and C1 be the midpoints of segments P A, P B and P C, respectively;
let Oa , Ob and Oc be the centers of the circumscribed circles of triangles BCP , ACP and
ABP , respectively. Points A1 , B1 and C1 are the bases of perpendiculars dropped from point
128
CHAPTER 5. TRIANGLES
P to sides of triangle Oa Ob Oc (or their extensions). Points A1 , B1 and C1 lie on one line,
hence, point P lies on the circumscribed circle of triangle Oa Oc Oc (cf. Problem 5.85, b).
5.87. Let the extension of the bisector AD intersect the circumscribed circle of triangle
ABC at point P . Let us drop from point P perpendiculars P A1 , P B1 and P C1 to lines
BC, CA and AB, respectively; clearly, A1 is the midpoint of segment BC. The homothety
centered at A that sends P to D sends points B1 and C1 to B ′ and C ′ and, therefore, it
sends point A1 to M , because M (???) lies on line B1 C1 and P A1 k DM .
5.88. a) The solution of Problem 5.85 can be adapted without changes to this case.
b) Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC and
CA, respectively, and let points A2 and B2 from lines BC and AC, respectively, be such
that ∠(P A2 , BC) = α = ∠(P B2 , AC). Then △P A1 A2 ∼ △P B1 B2 hence, points A1 and B1
turn under a rotational homothety centered at P into A2 and B2 and ∠A1 P A2 = 90◦ − α is
the angle of the rotation.
5.89. a) Let the angle between lines P C and AC be equal to ϕ. Then P A = 2R sin ϕ.
Since points A1 and B1 lie on the circle with diameter P C, the angle between lines P A1 and
A1 B1 is also equal to ϕ. Hence, P A1 = sind ϕ and, therefore, P A · P A1 = 2Rd.
b) Since P A1 ⊥ BC, it follows that cos α = sin ϕ = P dA1 . It remains to notice that
.
P A1 = 2Rd
PA
5.90. Points A1 and B1 lie on the circle with diameter P C, hence, A1 B1 = P C sin ∠A1 CB1 =
P C sin ∠C. Let the angle between lines AB and A1 B1 be equal to γ and C1 be the projection
of point P to line A1 B1 . Lines A1 B1 and B1 C1 coincide, hence, cos γ = P2RC (cf. Problem
5.89). Therefore, the length of the projection of segment AB to line A1 B1 is equal to
(2R sin ∠C)P C
= P C sin ∠C.
2R
5.91. Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC and
AC. Points A1 and B1 lie on the circle with diameter P C. Since sin ∠A1 CB1 = sin ∠ACB,
the chords A1 B1 of this circle are of the same length. Therefore, lines A1 B1 are tangent to
a fixed circle.
5.92. Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC
and CA. Then
⌣ BP
∠(A1 B1 , P B1 ) = ∠(A1 C, P C) =
.
2
It is also clear that for all points P lines P B1 have the same direction.
5.93. Let P1 and P2 be diametrically opposite points of the circumscribed circle of
triangle ABC; let Ai and Bi be the bases of perpendiculars dropped from point Pi to lines BC
and AC, respectively; let M and N be the midpoints of sides AC and BC, respectively; let
X be the intersection point of lines A1 B1 and A2 B2 , respectively. By Problem 5.92 A1 B1 ⊥
A2 B2 . It remains to verify that ∠(M X, XN ) = ∠(BC, AC). Since AB2 = B1 C, it follows
that XM is a median of right triangle B1 XB2 . Hence, ∠(XM, XB2 ) = ∠(XB2 , B2 M ).
Similarly, ∠(XA1 , XN ) = ∠(A1 N, XA1 ). Therefore,
AB cos γ =
∠(M X, XN ) = ∠(XM, XB2 ) + ∠(XB2 , XA1 ) + ∠(XA1 , XN ) =
∠(XB2 , B2 M ) + ∠(A1 N, XA1 ) + 90◦ .
Since
∠(XB2 , B2 M ) + ∠(AC, CB) + ∠(N A1 , A1 X) + 90◦ = 0◦ ,
we have: ∠(M N, XN ) + ∠(AC, CB) = 0◦ .
SOLUTIONS
129
−→ −→
5.94. If point R on the given circle is such that ∠(OP , OR) = 21 (β + γ), then OR ⊥ BC.
It remains to verify that ∠(OR, OQ) = ∠(P A1 , A1 B1 ). But ∠(OR, OQ) = 12 α and
−→ −→
∠(OP , OA)
α
∠(P A1 , A1 B1 ) = ∠(P B, BC1 ) =
= .
2
2
5.95. Let lines AC and P Q meet at point M . In triangle M P C draw heights P B1
and CA1 . Then A1 B1 is Simson’s line of point P with respect to triangle ABC. Moreover, by Problem 1.52 ∠(M B1 , B1 A1 ) = ∠(CP, P M ). It is also clear that ∠(CP, P M ) =
∠(CA, AQ) = ∠(M B1 , AQ). Hence, A1 B1 k AQ.
5.96. Let us draw chord P Q perpendicular to BC. Let points H ′ and P ′ be symmetric
to points H and P , respectively, through line BC; point H ′ lies on the circumscribed circle
of triangle ABC (Problem 5.9). First, let us prove that AQ k P ′ H. Indeed, ∠(AH ′ , AQ) =
∠(P H ′ , P Q) = ∠(AH ′ , P ′ H). Simson’s line of point P is parallel to AQ (Problem 5.95),
i.e., it passes through the midpoint of side P P ′ of triangle P P ′ H and is parallel to side P ′ H;
hence, it passes through the midpoint of side P H.
5.97. Let Ha , Hb , Hc and Hd be the orthocenters of triangles BCD, CDA, DAB and
ABC, respectively. Lines la , lb , lc and ld pass through the midpoints of segments AHa , BHb ,
CHc and DHd , respectively (cf. Problem 5.96). The midpoints of these segments coincide
−−→ −→ −−→ −→ −−→
with point H such that 2OH = OA + OB + OC + OD, where O is the center of the circle
(cf. Problem 13.33).
5.98. a) Let B1 , C1 and D1 be the projections of point P to lines AB, AC and AD,
respectively. Points B1 , C1 and D1 lie on the circle with diameter AP . Lines B1 C1 , C1 D1
and D1 B1 are Simson’s lines of point P with respect to triangles ABC, ACD and ADB,
respectively. Therefore, projections of point P to Simson’s lines of these triangles lie on one
line — Simson’s line of triangle B1 C1 D1 .
We similarly prove that any triple of considered points lies on one line.
b) Let P be a point of the circumscribed circle of n-gon A1 . . . An ; let B2 , B3 , . . . , Bn be
the projections of point P to lines A1 A2 , . . . , A1 An , respectively. Points B2 , . . . , Bn lie on
the circle with diameter A1 P .
Let us prove by induction that Simson’s line of point P with respect to n-gon A1 . . . An
coincides with Simson’s line of point P with respect to (n − 1)-gon B2 . . . Bn (for n = 4 this
had been proved in heading a)). By the inductive hypothesis Simson’s line of the (n − 1)-gon
A1 A3 . . . An coincides with Simson’s line of (n − 2)-gon B3 . . . Bn . Hence, the projections of
point P to Simson’s line of (n − 1)-gons whose vertices are obtained by consecutive deleting
points A2 , . . . , An from the collection A1 , . . . , An ????? lie on Simson’s line of the (n − 1)-gon
B2 . . . Bn . The projection of point P to Simson’s line of the (n − 1)-gon A2 . . . An lies on
the same line, because our arguments show that any n − 1 of the considered n points of
projections lie on one line.
5.99.
¡ ¢ Points B1 and C1 lie on the circle with diameter AP . Hence, B1 C1 = AP sin ∠B1 AC1 =
AP BC
.
2R
5.100. This problem is a particular case of Problem 2.43.
5.101. Clearly,
∠C1 AP = ∠C1 B1 P = ∠A2 B1 P = ∠A2 C2 P = ∠B3 C2 P = ∠B3 A3 P.
(The first, third and fifth equalities are obtained from the fact that the corresponding quadrilaterals are inscribed ones; the remaining equalities are obvious.) Similarly, ∠B1 AP =
∠C3 A3 P . Hence,
∠B3 A3 C3 = ∠B3 A3 P + ∠C3 A3 P = ∠C1 AP + ∠BAP = ∠BAC.
130
CHAPTER 5. TRIANGLES
Similarly, the equalities of the remaining angles of triangles ABC and A3 B3 C3 are similarly
obtained.
5.102. Let A1 , B1 and C1 be the bases of perpendiculars dropped from point P to lines
BC, CA and AB, respectively; let A2 , B2 and C2 be the intersection points of lines P A,
P B and P C, respectively, with the circumscribed circle of triangle ABC. Further, let S, S1
and S2 be areas of triangles ABC, A1 B1 C1 and A2 B2 C2 , respectively. It is easy to verify
2P
that a1 = a·AP
(Problem 5.99) and a2 = a·B
. Triangles A1 B1 C1 and A2 B2 C2 are similar
2R
CP
AP ·CP
. Since B2 P · BP = |d2 − R2 |, we
(Problem 5.100); hence, SS21 = k 2 , where k = aa12 = 2R·B
2P
have:
S1
(AP · BP · CP )2
.
=
S2
4R2 (d2 − R2 )2
b2 c 2
Triangles A2 B2 C2 and ABC are inscribed in one circle, hence, SS2 = a2abc
(cf. Problem
12.1). It is also clear that, for instance,
Therefore,
Hence,
B2 P
|d2 − R2 |
a2
=
=
.
a
CP
BP · CP
S2 : S = |d2 − R2 |3 : (AP · BP · CP )2 .
S1 S2
|d2 − R2 |
S1
=
=
.
·
S
S2 S
4R2
5.103. Points B1 and C1 lie on the circle with diameter P A and, therefore, the midpoint
of segment P A is the center of the circumscribed circle of triangle AB1 C1 . Consequenly, la
is the midperpendicular to segment B1 C1 . Hence, lines la , lb and lc pass through the center
of the circumscribed circle of triangle A1 B1 C1 .
5.104. a) Let us drop from points P1 and P2 perpendiculars P1 B1 and P2 B2 , respectively,
to AC and perpendiculars P1 C1 and P2 C2 to AB. Let us prove that points B1 , B2 , C1 and
C2 lie on one circle. Indeed,
∠P1 B1 C1 = ∠P1 AC1 = ∠P2 AB2 = ∠P2 C2 B2 ;
and, since ∠P1 B1 A = ∠P2 C2 A, it follows that ∠C1 B1 A = ∠B2 C2 A. The center of the
circle on which the indicated points lie is the intersection point of the midperpendiculars to
segments B1 B2 and C1 C2 ; observe that both these perpendiculars pass through the midpoint
O of segment P1 P2 , i.e., O is the center of this circle. In particular, points B1 and C1 are
equidistant from point O. Similarly, points A1 and B1 are equidistant from point O, i.e., O
is the center of the circumscribed circle of triangle A1 B1 C1 . Moreover, OB1 = OB2 .
b) The preceding proof passes virtually without changes in this case as well.
5.105. Let A1 , B1 and C1 be the midpoints of sides BC, CA and AB. Triangles A1 B1 C1
and ABC are similar and the similarity coefficient is equal to 2. The heights of triangle
A1 B1 C1 intersect at point O; hence, OA1 : HA = 1 : 2. Let M ′ be the intersection point of
segments OH and AA1 . Then OM ′ : M ′ H = OA1 : HA = 1 : 2 and AM ′ : M ′ A1 = OA1 :
HA = 1 : 2, i.e., M ′ = M .
5.106. Let A1 , B1 and C1 be the midpoints of sides BC, CA and AB, respectively; let
A2 , B2 and C2 the bases of heights; A3 , B3 and C3 the midpoints of segments that connect
the intersection point of heights with vertices. Since A2 C1 = C1 A = A1 B1 and A1 A2 k B1 C1 ,
point A2 lies on the circumscribed circle of triangle A1 B1 C1 . Similarly, points B2 and C2 lie
on the circumscribed circle of triangle A1 B1 C1 .
Now, consider circle S with diameter A1 A3 . Since A1 B3 k CC2 and A3 B3 k AB, it follows
that ∠A1 B3 A3 = 90◦ and, therefore, point B3 lies on S. We similarly prove that points C1 ,
SOLUTIONS
131
B1 and C3 lie on S. Circle S passes through the vertices of triangle A1 B1 C1 ; hence, it is its
circumscribed circle.
The homothety with center H and coefficient 12 sends the circumscribed circle of triangle ABC into the circumscribed circle of triangle A3 B3 C3 , i.e., into the circle of 9 points.
Therefore, this homothety sends point O into the center of the circle of nine points.
5.107. a) Let us prove that, for example, triangles ABC and HBC share the same circle
of nine points. Indeed, the circles of nine points of these triangles pass through the midpoint
of side BC and the midpoints of segments BH and CH.
b) Euler’s line passes through the center of the circle of 9 points and these triangles share
one circle of nine points.
c) The center of symmetry is the center of the circle of 9 points of these triangles.
5.108. Let AB > BC > CA. It is easy to verify that for an acute and an obtuse
triangles the intersection point H of heights and the center O of the circumscribed circle
are positioned precisely as on Fig. 59 (i.e., for an acute triangle point O lies inside triangle
BHC1 and for an acute triangle points O and B lie on one side of line CH).
Figure 59 (Sol. 5.108)
Therefore, in an acute triangle Euler’s line intersects the longest side AB and the shortest
side AC, whereas in an acute triangle it intersects the longest side AB, and side BC of
intermediate length.
5.109. a) Let Oa , Ob and Oc be the centers of the escribed circles of triangle ABC. The
vertices of triangle ABC are the bases of the heights of triangle Oa Ob Oc (Problem 5.2) and,
therefore, the circle of 9 points of triangle Oa Ob Oc passes through point A, B and C.
b) Let O be the intersection point of heights of triangle Oa Ob Oc , i.e., the intersection
point of the bisectors of triangle ABC. The circle of 9 points of triangle Oa Ob Oc divides
segment OOa in halves.
5.110. Let AA1 be an height, H the intersection point of heights. By Problem 5.45 b)
AH = 2R| cos ∠A|. The medians are divided by their intersection point in the ratio of 1:2,
−−→
hence, Euler’s line is parallel to BC if and only if AH : AA1 = 2 : 3 and vectors AH and
−−→
AA1 are codirected, i.e.,
2R cos ∠A : 2R sin ∠B sin ∠C = 2 : 3.
Taking into account that
we get
cos ∠A = − cos(∠B + ∠C) = sin ∠B sin ∠C − cos ∠B cos ∠C
sin ∠B sin ∠C = 3 cos ∠B cos ∠C.
5.111. Let CD be a height, O the center of the circumscribed circle, N the midpoint of
side AB and let point E divide the segment that connects C with the intersection point of the
132
CHAPTER 5. TRIANGLES
heights in halves. Then CEN O is a parallelogram, hence, ∠N ED = ∠OCH = |∠A − ∠B|
(cf. Problem 2.88). Points N , E and D lie on the circle of 9 points, hence, segment N D is
seen from its center under an angle of 2∠N ED = 2|∠A − ∠B|.
5.112. Let O and I be the centers of the circumscribed and inscribed circles, respectively,
of triangle ABC, let H be the intersection point of the heights; lines AI and BI intersect
the circumscribed circle at points A1 and B1 . Suppose that triangle ABC is not an isosceles
one. Then OI : IH = OA1 : AH and OI : IH = OB1 : BH. Since OB1 = OA1 , we see that
AH = BH and, therefore, AC = BC. Contradiction.
5.113. Let O and I be the centers of the circumscribed and inscribed circles, respectively,
of triangle ABC, H the orthocenter of triangle A1 B1 C1 . In triangle A1 B1 C1 , draw heights
A1 A2 , B1 B2 and C1 C2 . Triangle A1 B1 C1 is an acute one (e.g., ∠B1 A1 C1 = ∠B+∠C
< 90◦ ),
2
hence, H is the center of the inscribed circle of triangle A2 B2 C2 (cf. Problem 1.56, a).
The corresponding sides of triangles ABC and A2 B2 C2 are parallel (cf. Problem 1.54 a)
and, therefore, there exists a homothety that sends triangle ABC to triangle A2 B2 C2 . This
homothety sends point O to point I and point I to point H; hence, line IH passes through
point O.
5.114. Let H be be the intersection point of the heights of triangle ABC, let E and M
be the midpoints of segments CH and AB, see Fig. 60. Then C1 M C2 E is a rectangle.
Figure 60 (Sol. 5.114)
Let line CC2 meet line AB at point C3 . Let us prove that AC3 : C3 B = tan 2α : tan 2β.
It is easy to verify that
C3 M : C2 E = M C2 : EC, EC = R cos γ,
M C2 = C1 E = 2R sin α sin β − R cos γ
and C2 E = M C1 = R sin(β − α)
Hence,
C3 M =
R sin(β − α) cos(β − α)
R sin(β − α)(2 sin β sin α − cos γ)
=
.
cos γ
cos γ
Therefore,
AC3
AM + M C3
sin 2γ + sin 2(α − β)
tan 2α
=
=
=
.
sin 2γ − sin 2(α − β)
tan 2β
C3 B
C3 M + M B
Similar arguments show that
AC3 BA3 CB3
tan 2α tan 2β tan 2γ
·
·
= 1.
·
·
=
tan 2β tan 2γ tan 2α
C3 B A3 C B3 A
5.115. Let us solve a more general heading b). First, let us prove that lines AA1 , BB1
and CC1 meet at one point. Let the circumscribed circles of triangles A1 BC and AB1 C
SOLUTIONS
133
intersect at point O. Then
∠(BO, OA) = ∠(BO, OC) + ∠(OC, OA) = ∠(BA1 , A1 C) + ∠(CB1 , B1 A) =
= ∠(BA, AC1 ) + ∠(C1 B, BA) = ∠(C1 B, AC1 ),
i.e., the circumscribed circle of triangle ABC1 also passes through point O. Hence,
∠(AO, OA1 ) = ∠(AO, OB) + ∠(BO, OA1 ) = ∠(AC1 , C1 B + ∠(BC, CA1 ) = 0◦ ,
i.e., line AA1 passes through point O. We similarly prove that lines BB1 and CC1 pass
through point O.
Now, let us prove that point O coincides with point P we are looking for. Since ∠BAP =
∠A − ∠CAP , the equality ∠ABP = ∠CAP is equivalent to the equality ∠BAP + ∠ABP =
∠A, i.e., ∠AP B = ∠B + ∠C. For point O the latter equality is obvious since it lies on the
circumscribed circle of triangle ABC1 .
5.116. a) Let us prove that ⌣ AB =⌣ B1 C1 , i.e., AB = B1 C1 . Indeed, ⌣ AB =⌣
AC1 + ⌣ C1 B and ⌣ C1 B =⌣ AB1 ; hence, ⌣ AB =⌣ AC1 + ⌣ AB1 =⌣ B1 C1 .
b) Let us assume that triangles ABC and A1 B1 C1 are inscribed in one circle, where
triangle ABC is fixed and triangle A1 B1 C1 rotates. Lines AA1 , BB1 and CC1 meet at one
point for not more than one position of triangle A1 B1 C1 , see Problem 7.20 b). We can obtain
12 distinct families of triangles A1 B1 C1 : triangles ABC and A1 B1 C1 can be identified after
a rotation or an axial symmetry; moreover, there are 6 distinct ways to associate symbols
A1 , B1 and C1 to the vertices of the triangle.
From these 12 families of triangles 4 families can never produce the desired point P . For
similarly oriented triangles the cases
△ABC = △A1 C1 B1 ,
△ABC = △C1 B1 A1 ,
△ABC = △B1 A1 C1
are excluded: for example, if △ABC = △A1 C1 B1 , then point P is the intersection point of
line BC = B1 C1 with the tangent to the circle at point A = A1 ; in this case triangles ABC
and A1 B1 C1 coincide.
For differently oriented triangles the case △ABC = △A1 B1 C1 is excluded: in this case
AA1 k BB1 k CC1 .
Remark. Brokar’s points correspond to differently oriented triangles; for the first Brokar’s
point △ABC = △B1 C1 A1 and for the second Brokar’s point we have △ABC = △C1 A1 B1 .
∠CAP
∠CBP
5.117. a) Since P C = ACsinsin∠AP
and P C = BCsinsin∠BP
, it follows that
C
C
sin ϕ sin β
sin(β − ϕ) sin α
=
.
sin γ
sin β
Taking into account that
we get cot ϕ = cot β +
sin(β − γ) = sin β cos ϕ − cos β sin ϕ
sin β
.
sin α sin γ
It remains to notice that
sin β = sin(α + γ) = sin α cos γ + sin γ cos α.
b) For the second Brokar’s angle we get precisely the same expression as in heading a).
It is also clear that both Brokar’s angles are acute ones.
c) Since ∠A1 BC = ∠BCA and ∠BCA1 = ∠CAB, it follows that △CA1 B ∼ △ABC.
Therefore, Brokar’s point P lies on segment AA1 (cf. Problem 5.115 b)).
5.118. a) By Problem 10.38 a)
√
cot ϕ = cot α + cot β + cot γ ≥ 3 = cot 30◦ ;
hence, ϕ ≤ 30◦ .
134
CHAPTER 5. TRIANGLES
b) Let P be the first Brokar’s point of triangle ABC. Point M lies inside (or on the
boundary of) one of the triangles ABP , BCP and CAP . If, for example, point M lies inside
triangle ABP , then ∠ABM ≤ ∠ABP ≤ 30◦ .
5.119. Lines A1 B1 , B1 C1 and C1 A1 are the midperpendiculars to segments AQ, BQ and
CQ, respectively. Therefore, we have, for instance, ∠B1 A1 C1 = 180◦ − ∠AQC = ∠A. For
the other angles the proof is similar.
Moreover, lines A1 O, B1 O and C1 O are the midperpendiculars to segments CA, AB
and BC, respectively. Hence, acute angles ∠OA1 C1 and ∠ACQ, for example, have pairwise perpendicular sides and, consecutively, they are equal. Similar arguments show that
∠OA1 C1 = ∠OB1 A1 = ∠OC1 B1 = ϕ, where ϕ is the Brokar’s angle of triangle ABC.
5.120. By the law of sines
R1 =
AB
BC
, R2 =
2 sin ∠AP B
2 sin ∠BP C
and R3 =
CA
.
2 sin ∠CP A
It is also clear that
sin ∠AP B = sin ∠A, sin ∠BP C = sin ∠B and sin ∠CP A = sin ∠C.
5.121. Triangle ABC1 is an isosceles one and the angle at its base AB is equal to
Brokar’s angle ϕ. Hence, ∠(P C1 , C1 Q) = ∠(BC1 , C1 A) = 2ϕ. Similarly
∠(P A1 , A1 Q) = ∠(P B1 , B1 Q) = ∠(P C1 , C1 Q) = 2ϕ.
5.122. Since ∠CA1 B1 = ∠A + ∠AB1 A1 and ∠AB1 A1 = ∠CA1 C1 , we have ∠B1 A1 C1 =
∠A. We similarly prove that the remaining angles of triangles ABC and A1 B1 C1 are equal.
The circumscribed circles of triangles AA1 B1 , BB1 C1 and CC1 A1 meet at one point O.
(Problem 2.80 a). Clearly, ∠AOA1 = ∠AB1 A1 = ϕ. Similarly, ∠BOB1 = ∠COC1 = ϕ.
Hence, ∠AOB = ∠A1 OB1 = 180◦ −∠A. Similarly, ∠BOC = 180◦ −∠B and ∠COA = 180◦ −
∠C, i.e., O is the first Brokar’s point of both triangles. Hence, the rotational homothety by
angle ϕ with center O and coefficient AAO
sends triangle A1 B1 C1 to triangle ABC.
1O
AB
sin ∠AM B
AB
sin ∠AN B
5.123. By the law of sines BM = sin ∠BAM and BN
= sin
. Hence,
∠BAN
AB 2
sin ∠AM B sin ∠AN B
sin ∠AM C sin ∠AN C
AC 2
=
=
=
.
BM · BN
sin ∠BAM sin ∠BAN
sin ∠CAN sin ∠CAM
CM · CN
5.124. Since ∠BAS = ∠CAM , we have
BS
SBAS
AB · AS
=
,
=
CM
SCAM
AC · AM
AS
= 2b·BS
. It remains to observe that, as follows from Problems 5.123 and 12.11 a),
i.e., AM
ac
√
ac2
BS = b2 +c2 and 2AM = 2b2 + 2c2 − a2 .
5.125. The symmetry through the bisector of angle A sends segment B1 C1 into a segment
parallel to side BC, it sends line AS to line AM , where M is the midpoint of side BC.
5.126. On segments BC and BA, take points A1 and C1 , respectively, so that A1 C1 k
BK. Since ∠BAC = ∠CBK = ∠BA1 C1 , segment A1 C1 is antiparallel to side AC. On the
other hand, by Problem 3.31 b) line BD divides segment A1 C1 in halves.
5.127. It suffices to make use of the result of Problem 3.30.
5.128. Let AP be the common chord of the considered circles, Q the intersection point
of lines AP and BC. Then
sin ∠BAQ
AC
sin ∠AQC
BQ
=
and
=
.
AB
sin ∠AQB
CQ
sin ∠CAQ
SOLUTIONS
135
sin ∠BAP
= AB
. Since AC and AB are tangents to circles S1 and S2 , it follows that
Hence, BQ
CQ
AC sin ∠CAP
∠CAP = ∠ABP and ∠BAP = ∠ACP and, therefore, ∠AP B = ∠AP C. Hence,
AB AP
sin ∠AP B sin ∠ACP
sin ∠ACP
sin ∠BAP
AB
=
·
=
·
=
=
.
AC
AP AC
sin ∠ABP sin ∠AP C
sin ∠ABP
sin ∠CAP
2
= AB
.
It follows that BQ
CQ
AC 2
AS
CS
5.129. Let S be the intersection point of lines AX and BC. Then AB
= CX
and
AS
= BSBX and, therefore,
AC
CS
AC XC
=
·
.
BS
AB XB
AC
XC
= AB
(see the solution of Problem 7.16 a)).
It remains to observe that XB
5.130. Let L, M and N be the midpoints of segments CA, CB and CH. Since △BAC ∼
△CAH, it follows that △BAM ∼ △CAN and, therefore, ∠BAM = ∠CAN . Similarly,
∠ABL = ∠CBN .
5.131. Let B1 C1 , C2 A2 and A3 B3 be given segments. Then triangles A2 XA3 , B1 XB3
and C1 XC2 are isosceles ones; let the lengths of their lateral sides be equal to a, b and c.
Line AX divides segment B1 C1 in halves if and only if this line contains a simedian. Hence,
if X is Lemoin’s point, then a = b, b = c and c = a. And if B1 C1 = C2 A2 = A3 B3 , then
b + c = c + a = a + b and, therefore, a = b = c.
5.132. Let M be the intersection point of medians of triangle ABC; let a1 , b2 , c1 and a2 ,
b2 , c2 be the distances from points K and M , respectively, to the sides of the triangle. Since
points K and M are isogonally conjugate, a1 a2 = b1 b2 = c1 c2 . Moreover, aa2 = bb2 = cc2
(cf. Problem 4.1). Therefore, aa1 = bb1 = cc1 . Making use of this equality and taking into
a1 b
b1 c b1 c 1 a
, 4R and c14R
,
account that areas of triangles A1 B1 K, B1 C1 K and C1 A1 K are equal to a14R
respectively, where R is the radius of the circumscribed circle of triangle ABC, we deduce
that the areas of these triangles are equal. Moreover, point K lies inside triangle A1 B1 C1 .
Therefore, K is the intersection point of medians of triangle A1 B1 C1 (cf. Problem 4.2).
5.133. Medians of triangle A1 B1 C1 intersect at point K (Problem 5.132); hence, the
sides of triangle ABC are perpendicular to the medians of triangle A1 B1 C1 . After a rotation
through an angle of 90◦ the sides of triangle ABC become pairwise parallel to the medians
of triangle A1 B1 C1 and, therefore, the medians of triangle ABC become parallel to the
corresponding sides of triangle A1 B1 C1 (cf. Problem 13.2). Hence, the medians of triangle
ABC are perpendicular to the corresponding sides of triangle A1 B1 C1 .
5.134. Let A2 , B2 and C2 be the projections of point K to lines BC, CA and AB,
respectively. Then △A1 B1 C1 ∼ △A2 B2 C2 (Problem 5.100) and K is the intersection point
of medians of triangle A2 B2 C2 (Problem 5.132). Hence, the similarity transformation that
sends triangle A2 B2 C2 to triangle A1 B1 C1 sends point K to the intersection point M of
medians of triangle A1 B1 C1 . Moreover, ∠KA2 C2 = ∠KBC2 = ∠B1 A1 K, i.e., points K and
M are isogonally conjugate with respect to triangle A1 B1 C1 and, therefore, K is Lemoin’s
point of triangle A1 B1 C1 .
5.135. Let K be Lemoin’s point of triangle ABC; let A1 , B1 and C1 be the projections
of point K on the sides of triangle ABC; let L be the midpoint of segment B1 C1 and N
the intersection point of line KL and median AM ; let O be the midpoint of segment AK
(Fig. 61). Points B1 and C1 lie on the circle with diameter AK, hence, by Problem 5.132
OL ⊥ B1 C1 . Moreover, AN ⊥ B1 C1 (Problem 5.133) and O is the midpoint of segment
AK, consequently, OL is the midline of triangle AKN and KL = LN . Therefore, K is
the midpoint of segment A1 N . It remains to notice that the homothety with center M that
sends N to A sends segment N A1 to height AH.
136
CHAPTER 5. TRIANGLES
Figure 61 (Sol. 5.135)
Chapter 6. POLYGONS
Background
1) A polygon is called a convex one if it lies on one side of any line that connects two of
its neighbouring vertices.
2) A convex polygon is called a circumscribed one if all its sides are tangent to a circle.
A convex quadrilateral is a circumscribed one if and only if AB + CD = BC + AD.
A convex polygon is called an inscribed one if all its vertices lie on one circle. A convex
quadrilateral is an inscribed one if and only if
∠ABC + ∠CDA = ∠DAB + ∠BCD.
3) A convex polygon is called a regular one if all its sides are equal and all its angles are
also equal.
with
A convex n-gon is a regular one if and only if under a rotation by the angle of 2π
n
center at point O it turns into itself. This point O is called the center of the regular polygon.
Introductory problems
1. Prove that a convex quadrilateral ABCD can be inscribed into a circle if and only if
∠ABC + ∠CDA = 180◦ .
2. Prove that a circle can be inscribed in a convex quadrilateral ABCD if and only if
AB + CD = BC + AD.
3. a) Prove that the axes of symmetry of a regular polygon meet at one point.
b) Prove that a regular 2n-gon has a center of symmetry.
4. a) Prove that the sum of the angles at the vertices of a convex n-gon is equal to
(n − 2) · 180◦ .
b) A convex n-gon is divided by nonintersecting diagonals into triangles. Prove that the
number of these triangles is equal to n − 2.
§1. The inscribed and circumscribed quadrilaterals
6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with the
intersection point of the quadrilateral’s diagonals, then this quadrilateral is a rhombus.
6.2. Quadrilateral ABCD is circumscribed about a circle centered at O. Prove that
∠AOB + ∠COD = 180◦ .
6.3. Prove that if there exists a circle tangent to all the sides of a convex quadrilateral
ABCD and a circle tangent to the extensions of all its sides then the diagonals of such a
quadrilateral are perpendicular.
6.4. A circle singles out equal chords on all the four sides of a quadrilateral. Prove that
a circle can be inscribed into this quadrilateral.
6.5. Prove that if a circle can be inscribed into a quadrilateral, then the center of this
circle lies on one line with the centers of the diagonals.
137
138
CHAPTER 6. POLYGONS
6.6. Quadrilateral ABCD is circumscribed about a circle centered at O. In triangle
AOB heights AA1 and BB1 are drawn. In triangle COD heights CC1 and DD1 are drawn.
Prove that points A1 , B1 , C1 and D1 lie on one line.
6.7. The angles at base AD of trapezoid ABCD are equal to 2α and 2β. Prove that the
BC
trapezoid is a circumscribed one if and only if AD
= tan α tan β.
6.8. In triangle ABC, segments P Q and RS parallel to side AC and a segment BM are
drawn as plotted on Fig. 62. Trapezoids RP KL and M LSC are circumscribed ones. Prove
that trapezoid AP QC is also a circumscribed one.
Figure 62 (6.8)
6.9. Given convex quadrilateral ABCD such that rays AB and CD intersects at a
point P and rays BC and AD intersect at a point Q. Prove that quadrilateral ABCD is a
circumscribed one if and only if one of the following conditions hold:
AB + CD = BC + AD,
AP + CQ = AQ + CP
BP + BQ = DP + DQ.
6.10. Through the intersection points of the extension of sides of convex quadrilateral
ABCD two lines are drawn that divide it into four quadrilaterals. Prove that if the quadrilaterals adjacent to vertices B and D are circumscribed ones, then quadrilateral ABCD is
also a circumscribed one.
6.11. Prove that the intersection point of the diagonals of a circumscribed quadrilateral
coincides with the intersection point of the diagonals of the quadrilateral whose vertices are
the tangent points of the sides of the initial quadrilateral with the inscribed circle.
***
6.12. Quadrilateral ABCD is an inscribed one; Hc and Hd are the orthocenters of
triangles ABD and ABC respectively. Prove that CDHc Hd is a parallelogram.
6.13. Quadrilateral ABCD is an inscribed one. Prove that the centers of the inscribed
circles of triangles ABC, BCD, CDA and DAB are the vertices of a rectangle.
6.14. The extensions of the sides of quadrilateral ABCD inscribed in a circle centered
at O intersect at points P and Q and its diagonals intersect at point S.
a) The distances from points P , Q and S to point O are equal to p, q and s, respectively,
and the radius of the circumscribed circle is equal to R. Find the lengths of the sides of
triangle P QS.
b) Prove that the heights of triangle P QS intersect at point O.
***
6.15. Diagonal AC divides quadrilateral ABCD into two triangles whose inscribed
circles are tangent to diagonal AC at one point. Prove that the inscribed circles of triangle
ABD and BCD are also tangent to diagonal BD at one point and their tangent points with
the sides of the quadrilateral lie on one circle.
§2. QUADRILATERALS
139
6.16. Prove that the projections of the intersection point of the diagonals of the inscribed
quadrilateral to its sides are vertices of a circumscribed quadrilateral only if the projections
do not lie on the extensions of the sides.
6.17. Prove that if the diagonals of a quadrilateral are perpendicular, then the projections of the intersection points of the diagonals on its sides are vertices of an inscribed
quadrilateral.
See also Problem 13.33, 13.34, 16.4.
§2. Quadrilaterals
6.18. The angle between sides AB and CD of quadrilateral ABCD is equal to ϕ. Prove
that
AD2 = AB 2 + BC 2 + CD2 − 2(AB · BC cos B + BC · CD cos C + CD · AB cos ϕ).
6.19. In quadrilateral ABCD, sides AB and CD are equal and rays AB and DC intersect
at point O. Prove that the line that connects the midpoints of the diagonals is perpendicular
to the bisector of angle AOD.
6.20. On sides BC and AD of quadrilateral ABCD, points M and N , respectively, are
taken so that BM : M C = AN : N D = AB : CD. Rays AB and DC intersect at point O.
Prove that line M N is parallel to the bisector of angle AOD.
6.21. Prove that the bisectors of the angles of a convex quadrilateral form an inscribed
quadrilateral.
6.22. Two distinct parallelograms ABCD and A1 B1 C1 D1 with corresponding parallel
sides are inscribed into quadrilateral P QRS (points A and A1 lie on side P Q, points B and
B1 lie on side QR, etc.). Prove that the diagonals of the quadrilateral are parallel to the
corresponding sides of the parallelograms.
6.23. The midpoints M and N of diagonals AC and BD of convex quadrilateral ABCD
do not coincide. Line M N intersects sides AB and CD at points M1 and N1 . Prove that if
M M1 = N N1 , then AD k BC.
6.24. Prove that two quadrilaterals are similar if and only if four of their corresponding
angles are equal and the corresponding angles between the diagonals are also equal.
6.25. Quadrilateral ABCD is a convex one; points A1 , B1 , C1 and D1 are such that
AB k C1 D1 and AC k B1 D1 , etc. for all pairs of vertices. Prove that quadrilateral A1 B1 C1 D1
is also a convex one and ∠A + ∠C1 = 180◦ .
6.26. From the vertices of a convex quadrilateral perpendiculars are dropped on the
diagonals. Prove that the quadrilateral with vertices at the basis of the perpendiculars is
similar to the initial quadrilateral.
6.27. A convex quadrilateral is divided by the diagonals into four triangles. Prove that
the line that connects the intersection points of the medians of two opposite triangles is
perpendicular to the line that connects the intersection points of the heights of the other
two triangles.
6.28. The diagonals of the circumscribed trapezoid ABCD with bases AD and BC
intersect at point O. The radii of the inscribed circles of triangles AOD, AOB, BOC and
COD are equal to r1 , r2 , r3 and r4 , respectively. Prove that r11 + r13 = r12 + r14 .
6.29. A circle of radius r1 is tangent to sides DA, AB and BC of a convex quadrilateral
ABCD; a circle of radius r2 is tangent to sides AB, BC and CD; the radii r3 and r4 are
+ CD
= BC
+ AD
.
similarly defined. Prove that AB
r1
r3
r2
r4
6.30. A quadrilateral ABCD is convex and the radii of the circles inscribed in triangles
ABC, BCD, CDA and DAB are equal. Prove that ABCD is a rectangle.
140
CHAPTER 6. POLYGONS
6.31. Given a convex quadrilateral ABCD and the centers A1 , B1 , C1 and D1 of the
circumscribed circles of triangles BCD, CDA, DAB and ABC, respectively. For quadrilateral A1 B1 C1 D1 points A2 , B2 , C2 and D2 are similarly defined. Prove that quadrilaterals
ABCD and A2 B2 C2 D2 are similar and their similarity coefficient is equal to
1
|(cot A + cot C)(cot B + cot D)| .
4
6.32. Circles whose diameters are sides AB and CD of a convex quadrilateral ABCD
are tangent to sides CD and AB, respectively. Prove that BC k AD.
6.33. Four lines determine four triangles. Prove that the orthocenters of these triangles
lie on one line.
§3. Ptolemy’s theorem
6.34. Quadrilateral ABCD is an inscribed one. Prove that
AB · CD + AD · BC = AC · BD
(Ptolemy’s theorem).
6.35. Quadrilateral ABCD is an inscribed one. Prove that
AC
AB · AD + CB · CD
=
.
BD
BA · BC + DA · DC
6.36. Let α = π7 . Prove that
1
1
1
=
+
.
sin α
sin 2α sin 3α
6.37. The distances from the center of the circumscribed circle of an acute triangle to
its sides are equal to da , db and dc . Prove that da + db + dc = R + r.
6.38. The bisector of angle ∠A of triangle ABC intersects the circumscribed circle at
point D. Prove that AB + AC ≤ 2AD.
6.39. On arc ⌣
√ CD of the circumscribed circle of square ABCD point P is taken. Prove
that P A + P C = 2P B.
6.40. Parallelogram ABCD is given. A circle passing through point A intersects segments AB, AC and AD at points P , Q and R, respectively. Prove that
AP · AB + AR · AD = AQ · AC.
6.41. On arc ⌣ A1 A2n+1 of the circumscribed circle S of a regular (2n + 1)-gon
A1 . . . A2n+1 a point A is taken. Prove that:
a) d1 + d3 + · · · + d2n+1 = d2 + d4 + · · · + d2n , where di = AAi ;
b) l1 + · · · + l2n+1 = l2 + · · · + l2n , where li is the length of the tangent drawn from point A
to the circle of radius r tangent to S at point Ai (all the tangent points are simultaneously
either inner or outer ones).
6.42. Circles of radii x and y are tangent to a circle of radius R and the distance between
the tangent points is equal to a. Calculate the length of the following common tangent to
the first two circles:
a) the outer one if both tangents are simultaneously either outer or inner ones;
b) the inner one if one tangent is an inner one and the other one is an outer one.
6.43. Circles α, β, γ and δ are tangent to a given circle at vertices A, B, C and D,
respectively, of convex quadrilateral ABCD. Let tαβ be the length of the common tangent
to circles α and β (the outer one if both tangent are simultaneously either inner or outer
§5. HEXAGONS
141
ones and the inner one if one tangent is an inner one and the other one is an outer one); tβγ ,
tγδ , etc. are similarly determined. Prove that
tαβ tγδ + tβγ tδα = tαγ tβδ
(The generalized Ptolemy’s theorem)
See also Problem 9.67.
§4. Pentagons
6.44. In an equilateral (non-regular) pentagon ABCDE we have angle ∠ABC =
2∠DBE. Find the value of angle ∠ABC.
6.45. a) Diagonals AC and BE of a regular pentagon ABCDE intersect at point K.
Prove that the inscribed circle of triangle CKE is tangent to line BC.
b) Let a be the length of the side of a regular pentagon, d the length of its diagonal.
Prove that d2 = a2 + ad.
6.46. Prove that a square can be inscribed in a regular pentagon so that the vertices of
the square would lie on four sides of the pentagon.
Figure 63 (6.46)
6.47. Regular pentagon ABCDE with side a is inscribed in circle S. The lines that
pass through the pentagon’s vertices perpendicularly to the sides form a regular pentagon
with side b (Fig. 63). A side of a regular pentagon circumsribed about circle S is equal to
c. Prove that a2 + b2 = c2 .
See also Problems 2.59, 4.9, 9.23, 9.44, 10.63, 10.67, 13.10, 13.56, 20.11.
§5. Hexagons
6.48. The opposite sides of a convex hexagon ABCDEF are pairwise parallel. Prove
that:
a) the area of triangle ACE constitutes not less than a half area of the hexagon.
b) the areas of triangles ACE and BDF are equal.
6.49. All the angles of a convex hexagon ABCDEF are equal. Prove that
|BC − EF | = |DE − AB| = |AF − CD|.
6.50. The sums of the angles at vertices A, C, E and B, D, F of a convex hexagon
ABCDEF with equal sides are equal. Prove that the opposite sides of this hexagon are
parallel.
6.51. Prove that if in a convex hexagon each of the three diagonals that connect the
opposite vertices divides the area in halves then these diagonals intersect at one point.
6.52. Prove that if in a convex hexagon each of the three segments that connect the
midpoints of the opposite sides divides the area in halves then these segments intersect at
one point.
142
CHAPTER 6. POLYGONS
See also problems 2.11, 2.20, 2.46, 3.66, 4.6, 4.28, 4.31, 5.80, 9.45 a), 9.76–9.78, 13.3,
14.6, 18.22, 18.23.
§6. Regular polygons
6.53. The number of sides of a polygon A1 . . . An is odd. Prove that:
a) if this polygon is an inscribed one and all its angles are equal, then it is a regular
polygon;
b) if this polygon is a circumscribed one and all its sides are equal, then it is a regular
polygon.
6.54. All the angles of a convex polygon A1 . . . An are equal; an inner point O of the
polygon is the vertex of equal angles that subtend all the polygon’s sides. Prove that the
polygon is a regular one.
6.55. A paper band of constant width is tied in a simple knot and then tightened in
order to make the knot flat, cf. Fig. 64. Prove that the knot is of the form of a regular
pentagon.
Figure 64 (6.55)
6.56. On sides AB, BC, CD and DA of square ABCD equilateral triangles ABK,
BCL, CDM and DAN are constructed inwards. Prove that the midpoints of sides of these
triangles (which are not the sides of a square) and the midpoints of segments KL, LM , M N
and N K form a regular 12-gon.
***
6.57. Does there exist a regular polygon the length of one of whose diagonal is equal to
the sum of lengths of some other two diagonals?
6.58. A regular (4k + 2)-gon is inscribed in a circle of radius R centered at O. Prove
that the sum of the lengths of segments singled out by the legs of angle ∠Ak OAk+1 on lines
A1 A2k , A2 A2k−1 , . . . , Ak Ak+1 is equal to R.
6.59. In regular 18-gon A1 . . . A18 , diagonals Aa Ad , Ab Ae and Ac Af are drawn. Let
k = a − b, p = b − c, m = c − d, q = d − e, n = e − f and r = f − a. Prove that the indicated
diagonals intersect at one point in any of the following cases and only in these cases:
−−−−→
−→r;
a) k, m, n = −
p,−q,
−−−−→ −−−→
−−→
−→r = −
b) k, m, n = 1, 2, 7 and −
p,−q,
1, 3, 4;
−−−−→ −−−→
−−→
−→r = −
c) k, m, n = 1, 2, 8 and −
p,−q,
2, 2, 3.
−−−−→
−→
Remark. The equality k, m, n = −
x,−y,
z means that the indicated tuples of numbers
coincide; the order in which they are written in not taken into account.
−−−−→
6.60. In a regular 30-gon three diagonals are drawn. For them define tuples k, m, n and
−−−→ −−−−→
−−→
−
−→r as in the preceding problem. Prove that if −
−→r = −
p,−q,
k, m, n = 1, 3, 14 and −
p,−q,
2, 2, 8, then
the diagonals intersect at one point.
* * *
143
6.61. In a regular n-gon (n ≥ 3) the midpoints of all its sides and the diagonals are
marked. What is the greatest number of marked points that lie on one circle?
6.62. The vertices of a regular n-gon are painted several colours so that the points of
one colour are the vertices of a regular polygon. Prove that among these polygons there are
two equal ones.
6.63. Prove that for n ≥ 6 a regular (n − 1)-gon is impossible to inscribe in a regular
n-gon so that on every side of the n-gon except one there lies exactly one vertex of the
(n − 1)-gon.
***
6.64. Let O be the center of a regular n-gon A1 . . . An and X an arbitrary point. Prove
that
−−→
−−→ −
−−→
−−→
−−→
→
OA1 + · · · + OAn = 0 and XA1 + · · · + XAn = nXO.
6.65. Prove that it is possible to place real numbers x1 , . . . , xn all distinct from zero in
the vertices of a regular n-gon so that for any regular k-gon all vertices of which are vertices
of the initial n-gon the sum of the numbers at the vertices of the k-gon is equal to zero.
6.66. Point A lies inside regular 10-gon X1 . . . X10 and point B outside it. Let a =
−−→
−−−→
−−→
−−−→
AX1 + . . . AX10 and b = BX1 + . . . BX10 . Is it possible that |a| > |b|?
6.67. A regular polygon A1 . . . An is inscribed in the circle of radius R centered at O;
let X be an arbitrary point. Prove that
A1 X 2 + · · · + An X 2 = n(R2 + d2 ),
where d = OX.
6.68. Find the sum of squares of the lengths of all the sides and diagonals of a regular
n-gon inscribed in a circle of radius R.
6.69. Prove that the sum of distances from an arbitrary X to the vertices of a regular
n-gon is the least if X is the center of the n-gon.
6.70. A regular n-gon A1 . . . An is inscribed in the circle of radius R centered at O; let
−−→
−−→
ei = OAi and x = OX be an arbitrary vector. Prove that
X
nR2 · OX 2
(ei , x)2 =
.
2
6.71. Find the sum of the squared distances from the vertices of a regular n-gon inscribed
in a circle of radius R to an arbitrary line that passes through the center of the n-gon.
6.72. The distance from point X to the center of a regular n-gon is equal to d and r is
the radius of the inscribed circle of the n-gon. Prove that the sum of squared
distances
from
´
³
d2
2
point X to the lines that contain the sides of the n-gon is equal to n r + 2 .
6.73. Prove that the sum of squared lengths of the projections of the sides of a regular
n-gon to any line is equal to 21 na2 , where a is the length of the side of the n-gon.
6.74. A regular n-gon A1 . . . An is inscribed in a circle of radius R; let X be a point on
this circle. Prove that
XA41 + · · · + XA4n = 6nR4 .
6.75. a) A regular n-gon A1 . . . An is inscribed in the circle of radius 1 centered at 0, let
P
−−→
ei = OAi and u an arbitrary vector. Prove that (u, ei )ei = 21 nu.
b) From an arbitrary point X perpendiculars XA1 , . . . , XAn are dropped to the sides
P −−→ 1 −−→
(or their extensions) of a regular n-gon. Prove that
XAi = 2 nXO, where O is the center
of the n-gon.
6.76. Prove that if the number n is not a power of a prime, then there exists a convex
n-gon with sides of length 1, 2, . . . , n, all the angles of which are equal.
144
CHAPTER 6. POLYGONS
See also Problems 2.9, 4.59, 4.62, 6.36, 6.41, 6.45–6.47, 9.83, 9.84, 11.46, 11.48, 17.31,
18.30, 19.47, 23.8, 24.2.
§7. The inscribed and circumscribed polygons
6.77. On the sides of a triangle three squares are constructed outwards. What should be
the values of the angles of the triangle in order for the six vertices of these squares distinct
from the vertices of the triangle belong to one circle?
6.78. A 2n-gon A1 . . . A2n is inscribed in a circle. Let p1 , . . . , p2n be the distances
from an arbitrary point M on the circle to sides A1 A2 , A2 A3 , . . . , A2n A1 . Prove that
p1 p3 . . . p2n−1 = p2 p4 . . . p2n .
6.79. An inscribed polygon is divided by nonintersecting diagonals into triangles. Prove
that the sum of radii of all the circles inscribed in these triangles does not depend on the
partition.
6.80. Two n-gons are inscribed in one circle and the collections of the length of their
sides are equal but the corresponding sides are not necessarily equal. Prove that the areas
of these polygons are equal.
6.81. Positive numbers a1 , . . . , an are such that 2ai < a1 + · · · + an for all i = 1, . . . ,
n. Prove that there exists an inscribed n-gon the lengths of whose sides are equal to a1 , . . . ,
an .
***
6.82. A point inside a circumscribed n-gon is connected by segments with all the vertices
and tangent points. The triangles formed in this way are alternately painted red and blue.
Prove that the product of areas of red triangles is equal to the product of areas of blue
triangles.
6.83. In a 2n-gon (n is odd) A1 . . . A2n circumscribed about a circle centered at O the
diagonals A1 An+1 , A2 An+2 , . . . , An−1 A2n−1 pass through point O. Prove that the diagonals
An A2n also passes through point O.
6.84. A circle of radius r is tangent to the sides of a polygon at points A1 , . . . , An and
the length of the side on which point Ai lies is equal to ai . The distance from point X to
the center of the circle is equal to d. Prove that
a1 XA21 + · · · + an XA2n = P (r2 + d2 ),
where P is the perimeter of the polygon.
6.85. An n-gon A1 . . . An is circumscribed about a circle; l is an arbitrary tangent to
the circle that does not pass through any vertex of the n-gon. Let ai be the distance from
vertex Ai to line l and bi the distance from the tangent point of side Ai Ai+1 with the circle
to line l. Prove that:
...bn
a) the value ab11 ...a
does not depend on the choice of line l;
n
2m−1
does not depend on the choice of line l if n = 2m.
b) the value a1aa23a...a
4 ...a2m
6.86. Certain sides of a convex polygon are red; the other ones are blue. The sum of the
lengths of the red sides is smaller than the semiperimeter and there is no pair of neighbouring
blue sides. Prove that it is impossible to inscribe this polygon in a circle.
See also Problems 2.12, 4.39, 19.6.
§8. Arbitrary convex polygons
6.87. What is the greatest number of acute angles that a convex polygon can have?
PROBLEMS FOR INDEPENDENT STUDY
145
6.88. How many sides whose length is equal to the length of the longest diagonal can a
convex polygon have?
6.89. For which n there exists a convex n-gon one side of which is of length 1 and the
lengths of the diagonals are integers?
6.90. Can a convex non-regular pentagon have exactly four sides of equal length and
exactly four diagonals of equal lengths? Can the fifth side of such a pentagon have a common
point with the fifth diagonal?
6.91. Point O that lies inside a convex polygon forms, together with each two of its
vertices, an isosceles triangle. Prove that point O is equidistant from the vertices of this
polygon.
See also Problems 4.49, 4.50, 9.82, 9.85, 9.86, 11.35, 13.14, 14.26, 16.8, 17.33, 17.34, 19.9,
23.13, 23.15.
§9. Pascal’s theorem
6.92. Prove that the intersection points of the opposite sides (if these sides are not
parallel) of an inscribed hexagon lie on one line. (Pascal’s theorem.)
6.93. Point M lies on the circumscribed cirlce of triangle ABC; let R be an arbitrary
point. Lines AR, BR and CR intersect the circumscribed circle at points A1 , B1 and C1 ,
respectively. Prove that the intersection points of lines M A1 and BC, M B1 and CA, M C1
and AB lie on one line and this line passes through point R.
6.94. In triangle ABC, heights AA1 and BB1 and bisectors AA2 and BB2 are drawn;
the inscribed circle is tangent to sides BC and AC at points A3 and B3 , respectively. Prove
that lines A1 B1 , A2 B2 and A3 B3 either intersect at one point or are parallel.
6.95. Quadrilateral ABCD is inscribed in circle S; let X be an arbitrary point, M and
N be the other intersection points of lines XA and XD with circle S. Lines DC and AX,
AB and DX intersect at points E and F , respectively. Prove that the intersection point of
lines M N and EF lies on line BC.
6.96. Points A and A1 that lie inside a circle centered at O are symmetric through point
O. Rays AP and A1 P1 are codirected, rays AQ and A1 Q1 are also codirected. Prove that
the intersection point of lines P1 Q and P Q1 lies on line AA1 . (Points P , P1 , Q and Q1 lie
on the circle.)
6.97. On a circle, five points are given. With the help of a ruler only construct a sixth
point on this circle.
6.98. Points A1 , . . . , A6 lie on one circle and points K, L, M and N lie on lines A1 A2 ,
A3 A4 , A1 A6 and A4 A5 , respectively, so that KL k A2 A3 , LM k A3 A6 and M N k A6 A5 .
Prove that N K k A5 A2 .
Problems for independent study
6.99. Prove that if ABCD is a rectangle and P is an arbitrary point, then AP 2 + CP 2 =
DP 2 + BP 2 .
6.100. The diagonals of convex quadrilateral ABCD are perpendicular. On the sides
of the quadrilateral, squares centered at P , Q, R and S are constructed outwards. Prove
that segment P R passes through the intersection point of diagonals AC and BD so that
P R = 21 (AC + BD).
6.101. On the longest side AC of triangle ABC, points A1 and C1 are taken so that
AC1 = AB and CA1 = CB and on sides AB and BC points A2 and C2 are taken so that
AA1 = AA2 and CC1 = CC2 . Prove that quadrilateral A1 A2 C2 C1 is an inscribed one.
146
CHAPTER 6. POLYGONS
6.102. A convex 7-gon is inscribed in a circle. Prove that if certain three of its angles
are equal to 120◦ each, then some two of its sides are equal.
6.103. In plane, there are given a regular n-gon A1 . . . An and point P . Prove that from
segments A1 P , . . . , An P a closed broken line can be constructed.
6.104. Quadrilateral ABCD is inscribed in circle S1 and circumscribed about circle S2 ;
let K, L, M and N be tangent points of its sides with circle S2 . Prove that KM ⊥ LN .
6.105. Pentagon ABCDE the lengths of whose sides are integers and AB = CD = 1 is
circumscribed about a circle. Find the length of segment BK, where K is the tangent point
of side BC with the circle.
6.106. Prove that in a regular 2n-gon A1 . . . A2n the diagonals A1 An+2 , A2n−1 A3 and
A2n A5 meet at one point.
6.107. Prove that in a regular 24-gon A1 . . . A24 diagonals A1 A7 , A3 A11 and A5 A21
intersect at a point that lies on diameter A4 A16 .
Solutions
6.1. Let O be the center of the inscribed circle and the intersection point of the diagonals
of quadrilateral ABCD. Then ∠ACB = ∠ACD and ∠BAC = ∠CAD. Hence, triangles
ABC and ADC are equal, since they have a common side AC. Therefore, AB = DA.
Similarly, AB = BC = CD = DA.
6.2. Clearly,
∠AOB = 180◦ − ∠BAO − ∠ABO = 180◦ −
and ∠COD = 180◦ −
∠C+∠D
.
2
∠A + ∠B
2
Hence,
∠A + ∠B + ∠C + ∠D
= 180◦ .
2
6.3. Let us consider two circles tangent to the sides of the given quadrilateral and their
extensions. The lines that contain the sides of the quadrilateral are the common inner and
outer tangents to these circles. The line that connects the midpoints of the circles contains
a diagonal of the quadrilateral and besides it is an axis of symmetry of the quadrilateral.
Hence, the other diagonal is perpendicular to this line.
6.4. Let O be the center of the given circle, R its radius, a the length of chords singled
out by the circle on the sides of the quadrilateral.
Then the distances from point O to the
q
∠AOB + ∠COD = 360◦ −
2
sides of the quadrilateral are equal to R2 − a4 , i.e., point O is equidistant from the sides
of the quadrilateral and is the center of the inscribed circle.
6.5. For a parallelogram the statement of the problem is obvious therefore, we can
assume that lines AB and CD intersect. Let O be the center of the inscribed circle of
quadrilateral ABCD; let M and N be the midpoints of diagonals AC and BD. Then
SAN B + SCN D = SAM B + SCM D = SAOB + SCOD =
SABCD
.
2
It remains to make use of the result of Problem 7.2.
6.6. Let the inscribed circle be tangent to sides DA, AB and BC at points M , H and
N , respectively. Then OH is a height of triangle AOB and the symmetries through lines AO
and BO sends point H into points M and N , respectively. Hence, by Problem 1.57 points
A1 and B1 lie on line M N . Similarly, points C1 and D1 lie on line M N .
6.7. Let r be the distance from the intersection point of bisectors of angles A and D to
the base AD, let r′ be the distance from the intersection point of bisectors of angles B and
SOLUTIONS
147
C to base BC. Then AD = r(cot α + cot β) and BC = r′ (tan α + tan β). Hence, r = r′ if
and only if
tan α + tan β
BC
=
= tan α · tan β.
AD
cot α cot β
6.8. Let ∠A = 2α, ∠C = 2β and ∠BM A = 2ϕ. By Problem 6.7,
LS
K
RS
LS
= cot ϕ tan β. Since PRSQ = PRL
and AC
=M
, it follows that
MC
C
PK
RL
=
tan α
tan ϕ
and
PQ
P K LS
=
= tan α tan β.
AC
RL M C
Hence, trapezoid AP QC is a circumscribed one.
6.9. First, let us prove that if quadrilateral ABCD is a circumscribed one, then all the
conditions take place. Let K, L, M and N be the tangent points of the inscribed circle with
sides AB, BC, CD and DA. Then
AB + CD = AK + BK + CM + DM = AN + BL + CL + DN = BC + AD,
AP + CQ = AK + P K + QL − CL = AN + P M + QN − CM = AQ + CP,
BP + BQ = AP − AB + BC + CQ = (AP + CQ) + (BC − AB) =
AQ + CP + CD − AD = DP + DQ.
Now, let us prove, for instance, that if BP + BQ = DP + DQ, then quadrilateral ABCD
is a circumscribed one. For this let us consider the circle tangent to side BC and rays BA
and CD. Assume that line AD is not tangent to this circle; let us shift this line in order for
it to touch the circle (Fig. 65).
Figure 65 (Sol. 6.9)
Let S be a point on line AQ such that Q′ S k DD′ . Since BP + BQ = DP + DQ and
BP + BQ′ = D′ P + D′ Q′ , it follows that QS + SQ′ = QQ′ . Contradiction.
In the other two cases the proof is similar.
6.10. Let rays AB and DC intersect at point P , let rays BC and AD intersect at point
Q; let the given lines passing through points P and Q intersect at point O. By Problem 6.9
we have BP + BQ = OP + OQ and OP + OQ = DP + DQ. Hence, BP + BQ = DP + DQ
and, therefore, quadrilateral ABCD is a circumscribed one.
6.11. Let sides AB, BC, CD and DA of quadrilateral ABCD be tangent of the inscribed
circle at points E, F , G and H, respectively. First, let us show that lines F H, EG and AC
intersect at one point. Denote the points at which lines F H and EG intersect line AC by
M and M ′ , respectively. Since ∠AM H = ∠BF M as angles between the tangents and chord
HF , it follows that sin ∠AHM = sin ∠CF M . Hence,
SAM H
AH · M H
AM · M H
=
,
=
F M · MC
SF M C
FC · FM
148
i.e.,
CHAPTER 6. POLYGONS
AM
MC
=
AH
.
FC
Similarly,
AE
AH
AM
AM ′
=
=
=
;
M ′C
CG
FC
MC
hence, M = M ′ , i.e., lines F H, EG and AC intersect at one point.
Similar arguments show that lines F H, EG and BD intersect at one point and therefore,
lines AC, BD, F H and EG intersect at one point.
6.12. Segments CHd and DHc are parallel because they are perpendicular to line BC.
Moreover, since ∠BCA = ∠BDA = ϕ, the lengths of these segments are equal to AB| cot ϕ|,
cf. Problem 5.45 b).
6.13. Let Oa , Ob , Oc and Od be the centers of the inscribed circles of triangles BCD,
ACD, ABD and ABC, respectively. Since ∠ADB = ∠ACB, it follows that
∠ADB
∠ACB
∠AOc B = 90◦ +
= 90◦ +
= ∠AOd B,
2
2
cf. Problem 5.3. Therefore, quadrilateral ABOd Oc is an inscribed one, i.e.,
∠A
.
∠Oc Od B = 180◦ − ∠Oc AB = 180◦ −
2
. Since ∠A + ∠C = 180◦ , it follows that ∠Oc Od B +
Similarly, ∠Oa Od B = 180◦ − ∠C
2
∠Oa Ob B = 270◦ and, therefore, ∠Oa Od Oc = 90◦ . We similarly prove that the remaining
angles of quadrilateral Oa Ob Oc Od are equal to 90◦ .
6.14. a) Let rays AB and DC intersect at point P and rays BC and AD intersect at
point Q. Let us prove that point M at which the circumscribed circles of triangles CBP
and CDQ intersect lies on segment P Q. Indeed,
∠CM P + ∠CM Q = ∠ABC + ∠ADC = 180◦ .
Hence, P M + QM = P Q and since
P M · P Q = P D · P C = p2 − R 2
and QM · P Q = QD · QA = q 2 − R2 ,
it follows that P Q2 = P M · P Q + QM · P Q = p2 + q 2 − 2R2 . Let N be the intersection point
of the circumscribed circles of triangles ACP and ABS. Let us prove that point S lies on
segment P N . Indeed,
∠AN P = ∠ACP = 180◦ − ∠ACD = 180◦ − ∠ABD = ∠AN S.
Hence, P N − SN = P S and since
P N · P S = P A · P B = p2 − R 2
it follows that
and SN · P S = SA · SC = R2 − s2 ,
P S 2 = P N · P S − SN · P S = p2 + s2 − 2R2 .
Similarly, QS 2 = q 2 + s2 − 2R2 .
b) By heading a)
P Q2 − P S 2 = q 2 − s2 = OQ2 − OS 2 .
Hence, OP ⊥ QS, cf. Problem 7.6. We similarly prove that OQ ⊥ P S and OS ⊥ P Q.
6.15. Let the inscribed circles of triangles ABC and ACD be tangent to diagonal AC
at points M and N , respectively. Then
AC + AB − BC
AC + AD − CD
AM =
andquadAN =
,
2
2
cf. Problem 3.2. Points M and N coincide if and only if AM = AN , i.e., AB + CD =
BC + AD. Thus, if points M and N coincide, then quadrilateral ABCD is a circumscribed
SOLUTIONS
149
one and similar arguments show that the tangent points of the inscribed circles of triangles
ABD and BCD with the diagonal BD coincide.
Let the inscribed circle of triangle ABC be tangent to sides AB, BC and CA at points P ,
Q and M , respectively and the inscribed circle of triangle ACD be tangent to sides AC, CD
and DA at points M , R and S, respectively. Since AP = AM = AS and CQ = CM = CR,
it follows that triangles AP S, BP Q, CQR and DRS are isosceles ones; let α, β, γ and δ be
the angles at the bases of these isosceles triangles. The sum of the angles of these triangles
is equal to
2(α + β + γ + δ) + ∠A + ∠B + ∠C + ∠D;
hence, α + β + γ + δ = 180◦ . Therefore,
∠SP Q + ∠SRQ = 360◦ − (α + β + γ + δ) = 180◦ ,
i.e., quadrilateral P QRS is an inscribed one.
6.16. Let O be the intersection point of diagonals AC and BD; let A1 , B1 , C1 and D1
be the projections of O to sides AB, BC, CD and DA, respectively. Points A1 and D1 lie
on the circle with diameter AO, hence, ∠OA1 D1 = ∠OAD1 . Similarly, ∠OA1 B1 = ∠OBB1 .
Since ∠CAD = ∠CBD, we have: ∠OA1 D1 = ∠OA1 B1 .
We similarly prove that B1 O, C1 O and D1 O are the bisectors of the angles of quadrilateral
A1 B1 C1 D1 , i.e., O is the center of its inscribed circle.
Figure 66 (Sol. 6.17)
6.17. Let us make use of the notations on Fig. 66. The condition that quadrilateral
A1 B1 C1 D1 is an inscribed one is equivalent to the fact that (α+β)+(γ+δ) = 180◦ and the the
fact that AC and BD are perpendicular is equivalent to the fact that (α1 + δ1 ) + (β1 + γ1 ) =
180◦ . It is also clear that α = α1 , β = β1 , γ = γ1 and δ = δ1 .
6.18. By the law of cosines
AD2 = AC 2 + CD2 − 2AC · CD cos ACD,
AC 2 = AB 2 + BC 2 − 2AB · BC cos B.
Since the length of the projection of segment AC to line l perpendicular to CD is equal to
the sum of the lengths of projections of segments AB and BC to line l,
AC cos ACD = AB cos ϕ + BC cos C.
6.19. Let ∠AOD = 2α; then the distances from point O to the projections of the
midpoints of diagonals AC and BD to the bisector of angle ∠AOD are equal to OA+OC
cos α
2
OB+OD
and
cos α, respectively. Since
2
OA + OC = AB + OB + OC = CD + OB + OC = OB + OD,
these projections coincide.
150
CHAPTER 6. POLYGONS
6.20. Let us complement triangles ABM and DCM to parallelograms ABM M1 and
DCM M2 . Since AM1 : DM2 = BM : M C = AN : DN , it follows that △AN M1 ∼
△DN M2 . Hence, point N lies on segment M1 M2 and
M M1 : M M2 = AB : CD = AN : N D = M1 N : M2 N,
i.e., M N is the bisector of angle M1 M M2 .
6.21. Let a, b, c and d be (the lengths of) the bisectors of the angles at vertices A, B,
C and D. We have to verify that ∠(a, b) + ∠(c, d) = 0◦ . Clearly,
∠(a, b) = ∠(a, AB) + ∠(AB, b) and ∠(c, d) = ∠(c, CD) + ∠(CD, d).
Since quadrilateral ABCD is a convex one and
∠(a, AB) =
∠(AD, AB)
∠(AB, BC)
, ∠(AB, b) =
,
2
2
∠(CB, CD)
∠(CD, DA)
∠(c, CD) =
, ∠(CD, d) =
,
2
2
it follows that
∠(a, b) + ∠(c, d) =
∠(AD, AB) + ∠(AB, BC) + ∠(CB, CD) + ∠(CD, DA)
=
2
360◦
= 0◦
2
(see Background to Chapter 2).
−−→
6.22. Let, for definiteness, AB > A1 B1 . The parallel translation by vector CB sends
triangle SD1 C1 to S ′ D1′ C1′ and segment CD to BA. Since QA1 : QA = A1 B1 : AB = S ′ D1′ :
S ′ A, we see that QS ′ k A1 D1′ . Hence, QS k AD. Similarly, P R k AB.
6.23. Suppose that lines AD and BC are not parallel. Let M2 , K, N2 be the midpoints
of sides AB, BC, CD, respectively. If M N k BC, then BC k AD, because AM = M C and
BN = N D. Therefore, let us assume that lines M N and BC are not parallel, i.e., M1 6= M2
−−−→
−−→
−−→
−−−→
−−→
and N1 6= N2 . Clearly, M2 M = 21 BC = N N2 and M1 M = N N1 . Hence, M1 M2 k N1 N2 .
Therefore, KM k AB k CD k KN , i.e., M = N . Contradiction.
6.24. By a similarity transformation we can identify one pair of the corresponding sides
of quadrilaterals, therefore, it suffices to consider quadrilaterals ABCD and ABC1 D1 whose
points C1 and D1 lie on rays BC and AD and such that CD k C1 D1 . Denote the intersection
points of diagonals of quadrilaterals ABCD and ABC1 D1 by O and O1 , respectively.
Suppose that points C and D lie closer to points B and A, then points C1 and D1 ,
respectively. Let us prove then that ∠AOB > ∠AO1 B. Indeed, ∠C1 BA > ∠CAB and
∠D1 BA > ∠DBA, hence,
∠AO1 B = 180◦ − ∠C1 AB − ∠D1 BA < 180◦ − ∠CAB − ∠DBA = ∠AOB.
We have obtained a contradiction and, therefore, C1 = C, D1 = D.
6.25. Any quadrilateral is determined up to similarity by the directions of its sides and
diagonals. Therefore, it suffices to construct one example of a quadrilateral A1 B1 C1 D1 with
the required directions of sides and diagonals. Let O be the intersection point of diagonals
AC and BD. On ray OA, take an arbitrary point D1 and draw D1 A1 k BC, A1 B1 k CD
and B1 C1 k DA (Fig. 67).
SOLUTIONS
151
Figure 67 (Sol. 6.25)
Since
OC1 : OB1 = OD : OA,
OB1 : OA1 = OC : OD
and OA1 : OD1 = OB : OC,
it follows that OC1 : OD1 = OB : OA, consequently, C1 D1 k AB. The obtained plot shows
that ∠A + ∠C1 = 180◦ .
6.26. Let O be the intersection point of the diagonals of quadrilateral ABCD. Without
loss of generality we may assume that α = ∠AOB < 90◦ . Let us drop perpendiculars AA1 ,
BB1 , CC1 , DD1 to the diagonals of quadrilateral ABCD. Since
OA1 = OA cos α,
OB1 = OB cos α,
OC1 = OC cos α,
OD1 = OD cos α,
it follows that the symmetry through the bisector of angle AOB sends quadrilateral ABCD
−−→
into a quadrilateral homothetic to quadrilateral A1 B1 C1 D1 with coefficient BC1cos α.
6.27. Let the diagonals of quadrilateral ABCD intersect at point O; let Ha and Hb
be the orthocentres of triangles AOB and COD; let Ka and Kb be the midpoints of sides
BC and AD; let P be the midpoint of diagonal AC. The intersection point of medians of
triangles AOD and BOC divide segments Ka O and Kb O in the ratio of 1 : 2 and, therefore,
we have to prove that Ha Hb ⊥ Ka Kb .
Since OHa = AB| cot ϕ| and OHb = CD| cot ϕ|, where ϕ = ∠AOB, cf. Problem 5.45 b),
then OHa : OHb = P Ka : P Kb . The correspondiong legs of angles ∠Ha OHb and ∠Ka P Kb
−−→
−−→
are perpendicular; moreover, vectors OHa and OHb are directed towards lines AB and CD
for ϕ < 90◦ and away from these lines for ϕ > 90◦ . Hence, ∠Ha OHb = ∠Ka P Kb and
△Ha OHb ∼ △Ka P Kb . It follows that Ha Hb ⊥ Ka Kb .
6.28. Let S = SAOD , x = AO, y = DO, a = AB, b = BC, c = CD, d = DA and k the
similarity coefficient of triangles BOC and AOD. Then
2
µ
1
1
+
r1 r3
¶
=
d + x + y kd + kx + ky
+
,
S
k2S
2
µ
1
1
+
r2 r4
¶
=
a + x + ky c + kx + y
+
kS
kS
because SBOC = k 2 S and SAOB = SCOD = kS. Since
x + ky kx + y
x+y x+y
+ 2 =
+
,
S
k S
kS
kS
it remains to notice that a + c = b + d = kd + d.
6.29. It is easy to verify that
µ
¶
¶
µ
A
C
B
D
AB = r1 cot + cot
and CD = r3 cot + cot
.
2
2
2
2
Hence,
B
C
D
BC AD
A
AB CD
+
= cot + cot + cot + cot =
+
.
r1
r3
2
2
2
2
r2
r4
152
CHAPTER 6. POLYGONS
6.30. Let us complete triangles ABD and DBC to parallelograms ABDA1 and DBCC1 .
The segments that connect point D with the vertices of parallelogram ACC1 A1 divide it into
four triangles equal to triangles DAB, CDA, BCD and ABC and, therefore, the radii of
the inscribed circles of these triangles are equal.
Let us prove that point D coincides with the intersection point O of the diagonals of
the parallelogram. If D 6= 0, then we may assume that point D lies inside triangle AOC.
Then rADC < rAOC = rA1 OC1 < rA1 DC1 = rABC , cf. Problem 10.86. We have obtained a
contradiction, hence, D = O.
−−→
Since p = BCSr and the areas and radii of the inscribed circles of triangles into which the
diagonals divide the parallelogram ACC1 A1 are equal, the triangles’ perimeters are equal.
Hence, ACC1 A1 is a rhombus and ABCD is a rectangular.
6.31. Points C1 and D1 lie on the midperpendicular to segment AB, hence, AB ⊥ C1 D1 .
Similarly, C1 D1 ⊥ A2 B2 and, therefore, AB k A2 B2 . We similarly prove that the remaining
corresponding sides and the diagonals of quadrilaterals ABCD and A2 B2 C2 D2 are parallel.
Therefore, these quadrilaterals are similar.
Let M be the midpoint of segment AC. Then B1 M = |AM cot D| and D1 M = |AM cot B|,
where B1 D1 = | cot B + cot D| · 21 AC. Let us rotate quadrilateral A1 B1 C1 D1 by 90◦ . Then
making use of the result of Problem 6.25 we see that this quadrilateral is a convex one and
cot A = − cot C1 , etc. Therefore,
1
1
A2 C2 = | cot A + cot C| · B1 D1 = |(cot A + cot C)(cot B + cot D)| · AC.
2
4
6.32. Let M and N be the midpoints of sides AB and CD, respectively. Let us drop
from point D perpendicular DP to line M N and from point M perpendicular M Q to line
CD. Then Q is the tangent point of line CD and a circle with diameter AB. Right triangles
P DN and OM N are similar, hence,
ND · MA
ND · MQ
=
.
DP =
MN
MN
−−→
Similarly, the distance from point A to line M N is equal to BCN D · M AM N . Therefore,
AD k M N . Similarly, BC k M N .
6.33. It suffices to verify that the orthocentres of any three of the four given triangles
lie on one line. Let a certain line intersect lines B1 C1 , C1 A1 and A1 B1 at points A, B and
C, respectively; let A2 , B2 and C2 be the orthocentres of triangles A1 BC, AB1 C and ABC1 ,
respectively. Lines AB2 and A2 B are perpendicular to line A1 B1 and, therefore, they are
parallel. Similarly, BC2 k B2 C and CA2 k C2 A. Points A, B and C lie on one line and,
therefore, points A2 , B2 and C2 also lie on one line, cf. Problem 1.12 b).
6.34. On diagonal BD, take point M so that ∠M CD = ∠BCA. Then △ABC ∼
△DM C, because angles ∠BAC and ∠BDC subtend the same arc. Hence, AB · CD =
AC · M D. Since ∠M CD = ∠BCA, then ∠BCM = ∠ACD and △BCM ∼ △ACD because
angles ∠CBD and ∠CAD subtend one arc. Hence, BC · AD = AC · BM . It follows that
AB · CD + AD · BC = AC · M D + AC · BM = AC · BD.
6.35. Let S be the area of quadrilateral ABCD, let R be the radius of its circumscribed
circle. Then
AC(AB · BC + AD · DC)
S = SABC + SADC =
,
4R
cf. Problem 12.1. Similarly,
BD(AB · AD + BC · CD)
.
S=
4R
SOLUTIONS
153
By equating these equations for S we get the desired statement.
6.36. Let regular hexagon A1 . . . A7 be inscribed in a circle. By applying Ptolemey’s
theorem to qudrilateral A1 A3 A4 A5 we get
i.e.,
A1 A3 · A5 A4 + A3 A4 · A1 A5 = A1 A4 · A3 A5 ,
sin 2α sin α + sin α sin 3α = sin 3α sin 2α.
6.37. Let A1 , B1 and C1 be the midpoints of sides BC, CA and AB, respectively. By
Ptolemy’s theorem
AC1 · OB1 + AB1 · OC1 = AO · B1 C1 ,
where O is the center of the circumscribed circle. Hence, cdb +bdc = aR. Similarly, adc +cda =
bR and adb + bda = cR. Moreover, ada + bdb + cdc = 2S = (a + b + c)r. By adding all these
equalities and dividing by a + b + c we get the desired statement.
6.38. By Ptolemy’s theorem
AB · CD + AC · BD = AD · BC.
Taking into account that CD = BD ≥ 12 BC we get the desired statement.
6.39. By applying Ptolemy’s theorem to quadrilateral ABCP and dividing by the lengths
of the square’s side we get the desired statement.
6.40. By applying Ptolemy’s theorem to quadrilateral AP QR we get
AP · RQ + AR · QP = AQ · P R.
Since ∠ACB = ∠RAQ = ∠RP Q and ∠RQP = 180◦ − ∠P AR = ∠ABC, it follows that
△RQP ∼ △ABC and, therefore, RQ : QP : P R = AB : BC : CA. It remains to notice
that BC = AD.
6.41. a) Let us express Ptolemy’s theorem for all quadrilaterals with vertices at point
A and three consecutive vertices of the given polygon; then let us group in the obtained
equalities the factors in which di with even indices enter in the right-hand side. By adding
these equalities we get
(2a + b)(d1 + · · · + d2n+1 ) = (2a + b)(d2 + · · · + d2n ),
where a is the side of the given polygon and b is its q
shortest diagonal.
b) Let R be the radius of circle S. Then li = di R±r
, cf. Problem 3.20. It remains to
R
make use of the result of heading a).
Figure 68 (Sol. 6.42)
6.42. Let both tangent be exterior ones and x ≤ y. The line that passes through the
center O of the circle of radius x parallel to the segment that connects the tangent points
intersects the circle of radius y − x (centered in the center of the circle of radius y) at points
A and B (Fig. 68).
154
CHAPTER 6. POLYGONS
Then OA =
a(R+x)
R
and
a(R + y)
a(y − x)
=
.
R
R
The square of the length to be found of the common outer tangent is equal to
³ a ´2
OA · OB =
(R + x)(R + y).
R
Similar arguments show that ¡if both
¢ tangent are inner ones, then the square of the lengths
a 2
of the outer tangent is equal to R (R − x)(R − y) and if the circle of radius x is tangent
from the outside and the circle
¡ aof
¢2 radius y from the inside, then the square of the length of
the inner tangent is equal to R (R − y)(R + x).
OB = OA +
Remark. In the case of an inner tangency of the circles we assume that R > x and
R > y.
6.43. Let R be the radius of the circumscribed circle of quadrilateral ABCD;
let ra , rb ,
√
rc and rd be the radii of circles α, β, γ and δ, respectively. Further, let a = R ± ra , where
the plus sign is taken if the tangent is an outer one and the minus sign if it is an inner one;
, cf. Problem 6.42, etc. Therefore,
numbers b, c and d are similarly defined. Then tαβ = ab·AB
R
by multiplying the equality
AB · CD + BC · DA = AC · BD
abcd
R
we get the desired statement.
6.44. Since ∠EBD = ∠ABE + ∠CBD, it is possible to take a point P on side ED
so that ∠EBP = ∠ABE = ∠AEB, i.e., BP k AE. Then ∠P BD = ∠EBD − ∠EBP =
∠CBD = ∠BDC, i.e., BP k CD. Therefore, AE k CD and since AE = CD, CDEA is a
parallelogram. Hence, AC = ED, i.e., triangle ABC is an equilateral one and ∠ABC = 60◦ .
6.45. a) Let O be the center of the circumscribed circle of triangle CKE. It suffices to
verify that ∠COK = 2∠KCB. It is easy to calculate both these angles:
by
∠COK = 180◦ − 2∠OKC = 180◦ − ∠EKC = 180◦ − ∠EDC = 72◦
◦
and ∠KCB = 180 −∠ABC
= 36◦ .
2
b) Since BC is a tangent to the circumscribed circle of triangle CKE, then BE · BK =
BC 2 , i.e., d(d − a) = a2 .
6.46. Let the perpendiculars erected to line AB at points A and B intersect sides DE
and CD at points P and Q, respectively. Any point of segment CQ is a vertex of a rectangle
inscribed in pentagon ABCDE (the respective sides of this pentagon are parallel to AB and
AP ); as this point moves from Q to C the ratio of the lengths of the sides of the rectangles
AP
to 0. Since angle ∠AEP is an obtuse one, AP > AE = AB. Therefore, for
varies from AB
a certain point of segment QC the ratio of the lengths of the sides of the rectangle is equal
to 1.
6.47. Let points A1 , . . . , E1 be symmetric to points A, . . . , E through the center of
circle S; let P , Q and R be the intersection points of lines BC1 and AB1 , AE1 and BA1 ,
BA1 and CB1 , see Fig. 69.
Then P Q = AB = a and QR = b. Since P Q k AB and ∠ABA1 = 90◦ , it follows
that P R2 = P Q2 + QR2 = a2 + b2 . Line P R passes through the center of circle S and
∠AB1 C = 4 · 18◦ = 72◦ , hence, P R is a side of a regular pentagon circumscribed about the
circle with center B1 whose radius B1 O is equal to the radius of circle S.
SOLUTIONS
155
Figure 69 (Sol. 6.47)
Figure 70 (Sol. 6.48)
6.48. Through points A, C and E draw lines l1 , l2 and l3 parallel to lines BC, DE and
F A, respectively. Denote the intersection points of lines l1 and l2 , l2 and l3 , l3 and l1 by P ,
Q, R, see Fig. 70. Then
SACE =
SABCDEF + SP QR
SABCDEF
SABCDEF − SP QR
+ SP QR =
≥
.
2
2
2
Similarly, SBDF = 21 (SABCDEF + SP ′ Q′ R′ ). Clearly,
P Q = |AB − DE|, QR = |CD − AF |, P R = |EF − BC|,
hence, triangles P QR and P ′ Q′ R′ are equal. Therefore, SACE = SBDF .
6.49. Let us construct triangle P QR as in the preceding problem. This triangle is an
equilateral one and
P Q = |AB − DE|, QR = |CD − AF |, RP = |EF − BC|.
Hence, |AB − DE| = |CD − AF | = |EF − BC|.
6.50. The sum of the angles at vertices A, C and E is equal to 360◦ , hence, from isosceles
triangles ABF , CBD and EDF we can construct a triangle by juxtaposing AB to CB, ED
to CD and EF to AF . The sides of the obtained triangle are equal to the respective sides
of triangle BDF . Therefore, the symmetry through lines F B, BD and DF sends points A,
C and E, respectively, into the center O of the circumscribed circle of triangle BDF , and,
therefore, AB k OF k DE.
6.51. Let us suppose that the diagonals of the hexagon form triangle P QR. Denote the
vertices of the hexagon as follows: vertex A lies on ray QP , vertex B on RP , vertex C on
RQ, etc. Since lines AD and BE divide the area of the hexagon in halves, then
SAP EF + SP ED = SP DCB + SABP and SAP EF + SABP = SP DCB + SP ED .
156
CHAPTER 6. POLYGONS
Hence, SABP = SP ED , i.e.,
AP · BP = EP · DP = (ER + RP )(DQ + QP ) > ER · DQ.
Similarly, CQ · DQ > AP · F R and F R · ER > BP · CQ. By multiplying these inequalities
we get
AP · BP · CQ · DQ · F R · ER > ER · DQ · AP · F R · BP · CQ
which is impossible. Hence, the diagonals of the hexagon intersect at one point.
Figure 71 (Sol. 6.52)
6.52. Denote the midpoints of the sides of convex hexagon ABCDEF as plotted on Fig.
71. Let O be the intersection point of segments KM and LN . Let us denote the areas of
triangles into which the segments that connect point O with the vertices and the midpoints
of the sides divide the hexagon as indicated on the same figure. It is easy to verify that
SKON F = SLOM C , i.e., a + f = c + d. Therefore, the broken line P OQ divides the hexagon
into two parts of equal area; hence, segment P Q passes through point O.
6.53. a) Let O be the center of the circumscribed circle. Since
∠Ak OAk+2 = 360◦ − 2∠Ak Ak+1 Ak+2 = ϕ
is a constant, the rotation through an angle of ϕ with center O sends point Ak into Ak+2 .
For n odd this implies that all the sides of polygon A1 . . . An are equal.
b) Let a be the length of the side of the given polygon. If one of its sides is divided
by the tangent point with the inscribed circle into segments of length x and a − x, then its
neighbouring sides are also divided into segments of length x and a − x (the neighbouring
segments of neighbouring sides are equal), etc. For n odd this implies that all the sides
of polygon A1 . . . An are divided by the tangent points with the inscribed circle in halves;
therefore, all the angles of the polygon are equal.
6.54. The sides of polygon A1 . . . An are parallel to respective sides of a regular n-gon.
On rays OA1 , . . . , OAn mark equal segments OB1 , . . . , OBn . Then polygon B1 . . . Bn is a
regular one and the sides of polygon A1 . . . An form equal angles with the respective sides of
polygon B1 . . . Bn . Therefore,
i.e.,
OA1 : OA2 = OA2 : OA3 = · · · = OAn : OA1 = k,
OA1 = kOA2 = k 2 OA3 = · · · = k n OA1 ;
thus, k = 1.
6.55. Denote the vertices of the pentagon as indicated on Fig. 72. Notice that if in a
triangle two heights are equal, then the sides on which these heights are dropped are also
equal.
SOLUTIONS
157
Figure 72 (Sol. 6.55)
From consideration of triangles EAB, ABC and BCD we deduce that EA = AB,
AB = BC and BC = CD. Therefore, trapezoids EABC and ABCD are isosceles ones,
i.e., ∠A = ∠B = ∠C. By considering triangles ABD and BCE we get AD = BD and
BE = CE. Since triangles EAB, ABC, BCD are equal, it follows that BE = AC = BD.
Hence, AD = BE and BD = CE, i.e., trapezoids ABDE and CDEB are isosceles ones.
Therefore, ED = AB = BC = CD = AE and ∠E = ∠A = ∠B = ∠C = ∠D, i.e., ABCDE
is a regular pentagon.
6.56. Triangles BAM and BCN are isosceles ones with angle 15◦ at the base, cf. Problem
2.26, and, therefore, triangle BM N is an equilateral one. Let O be the centre of the square,
P and Q the midpoints of segments M N and BK (Fig. 73). Since OQ is the midline of
triangle M BK, it follows that OQ = 12 BM = M P = OP and ∠QON = ∠M BA = 15◦ .
Therefore, ∠P OQ = ∠P ON − ∠QON = 30◦ .
The remaining part of the proof is carried out similarly.
Figure 73 (Sol. 6.56)
6.57. Let us consider a regular 12-gon A1 . . . A12 inscribed in a circle of radius R. Clearly,
A1 A7 = 2R, A1 A3 = A1 A11 = R. Hence, A1 A7 = A1 A3 + A1 A11 .
6.58. For k = 3 the solution of the problem is clear from Fig. 74. Indeed, A3 A4 = OQ,
KL = QP and M N = P A14 and, therefore,
A3 A4 + KL + M N = OQ + QP + P A14 = OA14 = R.
Proof is carried out in a similar way for any k.
6.59. In the proof if suffices to apply the result of Problems 5.78 and 5.70 b) to triangle
Aa Ac Ae and lines Aa Ad , Ac Af and Ae Ab . Solvling heading b) we have to notice additionally
that
sin 40◦
= sin 30◦ sin 40◦
sin 20◦ sin 70◦ = sin 20◦ cos 20◦ =
2
and in the solution of heading c) that sin 10◦ sin 80◦ = sin 30◦ sin 20◦ .
158
CHAPTER 6. POLYGONS
Figure 74 (Sol. 6.58)
6.60. As in the preceding problem we have to verify the equality
180◦
sin 2α sin 2α sin 8α = sin α sin 3α sin 14α, where α =
= 6◦ .
30
Clearly, sin 14α = cos α, hence, 2 sin α sin 3α sin 14α = sin 2α sin 3α. It remains to verify
that
1
sin 3α = 2 sin 2α sin 8α = cos 6α − cos 10α = 1 − 2 sin2 3α − ,
2
2
◦
◦
i.e., 4 sin 18 + 2 sin 18 = 1, cf. Problem 5.46.
6.61. First, let n = 2m. The diagonals and sides of a regular 2m-gon have m distinct
lengths. Therefore, the marked points lie on m − 1 concentric circles (having n points each)
or in the common center of these circles. Since distinct circles have not more than two
common points, the circle that does not belong to this family of concentric circles contains
not more than 1 + 2(m − 1) = 2m − 1 = n − 1 of marked points.
Now, let n = 2m + 1. There are m distinct lengths among the lengths of the diagonals
and sides of a regular (2m + 1)-gon. Hence, the marked points lie on m concentric circles (n
points on each). A circle that does not belong to this family of concentric circles contains
not more than 2m = n − 1 marked points.
In either case the greatest number of marked points that lie on one circle is equal to n.
6.62. Denote the center of the polygon by O and the vertices of the polygon by
A1 , . . . , An . Suppose that there are no equal polygons among the polygons of the same
colour, i.e., they have m = m1 < m2 < m3 < · · · < mk sides, respectively. Let us consider a
transformation f defined on the set of vertices of the n-gon as the one that sends vertex Ak
to vertex Amk : f (Ak ) = Amk (we assume that Ap+qn = Ap ). This transformation sends the
−−−−→
vertices of a regular m-gon into one point, B, hence, the sum of vectors Of (Ai ), where Ai
−−→ −
→
are the vertices of an m-gon, is equal to mOB 6= 0 .
Since ∠Ami OAmj = m∠Ai OAj , the vertices of any regular polygon with the number of
sides greater than m pass under the considered transformation into the vertices of a regular
−−−−→
polygon. Therefore, the sum of vectors Of (Ai ) over all vertices of an n-gon and similar
sums over the vertices of m2 -, m3 -, . . . , mk -gons are equal to zero. We have obtained a
−−−−→
contradiction with the fact that the sum of vectors Of (Ai ) over the vertices of an m-gon is
not equal to zero.
Therefore, among the polygons of one color there are two equal ones.
6.63. Let a regular (n − 1)-gon B1 . . . Bn−1 be inscribed into a regular n-gon A1 . . . An .
We may assume that A1 and B1 are the least distant from each other vertices of these
polygons and points B2 , B3 , B4 and B5 lie on sides A2 A3 , A3 A4 , A4 A5 and A5 A6 . Let
αi = ∠Ai+1 Bi Bi+1 and βi = ∠Bi Bi+1 Ai+1 , where i = 1, 2, 3, 4. By the sine theorem
SOLUTIONS
159
A2 B2 : B1 B2 = sin α1 : sin ϕ and B2 A3 : B2 B3 = sin β2 : sin ϕ, where ϕ is the angle at a
sin ϕ
vertex of a regular n-gon. Therefore, sin α1 + sin β2 = anan−1
, where an and an−1 are the
(lengths of the) sides of the given polygons.
Similar arguments show that
sin α1 + sin β2 = sin α2 + sin β3 = sin α3 + sin β4 .
Now, observe that
αi − βi+1
αi + βi+1
cos
2
2
2π
and compute αi + βi+1 and αi − βi+1 . Since αi + βi = n and αi+1 + βi =
2π
2π
αi+1 = αi + n(n−1)
and βi+1 = βi − n(n−1)
; therefore,
sin αi + sin βi+1 = 2 sin
αi + βi+1 =
is a constant and
Hence,
2π
n(n − 1)
¶
it follows that
2π
2π
−
n
n(n − 1)
αi − βi+1 = αi−1 − βi +
µ
cos θ = cos θ +
2π
,
n−1
4π
.
n(n − 1)
µ
4π
= cos θ +
(n − 1)n
¶
for θ =
α1 − β2
.
2
We have obtained a contradiction because on an interval shorter than 2π the cosine cannot
attain the same value at three distinct points.
Remark. A square can be inscribed in a regular pentagon, cf. Problem 6.46.
−−→
−−→
◦
sends point Ai to
6.64. Let a = OA1 + · · · + OAn . A rotation about point O by 360
n
Ai+1 and, therefore, sends vector a into itself, i.e., a = 0.
−−→ −−→ −−→
−−→
−−→ −
−−→
−−→
→
Since XAi = XO + OAi and OA1 + · · · + OAn = 0 , it follows that XA1 + · · · + XAn =
−−→
nXO.
6.65. Through the center of a regular polygon A1 . . . An , draw line l that does not pass
through the vertices of the polygon. Let xi be equal to the length of the projection of vector
−−→
OAi to a line perpendicular to l. Then all the xi are nonzero and the sum of numbers xi
assigned to the vertices of a regular k-gon is equal to zero since the corresponding sum of
−−→
vectors OAi vanishes, cf. Problem 6.64.
−→
−−→
6.66. By Problem 6.64 a = 10AO and b = 10BO, where O is the center of polygon
X1 . . . X10 . Clearly, if point A is situated rather close to a vertex of the polygon and point
B rather close to the midpoint of a side, then AO > BO.
6.67. Since
−−→ −−→
−−→ −−→
Ai X 2 = |Ai O + OX|2 = Ai O2 + OX 2 + 2(Ai O, OX) =
−−→ −−→
R2 + d2 + 2(Ai O, OX),
it follows that
X
cf. Problem 6.64.
X −−→ −−→
Ai X 2 = n(R2 + d2 ) + 2(
Ai O, OX) = n(R2 + d2 ),
160
CHAPTER 6. POLYGONS
6.68. Denote by Sk the sum of squared distances from vertex Ak to all the other vertices.
Then
−−→ −−→
Sk = Ak A21 + Ak A22 + · · · + Ak A2n = Ak O2 + 2(Ak O, OA1 ) + A1 O2 + . . .
−−→ −−→
+ Ak O2 + 2(Ak O, OAn ) + An O2 = 2nR2
P
P −−→
−
→
because ni=1 OAi = 0 . Hence, ni=1 Sk = 2n2 R2 . Since each squared side and diagonal
enters this sum twice, the sum to be found is equal to n2 R2 .
6.69. Consider the rotation of the given n-gon about the n-gon’s center O that sends
Ak to A1 . Let Xk be the image of point X under the rotation. This rotation sends segment
Ak X to A1 Xk . Therefore,
A1 X + · · · + An X = A1 X1 + · · · + A1 Xn .
Since n-gon X1 . . . Xn is a regular one,
−−−→
−−−→
−−→
A1 X1 + · · · + A1 Xn = nA1 O,
−−→
cf. Problem 6.64. Therefore, A1 X1 + · · · + An Xn ≥ nA1 O.
6.70. Let Bi be the projection of point X to line OAi . Then
−−→ −−→
−−→ −−→ −−→
(ei , x) = (OAi , OBi + Bi X) = (OAi , OBi ) = ±R · OBi .
Points B1 , . . . , Bn lie on the circle with diameter OX and are vertices of a regular n-gon
for n odd and vertices of an n2 -gon counted twice for n even, cf. Problem 2.9. Therefore,
P
OBi2 = 21 n · OX 2 , cf. Problem 6.67.
6.71. Let e1 , . . . , en be the vectors that go from the center of the given n-gon into
its
x a unit vector perpendicular to line l. The sum to be found is equal to
P vertices;
1
2
(ei , x) = 2 n · R2 , cf. Problem 6.70.
6.72. Let e1 , . . . , en be the unit vectors directed from the center O of a regular n-gon
−−→
into the midpoints of its sides; x = OX. Then the distance from point X to the i-th side is
equal to |(x, ei ) − r|. Hence, the sum to be found is equal to
X
X
((x, ei )2 − 2r(x, ei ) + r2 ) =
(x, ei )2 + nr2 .
P
By Problem 6.70 (x, ei )2 = 12 nd2 .
−−−−→
6.73. Let x be the unit vector parallel to line l and ei = Ai Ai+1 . Then the squared
P length
of the projection of side Ai Ai+1 to line l is equal to (x, ei )2 . By Problem 6.70 (x, ei )2 =
1
na2 .
2
−−→
−−→
6.74. Let a = OX, ei = OAi . Then
XA4i = |a + ei |4 = (|a|2 + 2(a, ei ) + |ei |2 )2 =
4(R2 + (a, ei ))2 = 4(R4 + 2R2 (a, ei ) + (a, ei )2 ).
P
P
P
(a, ei ) = (a,³ ei ) = 0.´ By Problem 6.70 (a, ei )2 = 12 nR4 ; hence, the sum to
4
be found is equal to 4 nR4 + nR2 = 6nR4 .
Clearly,
SOLUTIONS
161
6.75. a) First, let us prove the required relation for u = e1 . Let ei = (sin ϕi , cos ϕi ),
where cos ϕ1 = 1. Then
X
(e1 , ei )ei =
X
cos ϕi ei =
X
(sin ϕi cos ϕi , cos2 ϕi ) =
X µ sin 2ϕi 1 + cos 2ϕi ¶ ³ n ´ ne1
= 0,
,
.
=
2
2
2
2
For u = e2 the proof is similar.
It remains to notice that any vector u can be represented in the form u = λe1 + µe2 .
−−→
−−→
i
b) Let B1 , . . . , Bn be the midpoints of sides of the given polygon, ei = OB
, u = XO.
OBi
P −−→ −
−−→ −−→
→
Then XAi = OBi + (u, ei )ei . Since
OBi = 0 , it follows that
−−→
X −−→ X
nXO
nu
=
.
XAi =
(u, ei )ei =
2
2
6.76. Let e0 , . . . , en−1 be the vectors of sides of a regular n-gon. It sufficesPto prove that
n
−−−−−→
by reordering these vectors we can get a set of vectors −
a−
1 , . . . , an such that
k=1 kak = 0.
A number n which is not a power of a prime can be represented in the form n = pq, where
p and q are relatively prime. Now, let us prove that the collection
e0 , ep , . . . , e(q−1)p ; eq , eq+p , . . . , eq+(q−1)p ;
..............................
e(p−1)q , e(p−1)q+p , . . . , e(p−1)q+(q−1)p
is the one to be found. First, notice that if
x1 q + y1 p ≡ x2 q + y2 p ( mod pq),
then x1 ≡ x2 ( mod p) and y1 ≡ y2 ( mod q); therefore, in the considered collection each
of the vectors e0 , . . . , en−1 is encountered exactly once.
The endpoints of vectors eq , eq+p , . . . , eq+(q−1)p with a common beginning point destinguish a regular q-gon and, therefore, their sum is equal to zero. Moreover, vectors e0 ,
ep , . . . , e(q−1)p turn into eq , eq+p , . . . , eq+(p−1)q under the rotation by an angle of ϕ = 2π
.
p
Hence, if e0 + 2ep + · · · + qe(q−1)p = b, then
(q + 1)eq + (q + 2)eq+p + · · · + 2qeq+(q−1)p =
q(eq + · · · + eq+(q−1)p ) + eq + 2eq+p + · · · + qeq+(q−1)p = Rϕ b,
where Rϕ b is the vector obtained from b after the rotation by ϕ =
show that for the considered set of vectors we have
n
X
kak = b + Rϕ b + · · · + R(p−1)ϕ b = 0.
2π
.
p
Similar arguments
k=1
6.77. Suppose that on the sides of triangle ABC squares ABB1 A1 , BCC2 B2 , ACC3 A3
are constructed outwards and vertices A1 , B1 , B2 , C2 , C3 , A3 lie on one circle S. The midperpendiculars to segments A1 B1 , B2 C2 , A3 C3 pass through the center of circle S. It is clear
162
CHAPTER 6. POLYGONS
that the midperpendiculars to segments A1 B1 , B2 C2 , A3 C3 coincide with the midperpendiculars to sides of triangle ABC and therefore, the center of circle S coincides with the center
of the circumscribed circle of the triangle.
Denote the center of the circumscribed circle of triangle ABC by O. The distance from
O to line B2 C2 is equal to R cos ∠A + 2R sin ∠A, where R is the radius of the circumscribed
circle of triangle ABC. Hence,
OB22 = (R sin ∠A)2 + (R cos ∠A + 2R sin ∠A)2 =
√
R2 (3 + 2(sin ∠2A − cos 2∠A)) = R2 (3 − 2 2 cos(45◦ + 2∠A)).
Clearly, in order for the triangle to possess the desired property, it is necessary and sufficient
that OB22 = OC32 = OA21 , i.e.,
cos(45◦ + 2∠A) = cos(45◦ + 2∠B) = cos(45◦ + 2∠C).
This equality holds for ∠A = ∠B = ∠C = 60◦ . If, contrarywise, ∠A 6= ∠B, then (45◦ +
2∠A) + (45◦ + 2∠B) = 360◦ , i.e., ∠A + ∠B = 135◦ . Hence, ∠C = 45◦ and ∠A = ∠C = 45◦ ,
∠B = 90◦ (or ∠B = 45◦ , ∠A = 90◦ ). We see that the triangle should be either an equilateral
or an isosceles one.
Ak+1
ab
(Problem 12.33); hence, pk = M Ak ·M
. There6.78. In any triangle we have hc = 2R
2R
fore,
M A1 · M A2 . . . M A2n
p1 p3 . . . p2n−1 =
= p2 p4 . . . p2n .
(2R)n
6.79. Let ABC be a triangle inscribed in circle S. Denote the distances from the center
O of S to sides BC, CA and AB by a, b and c, respectively. Then R + r = a + b + c if point
O lies inside triangle ABC and R + r = −a + b + c if points A and O lie on various sides of
line BC, cf. Problem 12.38.
Each of the diagonals of the partition belongs to two triangles of the partition. For one
of these triangles point O and the remaining vertex lie on one side of the diagonal, for the
other one the points lie on different sides.
A partition of an n-gon by nonintersecting diagonals into triangles consists of n − 2
triangles. Therefore, the sum (n − 2)R + r1 + · · · + rn−2 is equal to the sum of distances from
point O to the sides of an n-gon (the distances to the sides are taken with the corresponding
signs). This implies that the sum r1 + · · · + rn−2 does not depend on the partition.
6.80. Let polygon A1 . . . An be inscribed in a circle. Let us consider point A′2 symmetric
to point A2 through the midperpendicular to segment A1 A3 . Then polygon A1 A′2 A3 . . . An
is an inscribed one and its area is equal to the area of polygon A1 . . . An . Therefore, we can
transpose any two sides. Therefore, we can make any side, call it X, a neighbouring side of
any given side, Y ; next, make any of the remaining sides a neighbour of X, etc. Therefore,
the area of an n-gon inscribed into the given circle only depends on the set of lengths of the
sides but not on their order.
6.81. Without loss of generality we may assume that an is the greatest of the numbers
a1 , . . . , an . Let n-gon A1 . . . An be inscribed into a circle centered at O. Then
Ai Ai+1 : A1 An = sin
∠A1 OAn
∠Ai OAi+1
: sin
.
2
2
Therefore, let us proceed as follows. From the relation sin ϕ2i : sin ϕ2 = ai : an the angle ϕi
is uniquely determined in terms of ϕ if ϕi < π. On a circle of radius 1, fix a point An and
SOLUTIONS
163
consider variable points A1 , . . . , An−1 , A′n such that
⌣ An A1 = ϕ, ⌣ A1 A2 = ϕ1 , . . . , ⌣ An−2 An−1 = ϕn−2 and ⌣ An−1 A′n = ϕn−1 .
Denote these points in two distinct ways as plotted on Fig. 75. (The first way — Fig. 75 a)
— corresponds to an n-gon that contains the center of the circle, and the second way — Fig.
75 b) — corresponds to an n-gon that does not contain the center of the circle). It remains
to prove that as ϕ varies from 0 to π, then in one of these cases point A′n coincides with An
(indeed, then up to a similarity we get the required n-gon). Suppose that in the first case
points A′n and An never coincide for 0 ≤ ϕ ≤ π, i.e., for ϕ = π we have ϕ1 + · · · + ϕn−1 < π.
Figure 75 (Sol. 6.81)
Fig. 75 b) requires certain comments: sin α ≈ α for small values of α; hence, the
conditions of the problem imply that for small angles point An does indeed lie on arc ⌣ A1 A′n
because ϕ1 + · · · + ϕn−1 > ϕ. Thus, for small angles ϕ1 + · · · + ϕn−1 > ϕ and if ϕ = π, then
by the hypothesis ϕ1 + · · · + ϕn−1 < π = ϕ. Hence, at certain moment ϕ = ϕ1 + · · · + ϕn−1 ,
i.e., points An and A′n coincide.
6.82. Let h1 , . . . , hn be the distances from the given point to the corresponding sides;
let a1 , . . . , an be the distances from the vertices of the polygon to tangent points. Then the
product of areas of red as well as blue triangles is equal to a1 ...an2nh1 ...hn .
6.83. Let OHi be a height of triangle OAi Ai+1 . Then ∠Hi−1 OAi = ∠Hi OAi = ϕi . The
conditions of the problem imply that
ϕ1 + ϕ2 = ϕn+1 + ϕn+2 ,
ϕn+2 + ϕn+3 = ϕ2 + ϕ3 ,
ϕ3 + ϕ4 = ϕn+3 + ϕn+4 ,
..............................,
ϕn−2 + ϕn−1 = ϕ2n−2 + ϕ2n−1
(expressing the last equality we have taken into account that n is odd) and
ϕn−1 + 2ϕn + ϕn+1 = ϕ2n−1 + 2ϕ2n + ϕ1 .
Adding all these equalities we get
ϕn−1 + ϕn = ϕ2n−1 + ϕ2n ,
as required.
−−→ −−→ −−→
6.84. Let O be the center of the given circle. Then XAi = XO + OAi and, therefore,
−−→ −−→
−−→ −−→
XA2i = XO2 + OA2i + 2(OX, OAi ) = d2 + r2 + 2(XO, OAi ).
−−→
−−→ −
→
Since a1 OA1 + · · · + An OAn = 0 (cf. Problem 13.4), it follows that
a1 XA21 + · · · + an XA2n = (a1 + · · · + an )(d2 + r2 ).
164
CHAPTER 6. POLYGONS
bi
i
6.85. By Problem 5.8 bi−1
= sin2 ∠A
. To solve heading a) it suffices to multiply all
2
a2i
these equalities and to solve heading b) we have to divide the product of all equalities with
even index i by the product of all equalities with odd index i.
6.86. Let BC be a blue side, AB and CD be the sides neighbouring with BC. By the
hypothesis sides AB and CD are red ones. Suppose that the polygon is a circumscribed one;
let P , Q, R be the tangent points of sides AB, BC, CD, respectively, with the inscribed
circle. Clearly, BP = BQ, CR = CQ and segments BP , CR only neighbour one blue
segment. Therefore, the sum of the lengths of the red sides is not smaller than the sum of
the lengths of the blue sides. We have obtained a contradiction with the fact that the sum
of the lengths of red sides is smaller than the semiperimeter. Therefore, a circle cannot be
inscribed into the polygon.
6.87. Let the given n-gon have k acute angles. Then the sum of its angles is smaller
than k · 90◦ + (n − k) · 180◦ . On the other hand, the sum of the angles of the n-gon is equal
to (n − 2) · 180◦ . Hence,
(n − 2) · 180◦ < k · 90◦ + (n − k) · 180◦ ,
i.e., k < 4.
Since k is an integer, k ≤ 3.
Figure 76 (Sol. 6.87)
For any n ≥ 3 there exists a convex n-gon with three acute angles (Fig. 76).
6.88. Suppose that the lengths of nonadjacent sides AB and CD are equal to the
length of the greatest diagonal. Then AB + CD ≥ AC + BD. But by Problem 9.14
AB + CD < AC + BD. We have obtained a contradiction and therefore, the sides whose
length is equal to the length of the longest diagonal should be adjacent ones, i.e., there are
not more than two of such sides.
Figure 77 (Sol. 6.88)
An example of a polygon with two sides whose lengths are equal to the length of the
longest diagonal is given on Fig. 77. Clearly, such an n-gon exists for any n > 3.
6.89. Let us prove that n ≤ 5. Let AB = 1 and C the vertex not adjacent to either A or
B. Then |AC −BC| < AB = 1. Hence, AC = BC, i.e., point C lies on the midperpendicular
SOLUTIONS
165
to side AB. Therefore, in addition to vertices A, B, C the polygon can have only two more
vertices.
Figure 78 (Sol. 6.89)
An example of a pentagon with the required property is given on Fig. 78. Let us elucidate
its construction. Clearly, ACDE is a rectangle, AC = ED = 1 and ∠CAD = 60◦ . Point B
is determined from the condition BE = BD = 3.
An example of a quadrilateral with the desired property is rectangle ACDE on the same
figure.
6.90. An example of a pentagon satisfying the conditions of the problem is plotted on
Fig. 79. Let us clarify its construction. Take an equilateral right triangle EAB and draw
midperpendiculars to sides EA, AB; on them construct points C and D, respectively, so
that ED = BC = AB (i.e., lines BC and ED form angles of 30◦ with the corresponding
midperpendiculars). Clearly,
DE = BC = AB = EA < EB < DC
and DB = DA = CA = CE > EB.
Now, let us prove that the fifth side and the fifth diagonal cannot have a common point.
Suppose that the fifth side AB has a common point A with the fifth diagonal. Then the
fifth diagonal is either AC or AD. Let us consider these two cases.
Figure 79 (Sol. 6.90)
In the first case △AED = △CDE; hence, under the symmetry through the midperpendicular to segment ED point A turns into point C. This symmetry preserves point B
because BE = BD. Therefore, segment AB turns into CB, i.e., AB = CB. Contradiction.
166
CHAPTER 6. POLYGONS
In the second case △ACE = △EBD; hence, under the symmetry through the bisector
of angle ∠AED segment AB turns into DC, i.e., AB = CD. Contradiction.
6.91. Let us consider two neighbouring vertices A1 and A2 . If ∠A1 OA2 ≥ 90◦ , then
OA1 = OA2 because neither right nor acute angle can be adjacent to the base of an isosceles
triangle.
Figure 80 (Sol. 6.91)
Now, let ∠A1 OA2 < 90◦ . Let us draw through point O lines l1 and l2 perpendicular
to lines OA1 and OA2 , respectively. Denote the regions into which these lines divide the
plane as indicated on Fig. 80. If in region 3 there is a vertex, Ak , then A1 O = Ak O = A2 O
because ∠A1 OAk ≥ 90◦ and ∠A2 OAk ≥ 90◦ . If region 3 has no vertices of the polygon, then
in region 1 there is a vertex Ap and in region 2 there is a vertex Aq (if neither of the regions
1 or 2 would have contained vertices of the polygon, then point O would have been outside
the polygon). Since ∠A1 OAq ≥ 90◦ , ∠A2 OAp ≥ 90◦ and ∠Ap OAq ≥ 90◦ , it follows that
A1 O = Aq O = Ap O = A2 O.
It remains to notice that if the distances from point O to any pair of the neighbouring
vertices of the polygon are equal, then all the distances from point O to the vertices of the
polygon are equal.
6.92. Let us prove that if A, B, C, D, E, F are points on the circle placed in an
arbitrary order; lines AB and DE, BC and EF , CD and F A, intersect at points G, H, K,
respectively. Then points G, H and K lie on one line.
Let a, b, . . . , f be oriented angles between a fixed line and lines OA, OB, . . . , OF ,
respectively, where O is the center of the circumscribed circle of the hexagon. Then
∠(AB, DE) =
a+b−d−e
c+d−f −a
, ∠(CD, F A) =
,
2
2
∠(EF, BC) =
e+f −b−c
2
and, therefore, the sum of these angles is equal to 0.
Let Z be the intersection point of circumscribed circles of triangles BDG and DF K.
Let us prove that point B, F , Z and H lie on one circle. For this we have to verify that
∠(BZ, ZF ) = ∠(BH, HF ). Clearly,
∠(BZ, ZF ) = ∠(BZ, ZD) + ∠(DZ, ZF ),
∠(BZ, ZD) = ∠(BG, GD) = ∠(AB, DE),
∠(DZ, ZF ) = ∠(DK, KF ) = ∠(CD, F A)
and, as we have just proved,
∠(AB, DE) + ∠(CD, F A) = −∠(EF, BC) = ∠(BC, EF ) = ∠(BH, HF ).
SOLUTIONS
167
Now, let us prove that points H, Z and G lie on one line. For this it suffices to verify
that ∠(GZ, ZB) = ∠(HZ, ZB). Clearly,
∠(GZ, ZB) = ∠(GD, DB) = ∠(ED, DB), ∠(HZ, ZB) = ∠(HF, F B) = ∠(ED, DB).
We similarly prove that points K, Z and G lie on one line:
∠(DZ, ZG) = ∠(DB, BG) = ∠(DB, BA);
∠(DZ, ZK) = ∠(DF, F K) = ∠(DB, BA)
We have deduced that points H and K lie on line GZ, consequently, points G, H and K lie
on one line.
6.93. Let A2 , B2 and C2 be the indicated intersection points of lines. By applying
Pascal’s theorem to points M , A1 , A, C, B, B1 we deduce that A2 , B2 and R lie on one line.
Similarly, points A2 , C2 and R lie on one line. Hence, points A2 , B2 , C2 and R lie on one
line.
6.94. Points A1 and B1 lie on circle S of diameter AB. Let A4 and B4 be the intersection
points of lines AA2 and BB2 with line A3 B3 . By Problem 2.41 a) these points lie on circle
S. Lines A1 B and A4 A intersect at point A2 and lines BB4 and AB1 at point B2 . Therefore,
applying Pascal’s theorem to points B1 , A1 , B, B4 , A4 , A we see that the intersection point
of lines B1 A1 and B4 A4 (the latter line coincides with A3 B3 ) lies on line A2 B2 .
6.95. Let K be the intersection point of lines BC and M N . Apply Pascal’s theorem to
points A, M , N , D, C, B. We see that points E, K, F lie on one line and, therefore, K is
the intersection point of lines M N and EF .
6.96. Let rays P A and QA intersect the circle at points P2 and Q2 , i.e., P1 P2 and Q1 Q2
are diameters of the given circle. Let us apply Pascal’s theorem to hexagon P P2 P1 QQ2 Q1 .
Lines P P2 and QQ2 intersect at point A and lines P1 P2 and Q1 Q2 intersect at point O,
hence, the intersection point of lines P1 Q and Q1 P lies on line AO.
6.97. Let given points A, B, C, D, E lie on one line. Suppose that we have constructed
point F of the same circle. Denote by K, L, M the intersection points of lines AB and DE,
BC and EF , CD and F A, respectively. Then by Pascal’s theorem points K, L, M lie on
one line.
The above implies the following construction. Let us draw through point E an arbitrary
line a and denote its intersection point with line BC by L. Then construct the intersection
point K of lines AB and DE and the intersection point M of lines KL and CD. Finally, let
F be the intersection point of lines AM and a. Let us prove that F lies on our circle. Let
F1 be the intersection point of the circle and line a. From Pascal’s theorem it follows that
F1 lies on line AM , i.e., F1 is the intersection point of a and AM . Hence, F1 = F .
6.98. Let P and Q be the intersection points of line A3 A4 with A1 A2 and A1 A6 , respectively, and R and S be the intersection points of line A4 A5 with A1 A6 and A1 A2 , respectively.
Then
A2 K : A3 L = A2 P : A3 P, A3 L : A6 M = A3 Q : A6 Q, A6 M : A5 N = A6 R : A5 R.
Therefore, the desired relation A2 K : A5 N = A2 S : A5 S takes the form
A2 P A3 Q A6 R A5 S
·
·
·
= 1.
A3 P A6 Q A5 R A2 S
Let T be the intersection point of lines A2 A3 and A5 A6 ; by Pascal’s theorem points S, Q
and T lie on one line. By applying Menelau’s theorem (cf. Problem 5.58) to triangle P QS
and points T , A2 , A3 and also to triangle RQS and points T , A5 , A6 we get
T Q A5 S A6 R
A2 P A3 Q T S
·
·
= 1 and
·
·
= 1.
A2 S A3 P T Q
T S A5 R A6 Q
168
CHAPTER 6. POLYGONS
By multiplying these equalities we get the statement desired. (The ratio of segments should
be considered oriented ones.)
Chapter 7. LOCI
Background
1) A locus is a figure consisting of all points having a cirtain property.
2) A solution of a problem where a locus is to be found should contain the proof of the
following facts:
a) the points with a required property belong to figure Φ which is the answer to the
problem;
b) All points of Φ have the required property.
3) A locus possessing two properties is the intersection of two figures: (1) the locus of
points possessing the first property and (2) the locus of points possessing the other property.
4) Three most important loci:
a) The locus of points equidistant from points A and B is the midperpendicular to
segment AB;
b) The locus of points whose distance from a given point O is equal to R is the circle of
radius R centered at O;
c) The locus of vertices of a given angle that subtend given segment AB is the union of
two arcs of circles symmetric through line AB (points A and B do not belong to the locus).
Introductory problems
1. a) Find the locus of points equidistant from two parallel lines.
b) Find the locus of points equidistant from two intersecting lines.
2. Find the locus of the midpoints of segments with the endpoints on two given parallel
lines.
3. Given triangle ABC, find the locus of points X satisfying inequalities AX ≤ BX ≤CX.
4. Find the locus of points X such that the tangents drawn from X to the given circle
have a given length.
5. A point A on a circle is fixed. Find the locus of points X that divide chords with A
as an endpoint in the ratio of 1 : 2 counting from point A.
§1. The locus is a line or a segment of a line
7.1. Two wheels of radii r1 and r2 roll along line l. Find the set of intersection points
M of their common inner tangents.
7.2. Sides AB and CD of quadrilateral ABCD of area S are not parallel. Inside the
quadrilateral find the locus of points X for which SABX + SCDX = 12 S.
7.3. Given two lines that meet at point O. Find the locus of points X for which the sum
of the lengths of projections of segments OX to these lines is a constant.
7.4. Given rectangle ABCD, find the locus of points X for which AX +BX = CX +DX.
7.5. Find the locus of points M that lie inside rhombus ABCD and with the property
that ∠AM D + ∠BM C = 180◦ .
169
170
CHAPTER 7. LOCI
***
7.6. Given points A and B in plane, find the locus of points M for which the difference
of the squared lengths of segments AM and P M is a constant.
7.7. A circle S and a point M outside it are given. Through point M all possible circles
S1 that intersect S are drawn; X is the intersection point of the tangent at M to S1 with
the extension of the common chord of circles S and S1 . Find the locus of points X.
7.8. Given two nonintersecting circles, find the locus of the centers of circles that divide
the given circles in halves (i.e., that intersect the given circles in diametrically opposite
points).
7.9. A point A inside a circle is taken. Find the locus of the intersection points of
tangents to circles drawn through the endpoints of possible chords that contain point A.
7.10. a) Parallelogram ABCD is given. Prove that the quantity
AX 2 + CX 2 − BX 2 − DX 2
does not depend on the choice of point X.
b) Quadrilateral ABCD is not a parallelogram. Prove that all points X that satisfy
the relation AX 2 + CX 2 = BX 2 + DX 2 lie on one line perpendicular to the segment that
connects the midpoints of the diagonals.
See also Problems 6.14, 15.14.
§2. The locus is a circle or an arc of a circle
7.11. A segment moves along the plane so that its endpoints slide along the legs of a
right angle ∠ABC. What is the trajectory traversed by the midpoint of this segment? (We
naturally assume that the length of the segment does not vary while it moves.)
7.12. Find the locus of the midpoints of the chords of a given circle, provided the chords
pass through a given point.
7.13. Given two points, A and B and two circles that are tangent to line AB: one circle
is tangent at A and the other one at B, and the circles are tangent to each other at point
M . Find the locus of points M .
***
7.14. Two points, A and B in plane are given. Find the locus of points M for which
AM : BM = k. (Apollonius’s circle.)
7.15. Let S be Apollonius’s circle for points A and B where point A lies outside circle
S. From point A tangents AP and AQ to circle S are drawn. Prove that B is the midpoint
of segment P Q.
7.16. Let AD and AE be the bisectors of the inner and outer angles of triangle ABC
and Sa be the circle with diameter DE; circles Sb and Sc are similarly defined. Prove that:
a) circles Sa , Sb and Sc have two common points, M and N , such that line M N passes
through the center of the circumscribed circle of triangle ABC;
b) The projections of point M (and N ) to the sides of triangle ABC distinguish an
equilateral triangle.
7.17. Triangle ABC is an equilateral one, M is a point. Prove that if the lengths
of segments AM , BM and CM form a geometric progression, then the quotient of this
progression is smaller than 2.
See also Problems 14.19 a), 18.14.
§6. A METHOD OF LOCI
171
§3. The inscribed angle
7.18. Points A and B on a circle are fixed and a point C runs along the circle. Find the
set of the intersection points of a) heights; b) bisectors of triangles ABC.
7.19. Point P runs along the circumscribed circle of square ABCD. Lines AP and BD
intersect at point Q and the line that passes through point Q parallel to AC intersects line
BP at point X. Find the locus of points X.
7.20. a) Points A and B on a circle are fixed and points A1 and B1 run along the same
circle so that the value of arc ⌣ A1 B1 remains a constant; let M be the intersection point
of lines AA1 and BB1 . Find the locus of points M .
b) Triangles ABC and A1 B1 C1 are inscribed in a circle; triangle ABC is fixed and triangle
A1 B1 C1 rotates. Prove that lines AA1 , BB1 and CC1 intersect at one point for not more
than one position of triangle A1 B1 C1 .
7.21. Four points in the plane are given. Find the locus of the centers of rectangles
formed by four lines that pass through the given points.
7.22. Find the locus of points X that lie inside equilateral triangle ABC and such that
∠XAB + ∠XBC + ∠XCA = 90◦ .
See also Problems 2.5, 2.37.
§4. Auxiliary equal triangles
7.23. A semicircle centered at O is given. From every point X on the extension of the
diameter of the semicircle a ray tangent to the semicircle is drawn. On the ray segment XM
equal to segment XO is marked. Find the locus of points M obtained in this way.
7.24. Let A and B be fixed points in plane. Find the locus of points C with the following
property: height hb of triangle ABC is equal to b.
7.25. A circle and a point P inside it are given. Through every point Q on the circle
the tangent is drawn. The perpendicular dropped from the center of the circle to line P Q
and the tangent intersect at a point M . Find the locus of points M .
§5. The homothety
7.26. Points A and B on a circle are fixed. Point C runs along the circle. Find the set
of the intersection points of the medians of triangles ABC.
7.27. Triangle ABC is given. Find the locus of the centers of rectangles P QRS whose
vertices Q and P lie on side AC and vertices R and S lie on sides AB and BC, respectively.
7.28. Two circles intersect at points A and B. Through point A a line passes. It
intersects the circles for the second time at points P and Q. What is the line plotted by the
midpoint of segment P Q while the intersecting line rotates about point A.
7.29. Points A, B and C lie on one line; B is between A and C. Find the locus of points
M such that the radii of the circumscribed circles of triangles AM B and CM B are equal.
See also Problems 19.10, 19.21, 19.38.
§6. A method of loci
7.30. Points P and Q move with the same constant speed v along two lines that intersect
at point O. Prove that there exists a fixed point A in plane such that the distances from A
to P and Q are equal at all times.
7.31. Through the midpoint of each diagonal of a convex quadrilateral a line is drawn
parallel to the other diagonal. These lines meet at point O. Prove that segments that connect
172
CHAPTER 7. LOCI
O with the midpoints of the sides of the quadrilateral divide the area of the quadrilateral
into equal parts.
7.32. Let D and E be the midpoints of sides AB and BC of an acute triangle ABC and
point M lies on side AC. Prove that if M D < AD, then M E > EC.
7.33. Inside a convex polygon points P and Q are taken. Prove that there exists a vertex
of the polygon whose distance from Q is smaller than that from P .
7.34. Points A, B and C are such that for any fourth point M either M A ≤ M B or
M A ≤ M C. Prove that point A lies on segment BC.
7.35. Quadrilateral ABCD is given; in it AB < BC and AD < DC. Point M lies on
diagonal BD. Prove that AM < M C.
§7. The locus with a nonzero area
7.36. Let O be the center of rectangle ABCD. Find the locus of points M for which
AM ≥ OM , BM ≥ OM , CM ≥ OM and DM ≥ OM .
7.37. Find the locus of points X from which tangents to a given arc AB of a circle can
be drawn.
7.38. Let O be the center of an equilateral triangle ABC. Find the locus of points M
satisfying the following condition: any line drawn through M intersects either segment AB
or segment CO.
7.39. In plane, two nonintersecting disks are given. Does there necessarily exist a point
M outside these disks that satisfies the following condition: each line that passes through
M intersects at least one of these disks?
Find the locus of points M with this property.
See also Problem 18.11.
§8. Carnot’s theorem
7.40. Prove that the perpendiculars dropped from points A1 , B1 and C1 to sides BC,
CA, AB of triangle ABC intersect at one point if and only if
A1 B 2 + C1 A2 + B1 C 2 = B1 A2 + A1 C 2 + C1 B 2 .
(Carnot’s formula)
7.41. Prove that the heights of a triangle meet at one point.
7.42. Points A1 , B1 and C1 are such that AB1 = AC1 , BC = BA1 and CA1 = CB1 .
Prove that the perpendiculars dropped from points A1 , B1 and C1 to lines BC, CA and AB
meet at one point.
7.43. a) The perpendiculars dropped from the vertices of triangle ABC to the corresponding sides of triangle A1 B1 C1 meet at one point. Prove that the perpendiculars dropped
from the vertices of triangle A1 B1 C1 to the corresponding sides of triangle ABC also meet
at one point.
b) Lines drawn through vertices of triangle ABC parallelly to the corresponding sides of
triangle A1 B1 C1 intersect at one point. Prove that the lines drawn through the vertices of
triangle A1 B1 C1 parallelly to the corresponding sides of triangle ABC also intersect at one
point.
7.44. On line l points A1 , B1 and C1 are taken and from the vertices of triangle ABC
perpendiculars AA2 , BB2 and CC2 are dropped to this line. Prove that the perpendiculars
dropped from points A1 , B1 and C1 to lines BC, CA and AB, respectively, intersect at one
point if and only if
A1 B1 : B1 C1 = A2 B2 : B2 C2 .
The ratios of segments are oriented ones.
PROBLEMS FOR INDEPENDENT STUDY
173
7.45. Triangle ABC is an equilateral one, P an arbitrary point. Prove that the perpendiculars dropped from the centers of the inscribed circles of triangles P AB, P BC and P CA
to lines AB, BC and CA, respectively, meet at one point.
7.46. Prove that if perpendiculars raised at the bases of bisectors of a triangle meet at
one point, then the triangle is an isosceles one.
§9. Fermat-Apollonius’s circle
7.47. Prove that the set of points X such that
k1 A1 X 2 + · · · + kn An X 2 = c
is either
a) a circle or the empty set if k1 + · · · + kn 6= 0;
b) a line, a plane or the empty set if k1 + · · · + kn = 0.
7.48. Line l intersects two circles at four points. Prove that the quadrilateral formed by
the tangents at these points is a circumscribed one and the center of its circumscribed circle
lies on the line that connects the centers of the given circles.
7.49. Points M and N are such that AM : BM : CM = AN : BN : CN . Prove that
line M N passes through the center O of the circumscribed circle of triangle ABC.
See also Problems 7.6, 7.14, 8.59–8.63.
Problems for independent study
7.50. On sides AB and BC of triangle ABC, points D and E are taken. Find the locus
of the midpoints of segments DE.
7.51. Two circles are tangent to a given line at two given points A and B; the circles
are also tangent to each other. Let C and D be the tangent points of these circles with
another outer tangent. Both tangent lines to the circles are outer ones. Find the locus of
the midpoints of segments CD.
7.52. The bisector of one of the angles of a triangle has inside the triangle a common
point with the perpendicular erected from the midpoint of the side opposite the angle. Prove
that the triangle is an isosceles one.
7.53. Triangle ABC is given. Find the locus of points M of this triangle for which
the condition AM ≥ BM ≥ CM holds. When the obtained locus is a) a pentagon; b) a
triangle?
7.54. Square ABCD is given. Find the locus of the midpoints of the sides of the squares
inscribed in the given square.
7.55. An equilateral triangle ABC is given. Find the locus of points M such that
triangles AM B and BCM are isosceles ones.
7.56. Find the locus of the midpoints of segments of length √23 whose endpoints lie on
the sides of a unit square.
7.57. On sides AB, BC and CA of a given triangle ABC points P , Q and R, respectively,
are taken, so that P Q k AC and P R k BC. Find the locus of the midpoints of segments
QR.
7.58. Given a semicircle with diameter AB. For any point X on this semicircle, point
Y on ray XA is taken so that XY = XB. Find the locus of points Y .
7.59. Triangle ABC is given. On its sides AB, BC and CA points C1 , A1 and B1 ,
respectively, are selected. Find the locus of the intersection points of the circumscribed
circles of triangles AB1 C1 , A1 BC1 and A1 B1 C.
174
CHAPTER 7. LOCI
Solutions
7.1. Let O1 and O2 be the centers of the wheels of radii r1 and r2 , respectively. If M is
the intersection point of the inner tangents, then OM : O2 M = r1 : r2 . It is easy to derive
1 r2
from this condition that the distance from point M to line l is equal to r2r1 +r
. Hence, all
2
the intersection points of the common inner tangents lie on the line parallel to l and whose
1 r2
.
distance from l is equal to r2r1 +r
2
7.2. Let O be the intersection point of lines AB and CD. On rays OA and OD, mark
segments OK and OL equal to AB and CD, respectively. Then
SABX + SCDX = SKOX + SLOX = SKOL ± SKXL .
Therefore, the area of triangles KXL is a constant, i.e., point X lies on a line parallel to
KL.
−−→
7.3. Let a and b be unit vectors parallel to the given lines; x = OX. The sum of
the lengths of the projections of vector x to the given lines is equal to |(a, x)| + |(b, x)| =
|(a ± b, x)|, where the change of sign occurs on the perpendiculars to the given lines erected
at point O. Therefore, the locus to be found is a rectangle whose sides are parallel to
the bisectors of the angles between the given lines and the vertices lie on the indicated
perpendiculars.
7.4. Let l be the line that passes through the midpoints of sides BC and AD. Suppose
that point X does not lie on l; for instance, points A and X lie on one side of l. Then
AX < DX and BX < CX and, therefore, AX + BX < CX + DX. Hence, l is the locus to
be found.
−−→ −−→
7.5. Let N be a point such that M N = DA. Then ∠N AM = ∠DM A and ∠N BM =
∠BM C and, therefore, quadrilateral AM BN is an inscribed one. The diagonals of the
inscribed quadrilateral AM BN are equal, hence, either AM k BN or BM k AN . In the first
case ∠AM D = ∠M AN = ∠AM B and in the second case ∠BM C = ∠M BN = ∠BM A. If
∠AM B = ∠AM D, then ∠AM B + ∠BM C = 180◦ and point M lies on diagonal AC and if
∠BM A = ∠BM C, then point M lies on diagonal BD. It is also clear that if point M lies
on one of the diagonals, then ∠AM D + ∠BM C = 180◦ .
7.6. Introduce a coordinate system selecting point A as the origin and directing Ox-axis
along ray AB. Let (x, y) be the coordinates of M . Then AM 2 = x2 + y 2 and BM 2 =
(x − a)2 + y 2 , where a = AB. Hence, AM 2 − BM 2 = 2ax − a2 . This quantity is equal to k
2
for points M whose coordinates are ( a 2a+k , y). All such points lie on a line perpendicular to
AB.
7.7. Let A and B be the intersection points of circles S and S1 . Then XM 2 = XA·XB =
XO2 − R2 , where O and R are the center and the radius, respectively, of circle S. Hence,
XO2 −XM 2 = R2 and, therefore, points X lie on the perpendicular to line OM (cf. Problem
7.6).
7.8. Let O1 and O2 be the centers of the given circles, R1 and R2 their respective radii.
The circle of radius r centered at X intersects the first circle in the diametrically opposite
points if and only if r2 = XO12 + R12 ; hence, the locus to be found consists of points X such
that XO12 + R12 = XO22 + R22 . All such points X lie on a line perpendicular to O1 O2 , cf.
Problem 7.6.
7.9. Let O be the center of the circle, R its radius, M the intersection point of the
tangents drawn through the endpoints of the chord that contains point A, and P the midpoint
of this chord. Then OP · OM = R2 and OP = OA cos ϕ, where ϕ = ∠AOP . Hence,
AM 2 = OM 2 + OA2 − 2OM · OA cos ϕ = OM 2 + OA2 − 2R2 ,
SOLUTIONS
175
and, therefore, the quantity
OM 2 − AM 2 = 2R2 − OA2
is a constant. It follows that all points M lie on a line perpendicular to OA, cf. Problem
7.6.
7.10. Let P and Q be the midpoints of diagonals AC and BD. Then
AC 2
BD2
and BX 2 + DX 2 = 2QX 2 +
2
2
(cf. Problem 12.11 a)) and, therefore, in heading b) the locus to be found consists of points
X such that P X 2 − QX 2 = 41 (BD2 − AC 2 ) and in heading a) P = Q and, therefore, the
considered quantity is equal to 21 (BD2 − AC 2 ).
7.11. Let M and N be the midpoints of the given segment, O its midpoint. Point B lies
on the circle with diameter M N , hence, OB = 21 M N . The trajectory of point O is the part
of the circle of radius 12 M N centered at B confined inside angle ∠ABC.
7.12. Let M be the given point, O the center of the given circle. If X is the midpoint of
chord AB, then XO ⊥AB. Therefore, the locus to be found is the circle with diameter M O.
7.13. Let us draw through point M a common tangent to the circles. Let O be the
intersection point of this tangent with line AB. Then AO = M O = BO, i.e., O is the
midpoint of segment AB. Point M lies on the circle with center O and radius 12 AB. The
locus of points M is the circle with diameter AB (points A and B excluded).
7.14. For k = 1 we get the midperpendicular to segment AB. In what follows we will
assume that k 6= 1.
Let us introduce a coordinate system in plane so that the coordinates of A and B are
(−a, 0) and (a, 0), respectively. If the coordinates of point M are (x, y), then
AX 2 + CX 2 = 2P X 2 +
The equation
AM 2
BM 2
AM 2
y2
2
=
(x
+
a)
+
+ y2.
BM 2
(x − a)2
= k 2 takes the form
2ka 2
k2
2
+y =
.
x+1−
1 − k2a
1 − k2
2
2ka
k
This is an equation of the circle with center (−1 + 1−k
2 a , 0) and radius |1−k 2 | .
7.15. Let line AB intersect circle S at points E and F so that point E lies on segment
AB. Then P E is the bisector of triangle AP B, hence, ∠EP B = ∠EP A = ∠EF P . Since
∠EP F = 90◦ , it follows that P B ⊥ EF .
7.16. a) The considered circles are Apollonius’s circles for the pairs of vertices of triangle
ABC and, therefore, if X is a common point of circles Sa and Sb , then XB : XC = AB : AC
and XC : XA = BC : BA, i.e., XB : XA = CB : CA and, therefore, point X belongs to
circle Sc . It is also clear that if AB > BC, then point D lies inside circle Sb and point A
outside it. It follows that circles Sa and Sb intersect at two distinct points.
To complete the proof, it remains to make use of the result of Problem 7.49.
b) According to heading a) M A = λa , M B = λb and M C = λc . Let B1 and C1 be the
projections of point M on lines AC and AB, respectively. Points B1 and C1 lie on the circle
with diameter M A, hence,
λ
λ a
=
,
B1 C1 = M A sin ∠B1 AC1 =
a 2R
2R
where R is the radius of the circumscribed circle of triangle ABC. Similarly, A1 C1 = A1 B1 =
λ
.
2R
176
CHAPTER 7. LOCI
−−→
−→
−−→
−−→
7.17. Let O1 and O2 be points such that BO1 = 34 BA and CO2 = 43 CB. It is easy to
verify that if BM > 2AM , then point M lies inside circle S1 of radius 32 AB with center O1
(cf. Problem 7.14) and if CM > 2BM , then point M lies inside circle S2 of radius 32 AB
centered at O2 . Since O1 O2 > BO1 = 34 AB and the sum of the radii of circles S1 and S2 is
equal to 34 AB, it follows that these circles do not intersect. Therefore, if BM = qAM and
CM = qBM , then q < 2.
7.18. a) Let O be the intersection point of heights AA1 and BB1 . The points A1 and
B1 lie on the circle with diameter CO. Therefore, ∠AOB = 180◦ − ∠C. Hence, the locus to
be found is the circle symmetric to the given one through line AB (points A and B should
be excluded).
b) If O is the intersection point of the bisectors of triangle ABC, then ∠AOB = 90◦ +
1
∠C. On each of the two arcs ⌣ AB the angles C are constant and, therefore, the desired
2
locus of the vertices of angles of 90◦ + 12 ∠C that subtend segment AB is the union of two
arcs (points A and B should be excluded).
7.19. Points P and Q lie on the circle with diameter DX, hence,
∠(QD, DX) = ∠(QP, P X) = ∠(AP, P B) = 45◦ ,
i.e., point X lies on line CD.
7.20. a) If point A1 traverses along the circle an arc of value 2ϕ, then point B1 also
traverses an arc of value 2ϕ, consequently, lines AA1 and BB1 turn through an angle of ϕ
and the angle between them will not change.
Hence, point M moves along a circle that contains points A and B.
b) Let at some moment lines AA1 , BB1 and CC1 meet at point P . Then, for instance,
the intersection point of lines AA1 and BB1 moves along the circumscribed circle of triangle
ABP . It is also clear that the circumscribed circles of triangles ABP , BCP and CAP have
a unique common point, P .
7.21. Suppose that points A and C lie on opposite sides of a rectangle. Let M and N be
the midpoints of segments AC and BD, respectively. Let us draw through point M line l1
parallel to the sides of the rectangle on which points A and C lie and through point N line l2
parallel to the sides of the rectangle on which points B and D lie. Let O be the intersection
point of lines l1 and l2 . Clearly, point O lies on circle S constructed on segment M N as on
a diameter.
On the other hand, point O is the center of the rectangle. Clearly, the rectangle can be
constructed for any point O that lies on circle S.
It remains to notice that on the opposite sides of the rectangle points A and B or A and
D can also lie. Hence, the locus to be found is the union of three circles.
7.22. It is easy to verify that the points of heights of triangle ABC possess the required
property. Suppose that a point X not belonging to any of the heights of triangle ABC
possesses the required property. Then line BX intersects heights AA1 and CC1 at points
X1 and X2 . Since
∠XAB + ∠XBC + ∠XCA = 90◦ = ∠X1 AB + ∠X1 BC + ∠X1 CA,
it follows that
∠XAB − ∠X1 AB = ∠X1 CA − ∠XCA,
i.e., ∠(XA, AX1 ) = ∠(X1 C, CX). Therefore, point X lies on the circumscribed circle of
triangle AXC ′ , where point C ′ is symmetric to C through line BX. We similarly prove
that point X2 lies on the circle and, therefore, line BX intersects this circle at three distinct
points. Contradiction.
SOLUTIONS
177
7.23. Let K be the tangent point of line M X with the given semicircle and P the
projection of point M to the diameter. In right triangles M P X and OKX, the hypothenuses
are equal and ∠P XM = ∠OXK; hence, these triangles are equal. In particular, M P =
KO = R, where R is the radius of the given semicircle. It follows that point M lies on line
l parallel to the diameter of the semicircle and tangent to the semicircle. Let AB be the
segment of line l whose projection is the diameter of the semicircle. From a point on l that
does not belong to segment AB a tangent to the given semicircle cannot be drawn because
the tangent drawn to the circle should be tangent to the other semicircle as well.
The locus to be found is punctured segment AB: without points A, B, and the midpoint.
7.24. Let H be the base of height hb of triangle ABC and hb = b. Denote by B ′
the intersection point of the perpendicular to line AB drawn through point A and the
perpendicular to line AH drawn through point C. Right triangles AB ′ C and BAH are
equal, becuase ∠AB ′ C = ∠BAH and ∠AC = BH. Therefore, AB ′ = AB, i.e., point C lies
on the circle with diameter AB ′ .
Figure 81 (Sol. 7.24)
Let S1 and S2 be the images of circle S with diameter AB under the rotations through
angles of ±90◦ with center at A (Fig. 81). We have proved that point C 6= A belongs to the
union of circles S1 and S2 .
Conversely, let a point C, C 6= A, belong to either of the circles S1 or S2 ; let AB ′ be
a diameter of the corresponding circle. Then ∠AB ′ C = ∠HAB and A′ B = AB; hence,
AC = HB.
7.25. Let O be the center of the circle, N the intersection point of lines OM and QP .
Let us drop from point M perpendicular M S to line OP . Since △ON Q ∼ △OQM and
△OP N ∼ △OM S, we derive that
ON : OQ = OQ : OM
and OP : ON = OM : OS.
By multiplying these equalities we get OP : OQ = OQ : OS. Hence, OS = OQ2 : OP is a
constant. Since point S lies on line OP , its position does not depend on the choice of point
Q. The locus to be found is the line perpendicular to line OP and passing through point S.
7.26. Let O be the midpoint of segment AB, and M the intersection point of the medians
of triangle ABC. The homothety with center O and coefficient 31 sends point C to point M .
Therefore, the intersection point of the medians of triangle ABC lies on circle S which is
the image of the initial circle under the homothety with center O and coefficient 31 . To get
the desired locus we have to delete from S the images of points A and B.
178
CHAPTER 7. LOCI
7.27. Let O be the midpoint of height BH; let M , D and E be the midpoints of segment
AC, and sides RQ and P S, respectively (Fig. 82).
Figure 82 (Sol. 7.27)
Points D and E lie on lines AO and CO, respectively. The midpoint of segment DE is
the center of rectangle P QRS. Clearly, this midpoint lies on segment OM . The locus in
question is segment OM without its endpoints.
7.28. Let O1 and O2 be the centers of the given circles (point P lies on the circle centered
at O1 ); O the midpoint of segment O1 O2 ; P ′ , Q′ and O′ the projections of points O1 , O2 and
O to line P Q. As line P Q rotates, point O′ runs the circle S with diameter AO. Clearly, the
homothety with center A and coefficient 2 sends segment P ′ Q′ to segment P Q, i.e., point O′
turns into the midpoint of segment P Q. Hence, the locus in question is the image of circle
S under this homothety.
7.29. Let P and Q be the centers of the circumscribed circles of triangles AM B and
CM B. Point M belongs to the locus to be found if BP M Q is a rhombus, i.e., point M is
the image of the midpoint of segment P Q under the homothety with center B and coefficient
2. Since the projections of points P and Q to line AC are the midpoints of segments AB
and BC, respectively, the midpoints of all segments P Q lie on one line. (The locus to be
found is the above-obtained line without the intersection point with line AC.)
7.30. Point P passes through point O at time t1 , it passes point Q at time t2 . At
time 12 (t1 + t2 ) the distances from O to points P and Q are equal to 12 |t1 + t2 |v. At this
moment erect the perpendiculars to the lines at points P and Q. It is easy to verify that the
intersection point of these perpendiculars is the required one.
7.31. Denote the midpoints of diagonals AC and BD of quadrilateral ABCD by M and
N , respectively. Clearly, SAM B = SBM C and SAM D = SDM C , i.e., SDABM = SBCDM . Since
the areas of quadrilaterals DABM and BCDM do not vary as point M moves parallelly to
BD, it follows that SDABO = SCDAO . Similar arguments for point N show that SABCO =
SCDAO . Hence,
SADO + SABO = SBCO + SCDO and SABO + SBCO = SCDO + SADO
and, therefore,
SADO = SBCO = S1 and SABO = SCDO = S2 ,
i.e., the area of each of the four parts into which the segments that connect point O with
the midpoints of sides of the quadrilateral divide it is equal to 12 (S1 + S2 ).
7.32. Let us drop height BB1 from point B. Then AD = B1 D and CE = B1 E. Clearly,
if M D < AD, then point M lies on segment AB1 , i.e., outside segment B1 C. Therefore,
M E > EC.
7.33. Suppose that the distance from any vertex of the polygon to point Q is not shorter
than to point P . Then all the vertices of the polygon lie in the same half plane determined by
the perpendicular to segment P Q at point P ; point Q lies in the other half plane. Therefore,
point Q lies outside the polygon. This contradicts the hypothesis.
SOLUTIONS
179
7.34. Let us find the locus of points M for which M A > M B and M A > M C. Let us
draw midperpendiculars l1 and l2 to segments AB and AC. We have M A > M B for the
points that lie inside the half-plane bounded by line l1 and the one without point A. Therefore, the locus in question is the intersection of half-planes (without boundaries) bounded
by lines l1 and l2 and not containing point A.
If points A, B and C do not lie on one line, then this locus is always nonempty. If A, B,
C lie on one line but A does not lie on segment BC, then this locus is also nonempty. If point
A lies on segment BC, then this locus is empty, i.e., for any point M either M A ≤ M B or
M A ≤ M C.
7.35. Let O be the midpoint of diagonal AC. The projections of points B and D to line
AC lie on segment AO, hence, the projection of point M also lies on segment AO.
7.36. Let us draw the midperpendicular l to segment AO. Clearly, AM ≥ OM if and
only if point M lies on the same side of line l as O (or lies on line l itself). Therefore, the
locus in question is the rhombus formed by the midperpendiculars to segments OA, OB,
OC and OD.
7.37. The locus to be found is shaded on Fig. 83 (the boundary belongs to the locus).
Figure 83 (Sol. 7.37)
7.38. Let A1 and B1 be the midpoints of sides CB and AC, respectively. The locus to
be found is the interior of quadrilateral OA1 CB1 .
Figure 84 (Sol. 7.39)
7.39. Let us draw the common tangents to given disks (Fig. 84). It is easy to verify
that the points that belong to the shaded domains (but not to their boundaries) satisfy the
required condition and the points that do not belong to these domains do not satisfy this
condition.
7.40. Let the perpendiculars dropped from points A1 , B1 , C1 to lines BC, CA, AB,
respectively, intersect at point M . Since points B1 and M lie on one perpendicular to line
AC, we have
B1 A2 − B1 C 2 = M A2 − M C 2 .
180
CHAPTER 7. LOCI
Similarly,
C1 B 2 − C1 A2 = M B 2 − M A2 and A1 C 2 − A1 B 2 = M C 2 − M B 2 .
Adding these equalities we get
A1 B 2 + C1 A2 + B1 C 2 = B1 A2 + A1 C 2 + C1 B 2 .
(∗)
Conversely, let (∗) hold. Denote the intersection point of the perpendiculars dropped from
points A1 and B1 to lines BC and AC, respectively, by M . Let us draw through point M
line l perpendicular to line AB. If C1′ is a point on line l, then by the above
A1 B 2 + C1′ A2 + B1 C 2 = B1 A2 + A1 C 2 + C1′ B 2 .
Hence, C1′ A2 − C1′ B 2 = C1 A2 − C1 B 2 . By Problem 7.6 the locus of points X for which
XA2 − XB 2 = k is a line perpendicular to segment AB. Therefore, the perpendicular
dropped from point C1 to line AB passes through point M , as required.
7.41. Set A1 = A, B1 = B and C1 = C. From the obvious identity
AB 2 + CA2 + BC 2 = BA2 + AC 2 + CB 2
we derive that the heights dropped from points A, B and C to sides BC, CA and AB,
respectively, intersect at one point.
7.42. It suffices to make use of the result of Problem 7.40.
7.43. a) This problem is an obvious corollary of Problem 7.40.
b) Let the rotation by 90◦ about a point send triangle A1 B1 C1 to triangle A2 B2 C2 .
The perpendiculars to sides of triangle A2 B2 C2 are parallel to the corresponding sides of
triangle A1 B1 C1 and, therefore, the perpendiculars dropped from the vertices of triangle
ABC to the corresponding sides of triangle A2 B2 C2 intersect at one point. It follows that
the perpendiculars dropped from the vertices of triangle A2 B2 C2 to the corresponding sides
of triangle ABC intersect at one point. It remains to notice that the rotation by 90◦ that
sends triangle A2 B2 C2 to triangle A1 B1 C1 sends these perpendiculars to the lines that pass
through the sides of triangle A1 B1 C1 parallelly the corresponding sides of triangle ABC.
7.44. We have to find out when the identity
AB12 + BC12 + CA21 = BA21 + CB12 + AC12
holds. By subtracting AA22 + BB22 + CC22 from both sides of this identity we get
A2 B12 + B2 C12 + C2 A21 = B2 A21 + C2 B12 + A2 C12 ,
i.e.,
(b1 − a2 )2 + (c1 − b2 )2 + (a1 − c2 )2 = (a1 − b2 )2 + (b1 − c2 )2 + (c1 − a2 )2 ,
where ai , bi and ci are the coordinates of points Ai , Bi and Ci on line l. After simplification
we get
a2 b 1 + b 2 c 1 + c 2 a1 = a1 b 2 + b 1 c 2 + c 1 a2
and, therefore,
(b2 − a2 )(c1 − b1 ) = (b1 − a1 )(c2 − b2 ),
i.e., A2 B2 : B2 C2 = A1 B1 : B1 C1 .
7.45. We may assume that the length of a side of the given equilateral triangle is equal
to 2. Let P A = 2a, P B = 2b and P C = 2c; let A1 , B1 and C1 be the projections of the
SOLUTIONS
181
centers of the inscribed circles of triangles P BC, P CA and P AB to lines BC, CA and AB,
respectively. By Problem 3.2 we have
AB12 + BC12 + CA21 = (1 + a − c)2 + (1 + b − a)2 + (1 + c − b)2 =
3 + (a − c)2 + (b − a)2 + (c − b)2 = BA21 + CB12 + AC12 .
7.46. The segments into which the bisectors divide the sides of the triangle are easy to
calculate. As a result we see that if the perpendiculars raised from the bases of the bisectors
intersect, then
ac
ab
bc
ab
bc
ac
( + c)2 + ( + c)2 + ( + b)2 = ( + c)2 + ( + c)2 + ( + b)2 ,
b
a
a
b
a
a
i.e.,
a−c
b−a
a2 + b 2 + c 2
c−b
+ b2
+ c2
= −(b − a)(a − c)
.
0 = a2
b+c
a+c
a+b
(a + b)(a + c)(b + c)
7.47. Let (ai , bi ) be the coordinates of point Ai and (x, y) the coordinates of point X.
Then the equation satisfied by point X takes the form
c=
X
ki ((x − ai )2 + (x − bi )2 ) =
X
X
X
X
(
ki )(x2 + y 2 ) − (2
ki ai )x − (2
ki bi )y +
ki (a2i + b2i ).
If the coefficient of x2 + y 2 is nonzero, then this equation determines either a circle or the
empty set and if it is zero, then the equation determines either a line, or a plane, or the
empty set.
Remark. If in case a) points A1 , . . . , An lie on one line l, then this line can be taken for
Ox-axis. Then bi = 0 and, therefore, the coefficient of y is equal to zero, i.e., the center of
the circle lies on l.
7.48. Let line l cut on the given circles arcs ⌣ A1 B1 and ⌣ A2 B2 whose values are 2α1
and 2α2 , respectively; let O1 and O2 be the centers of the circles, R1 and R2 their respective
radii. Let K be the intersection point of the tangents at points A1 and A2 . By the law of
sines KA1 : KA2 = sin α2 : sin α1 , i.e., KA1 sin α1 = KA2 sin α2 . Since
KO12 = KA21 + R12
and KO22 = KA22 + R22 ,
it follows that
(sin2 α2 )KO12 − (sin2 α2 )KO22 = (R1 sin α1 )2 − (R2 sin α2 )2 = q.
We similarly prove that the other intersection points of the tangents belong to the locus of
points X such that
(sin2 α1 )XO12 − (sin2 α2 )XO22 = q.
This locus is a circle whose center lies on line O1 O2 (cf. Remark to Problem 7.47).
7.49. Let AM : BM : CM = p : q : r. All the points X that satisfy
(q 2 − r2 )AX 2 + (r2 − p2 )BX 2 + (p2 − q 2 )CX 2 = 0
lie on one line (cf. Problem 7.47) and points M , N and O satisfy this relation.
Chapter 8. CONSTRUCTIONS
§1. The method of loci
8.1. Construct triangle ABC given a, ha and R.
8.2. Inside triangle ABC construct point M so that SABM : SBCM : SACM = 1 : 2 : 3.
8.3. Through given point P inside a given circle draw a chord so that the difference of
the lengths of the segments into which P divides the chord would be equal to the given value
a.
8.4. Given a line and a circle without common points, construct a circle of a given radius
r tangent to them.
8.5. Given point A and circle S draw a line through point A so that the chord cut by
circle S on this line would be of given length d.
8.6. Quadrilateral ABCD is given. Inscribe in it a parallelogram with given directions
of sides.
§2. The inscribed angle
8.7. Given a, mc and angle ∠A, construct triangle ABC.
8.8. A circle and two points A and B inside it are given. Inscribe a right triangle in the
circle so that the legs would pass through the given points.
8.9. The extensions of sides AB and CD of rectangle ABCD intersect a line at points
M and N , respectively, and the extensions of sides AD and BC intersect the same line at
points P and Q, respectively. Construct rectangle ABCD given points M , N , P , Q and the
length a of side AB.
8.10. Construct a triangle given its bisector, median and height drawn from one vertex.
8.11. Construct triangle ABC given side a, angle ConstructatriangleA and the radius
r of the inscribed circle.
§3. Similar triangles and a homothety
8.12. Construct a triangle given two angles ∠A, ∠B and the perimeter P .
8.13. Construct triangle ABC given ma , mb and mc .
8.14. Construct triangle ABC given ha , hb and hc .
8.15. In a given acute triangle ABC inscribe square KLM N so that vertices K and N
lie on sides AB and AC and vertices L and M lie on side BC.
8.16. Construct triangle ABC given ha , b − c and r.
Cf. also Problems 19.15-19.20, 19.39, 19.40.
§4. Construction of triangles from various elements
In the problems of this section it is necessary to construct triangle ABC given the
elements indicated below.
8.17. c, ma and mb .
8.18. a, b and ha .
183
184
CHAPTER 8. CONSTRUCTIONS
8.19.
8.20.
8.21.
8.22.
8.23.
8.24.
8.25.
8.26.
hb , hc and ma .
∠A, hb and hc .
a, hb and mb .
ha , ma and hb .
a, b and mc .
ha , ma and ∠A.
a, b and lc .
∠A, ha and p.
See also Problems 17.6-17.8.
§5. Construction of triangles given various points
8.27. Construct triangle ABC given (1) line l containing side AB and (2) bases A1 and
B1 of heights dropped on sides BC and AC, respectively.
8.28. Construct an equilateral triangle given the bases of its bisectors.
8.29. a) Construct triangle ABC given three points A′ , B ′ , C ′ at which the bisectors of
the angles of triangle ABC intersect the circumscribed circle (both triangles are supposed
to be acute ones).
b) Construct triangle ABC given three points A′ , B ′ , C ′ at which the heights of the
triangle intersect the circumscribed circle (both triangles are supposed to be acute ones).
8.30. Construct triangle ABC given three points A′ , B ′ , C ′ symmetric to the center O
of the circumscribed circle of this triangle through sides BC, CA, AB, respectively.
8.31. Construct triangle ABC given three points A′ , B ′ , C ′ symmetric to the intersection
point of the heights of the triangle through sides BC, CA, AB, respectively (both triangles
are supposed to be acute ones).
8.32. Construct triangle ABC given three points P , Q, R at which the height, the
bisector and the median drawn from vertex C, respectively, intersect the circumscribed
circle.
8.33. Construct triangle ABC given the position of points A1 , B1 , C1 that are the
centeres of the escribed circles of triangle ABC.
8.34. Construct triangle ABC given the center of the circumscribed circle O, the intersection point of medians, M , and the base H of height CH.
8.35. Construct triangle ABC given the centers of the inscribed, the circumscribed, and
one of the escribed circles.
§6. Triangles
8.36. Construct points X and Y on sides AB and BC, respectively, of triangle ABC so
that AX = BY and XY k AC.
8.37. Construct a triangle from sides a and b if it is known that the angle opposite one
of the sides is three times the angle opposite the other side.
8.38. In given triangle ABC inscribe rectangle P RQS (vertices R and Q lie on sides
AB and BC and vertices P and S lie on side AC) so that its diagonal would be of a given
length.
8.39. Through given point M draw a line so that it would cut from the given angle with
vertex A a triangle ABC of a given perimeter 2p.
8.40. Construct triangle ABC given its median mc and bisector lc if ∠C = 90◦ .
8.41. Given triangle ABC such that AB < BC, construct on side AC point D so that
the perimeter of triangle ABD would be equal to the length of side BC.
§8. CIRCLES
185
8.42. Construct triangle ABC from the radius of its circumscribed circle and the bisector
of angle ∠A if it is known that ∠B − ∠C = 90◦ .
8.43. On side AB of triangle ABC point P is given. Draw a line (distinct from AB)
through point P that cuts rays CA and CB at points M and N , respectively, such that
AM = BN .
8.44. Construct triangle ABC from the radius of the inscribed circle r and (nonzero)
lengths of segments AO and AH, where O is the center of the inscribed circle and H the
orthocenter.
See also Problems 15.12 b), 17.12-17.15, 18.10, 18.29.
§7. Quadrilaterals
8.45. Construct a rhombus two sides of which lie on two given parallel lines and two
other sides pass through two given points.
8.46. Construct quadrilateral ABCD given the lengths of the four sides and the angle
between AB and CD.
8.47. Through vertex A of convex quadrilateral ABCD draw a line that divides ABCD
into two parts of equal area.
8.48. In a convex quadrilateral three sides are equal. Given the midpoints of the equal
sides construct the quadrilateral.
8.49. A quadrilateral is both inscribed and circumscribed. Given three of its vertices,
construct its fourth vertex.
8.50. Given vertices A and C of an isosceles circumscribed trapezoid ABCD (AD k BC)
and the directions of its bases, construct vertices B and D.
8.51. On the plane trapezoid ABCD is drawn (AD k BC) and perpendicular OK from
the intersection point O is dropped on base AD; the midpoint EF is drawn. Then the
trapezoid itself was erased. How to recover the plot of the trapezoid from the remaining
segments OK and EF ?
8.52. Construct a convex quadrilateral given the lengths of all its sides and one of the
midlines.
8.53. (Brachmagupta.) Construct an inscribed quadrilateral given its four sides.
See also Problems 15.10, 15.13, 16.17, 17.4, 17.5.
§8. Circles
8.54. Inside an angle two points A and B are given. Construct a circle that passes
through these points and intercepts equal segments on the sides of the angle.
8.55. Given circle S, point A on it and line l. Construct a circle tangent to the given
circle at point A and tangent to the given line.
8.56. a) Two points, A, B and line l are given. Construct a circle that passes through
point A, B and is tangent to l.
b) Two points A, B and circle S are given. Construct a circle that passes through points
A and B and is tangent to S.
8.57. Three points A, B and C are given. Construct three circles that are pairwise
tangent at these points.
8.58. Construct a circle the tangents to which drawn from three given points A, B and
C have given lengths a, b and c, respectively.
See also Problems 15.8, 15.9, 15.11, 15.12 a), 16.13, 16.14, 16.18–16.20, 18.24.
186
CHAPTER 8. CONSTRUCTIONS
§9. Apollonius’ circle
8.59. Construct triangle ABC given a, ha and cb .
8.60. Construct triangle ABC given the length of bisector CD and the lengths of
segments AD and BD into which the bisector divides side AB.
8.61. On a line four points A, B, C, D are given in the indicated order. Construct point
M — the vertex of equal angles that subtend segments AB, BC, CD.
8.62. Two segments AB and A′ B ′ are given in plane. Construct point O so that triangles
AOB and A′ OB ′ would be similar (equal letters stand for the corresponding vertices of
similar triangles).
8.63. Points A and B lie on a diameter of a given circle. Through A and B draw two
equal chords with a common endpoint.
§10. Miscellaneous problems
8.64. a) On parallel lines a and b, points A and B are given. Through a given point
C draw line l that intersects lines a and b at points A1 and B1 , respectively, and such that
AA1 = BB1 .
b) Through point C draw a line equidistant from given points A and B.
8.65. Construct a regular decagon.
8.66. Construct a rectangle with the given ratio of sides knowing one point on each of
its sides.
8.67. Given diameter AB of a circle and point C on the diameter. On this circle,
construct points X and Y symmetric through line AB and such that lines AX and Y C are
perpendicular.
See also Problems 15.7, 16.15, 16.16, 16.21, 17.9–17.11, 17.27–17.29, 18.41.
§11. Unusual constructions
8.68. With the help of a ruler and a compass divide the angle of 19◦ into 19 equal parts.
8.69. Prove that an angle of value n◦ , where n is an integer not divisible by 3, can be
divided into n equal parts with the help of a compass and ruler.
8.70. On a piece of paper two lines are drawn. They form an angle whose vertex lies
outside this piece of paper. With the help of a ruler and a compass draw the part of the
bisector of the angle that lies on this piece of paper.
8.71. With the help of a two-sided ruler construct the center of the given circle whose
diameter is greater than the width of the ruler.
8.72. Given points A and B; the distance between them is greater than 1 m. The length
of a ruler is 10 cm. With the help of the ruler only construct segment AB. (Recall that with
the help of a ruler one can only draw straight lines.)
8.73. On a circle of radius a a point is given. With the help of a coin of radius a
construct the point diametrically opposite to the given one.
§12. Construction with a ruler only
In the problems of this section we have to perform certain constructions with the help of
a ruler only, without a compass or anything else. With the help of one ruler it is almost impossible to construct anything. For example, it is even impossible to construct the midpoint
of a segment (Problem 30.59).
But if certain additional lines are drawn on the plane, it is possible to perform certain
constructions. In particular, if an additional circle is drawn on the plane and its center is
§13. CONSTRUCTIONS WITH THE HELP OF A TWO-SIDED RULER
187
marked, then with the help of a ruler one can perform all the constructions that can be
performed with the help of a ruler and a compass. One has, however, to convene that a
circle is “constructed” whenever its center and one of its points are marked.
Remark. If a circle is drawn on the plane but its center is not marked then to construct
its center with the help of a ruler only is impossible (Problem 30.60).
8.74. Given two parallel lines and a segment that lies on one of the given lines. Divide
the segment in halves.
8.75. Given two parallel lines and a segment that lies on one of the given lines. Double
the segment.
8.76. Given two parallel lines and a segment that lies on one of the given lines. Divide
the segment into n equal parts.
8.77. Given two parallel lines and point P , draw a line through P parallel to the given
lines.
8.78. A circle, its diameter AB and point P are given. Through point P draw the
perpendicular to line AB.
8.79. In plane circle S and its center O are given. Then with the help of a ruler only
one can:
a) additionally given a line, draw a line through any point parallel to the given line and
drop the perpendicular to the given line from this point;
b) additionally given a line a point on it and a length of a segment, on the given line,
mark a segment of length equal to the given one and with one of the endpoints in the given
point;
c) additionally given lengths of a, b, c of segments, construct a segment of length abc ;
d) additionally given line l, point A and the length r of a segment, construct the intersection points of line l with the circle whose center is point A and the radius is equal to
r;
e) additionally given two points and two segments, construct the intersection points of
the two circles whose centers are the given points and the radii are the given segments.
See also Problem 6.97.
§13. Constructions with the help of a two-sided ruler
In problems of this section we have to perform constructions with the help of a ruler
with two parallel sides (without a compass or anything else). With the help of a two-sided
ruler one can perform all the constructions that are possible to perform with the help of a
compass and a ruler.
Let a be the width of a two-sided ruler. By definition of the two-sided ruler with the
help of it one can perform the following elementary constructions:
1) draw the line through two given points;
2) draw the line parallel to a given one and with the distance between the lines equal to
a;
3) through two given points A and B, where AB ≥ a, draw a pair of parallel lines the
distance between which is equal to a (there are two pairs of such lines).
8.80. a) Construct the bisector of given angle ∠AOB.
b) Given acute angle ∠AOB, construct angle ∠BOC whose bisector is ray OA.
8.81. Erect perpendicular to given line l at given point A.
8.82. a) Given a line and a point not on the line. Through the given point draw a line
parallel to the given line.
188
CHAPTER 8. CONSTRUCTIONS
b) Construct the midpoint of a given segment.
8.83. Given angle ∠AOB, line l and point P on it, draw through P lines that form
together with l an angle equal to angle ∠AOB.
8.84. Given segment AB, a non-parallel to it line l and point M on it, construct the
intersection points of line l with the circle of radius AB centered at M .
8.85. Given line l and segment OA, parallel to l, construct the intersection points of l
with the circle of radius OA centered at O.
8.86. Given segments O1 A1 and O2 A2 , construct the radical axis of circles of radii O1 A1
and O2 A2 centered at O1 and O2 , respectively.
§14. Constructions using a right angle
In problems of this section we have to perform the constructions indicated using a right
angle. A right angle enables one to perform the following elementary constructions:
a) given a line and a point not on it, place the right angle so that one of its legs lies on
the given line and the other leg runs through the given point;
b) given a line and two points not on it, place the right angle so that its vertex lies on
the given line and thelegs pass through two given points (if, certainly, for the given line and
points such a position of the right angle exists).
Placing the right angle in one of the indicated ways we can draw rays corresponding to
its sides.
8.87. Given line l and point A not on it, draw a line parallel to l.
8.88. Given segment AB, construct
a) the midpoint of AB;
b) segment AC whose midpoint is point B.
8.89. Given angle ∠AOB, construct
a) an angle of value 2∠AOB;
b) an angle of value 12 ∠AOB.
8.90. Given angle ∠AOB and line l, draw line l1 so that the angle between lines l and
l1 is equal to ∠AOB.
8.91. Given segment AB, line l and point O on it, construct on l point X such that
OX = AB.
8.92. Given segment OA parallel to line l, construct the locus of points in which the
disc segment of radius OA centered at O intersects l.
Problems for independent study
8.93. Construct a line tangent to two given circles (consider all the possible cases).
8.94. Construct a triangle given (the lengths of) the segments into which a height divides
the base and a median drawn to a lateral side.
8.95. Construct parallelogram ABCD given vertex A and the midpoints of sides BC
and CD.
8.96. Given 3 lines, a line segment and a point. Construct a trapezoid whose lateral
sides lie on the given lines, the diagonals intersect at the given point and one of the bases is
of the given length.
8.97. Two circles are given. Draw a line so that it would be tangent to one of the circles
and the other circle would intersept on it a chord of a given length.
8.98. Through vertex C of triangle ABC draw line l so that the areas of triangles AA1 C
and BB1 C, where A1 and B1 are projections of points A and B on line l, are equal.
SOLUTIONS
189
8.99. Construct triangle ABC given sides AB and AC if it is given that bisector AD,
median BM , and height CH meet at one point.
8.100. Points A1 , B1 and C1 that divide sides BC, CA and AB, respectively, of triangle
ABC in the ratio of 1 : 2 are given. Recover triangle ABC from this data.
Solutions
8.1. Let us construct segment BC of length a. The center O of the circumscribed circle
of triangle ABC is the intersection point of two circles of radius R centered at B and C.
Select one of these intersection points and construct the circumscribed circle S of triangle
ABC. Point A is the intersection point of circle S and a line parallel line BC and whose
distance from BC is equal to ha (there are two such lines).
8.2. Let us construct points A1 and B1 on sides BC and AC, respectively, so that
BA1 : A1 C = 1 : 3 and AB1 : B1 C = 1 : 2. Let point X lie inside triangle ABC. Clearly,
SABX : SBCX = 1 : 2 if and only if point X lies on segment BB1 and SABX : SACX = 1 : 3
if and only if point X lies on segment AA1 . Therefore, the point M to be constucted is the
intersection point of segments AA1 and BB1 .
8.3. Let O be the center of the given circle, AB a chord that passes through point P
and M the midpoint of AB. Then |AP − BP | = 2P M . Since ∠P M O = 90◦ , point M lies
on circle S with diameter OP . Let us construct chord P M of circle S so that P M = 12 a
(there are two such chords). The chord to be constructed is determined by line P M .
8.4. Let R be the radius of the given circle, O its center. The center of the circle to be
constructed lies on circle S of radius R + r centered at O. On the other hand, the center
to be constructed lies on line l passing parallelly to the given line at distance r (there are
two such lines). Any intersection point of S with l can serve as the center of the circle to be
constructed.
8.5. Let R be the radius of circle S and O its center. If circle S intersepts on the line
that passes through point A chord P Q and M is the midpoint of P Q, then
OM 2 + OQ2 − N Q2 = R2 −
d2
.
4
q
2
Therefore, the line to be constructed is tangent to the circle of radius R2 − d4 centered at
O.
8.6. On lines AB and CD take points E and F so that lines BF and CE would have had
prescribed directions. Let us considered all possible parallelograms P QRS with prescribed
directions of sides whose vertices P and R lie on rays BA and CD and vertex Q lies on side
BC (Fig. 85).
Figure 85 (8.6)
190
CHAPTER 8. CONSTRUCTIONS
PQ
SR
= EC
= BQ
= FF CR ,
Let us prove that the locus of vertices S is segment EF . Indeed, EC
BC
′
i.e., point S lies on segment EF . Conversely if point S lies on segment EF then let us draw
lines S ′ P ′ , P ′ Q′ and Q′ R′ so that S ′ P ′ k BF , P ′ Q′ k EC and Q′ R′ k BF , where P ′ , Q′ and
′
′ ′
′ ′
′
R′ are some points on lines AB, BC, CD, respectively. Then SBFP = PBEE = QBCC = QBFR , i.e.,
S ′ P ′ = Q′ R′ and P ′ Q′ R′ S ′ is a parallelogram.
This implies the following construction. First, construct points E and F . Vertex S is the
intersection point of segments AD and EF . The continuation of construction is obvious.
8.7. Suppose that triangle ABC is constructed. Let A1 and C1 be the midpoints of sides
CD and AB, respectively. Since C1 A1 k AC, it follows that ∠A1 C1 B = ∠A. This implies
the following construction.
First, let us construct segment CD of length a and its midpoint, A1 . Point C1 is the
intersection point of the circle of radius mc centered at C and the arcs of the circles whose
points are vertices of the angles equal to ∠A that segment A1 B subtends. Construct point
C1 , then mark on ray BC1 segment BA = 2BC1 . Then A is the vertex of the triangle to be
constructed.
8.8. Suppose that the desired triangle is constructed and C is the vertex of its right
angle. Since ∠ACB = 90◦ , point C lies on circle S with diameter AB. Hence, point C is
the intersection point of circle S and the given circle. Constructing point C and drawing
lines CA and AB, we find the remaining vertices of the triangle to be constructed.
8.9. Suppose that rectangle ABCD is constructed. Let us drop perpendicular P R from
point P to line BC. Point R can be constructed because it lies on the circle with diameter
P Q and P R = AB = a. Constructing point R, let us construct lines BC and AD and drop
on them perpendiculars from points M and N , respectively.
8.10. Suppose that triangle ABC is constructed, AH is its height, AD its bisector, AM
its median. By Problem 2.67 point D lies between M and H. Point E, the intersection point
of line AD with the perpendicular drawn from point M to side BC, lies on the circumscribed
circle of triangle ABC. Hence, the center O of the circumscribed circle lies on the intersection
of the midperpendicular to segment AE and the perpendicular to side BC drawn through
point M .
The sequence of constructions is as follows: on an arbitrary line (which in what follows
turns out to be line BC) construct point H, then consecutively construct points A, D, M ,
E, O. The desired vertices B and C of triangle ABC are intersection points of the initial
line with the circle of radius OA centered at O.
8.11. Suppose that triangle ABC is constructed and O is the center of its inscribed
circle. Then ∠BOC = 90◦ + 12 ∠A (Problem 5.3). Point O is the vertex of an angle of
90◦ + 21 ∠A that subtends segment BC; the distance from O to line BC is equal to r, hence,
BC(??) can be constructed. Further, let us construct the inscribed circle and draw the
tangents to it from points B and C.
8.12. Let us construct any triangle with angles ∠A and ∠B and find its perimeter P1 .
The triangle to be found is similar to the constructed triangle with coefficient PP1 .
8.13. Suppose that triangle ABC is constructed. Let AA1 , BB1 and CC1 be its medians,
M their intersection point, M ′ the point symmetric to M through point A1 . Then M M ′ =
2
m , M C = 23 mc and M ′ C = 32 mb ; hence, triangle M M ′ C can be constructed. Point A is
3 a
symmetric to M ′ through point M and point B is symmetric to C through the midpoint of
segment M M ′ .
8.14. Clearly,
BC : AC : AB =
1 1 1
S S S
:
:
=
:
: .
ha hb hc
ha hb hc
SOLUTIONS
191
Let us take an arbitrary segment B ′ C ′ and construct triangle A′ B ′ C ′ so that B ′ C ′ : A′ C ′ =
hb : ha and B ′ C ′ : A′ B ′ = hc : ha . Let h′a be the height of triangle A′ B ′ C ′ dropped from
vertex A′ . The triangle to be found is similar to triangle A′ B ′ C ′ with coefficient hha′ .
a
8.15. On side AB, take an arbitrary point K ′ and drop from it perpendicular K ′ L′
to side BC; then construct square K ′ L′ M ′ N ′ that lies inside angle ∠ABC. Let line BN ′
intersect side AC at point N . Clearly, the square to be constructed is the image of square
K ′ L′ M ′ N ′ under the homothety with center B and coefficient BN : BN ′ .
8.16. Suppose that the desired triangle ABC is constructed. Let Q be the tangent point
of the inscribed circle with side BC; let P Q be a diameter of the circle, R the tangent point
of an escribed circle with side BC. Clearly,
a+b+c
a+b−c
a+c−b
BR =
−c=
and BQ =
.
2
2
2
Hence, RQ = |BR − BQ| = |b − c|. The inscribed circle of triangle ABC and the escribed
circle tangent to side BC are homothetic with A being the center of homothety. Hence,
point A lies on line P R (Fig. 86).
Figure 86 (Sol. 8.16)
This implies the following construction. Let us construct right triangle P QR from the
known legs P Q = 2r and RQ = |b − c|. Then draw two lines parallel to line RQ and whose
distances from RQ are equal to ha . Vertex A is the intersection point of one of these lines
with ray RP . Since the length of diameter P Q of the inscribed circle is known, it can be
constructed. The intersection points of the tangents to this circle drawn from point A with
line RQ are vertices B and C of the triangle.
8.17. Suppose that triangle ABC is constructed. Let M be the intersection point
of medians AA1 and BB1 . Then AM = 23 ma and BM = 23 mb . Triangle ABM can be
constructed from the lengths of sides AB = c, AM and BM . Then on rays AM and BM
segments AA1 = ma and BB1 = mb should be marked. Vertex C is the intersection point of
lines AB1 and A1 B.
8.18. Suppose triangle ABC is constructed. Let H be the base of the height dropped
from vertex A. Right triangle ACH can be constructed from its hypothenuse AC = b and
leg AH = ha . Then on line CH construct point B so that CB = a.
8.19. Suppose that triangle ABC is constructed. Let us draw from the midpoint A1 of
side BC perpendiculars A1 B ′ and A1 C ′ to lines AC and AB, respectively. Clearly, AA1 =
ma , A1 B ′ = 21 hb and A1 C ′ = 21 hc . This implies the following construction.
First, let us construct segment AA1 of length ma . Then construct right triangles AA1 B ′
and AA1 C ′ from the known legs and hypothenuse so that they would lie on distinct sides
of line AA1 . It remains to construct points B and C on sides AC ′ and AB ′ of angle C ′ AB ′
so that segment BC would be divided by points A1 in halves. For this let us mark on ray
AA1 segment AD = 2AA1 and then draw through point D the lines parallel to the legs of
192
CHAPTER 8. CONSTRUCTIONS
Figure 87 (Sol. 8.19)
angle ∠C ′ AB ′ . The intersection points of these lines with the legs of angle ∠C ′ AB ′ are the
vertices of the triangle to be constructed (Fig. 87).
8.20. Let us construct angle ∠B ′ AC ′ equal to ∠A. Point B is constructed as the
intersection of ray AB ′ with a line parallel to ray AC ′ and passsing at distance hb from it.
Point C is similarly constructed.
8.21. Suppose that triangle ABC is constructed. Let us drop height BH from point B
and draw median BB1 . In right triangles CBH and B1 BH, leg BH and hypothenuses CB
and B1 B are known; hence, these segments can be constructed. Then on ray CB1 we mark
segment CA = 2CB1 . The problem has two solutions because we can construct triangles
CBH and B1 BH either on one or on distinct sides of line BH.
8.22. Suppose that triangle ABC is constructed. Let M be the midpoint of segment
BC. From point A drop height AH and from point M drop perpendicular M D to side AC.
Clearly, M D = 12 hb . Hence, triangles AM D and AM H can be constructed.
Vertex C is the intersection point of lines AD and M H. On ray CM , mark segment
CB = 2CM . The problem has two solutions because triangles AM D and AM H can be
constructed either on one or on distinct sides of line AM .
8.23. Suppose that triangle ABC is constructed. Let A1 , B1 and C1 be the midpoints of
sides BC, CA and AB, respectively. In triangle CC1 B1 all the sides are known: CC1 = mc ,
C1 B1 = 21 a and CB1 = 12 b; hence, it can be constructed. Point A is symmetric to C through
point B1 and point B is symmetric to A through C1 .
8.24. Suppose that triangle ABC is constructed, AM is its median, AH its height. Let
point A′ be symmetric to A through point M .
Let us construct segment AA′ = 2ma . Let M be the midpoint of AA′ . Let us construct
right triangle AM H with hypothenuse AM and leg AH = ha . Point C lies on an arc of the
circle whose points are the vertices of the angles that subtend segment AA′ ; the values of
these angles are equal to 180◦ − ∠A because ∠ACA′ = 180◦ − ∠CAB. Hence, point C is the
intersection point of this arc and line M H. Point B is symmetric to C through point M .
8.25. Suppose triangle ABC is constructed. Let CD be its bisector. Let us draw line
M D parallel to side BC (point M lies on side AC). Triangle CM D is an isosceles one
because ∠M CD = ∠DCB = ∠M DC. Since
M C : AM = DB : AD = CB : AC = a : b and AM + M C = b,
ab
it follows that M C = a+b
. Let us construct an isosceles triangle CM D from its base CD = lc
ab
. Further, on ray CM , mark segment CA = b and on
and lateral sides M D = M C = a+b
the ray symmetric to ray CM through line CD mark segment CB = a.
8.26. Suppose that triangle ABC is constructed. Let S1 be the escribed circle tangent
to side BC. Denote the tangent points of circle S1 with the extensions of sides AB and AC
by K and L, respectively, and the tangent point of S1 with side BC by M . Since
AK = AL, AL = AC + CM
and AK = AB + BM,
SOLUTIONS
193
it follows that AK = AL = p. Let S2 be the circle of radius ha centered at A. Line BC is a
common inner tangent to circles S1 and S2 .
This implies the following construction. Let us construct angle ∠KAL whose value is
equal to that of A so that KA = LA = p. Next, construct circle S1 tangent to the sides of
angle ∠KAL at points K and L and circle S2 of radius ha centered at A. Then let us draw
a common inner tangent to circles S1 and S2 . The intersection points of this tangent with
the legs of angle ∠KAL are vertices B and C of the triangle to be constructed.
8.27. Points A1 and B1 lie on the circle S with diameter AB. The center O of this
circle lies on the midperpendicular to chord A1 B1 . This implies the following construction.
First, let us construct point O which is the intersection point of the midperpendicular to
segment A1 B1 with line l. Next, construct the circle of radius OA1 = OB1 centered at O.
The vertices A and B are the intersection points of circle S with line l. Vertex C is the
intersection point of lines AB1 and BA1 .
8.28. Let AB = BC and A1 , B1 , C1 the bases of the bisectors of triangle ABC. Then
∠A1 C1 C = ∠C1 CA = ∠C1 CA1 , i.e., triangle CA1 C1 is an isosceles one and A1 C = A1 C1 .
This implies the following construction.
Let us draw through point B1 line l parallel to A1 C1 . On l, construct point C such that
CA1 = C1 A1 and ∠C1 A1 C > 90◦ . Point A is symmetric to point C through point B1 and
vertex B is the intersection point of lines AC1 and A1 C.
8.29. a) By Problem 2.19 a) points A, B and C are the intersection points of the
extensions of heights of triangle A′ B ′ C ′ with its circumscribed circle.
b) By Problem 2.19 b) points A, B and C are the intersection points of the extensions
of bisectors of the angles of triangle A′ B ′ C ′ with its circumscribed circle.
8.30. Denote the midpoints of sides BC, CA, AB of the triangle by A1 , B1 , C1 , respectively. Since BC k B1 C1 k B ′ C ′ and OA1 ⊥ BC, it follows that OA′ ⊥ B ′ C ′ . Similarly,
OB ′ ⊥ A′ C ′ and OC ′ ⊥ A′ B ′ , i.e., O is the intersection point of the heights of triangle
A′ B ′ C ′ . Constructing point O, let us draw the midperpendiculars to segments OA′ , OB ′ ,
OC ′ . These lines form triangle ABC.
8.31. Thanks to Problem 5.9 our problem coincides with Problem 8.29 b).
8.32. Let O be the center of the circumscribed circle, M the midpoint of side AB and
H the base of the height dropped from point C. Point Q is the midpoint of arc ⌣ AB,
therefore, OQ ⊥ AB. This implies the following construction. First, the three given points
determine the circumscribed circle S of triangle P QR. Point C is the intersection point of
circle S and the line drawn parallelly to OQ through point P . Point M is the intersection
point of line OQ and line RC. Line AB passes through point M and is perpendicular to
OQ.
8.33. By Problem 5.2, points A, B and C are the bases of the heights of triangle A1 B1 C1 .
8.34. Let H1 be the intersection point of heights of triangle ABC. By Problem 5.105,
OM : M H1 = 1 : 2 and point M lies on segment OH1 . Therefore, we can construct point
H1 . Then let us draw line H1 H and erect at point H of this line perpendicular l. Dropping
perpendicular from point O to line l we get point C1 (the midpoint of segment AB). On ray
C1 M , construct point C so that CC1 : M C1 = 3 : 1. Points A and B are the intersection
points of line l with the circle of radius CO centered at O.
8.35. Let O and I be the centers of the circumscribed and inscribed circles, Ic the center
of the escribed circle tangent to side AB. The circumscribed circle of triangle ABC divides
segment IIc (see Problem 5.109 b)) in halves and segment IIc divides arc ⌣ AB in halves.
It is also clear that points A and B lie on the circle with diameter IIc . This implies the
following construction.
194
CHAPTER 8. CONSTRUCTIONS
Let us construct circle S with diameter IIc and circle S1 with center O and radius OD,
where D is the midpoint of segment IIc . Circles S and S1 intersect at points A and B. Now,
we can construct the inscribed circle of triangle ABC and draw tangents to it at points A
and B.
8.36. Suppose that we have constructed points X and Y on sides AB and BC, respectively, of triangle ABC so that AX = BY and XY k AC. Let us draw Y Y1 parallel to AB
and Y1 C1 parallel to BC (points Y1 and C1 lie on sides AC and AB, respectively). Then
Y1 Y = AX = BY , i.e., BY Y1 C is a rhombus and BY1 is the bisector of angle ∠B.
This implies the following construction. Let us draw bisector BY1 , then line Y1 Y parallel
to side AB (we assume that Y lies on BC). Now, it is obvious how to construct point X.
8.37. Let, for definiteness, a < b. Suppose that triangle ABC is constructed. On side
AC, take point D such that ∠ABD = ∠BAC. Then ∠BDC = 2∠BAC and
∠CBD = 3∠BAC − ∠BAC = 2∠BAC,
i.e., CD = CB = a. In triangle BCD all the sides are known: CD = CB = a and
DB = AD = b − a. Constructing triangle BCD, draw ray BA that does not intersect side
CD so that ∠DBA = 12 ∠DBC. Vertex A to be constructed is the intersection point of line
CD and this ray.
8.38. Let point B ′ lie on line l that passes through point B parallelly to AC. Sides of
triangles ABC and AB ′ C intersept equal segments on l. Hence, rectangles P ′ R′ Q′ S ′ and
P RQS inscribed in triangles ABC and AB ′ C, respectively, are equal if points R, Q, R′ and
Q′ lie on one line.
On line l, take point B ′ so that ∠B ′ AC = 90◦ . It is obvious how to inscribe rectangle
P ′ R′ Q′ S ′ with given diagonal P ′ Q′ in triangle AB ′ C (we assume that P ′ = A). Draw line
R′ Q′ ; we thus find vertices R and Q of the rectangle to be found.
8.39. Suppose that triangle ABC is constructed. Let K and L be points at which
the escribed circle tangent to side BC is tangent to the extensions of sides AB and AC,
respectively. Since AK = AL = p, this escribed circle can be constructed; it remains to
draw the tangent through the given point M to the constructed circle.
8.40. Let the extension of the bisector CD intersect the circumscribed circle of triangle
ABC (with right angle ∠C) at point P , let P Q be a diameter of the circumscribed circle
and O its center. Then P D√: P O = P Q : P C, i.e., P D · P C = 2R2 = 2m2c . Therefore,
drawing a tangent of length 2mc to the circle with diameter CD, it is easy to construct a
segment of length P C. Now, the lengths of all the sides of triangle OP C are known.
8.41. Let us construct point K on side AC so that AK = BC − AB. Let point D
lie on segment AC. The equality AD + BD + AB = BC is equivalent to the equality
AD + BD = AK. For point D that lies on segment AK the latter equality takes the
form AD + BD = AD + DK and for point D outside segment AK it takes the form
AD + BD = AD − DK. In the first case BD = DK and the second case is impossible.
Hence, point D is the intersection point of the midperpendicular to segment BK and segment
AC.
8.42. Suppose that triangle ABC is constructed. Let us draw diameter CD of the
circumscribed circle. Let O be the center of the circumscribed circle, L the intersection
point of the extension of the bisector AK with the circumscribed circle (Fig. 88). Since
∠ABC − ∠ACB = 90◦ , it follows that ∠ABD = ∠ACB; hence, ⌣ DA =⌣ AB. It is also
clear that ⌣ BL =⌣ LC. Therefore, ∠AOL = 90◦ .
This implies the following construction. Let us construct circle S with center O and a
given radius. On circle S select an arbitrary point A. Let us construct a point L on circle
S so that ∠AOL = 90◦ . On segment AL, construct segment AK whose length is equal to
SOLUTIONS
195
Figure 88 (Sol. 8.42)
that of the given bisector. Let us draw through point K line l perpendicular to OL. The
intersection points of l with circle S are vertices B and C of triangle ABC to be constructed.
8.43. On sides BC and AC, take points A1 and B1 such that P A1 k AC and P B1 k BC.
Next, on rays A1 B and B1 A mark segments A1 B2 = AB1 and B1 A2 = BA1 . Let us prove
AP
that line A2 B2 is the one to be found. Indeed, let k = AB
. Then
(1 − k)a
(1 − k)a + (1 − k)b
CA2
BA2
=
=
=
,
BP
ka
ka + kb
CB2
i.e., △A2 B1 P ∼ △A2 CB2 and line A2 B2 passes through point P . Moreover, AA2 = |(1 −
k)a − kb| = BB2 .
8.44. Suppose that triangle ABC is constructed. Let B1 be the tangent point of the
inscribed circle with side AC. In right triangle AOB1 leg OB1 = r and hypothenuse AO
are known, therefore, we can construct angle ∠OAB1 , hence, angle ∠BAC. Let O1 be the
center of the circumscribed circle of triangle ABC, let M be the midpoint of side BC. In
right triangle BO1 M leg O1 M = 12 AH is known (see solution to Problem 5.105) and angle
∠BO1 M is known (it is equal to either ∠A or 180◦ − ∠A);
p hence, it can be constructed.
Next, we can determine the length of segment OO1 = R(R − 2r), cf. Problem 5.11 a).
Thus, we can construct segments of length R and OO1 = d.
After this take segment AO and construct point O1 for which AO1 = R and OO1 = d
(there could be two such points). Let us draw from point A tangents to the circle of radius
r centered at O. Points B and C to be found lie on these tangents and their distance from
point O1 is equal to R; obviously, points B and C are distinct from point A.
8.45. Let the distance between the given parallel lines be equal to a. We have to draw
parallel lines through points A and B so that the distance between the lines is equal to
a. To this end, let us construct the circle with segment AB as its diameter and find the
intersection points C1 and C2 of this circle with the circle of radius a centered at B. A side
of the rhombus to be constructed lies on line AC1 (another solution: it lies on AC2 ). Next,
let us draw through point B the line parallel to AC1 (resp. AC2 ).
8.46. Suppose that quadrilateral ABCD is constructed. Let us denote the midpoints of
sides AB, BC, CD and DA by P , Q, R and S, respectively, and the midpoints of diagonals
AC and BD by K and L, respectively. In triangle KSL we know KS = 21 CD, LS = 21 AB
and angle ∠KSL equal to the angle between the sides AB and CD.
Having constructed triangle KSL, we can construct triangle KRL because the lengths of
all its sides are known. After this we complement triangles KSL and KRL to parallelograms
KSLQ and KRLP , respectively. Points A, B, C, D are vertices of parallelograms P LSA,
QKP B, RLQC, SKRD (Fig. 89).
8.47. Let us drop perpendiculars BB1 and DD1 from vertices B and D, respectively,
to diagonal AC. Let, for definiteness, DD1 > BB1 . Let us construct a segment of length
196
CHAPTER 8. CONSTRUCTIONS
Figure 89 (Sol. 8.46)
a = DD1 − BB1 ; draw a line parallel to line AC and such that the distance between this
line and AC is equal to a and which intersects side CD at a point, E. Clearly,
BB1
ED
SACD =
SACD = SABC .
SAED =
CD
DD1
Therefore, the median of triangle AEC lies on the line to be constructed.
8.48. Let P , Q, R be the midpoints of equal sides AB, BC, CD of quadrilateral
ABCD. Let us draw the midperpendiculars l1 and l2 to segments P Q and QR. Since
AB = BC = CD, it is clear that points B and C lie on lines l1 and l2 and BQ = QC.
This implies the following construction. Let us draw the midperpendiculars l1 and l2
to segments P Q and QR, respectively. Then through point Q we draw a segment with
endpoints on lines l1 and l2 so that Q were its midpoint, cf. Problem 16.15.
8.49. Let vertices A, B and C of quadrilateral ABCD which is both inscribed and
circumscribed be given and AB ≥ BC. Then AD − CD = AB − BC ≥ 0. Hence, on side
AD we can mark segment DC1 equal to DC. In triangle AC1 C the lengths of sides AC and
AC1 = AB − BC are known and ∠AC1 C = 90◦ + 12 ∠D = 180◦ − 12 ∠B. Since angle ∠AC1 C
is an obtuse one, triangle AC1 C is uniquely recoverable from these elements. The remaining
part of the construction is obvious.
8.50. Let ABCD be a circumscribed equilateral trapezoid with bases AD and BC such
that AD > BC; let C1 be the projection of point C to line AD. Let us prove that AB = AC1 .
Indeed, if P and Q are the tangent points of sides AB and AD with the inscribed circle,
then AB = AP + P B = AQ + 12 BC = AQ + QC1 = AC1 .
This implies the following construction. Let C1 be the projection of point C to base AD.
Then B is the intersection point of line BC and the circle of radius AC1 centered at A. A
trapezoid with AD < BC is similarly constructed.
8.51. Let us denote the midpoints of bases AD and BC by L and N and the midpoint
of segment EF by M . Points L, O, N lie on one line (by Problem 19.2). Clearly, point M
also lies on this line. This implies the following construction.
Let us draw through point K line l perpendicular to line OK. Base AD lies on l. Point
L is the intersection point of l and line OM . Point N is symmetric to point L through point
M . Let us draw lines through point O parallel to lines EN and F N . The intersection points
of the lines we have just drawn are vertices A and D of the trapezoid. Vertices B and C are
symmetric to vertices A and D through points E and F , respectively.
8.52. Suppose that we have constructed quadrilateral ABCD with given lengths of sides
and a given midline KP (here K and P are the midpoints of sides AB and CD, respectively).
Let A1 and B1 be the points symmetric to points A and B, respectively, through point P .
Triangle A1 BC can be constructed because its sides BC, CA1 = AD and BA1 = 2KP are
SOLUTIONS
197
known. Let us complement triangle A1 BC to parallelogram A1 EBC. Now we can construct
−−→ −−→
point D because CD and ED = BA are known. Making use of the fact that DA = A1 C we
construct point A.
8.53. Making use of the formulas of Problems 6.34 and 6.35 it is easy to express the
lengths of the diagonals of the inscribed triangle in terms of the lengths of its sides. The
obtained formulas can be applied for the construction of the diagonals (for convenience it is
advisable to introduce an arbitrary segment e as the measure of unit length and construct
√
√
segments of length pq, pq and p as pqe , pe
and pe).
q
8.54. A circle intercepts equal segments on the legs of an angle if and only if the center
of the circle lies on the bisector of the angle. Therefore, the center of the circle to be found is
the intersection point of the midperpendicular to segment AB and the bisector of the given
angle.
8.55. Let us suppose that we have constructed circle S ′ tangent to the given circle S at
point A and the given line l at a point, B. Let O and O′ be the centers of circles S and S ′ ,
respectively (Fig. 90). Clearly, points O, O′ and A lie on one line and O′ B = O′ A. Hence,
we have to construct point O′ on line OA so that O′ A = O′ B, where B is the base of the
perpendicular dropped from point O′ to line l.
Figure 90 (Sol. 8.55)
To this end let us drop perpendiculr OB ′ on line l. Next, on line AO mark segment OA′
of length OB ′ . Let us draw through point A line AB parallel to A′ B ′ (point B lies on line
l). Point O′ is the intersection point of line OA and the perpendicular to l drawn through
point B.
8.56. a) Let l1 be the midperpendicular to segment AB, let C be the intercection point
of lines l1 and l; let l′ be the line symmetric to l through line l1 . The problem reduces to
the necessity to construct a circle that passes through point A and is tangent to lines l and
l′ , cf. Problem 19.15.
b) We may assume that the center of circle S does not lie on the midperpendicular to
segment AB (otherwise the construction is obvious). Let us take an arbitrary point C on
circle S and construct the circumscribed circle of triangle ABC; this circle intersects S at a
point D. Let M be the intersection point of lines AB and CD. Let us draw tangents M P
and M Q to circle S. Then the circumscribed circles of triangles ABP and ABQ are the
ones to be found since M P 2 = M Q2 = M A · M B.
8.57. Suppose we have constructed circles S1 , S2 and S3 tangent to each other pairwise
at given points: S1 and S2 are tangent at point C; circles S1 and S3 are tangent at point
B; circles S2 and S3 are tangent at point A. Let O1 , O2 and O3 be the centers of circles
S1 , S2 and S3 , respectively. Then points A, B and C lie on the sides of triangle O1 O2 O3
and O1 B = O1 C, O2 C = O2 A and O3 A = O3 B. Hence, points A, B and C are the tangent
points of the inscribed circle of triangle O1 O2 O3 with its sides.
198
CHAPTER 8. CONSTRUCTIONS
This implies the following construction. First, let us construct the circumscribed circle
of triangle ABC and draw tangents to it at points A, B and C. The intersection points of
these tangents are the centers of circles to be found.
8.58. Suppose that we have constructed circle S whose tangents AA1 , BB1 and CC1 ,
where A1 , B1 and C1 are the tangent points, are of length a, b and c, respectively. Let us
construct circles Sa , Sb and Sc with the centers A, B and C and radii a, b and c, respectively
(Fig. 91). If O is the center of circle S, then segments OA1 , OB1 and OC1 are radii of circle
S and tangents to circles Sa , Sb and Sc as well. Hence, point O is the radical center (cf.
§3.10) of circles Sa , Sb and Sc .
Figure 91 (Sol. 8.58)
This implies the following construction. First, construct circles Sa , Sb and Sc . Then let
us construct their radical center O. The circle to be found is the circle with center O and
the radius whose length is equal to that of the tangent drawn from point O to circle Sa .
8.59. First, let us construct segment BC of length a. Next, let us construct the locus
of points X for which CX : BX = b : c, cf. Problem 7.14. For vertex A we can take any of
the intersection points of this locus with a line whose distance from line BC is equal to ha .
8.60. Given the lengths of segments AD′ and BD, we can construct segment AB and
point D on this segment. Point C is the intersection point of the circle of radius CD centered
at D and the locus of points X for which AX : BX = AD : BD.
8.61. Let X be a point that does not lie on line AB. Clearly, ∠AXB = ∠BCX if and
only if AX : CX = AB : CB. Hence, point M is the intersection point of the locus of points
X for which AX : CX = AB : CB and the locus of points Y for which BY : DY = BC : DC
(it is possible for these loci not to intersect).
8.62. We have to construct a point O for which AO : A′ O = AB : A′ B ′ and BO :
′
B O = AB : A′ B ′ . Point O is the intersection point of the locus of points X for which
AX : A′ X = AB : A′ B ′ and the locus of points Y for which BY : B ′ Y = AB : A′ B ′ .
8.63. Let O be the center of the given circle. Chords XP and XQ that pass through
points A and B are equal if and only if XO is the bisector of angle P XO, i.e., AX :
BX = AO : BO. The point X to be found is the intersection point of the corresponding
Apollonius’s circle with the given circle.
8.64. a) If line l does not intersect segment AB, then ABB1 A1 is a parallelogram and
l k AB. If line l intersects segment AB, then AA1 BB1 is a parallelogram and l passes
through the midpoint of segment AB.
b) One of the lines to be found is parallel to line AB and another one passes through the
midpoint of segment AB.
8.65. Let us construct a circle of radius 1 and in it draw two perpendicular diameters, AB
and CD. Let O be the center of the circle, M the midpoint of segment OC, P the intersection
SOLUTIONS
199
point of line AM and the circle√with diameter OC (Fig. 92). Then AM 2 = 1 + 14 = 54 and,
= 2 sin 18◦ (cf. Problem 5.46), i.e., AP is the length of
therefore, AP = AM − P M = 5−1
2
a side of a regular decagon inscribed in the given circle.
Figure 92 (Sol. 8.65)
8.66. Suppose we have constructed rectangle P QRS so that the given points A, B, C,
D lie on sides P Q, QR, RS, SP , respectively, and P Q : QR = a, where a is the given ratio
of sides. Let F be the intersection point of the line drawn through point D perpendicularly
to line AC and line QR. Then DF : AC = a.
This implies the following construction. From point D draw a ray that intersects segment
AC at a right angle and on this ray construct a point F so that DF = a · AC. Side QR lies
on line BF . The continuation of the construction is obvious.
8.67. Suppose that points X and Y with the required properties are constructed. Denote
the intersection point of lines AX and Y C by M , that of lines AB and XY by K. Right
triangles AXK and Y XM have a common acute angle ∠X, hence, ∠XAK = ∠XY M .
Angles ∠XAB and ∠XY B subtend the same arc, hence, ∠XAB = ∠XY B. Therefore,
∠XY M = ∠XY B. Since XY ⊥ AB, it follows that K is the midpoint of segment CB.
Conversely, if K is the midpoint of segment CB, then ∠M Y X = ∠BY X = ∠XAB.
Triangles AXK and Y XM have a common angle ∠X and ∠XAK = ∠XY M ; hence,
∠Y M X = ∠AKX = 90◦ .
This implies the following construction. Through the midpoint K of segment CB draw
line l perpendicular to line AB. Points X and Y are the intersection points of line l with
the given circle.
8.68. If we have an angle of value α, then we can construct angles of value 2α, 3α, etc.
Since 19 · 19◦ = 361◦ , we can construct an angle of 361◦ that coincides with the angle of 1◦ .
8.69. First, let us construct an angle of 36◦ , cf. Problem 8.65. Then we can construct
◦
◦
the angle of 36 −30
= 3◦ . If n is not divisible by 3, then having at our disposal angles of n◦
2
and 3◦ we can construct an angle of 1◦ . Indeed, if n = 3k + 1, then 1◦ = n◦ − k · 3◦ and if
n = 3k + 2, then 1◦ = 2n◦ − (2k + 1) · 3◦ .
8.70. The sequence of constructions is as follows. On the piece of paper take an arbitrary
point O and perform the homothety with center O and sufficiently small coefficient k so that
this homothety sends the image of the intersection point of the given lines on the piece of
paper. Then we can construct the bisector of the angle between the images of the lines.
Next, let us perform the homothety with the same center and coefficient k1 which yields the
desired segment of the bisector.
8.71. Let us construct with the help of a two-sided ruler two parallel chords AB and CD.
Let P and Q be the intersection points of lines AC with BD and AD with BC, respectively.
Then line P Q passes through the center of the given circle. Constructing similarly one more
such line we find the center of the circle.
200
CHAPTER 8. CONSTRUCTIONS
8.72. Let us draw through point A two rays p and q that form a small angle inside
which point B lies (the rays can be constructed by replacing the ruler). Let us draw through
point B segments P Q1 and P1 Q (Fig. 93). If P Q < 10 cm and P1 Q1 < 10 cm, then we can
construct point O at which lines P Q and P1 Q1 intersect.
Figure 93 (Sol. 8.72)
Through point O draw line P2 Q2 . If P Q2 < 10 cm and P2 Q < 10 cm; then we can
construct point B ′ at which lines P Q2 and P2 Q intersect. If BB ′ < 10 cm, then by Problem
5.67 we can construct line BB ′ ; this line passes through point A.
8.73. The construction is based on the fact that if A and B are the intersection points of
−→ −−→
equal circles centered at P and Q, then P A = BQ. Let S1 be the initial circle, A1 the given
point. Let us draw circle S2 through point A1 and circle S3 through the intersection point
A2 of circles S1 and S2 ; circle S4 through the intersection point A3 of circles S2 and S3 and,
finally, circle S5 through the intersection points B1 and A4 of circles S1 and S3 , respectively,
with circle S4 . Let us prove that the intersection point B2 of circles S5 and S1 is the one to
be found.
Let Oi be the center of circle Si . Then
−−−→ −−−→ −−−→ −−−→ −−−→ −−−→
A1 O1 = O2 A2 = A3 O3 = O4 A4 = B1 O5 = O1 B2 .
Remark. There are two intersection points of circles S1 and S4 ; for point B1 we can
take any of them.
8.74. Let AB be the given segment, P an arbitrary point not on the given lines. Let
us construct the intersection points C and D of the second of the given lines with lines P A
and P B, respectively, and the intersection point Q of lines AD and BC. By Problem 19.2
line P Q passes through the midpoint of segment AB.
8.75. Let AB be the given segment; let C and D be arbitrary points on the second of
given lines. By the preceding problem we can construct the midpoint, M , of segment CD.
Let P be the intersection point of lines AM and BD; let E be the intersection point of lines
P C and AB. Let us prove that EB is the segment to be found.
Since △P M C ∼ △P AE and △P M D ∼ △P AB, it follows that
AB
AB AE
MD MC
MD
=
:
=
:
=
= 1.
AE
AP AP
MP MP
MC
8.76. Let AB be the given segment; let C and D be arbitrary points on the second of
the given lines. By the preceding problem we can construct points D1 = D, D2 , . . . , Dn such
that all the segments Di Di+1 are equal to segment CD. Let P be the intersection point
of lines AC and BDn and let B1 , . . . , Bn−1 be the intersection points of line AB with lines
P D1 , . . . , P Dn−1 , respectively. Clearly, points B1 , . . . , Bn−1 divide segment AB in n equal
parts.
SOLUTIONS
201
8.77. On one of the given lines take segment AB and construct its midpoint, M (cf.
Problem 8.74). Let A1 and M1 be the intersection points of lines P A and P M with the
second of the given lines, Q the intersection point of lines BM1 and M A1 . It is easy to verify
that line P Q is parallel to the given lines.
8.78. In the case when point P does not lie on line AB, we can make use of the solution
of Problem 3.36. If point P lies on line AB, then we can first drop perpendiculars l1 and
l2 from some other points and then in accordance with Problem 8.77 draw through point P
the line parallel to lines l1 and l2 .
8.79. a) Let A be the given point, l the given line. First, let us consider the case
when point O does not lie on line l. Let us draw through point O two arbitrary lines that
intersect line l at points B and C. By Problem 8.78, in triangle OBC, heights to sides OB
and OC can be dropped. Let H be their intersection point. Then we can draw line OH
perpendicular to l. By Problem 8.78 we can drop the perpendicular from point A to OH.
This is the line to be constructed that passes through A and is parallel to l. In order to drop
the perpendicular from A to l we have to erect perpendicular l′ to OH at point O and then
drop the perpendicular from A to l′ .
If point O lies on line l, then by Problem 8.78 we can immediately drop the perpendicular
′
l from point A to line l and then erect the perpendicular to line l′ from the same point A.
b) Let l be the given line, A the given point on it and BC the given segment. Let us
draw through point O lines OD and OE parallel to lines l and BC, respectively (D and E
are the intersection points of these lines with circle S). Let us draw through point C the
line parallel to OB to its intersection with line OE at point F and through point F the line
parallel to ED to its intersection with OD at point G and, finally, through point G the line
parallel to OA to its intersection with l at point H. Then AH = OG = OF = BC, i.e., AH
is the segment to be constructed.
c) Let us take two arbitrary lines that intersect at point P . Let us mark on one of them
segment P A = a and on the other one segments P B = b and P C = c. Let D be the
intersection point of line P A with the line that passes through B and is parallel to AC.
Clearly, P D = abc .
d) Let H be the homothety (or the parallel translation) that sends the circle with center
A and radius r to circle S (i.e., to the given circle with the marked center O). Since the radii
of both circles are known, we can construct the image of any point X under the mapping
H. For this we have to draw through point O the line parallel to line AX and mark on it a
, where rs is the radius of circle S.
segment equal to rs ·AX
r
We similarly construct the image of any point under the mapping H −1 . Hence, we can
construct the line l′ = H(l) and find its intersection points with circle S and then construct
the images of these points under the map H −1 .
e) Let A and B be the centers of the given circles, C one of the points to be constructed,
CH the height of triangle ABC. From Pythagoras theorem for triangles ACH and BCH we
2
2
2
deduce that AH = b +c2c−a . The quantities a, b and c are known, hence, we can construct
point H and the intersection points of line CH with one of the given circles.
8.80. a) Let us draw lines parallel to lines OA and OB, whose distance from the latter
lines is equal to a and which intersect the legs of the angles. The intersection point of these
lines lies on the bisector to be constructed.
b) Let us draw the line parallel to OB, whose distance from OB is equal to a and which
intersects ray OA at a point M . Let us draw through points O and M another pair of
parallel lines the distance between which is equal to a; the line that passes through point O
contains the leg of the angle to be found.
202
CHAPTER 8. CONSTRUCTIONS
8.81. Let us draw through point A an arbitrary line and then draw lines l1 and l2 parallel
to it and whose distance from this line is equal to a; these lines intersect line l at points M1
and M2 . Let us draw through points A and M1 one more pair of parallel lines, la and lm ,
the distance between which is equal to a. The intersection point of lines l2 and lm belongs
to the perpendicular to be found.
8.82. Let us draw a line parallel to the given one at a distance of a. Now, we can make
use of the results of Problems 8.77 and 8.74.
8.83. Let us draw through point P lines P A1 and P B1 so that P A1 k OA and P B1 k OB.
Let line P M divide the angle between lines l and P A1 in halves. The symmetry through
line P M sends line P A1 to line l and, therefore, line P B1 turns under this symmetry into
one of the lines to be constructed.
8.84. Let us complement triangle ABM to parallelogram ABM N . Through point N
draw lines parallel to the bisectors of the angles between lines l and M N . The intersection
points of these lines with line l are the ones to be found.
8.85. Let us draw line l1 parallel to line OA at a distance of a. On l, take an arbitrary
point B. Let B1 be the intersection point of lines OB and l1 . Through point B1 draw the
line parallel to AB; this line intersects line OA at point A1 . Now, let us draw through points
O and A1 a pair of parallel lines the distance between which is equal to a.
There could be two pairs of such lines. Let X and X1 be the intersection points of the line
that passes through point O with lines l and l1 . Since OA1 = OX1 and △OA1 X1 ∼ △OAX,
point X is the one to be found.
8.86. Let us erect perpendiculars to line O1 O2 at points O1 and O2 and on the perpendiculars mark segments O1 B1 = O2 A2 and O2 B2 = O1 A1 . Let us construct the midpoint
M of segment B1 B2 and erect the perpendicular to B1 B2 at point M . This perpendicular intersects line O1 O2 at point N . Then O1 N 2 + O1 B12 = O2 N 2 + O2 B22 and, therefore,
O1 N 2 − O1 A21 = O2 N 2 − O2 A22 , i.e., point N lies on the radical axis. It remains to erect the
perpendicular to O1 O2 at point N .
8.87. First, let us construct an arbitrary line l1 perpendicular to line l and then draw
through point A the line perpendicular to l1 .
8.88. a) Let us draw through points A and B lines AB and BQ perpendicular to line
AB and then draw an arbitrary perpendicular to line AP . As a result we get a rectangle.
It remains to drop from the intersection point of its diagonals the perpendicular to line AB.
b) Let us raise from point B perpendicular l to line AB and draw through point A
two perpendicular lines; they intersect line l at points M and N . Let us complement right
triangle M AN to rectangle M AN R. The base of the perpendicular dropped from point R
to line AB is point C to be found.
8.89. a) Let us drop perpendicular AP from point A to line OB and construct segment
AC whose midpoint is points P . Then angle ∠AOC is the one to be found.
b) On line OB, take points B and B1 such that OB = OB1 . Let us place the right angle
so that its sides would pass through points B and B1 and the vertex would lie on ray OA.
If A is the vertex of the right angle, then angle ∠AB1 B is the one to be found.
8.90. Let us draw through point O line l′ parallel to line l. Let us drop perpendiculars
BP and BQ from point B to lines l′ and OA, respectively, and then drop perpendicular OX
from point O to line P Q. Then line XO is the desired one (cf. Problem 2.3); if point Y is
symmetric to point X through line l′ , then line Y O is also the one to be found.
8.91. Let us complement triangle OAB to parallelogram OABC and then construct
segment CC1 whose midpoint is point O. Let us place the right angle so that its legs pass
through points C and C1 and the vertex lies on line l. Then the vertex of the right angle
coincides with point X to be found.
SOLUTIONS
203
8.92. Let us construct segment AB whose midpoint is point O and place the right angle
so that its legs passes through points A and B and the vertex lies on line l. Then the vertex
of the right angle coincides with the point to be found.
Chapter 9. GEOMETRIC INEQUALITIES
Background
1) For elements of a triangle the following notations are used:
a, b, c are the lengths of sides BC, CA, AB, respectively;
α, β, γ the values of the angles at vertices A, B, C, respectively;
ma , mb , mc are the lengths of the medians drawn from vertices A, B, C, respectively;
ha , hb , hc are the lengths of the heights dropped from vertices A, B, C, respectively;
la , lb , lc are the lengths of the bisectors drawn from vertices A, B, C, respectively;
r and R are the radii of the inscribed and circumscribed circles, respectively.
2) If A, B, C are arbitrary points, then AB ≤ AC + CB and the equality takes place
only if point C lies on segment AB (the triangle inequality).
3) The median of a triangle is shorter than a half sum of the sides that confine it:
1
(Problem 9.1).
ma < 2(b+c)
4) If one convex polygon lies inside another one, then the perimeter of the outer polygon
is greater than the perimeter of the inner one (Problem 9.27 b).
5) The sum of the lengths of the diagonals of a convex quadrilateral is greater than the
sum of the length of any pair of the opposite sides of the quadrilateral (Problem 9.14).
6) The longer side of a triangle subtends the greater angle (Problem 10.59).
7) The length of the segment that lies inside a convex polygon does not exceed either
that of its longest side or that of its longest diagonal (Problem 10.64).
Remark. While solving certain problems of this chapter we have to know various algebraic inequalities. The data on these inequalities and their proof are given in an appendix to
this chapter; one should acquaint oneself with them but it should be taken into account that
these inequalities are only needed in the solution of comparatively complicated
problems;
√
1
in order to solve simple problems we will only need the inequality ab ≤ 2 a + b and its
corollaries.
Introductory problems
1. Prove that SABC ≤ 12 AB · BC.
2. Prove that SABCD ≤ 21 (AB · BC + AD · DC).
3. Prove that ∠ABC > 90◦ if and only if point B lies inside the circle with diameter
AC.
4. The radii of two circles are equal to R and r and the distance between the centers of
the circles is equal to d. Prove that these circles intersect if and only if |R − r| < d < R + r.
5. Prove that any diagonal of a quadrilateral is shorter than the quadrilateral’s semiperimeter.
§1. A median of a triangle
9.1. Prove that 21 (a + b − c) < mc < 21 (a + b).
205
206
CHAPTER 9. GEOMETRIC INEQUALITIES
9.2. Prove that in any triangle the sum of the medians is greater than 43 of the perimeter
but less than the perimeter.
9.3. Given n points A1 , . . . , An and a unit circle, prove that it is possible to find a point
M on the circle so that M A1 + · · · + M An ≥ n.
9.4. Points A1 , . . . , An do not lie on one line. Let two distinct points P and Q have the
following property
A1 P + · · · + An P = A1 Q + · · · + An Q = s.
Prove that A1 K + · · · + An K < s for a point K.
9.5. On a table lies 50 working watches (old style, with hands); all work correctly.
Prove that at a certain moment the sum of the distances from the center of the table to the
endpoints of the minute’s hands becomes greater than the sum of the distances from the
center of the table to the centers of watches. (We assume that each watch is of the form of
a disk.)
§2. Algebraic problems on the triangle inequality
In problems of this section a, b and c are the lengths of the sides of an arbitrary triangle.
9.6. Prove that a = y + z, b = x + z and c = x + y, where x, y and z are positive
numbers.
9.7. Prove that a2 + b2 + c2 < 2(ab + bc + ca).
9.8. For any positive integer n, a triangle can be composed of segments whose lengths
are an , bn and cn . Prove that among numbers a, b and c two are equal.
9.9. Prove that
a(b − c)2 + b(c − a)2 + c(a − b)2 + 4abc > a3 + b3 + c3 .
9.10. Let p = ab + cb + ac and q = ac + cb + ab . Prove that |p − q| < 1.
9.11. Five segments are such that from any three of them a triangle can be constructed.
Prove that at least one of these triangles is an acute one.
9.12. Prove that
(a + b − c)(a − b + c)(−a + b + c) ≤ abc.
9.13. Prove that
a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0.
§3. The sum of the lengths of quadrilateral’s diagonals
9.14. Let ABCD be a convex quadrilateral. Prove that AB + CD < AC + BD.
9.15. Let ABCD be a convex quadrilateral and AB + BD ≤ AC + CD. Prove that
AB < AC.
9.16. Inside a convex quadrilateral the sum of lengths of whose diagonals is equal to d,
a convex quadrilateral the sum of lengths of whose diagonals is equal to d′ is placed. Prove
that d′ < 2d.
9.17. Given closed broken line has the property that any other closed broken line with
the same vertices (?) is longer. Prove that the given broken line is not a self-intersecting
one.
9.18. How many sides can a convex polygon have if all its diagonals are of equal length?
9.19. In plane, there are n red and n blue dots no three of which lie on one line. Prove
that it is possible to draw n segments with the endpoints of distinct colours without common
points.
9.20. Prove that the mean arithmetic of the lengths of sides of an arbitrary convex
polygon is less than the mean arithmetic of the lengths of all its diagonals.
THE AREA OF A TRIANGLE
207
9.21. A convex (2n + 1)-gon A1 A3 A5 . . . A2n+1 A2 . . . A2n is given. Prove that among all
the closed broken lines with the vertices in the vertices of the given (2n + 1)-gon the broken
line A1 A2 A3 . . . A2n+1 A1 is the longest.
§4. Miscellaneous problems on the triangle inequality
9.22. In a triangle, the lengths of two sides are equal to 3.14 and 0.67. Find the length
of the third side if it is known that it is an integer.
9.23. Prove that the sum of lengths of diagonals of convex pentagon ABCDE is greater
than its perimeter but less than the doubled perimeter.
9.24. Prove that if the lengths of a triangle’s sides satisfy the inequality a2 + b2 > 5c2 ,
then c is the length of the shortest side.
9.25. The lengths of two heights of a triangle are equal to 12 and 20. Prove that the
third height is shorter than 30.
9.26. On sides AB, BC, CA of triangle ABC, points C1 , A1 , B1 , respectively, are taken
so that BA1 = λ · BC, CB1 = λ · CA and AC1 = λ · AB, where 21 < λ < 1. Prove that the
perimeter P of triangle ABC and the perimeter P1 of triangle A1 B1 C1 satisfy the following
inequality: (2λ − 1)P < P1 < λP .
***
9.27. a) Prove that under the passage from a nonconvex polygon to its convex hull the
perimeter diminishes. (The convex hull of a polygon is the smallest convex polygon that
contains the given one.)
b) Inside a convex polygon there lies another convex polygon. Prove that the perimeter
of the outer polygon is not less than the perimeter of the inner one.
9.28. Inside triangle ABC of perimeter P , a point O is taken. Prove that 21 P <
AO + BO + CO < P .
9.29. On base AD of trapezoid ABCD, a point E is taken such that the perimeters of
triangles ABE, BCE and CDE are equal. Prove that BC = 21 AD.
See also Problems 13.40, 20.11.
§5. The area of a triangle does not exceed a half product of two sides
9.30.
Given a triangle of area 1 the lengths of whose sides satisfy a ≤ b ≤ c. Prove that
√
b ≥ 2.
9.31. Let E, F , G and H be the midpoints of sides AB, BC, CD and DA of quadrilateral
ABCD. Prove that
(AB + CD)(AD + BC)
SABCD ≤ EG · HF ≤
.
4
9.32. The perimeter of a convex quadrilateral is equal to 4. Prove that its area does not
exceed 1.
9.33. Inside triangle ABC a point M is taken. Prove that
4S ≤ AM · BC + BM · AC + CM · AB,
where S is the area of triangle ABC.
9.34. In a circle of radius R a polygon of area S is inscribed; the polygon contains the
center of the circle and on each of its sides a point is chosen. Prove that the perimeter of
the convex polygon with vertices in the chosen points is not less than 2S
.
R
9.35. Inside a convex quadrilateral ABCD of area S point O is taken such that AO2 +
BO2 + CO2 + DO2 = 2S. Prove that ABCD is a square and O is its center.
208
CHAPTER 9. GEOMETRIC INEQUALITIES
§6. Inequalities of areas
9.36. Points M and N lie on sides AB and AC, respectively, of triangle ABC, where
AM = CN and AN = BM . Prove that the area of quadrilateral BM N C is at least three
times that of triangle AM N .
9.37. Areas of triangles ABC, A1 B1 C1 , A2 B2 C2 are equal to S, S1 , S2 , respectively,
and
√
AB = A1 B1 + A2 B2 , AC = A1 C1 + A2 B2 , BC = B1 C1 + B2 C2 . Prove that S ≤ 4 S1 S2 .
9.38. Let ABCD be a convex quadrilateral of area S. The angle between lines AB and
CD is equal to α and the angle between AD and BC is equal to β. Prove that
AB · CD sin α + AD · BC sin β ≤ 2S ≤ AB · CD + AD · BC.
9.39. Through a point inside a triangle three lines parallel to the triangle’s sides are
drawn.
Figure 94 (9.39)
Denote the areas of the parts into which these lines divide the triangle as plotted on Fig.
94. Prove that αa + βb + γc ≥ 32 .
9.40. The areas of triangles ABC and A1 B1 C1 are equal to S and S1 , respectively, and
we know that triangle ABC is not an obtuse one. The greatest of the ratios aa1 , bb1 and cc1 is
equal to k. Prove that S1 ≤ k 2 S.
9.41. a) Points B, C and D divide the (smaller) arc ⌣ AE of a circle into four equal
parts. Prove that SACE < 8SBCD .
b) From point A tangents AB and AC to a circle are drawn. Through the midpoint D
of the (lesser) arc ⌣ BC the tangent that intersects segments AB and AC at points M and
N , respectively is drawn. Prove that SBCD < 2SM AN .
9.42. All sides of a convex polygon are moved outwards at distance h and extended
to form a new polygon. Prove that the difference of areas of the polygons is more than
P h + πh2 , where P is the perimeter.
9.43. A square is cut into rectangles. Prove that the sum of areas of the disks circumscribed about all these rectangles is not less than the area of the disk circumscribed about
the initial square.
9.44. Prove that the sum of areas of five triangles formed by the pairs of neighbouring
sides and the corresponding diagonals of a convex pentagon is greater than the area of the
pentagon itself.
9.45. a) Prove that in any convex hexagon of area S there exists a diagonal that cuts
off the hexagon a triangle whose area does not exceed 16 S.
b) Prove that in any convex 8-gon of area S there exists a diagonal that cuts off it a
triangle of area not greater than 81 S.
See also Problem 17.19.
§8. BROKEN LINES INSIDE A SQUARE
209
§7. Area. One figure lies inside another
9.46. A convex polygon whose area is greater than 0.5 is placed in a unit square. Prove
that inside the polygon one can place a segment of length 0.5 parallel to a side of the square.
9.47. Inside a unit square n points are given. Prove that:
a) the area of one of the triangles some of whose vertices are in these points and some in
1
vertices of the square does not exceed 2(n+1)
;
1
.
b) the area of one of the triangles with the vertices in these points does not exceed n−2
9.48. a) In a disk of area S a regular n-gon of area S1 is inscribed and a regular n-gon
of area S2 is circumscribed about the disk. Prove that S 2 > S1 S2 .
b) In a circle of length L a regular n-gon of perimeter P1 is inscribed and another regular
n-gon of perimeter P2 is circumscribed about the circle. Prove that L2 < P1 P2 .
9.49. A polygon of area B is inscribed in a circle of area A and circumscribed about a
circle of area C. Prove that 2B ≤ A + C.
9.50. In a unit disk two triangles the area of each of which is greater than 1 are placed.
Prove that these triangles intersect.
9.51. a) Prove that inside a convex polygon of area S and perimeter P one can place a
disk of radius PS .
b) Inside a convex polygon of area S1 and perimeter P1 a convex polygon of area S2 and
1
> PS22 .
perimeter P2 is placed. Prove that 2S
P1
9.52. Prove that the area of a parallelogram that lies inside a triangle does not exceed
a half area of the triangle.
9.53. Prove that the area of a triangle whose vertices lie on sides of a parallelogram does
not exceed a half area of the parallelogram.
***
√ 9.54. Prove that any acute triangle of area 1 can be placed in a right triangle of area
3.
9.55. a) Prove that a convex polygon of area S can be placed in a rectangle of area not
greater than 2S.
b) Prove that in a convex polygon of area S a parallelogram of area not less than 12 S can
be inscribed.
9.56. Prove that in any convex polygon of area 1 a triangle whose area is not less than
1
a) 4 ; b) 38 can be placed.
9.57. A convex n-gon is placed in a unit square. Prove that there are three vertices A, B
and C of this n-gon, such that the area of triangle ABC does not exceed a) n82 ; b) 16π
.
n3
See also Problem 15.6.
§8. Broken lines inside a square
9.58. Inside a unit square a non-self-intersecting broken line of length 1000 is placed.
Prove that there exists a line parallel to one of the sides of the square that intersects this
broken line in at least 500 points.
9.59. In a unit square a broken line of length L is placed. It is known that each point of
1
− 12 πε.
the square is distant from a point of this broken line less than by ε. Prove that L ≥ 2ε
9.60. Inside a unit square n2 points are placed. Prove that there exists a broken line
that passes through all these points and whose length does not exceed 2n.
9.61. Inside a square of side 100 a broken line L is placed. This broken line has the
following property: the distance from any point of the square to L does not exceed 0.5.
210
CHAPTER 9. GEOMETRIC INEQUALITIES
Prove that on L there are two points the distance between which does not exceed 1 and the
distance between which along L is not less than 198.
§9. The quadrilateral
9.62. In quadrilateral ABCD angles ∠A and ∠B are equal and ∠D > ∠C. Prove that
AD < BC.
9.63. In trapezoid ABCD, the angles at base AD satisfy inequalities ∠A < ∠D < 90◦ .
Prove that AC > BD.
9.64. Prove that if two opposite angles of a quadrilateral are obtuse ones, then the
diagonal that connects the vertices of these angles is shorter than the other diagonal.
9.65. Prove that the sum of distances from an arbitrary point to three vertices of an
isosceles trapezoid is greater than the distance from this point to the fourth vertex.
9.66. Angle ∠A of quadrilateral ABCD is an obtuse one; F is the midpoint of side BC.
Prove that 2F A < BD + CD.
9.67. Quadrilateral ABCD is given. Prove that AC · BD ≤ AB · CD + BC · AD.
(Ptolemy’s inequality.)
9.68. Let M and N be the midpoints of sides BC and CD, respectively, of a convex
quadrilateral ABCD. Prove that SABCD < 4SAM N .
9.69. Point P lies inside convex quadrilateral ABCD. Prove that the sum of distances
from point P to the vertices of the quadrilateral is less than the sum of pairwise distances
between the vertices of the quadrilateral.
9.70. The diagonals divide a convex quadrilateral ABCD into four triangles. Let P be
the perimeter of ABCD and Q the perimeter of the quadrilateral formed by the centers of
the inscribed circles of the obtained triangles. Prove that P Q > 4SABCD .
9.71. Prove that the distance from one of the vertices of a convex quadrilateral to the
opposite diagonal does not exceed a half length of this diagonal.
9.72. Segment KL passes through the intersection point of diagonals of quadrilateral
ABCD and the endpoints of KL lie on sides AB and CD of the quadrilateral. Prove that the
length of segment KL does not exceed the length of one of the diagonals of the quadrilateral.
9.73. Parallelogram P2 is inscribed in parallelogram P1 and parallelogram P3 whose sides
are parallel to the corresponding sides of P1 is inscribed in parallelogram P2 . Prove that the
length of at least one of the sides of P1 does not exceed the doubled length of a parallel to
it side of P3 .
See also Problems 13.19, 15.3 a).
§10. Polygons
9.74. Prove that if the angles of a convex pentagon form an arithmetic progression, then
each of them is greater than 36◦ .
9.75. Let ABCDE is a convex pentagon inscribed in a circle of radius 1 so that AB = A,
BC = b, CD = c, DE = d, AE = 2. Prove that
a2 + b2 + c2 + d2 + abc + bcd < 4.
9.76. Inside a regular hexagon with side 1 point P is taken. Prove that the sum of the
distances from point P to certain three vertices of the hexagon is not less than 1.
9.77. Prove that if the sides of convex hexagon ABCDEF are equal to 1, then the
radius of the circumscribed circle of one of triangles ACE and BDF does not exceed 1.
9.78. Each side of convex hexagon ABCDEF is shorter than 1. Prove that one of the
diagonals AD, BE, CF is shorter than 2.
* * *
211
9.79. Heptagon A1 . . . A7 is inscribed in a circle. Prove that if the center of this circle
lies inside it, then the value of any angle at vertices A1 , A3 , A5 is less than 450◦ .
***
9.80. a) Prove that if the lengths of the projections of a segment to two perpendicular
√ .
lines are equal to a and b, then the segment’s length is not less than a+b
2
b) The lengths of the projections of a√polygon to coordinate axes are equal to a and b.
Prove that its perimeter is not less than 2(a + b).
9.81. Prove that from the sides of a convex polygon of perimeter P two segments whose
lengths differ not more than by 13 P can be constructed.
9.82. Inside a convex polygon A1 . . . An a point O is taken. Let αk be the value of the
angle
at vertexPAk , xk = P
OAk and dk the distance from point O to line Ak Ak+1 . Prove that
P
αk
xk sin 2 ≥
dk and
xk cos α2k ≥ p, where p is the semiperimeter of the polygon.
9.83. Regular 2n-gon M1 with side a lies inside regular 2n-gon M2 with side 2a. Prove
that M1 contains the center of M2 .
9.84. Inside regular polygon A1 . . . An point O is taken.
¢
¡
Prove that at least one of the angles ∠Ai OAj satisfies the inequalities π 1 − n1 ≤
∠Ai OAj ≤ π.
9.85. Prove that for n ≥ 7 inside a convex n-gon there is a point the sum of distances
from which to the vertices is greater than the semiperimeter of the n-gon.
9.86. a) Convex polygons A1 . . . An and B1 . . . Bn are such that all their corresponding
sides except for A1 An and B1 Bn are equal and ∠A2 ≥ ∠B2 , . . . , ∠An−1 ≥ ∠Bn−1 , where at
least one of the inequalities is a strict one. Prove that A1 An > B1 Bn .
b) The corresponding sides of nonequal polygons A1 . . . An and B1 . . . Bn are equal.
Let us write beside each vertex of polygon A1 . . . An the sign of the difference ∠Ai − ∠Bi .
Prove that for n ≥ 4 there are at least four pairs of neighbouring vertices with distinct signs.
(The vertices with the zero difference are disregarded: two vertices between which there only
stand vertices with the zero difference are considered to be neighbouring ones.)
See also Problems 4.37, 4.53, 13.42.
§11. Miscellaneous problems
9.87. On a segment of length 1 there are given n points. Prove that the sum of distances
from a point of the segment to these points is not less than 21 n.
9.88. In a forest, trees of cylindrical form grow. A communication service person has to
connect a line from point A to point B through this forest the distance between the points
being equal to l. Prove that to acheave the goal a piece of wire of length 1.6l will be sufficient.
9.89. In a forest, the distance between any two trees does not exceed the difference of
their heights. Any tree is shorter than 100 m. Prove that this forest can be fenced by a
fence of length 200 m.
9.90. A (not necessarily convex) paper polygon is folded along a line and both halves
are glued together. Can the perimeter of the obtained lamina be greater than the perimeter
of the initial polygon?
***
9.91. Prove that a closed broken line of length 1 can be placed in a disk of radius 0.25.
9.92. An acute triangle is placed inside a circumscribed circle. Prove that the radius of
the circle is not less than the radius of the circumscribed circle of the triangle.
212
CHAPTER 9. GEOMETRIC INEQUALITIES
Is a similar statement true for an obtuse triangle?
9.93. Prove that the perimeter of an acute triangle is not less than 4R.
See also problems 14.23, 20.4.
Problems for independent study
9.94. Two circles divide rectangle ABCD into four rectangles. Prove that the area of
one of the rectangles, the one adjacent to vertices A and C, does not exceed a quarter of the
area of ABCD.
9.95. Prove that if AB + BD = AC + CD, then the midperpendicular to side BC of
quadrilateral ABCD intersects segment AD.
9.96. Prove that if diagonal BD of convex quadrilateral ABCD divides diagonal AC in
halves and AB > BC, then AD < DC.
9.97. The lengths of bases of a circumscribed trapezoid are equal to 2 and 11. Prove
that the angle between the extensions of its lateral sides is an acute one.
9.98. The bases of a trapezoid are equal to a and q
b and its height is equal to h. Prove
2
2
.
that the length of one of its diagonals is not less than h +(b+a)
4
9.99. The vertices of an n-gon M1 are the midpoints of sides of a convex n-gon M . Prove
that for n ≥ 3 the perimeter of M1 is not less than the semiperimeter of M and for n ≥ 4
the area of M1 is not less than a half area of M .
√ 9.100. In a unit circle a polygon the lengths of whose sides are confined between 1 and
2 is inscribed. Find how many sides does the polygon have.
Supplement. Certain inequalities
inequality between the mean arithmetic and the mean geometric of two numbers
√ 1. The
ab ≤ 12 (a + b), where a and b are positive numbers, is often encountered. This inequality
√
√
√
follows from the fact that a − 2 ab + b = ( a − b)2 ≥ 0, where the equality takes place
only if a = b.
This inequality implies several useful inequalities, for example:
¡ x+a−x ¢2 a2
= 4;
x(a − x)
≤
2
q
a + a1
≥ 2 a · a1 = 2 fora > 0.
2. The inequality between the mean arithmetic and the mean geometric of n positive
1
n
is sometimes used. In this inequality the equality takes
numbers (a1 a2 . . . an ) n ≤ a1 +···+a
n
place only if a1 = · · · = an .
First, let us prove this inequality for the numbers of the form n = 2m by induction on
m. For m = 1 the equality was proved above.
Suppose that it is proved for m and let us prove it for m + 1. Clearly, ak ak+2m ≤
¡ ak +ak+2m ¢2
. Therefore,
2
1
1
(a1 a2 . . . a2m+1 ) 2m+1 ≤ (b1 b2 . . . b2m ) 2m ,
where bk = 21 (ak + ak+2m ) and by the inductive hypothesis
1
1
1
(b1 . . . b2m ) 2m ≤ m (b1 + · · · + b2m ) = m+1 (a1 + · · · + a2m+1 ).
2
2
m
Now, let n be an arbitrary number. Then n < 2 for some m. Suppose an+1 = · · · = a2m =
a1 +···+an
= A. Clearly,
n
(a1 + · · · + an ) + (an+1 + · · · + a2m ) = nA + (2m − n)A = 2m A
SOLUTIONS
213
m −n
and a1 . . . a2m = a1 . . . an · A2
. Hence,
µ m ¶2m
2 A
m
2m −n
= A2 , i.e. a1 . . . an ≤ An .
a1 . . . a n · A
≤
m
2
The equality is attained only for a1 = · · · = an .
3. For arbitrary numbers a1 , . . . , an we have
(a + · · · + an )2 ≤ n(a21 + · · · + a2n ).
Indeed,
(a1 + · · · + an )2 =
Rα
X
a2i + 2
X
i<j
Rα
ai aj ≤
X
a2i +
X
X
(a2i + a2j ) = n
a2i .
i<j
4. Since 0 cos t dt = sin α and 0 sin t dt = 1 − cos α, it follows that starting from the
2
2
inequality cos t ≤ 1 we get: first, sin α ≤ α, then 1 − cos α ≤ α2 (i.e. cos α ≥ 1 − α2 ), next,
3
2
4
sin α ≥ α − α6 , cos α ≤ 1 − α2 + α24 , etc. (the inequalities are true for all α ≥ 0).
5. Let us prove that tan α ≥ α for 0 ≤ α < π2 . Let AB be the tangent to the unit circle
centered at O; let B be the tangent point, C the intersection point of ray OA with the circle
and S the area of the disk sector BOC. Then α = 2S < 2SAOB = tan α.
6. On the segment [0, π2 ] the function
f (x) = sinx x monotonously grows because f ′ (x) =
¡
¢
tan x−x
> 0. In particular, f (α) ≤ f π2 , i.e.,
cos x sin2 x
π
π
α
≤
for 0 < α < .
sin α
2
2
√
7. If f (x) = a cos x + b sin x, then f (x) ≤ a2 + b2 . Indeed, there exists an angle ϕ such
that cos ϕ = √a2a+b2 and sin ϕ = √a2b+b2 ; hence,
√
√
f (x) = a2 + b2 cos(ϕ − x) ≤ a2 + b2 .
The equality takes place only if ϕ = x + 2kπ, i.e., cos x =
√ a
a2 +b2
and sin x =
√ b
.
a2 +b2
Solutions
9.1. Let C1 be the midpoint of side AB. Then CC1 + C1 A > CA and BC1 + C1 C > BC.
Therefore, 2CC1 + BA > CA + BC, i.e., mc > 21 (a + b − c).
Let point C ′ be symmetric to C through point C1 . Then CC1 = C1 C ′ and BC ′ = CA.
Hence, 2mc = CC ′ < CB + BC ′ = CB + CA, i.e., mc < 21 (a + b).
9.2. The preceding problem implies that ma < 12 (b + c), mb < 12 (a + c) and mc < 21 (a + b)
and, therefore, the sum of the lengths of medians does not exceed the perimeter.
Let O be the intersection point of medians of triangle ABC. Then BO + OA > BA,
AO + OC > AC and CO + OB > CB. Adding these inequalities and taking into account
that AO = 23 ma , BO = 23 mb , CO = 23 mc we get ma + mb + mc > 43 (a + b + c).
9.3. Let M1 and M2 be diametrically opposite points on a circle. Then M1 Ak + M2 Ak ≥
M1 M2 = 2. Adding up these inequalities for k = 1, . . . , n we get
(M1 A1 + · · · + M1 An ) + (M2 A1 + · · · + M2 An ) ≥ 2n.
Therefore, either M1 A1 +· · ·+M1 An ≥ n and then we set M = M1 or M2 A1 +· · ·+M2 An ≥ n
and then we set M = M2 .
9.4. For K we can take the midpoint of segment P Q. Indeed, then Ai K ≤ 12 (Ai P +Ai Q)
(cf. Problem 9.1), where at least one of the inequalities is a strict one because points Ai
cannot all lie on line P Q.
214
CHAPTER 9. GEOMETRIC INEQUALITIES
9.5. Let Ai and Bi be the positions of the minute hands of the i-th watch at times t
and t + 30 min, let Oi be the center of the i-th watch and O the center of the table. Then
OOi ≤ 21 (OAi + OBi ) for any i, cf. Problem 9.1. Clearly, at a certain moment points Ai
and Bi do not lie on line Oi O, i.e., at least one of n inequalities becomes a strict one. Then
either OO1 + · · · + OOn < OA1 + · · · + OAn or OO1 + · · · + OOn < OB1 + · · · + OBn .
9.6. Solving the system of equations
x + y = c,
x + z = b,
y+z =a
we get
a−b+c
a+b−c
−a + b + c
, y=
, z=
.
2
2
2
The positivity of numbers x, y and z follows from the triangle inequality.
9.7. Thanks to the triangle inequality we have
x=
a2 > (b − c)2 = b2 − 2bc + c2 ,
b2 > a2 − 2ac + c2 ,
c2 > a2 − 2ab + b2 .
Adding these inequalities we get the desired statement.
9.8. We may assume that a ≥ b ≥ c. Let us prove that a = b. Indeed, if b < a, then
b ≤ λa and c ≤ λa, where λ < 1. Hence, bn + cn ≤ 2λn an . For sufficiently large n we have
2λn < 1 which contradicts the triangle inequality.
9.9. Since c(a − b)2 + 4abc = c(a + b)2 , it follows that
a(b − c)2 + b(c − a)2 + c(a − b)2 + 4abc − a3 − b3 − c3 =
a((b − c)2 − a2 ) + +b((c − a)2 − b2 ) + c((a + b)2 − c2 ) =
(a + b − c)(a − b + c)(−a + b + c).
The latter equality is subject to a direct verification. All three factors of the latter expression
are positive thanks to the triangle inequality.
9.10. It is easy to verify that
abc|p − q| = |(b − c)(c − a)(a − b)|.
Since |b − c| < a, |c − a| < b and |a − b| < c, we have |(b − c)(c − a)(a − b)| < abc.
9.11. Let us index the lengths of the segments so that a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 . If all
the triangles that can be composed of these segments are not acute ones, then a23 ≥ a21 + a22 ,
a24 ≥ a22 + a23 and a25 ≥ a23 + a24 . Hence,
a25 ≥ a23 + a24 ≥ (a21 + a22 ) + (a22 + a23 ) ≥ 2a21 + 3a22 .
Since a21 + a22 ≥ 2a1 a2 , it follows that
2a21 + 3a22 > a21 + 2a1 a2 + a22 = (a1 + a2 )2 .
We get the inequality a25 > (a1 + a2 )2 which contradicts the triangle inequality.
9.12. First solution. Let us introduce new variables
x = −a + b + c,
y = a − b + c,
z = a + b − c.
Then a = 21 (y + z), b = 21 (x + z), c = 12 (x + y), i.e., we have to prove that either
1
xyz ≤ (x + y)(y + z)(x + z)
8
or
6xyz ≤ x(y 2 + z 2 ) + y(x2 + z 2 ) + z(x2 + y 2 ).
The latter inequality follows from the fact that 2xyz ≤ x(y 2 + z 2 ), 2xyz ≤ y(x2 + z 2 ) and
2xyz ≤ z(x2 + y 2 ), because x, y, z are positive numbers.
SOLUTIONS
215
Second solution. Since 2S = ab sin γ and sin γ =
Heron’s formula
c
,
2R
it follows that abc = 2SR. By
8S 2
.
p
≤ 4SR, i.e., 2S ≤ pR. Since S = pr, we infer that
(a + b − c)(a − b + c)(−a + b + c) =
2
Therefore, we have to prove that 8Sp
2r ≤ R, cf. Problem 10.26.
9.13. Let us introduce new variables
a−b+c
a+b−c
−a + b + c
, y=
, z=
.
x=
2
2
2
Then numbers x, y, z are positive and
a = y + z,
b = x + z,
c = x + y.
Simple but somewhat cumbersome calculations show that
a2 b(a − b) + b2 c(b − c) + c2 a(c − a) = 2(x3 z + y 3 x + z 3 y − xyz(x + y + z)) =
¶
µ 2
y2 z2
x
+
+
−x−y−z .
2xyz
y
z
x
Since 2 ≤
x
y
+ xy , it follows that
2x ≤ x
Similarly,
µ
y z
2y ≤ y
+
z y
Adding these inequalities we get
¶
=
µ
x y
+
y x
y2
+ z;
z
¶
=
x2
+ y.
y
2z ≤ z
³z
x
+
x ´ z2
+ x.
=
z
x
x2 y 2 z 2
+
+
≥ x + y + z.
y
z
x
9.14. Let O be the intersection point of the diagonals of quadrilateral ABCD. Then
AC + BD = (AO + OC) + (BO + OD) = (AO + OB) + (OC + OD) > AB + CD.
9.15. By the above problem AB + CD < AC + BD. Adding this inequality to the
inequality AB + BD ≤ AC + CD we get 2AB < 2AC.
9.16. First, let us prove that if P is the perimeter of convex quadrilateral ABCD and
d1 and d2 are the lengths of its diagonals, then P > d1 + d2 > 12 P . Clearly, AC < AB + BC
and AC < AD + DC; hence,
AB + BC + CD + AD
P
AC <
= .
2
2
1
Similarly, BD < 2 P . Therefore, AC + BD < P . On the other hand, adding the inequalities
AB + CD < AC + BD
and BC + AD < AC + BD
(cf. Problem 9.14) we get P < 2(AC + BD).
Let P be the perimeter of the outer quadrilateral, P ′ the perimeter of the inner one.
Then d > 21 P and since P ′ < P (by Problem 9.27 b)), we have d′ < P ′ < P < 2d.
9.17. Let the broken line of the shortest length be a self-intersecting one. Let us consider
two intersecting links. The vertices of these links can be connected in one of the following
three ways: Fig. 95. Let us consider a new broken line all the links of which are the same
216
CHAPTER 9. GEOMETRIC INEQUALITIES
as of the initial one except that the two solid intersecting links are replaced by the dotted
links (see Fig. 95).
Figure 95 (Sol. 9.17)
Then we get again a broken line but its length is less than that of the initial one because
the sum of the lengths of the opposite sides of a convex quadrilateral is less than the sum
of the length of its diagonals. We have obtained a contradiction and, therefore, the closed
broken line of the least length cannot have intersecting links.
9.18. Let us prove that the number of sides of such a polygon does not exceed 5. Suppose
that all the diagonals of polygon A1 . . . An are of the same length and n ≥ 6. Then segments
A1 A4 , A1 A5 , A2 A4 and A2 A5 are of equal length since they are the diagonals of this polygon.
But in convex quadrilateral A1 A2 A4 A5 segments A1 A5 and A2 A4 are opposite sides whereas
A1 A4 and A2 A5 are diagonals. Therefore, A1 A5 + A2 A4 < A1 A4 + A2 A5 . Contradiction.
It is also clear that a regular pentagon and a square satisfy the required condition.
9.19. Consider all the partitions of the given points into pairs of points of distinct
colours. There are finitely many such partitions and, therefore, there exists a partition for
which the sum of lengths of segments given by pairs of points of the partition is the least
one. Let us show that in this case these segments will not intersect. Indeed, if two segments
would have intersected, then we could have selected a partition with the lesser sum of lengths
of segments by replacing the diagonals of the convex quadrilateral by its opposite sides as
shown on Fig. 96.
Figure 96 (Sol. 9.19)
9.20. Let Ap Ap+1 and Aq Aq+1 be nonadjacent sides of n-gon A1 . . . An (i.e., |p − q| ≥ 2).
Then
Ap Ap+1 + Aq Aq+1 < Ap Aq + Ap+1 Aq+1 .
Let us write all such inequalities and add them. For each side there exist precisely n − 3
sides nonadjacent to it and, therefore, any side enters n − 3 inequality, i.e., in the left-hand
side of the obtained sum there stands (n − 3)p, where p is the sum of lengths of the n-gon’s
sides. Diagonal Am An enters two inequalities for p = n, q = m and for p = n − 1, q = m − 1;
hence, in the right-hand side stands 2d, where d is the sum of lengths of diagonals. Thus,
d
(n − 3)p < 2d. Therefore, np < n(n−3)/2
, as required.
9.21. Let us consider an arbitrary closed broken line with the vertices in vertices of the
given polygon. If we have two nonintersecting links then by replacing these links by the
SOLUTIONS
217
diagonals of the quadrilateral determined by them we enlarge the sum of the lengths of the
links. In this process, however, one broken line can get split into two nonintersecting ones.
Let us prove that if the number of links is odd then after several such operations we will
still get in the end a closed broken line (since the sum of lengths of the links increases each
time, there can be only a finite number of such operations). One of the obtained closed
broken lines should have an odd number of links but then any of the remaining links does
not intersect at least one of the links of this broken line (cf. Problem 23.1 a)); therefore, in
the end we get just one broken line.
Figure 97 (Sol. 9.21)
Now, let us successively construct a broken line with pairwise intersecting links (Fig.
97). For instance, the 10-th vertex should lie inside the shaded triangle and therefore,
the position of vertices is precisely as plotted on Fig. 97. Therefore, to convex polygon
A1 A3 A5 . . . A2n+1 A2 . . . A2n the broken line A1 A2 A3 . . . A2n+1 A1 corresponds.
9.22. Let the length of the third side be equal to n. From the triangle inequality we get
3.14 − 0.67 < n < 3.14 + 0.67. Since n is an integer, n = 3.
9.23. Clearly, AB + BC > AC, BC + CD > BD, CD + DE > CE, DE + EA > DA,
EA + AB > EB. Adding these inequalities we see that the sum of the lengths of the
pentagon’s diagonals is shorter than the doubled perimeter.
Figure 98 (Sol. 9.23)
The sum of the the diagonals’ lengths is longer than the sum of lengths of the sides of
the “rays of the star” and it, in turn, is greater than the perimeter of the pentagon (Fig.
98).
9.24. Suppose that c is the length of not the shortest side, for instance, a ≤ c. Then
a2 ≤ c2 and b2 < (a + c)2 ≤ 4c2 . Hence, a2 + b2 < 5c2 . Contradiction.
¯
¯
¯1
2S
1
1 ¯
,
c
=
,
it
follows
that
>
−
9.25. Since c > |b − a| and a = 2S
¯ ha hb ¯. Therefore, in
ha
hc
hc
our case hc <
20·12
8
= 30.
218
CHAPTER 9. GEOMETRIC INEQUALITIES
9.26. On sides AB, BC, CA take points C2 , A2 , B2 , respectively, so that A1 B2 k AB,
B1 C2 k BC, CA2 k CA (Fig. 99). Then
Similarly,
A1 B1 < A1 B2 + B2 B1 = (1 − λ)AB + (2λ − 1)CA.
BC1 < (1 − λ)BC + (2λ − 1)AB and C1 A1 < (1 − λ)CA + (2λ − 1)BC.
Adding these inequalities we get P1 < λP .
Figure 99 (Sol. 9.26)
Clearly, A1 B1 + AC > B1 C, i.e.,
Similarly,
A1 B1 + (1 − λ)BC > λ · CA.
B1 C1 + (1 − λ)CA > λ · AB
and C1 A1 + (1 − λ)AB > λ · BC.
Adding these inequalities we get P1 > (2λ − 1)P .
9.27. a) Passing from a nonconvex polygon to its convex hull we replace certain broken
lines formed by sides with segments of straight lines (Fig. 100). It remains to take into
account that any brokenline is longer than the line segment with the same endpoints.
Figure 100 (Sol. 9.27 a))
b) On the sides of the inner polygon construct half bands directed outwards; let the
parallel sides of half bands be perpendicular to the corresponding side of the polygon (Fig.
101).
Denote by P the part of the perimeter of the outer polygon corresponding to the boundary
of the polygon contained inside these half bands. Then the perimeter of the inner polygon
does not exceed P whereas the perimeter of the outer polygon is greater than P .
9.28. Since AO + BO > AB, BO + OC > BC and CO + OA > AC, it follows that
AB + BC + CA
AO + BO + CO >
.
2
Since triangle ABC contains triangle ABO, it follows that AB +BO +OA < AB +BC +CA
(cf. Problem 9.27 b)), i.e., BO + OA < BC + CA. Similarly,
AO + OC < AB + BC and CO + OB < CA + AB.
Adding these inequalities we get AO + BO + CO < AB + BC + CA.
SOLUTIONS
219
Figure 101 (Sol. 9.27 b))
9.29. It suffices to prove that ABCE and BCDE are parallelograms. Let us complement
triangle ABE to parallelogram ABC1 E. Then perimeters of triangles BC1 E and ABE are
equal and, therefore, perimeters of triangles BC1 E and BCE are equal. Hence, C1 = C
because otherwise one of the triangles BC1 E and BCE would have lied inside the other one
and their perimeters could not be equal. Hence, ABCE is a parallelogram. We similarly
prove that BCDE is a parallelogram.
√
9.30. Clearly, 2 = 2S = ab sin γ ≤ ab ≤ b2 , i.e., b ≥ 2.
9.31. Since EH is the midline of triangle ABD, it follows that SAEH = 41 SABD . Similarly,
SCF G = 41 SCBD . Therefore, SAEH + SCF G = 41 SABCD . Similarly, SBF E + SDGH = 41 SABCD .
It follows that
SABCD = 2SEF GH = EG · HF sin α,
where α is the angle between lines EG and HF . Since sin α ≤ 1, then SABCD ≤ EG · HF .
Adding equalities
−−→ −−→ −−→ −→
−−→ −→ −−→ −−→
EG = EB + BC + CG and EG = EA + AD + DG
we obtain
−−→
−−→ −→
−−→ −−→
−−→ −→
−−→ −−→
2EG = (EB + EA) + (BC + AD) + (DG + CG) = BC + AD.
Therefore, EG ≤ 21 (BC + AD). Similarly, HF ≤ 21 (AB + CD). It follows that
SABCD ≤ EG · HF ≤
(AB + CD)(BC + AD)
.
4
9.32. By Problem 9.31 SABCD ≤ 41 (AB + CD)(BC + AD). Since ab ≤ 41 (a + b)2 , it
1
follows that SABCD ≤ 16
(AB + CD + AD + BC)2 = 1.
9.33. From points B and C drop perpendiculars BB1 and CC1 to line AM . Then
2SAM B + 2SAM C = AM · BB1 + AM · CC1 ≤ AM · BC
because BB1 + CC1 ≤ BC. Similarly,
2SBM C + 2SBM A ≤ BM · AC and 2SCM A + 2SCM B ≤ CM · AB.
Adding these inequalities we get the desired statement.
9.34. Let on sides A1 A2 , A2 A3 , . . . , An A1 points B1 , . . . , Bn , respectively, be selected;
let O be the center of the circle. Further, let
Sk = SOBk Ak+1 Bk+1 =
OAk+1 · Bk Bk+1 sin ϕ
,
2
220
CHAPTER 9. GEOMETRIC INEQUALITIES
where ϕ is the angle between OAk+1 and Bk Bk+1 . Since OAk+1 = R and sin ϕ ≤ 1, it follows
that Sk ≤ 12 R · Bk Bk+1 . Hence,
S = S1 + · · · + S n ≤
R(B1 B2 + · · · + Bn B1 )
,
2
i.e., the perimeter of polygon B1 B2 . . . Bn is not less than 2S
.
R
1
2
2
9.35. We have 2SAOB ≤ AO · OB ≤ 2 (AO + BO ), where the equality is only possible
if ∠AOB = 90◦ and AO = BO. Similarly,
2SBOC ≤
CO2 + DO2
DO2 + AO2
BO2 + CO2
, 2SCOD ≤
and 2SDOA ≤
.
2
2
2
Adding these inequalities we get
2S = 2(SAOB + SBOC + SCOD + SDOA ) ≤ AO2 + BO2 + CO2 + DO2 ,
where the equality is only possible if AO = BO = CO = DO and ∠AOB = ∠BOC =
∠COD = ∠DOA = 90◦ , i.e., ABCD is a square and O is its center.
ABC
9.36. We have to prove that SSAM
≥ 4. Since AB = AM + M B = AM + AN =
N
AN + N C = AC, it follows that
AB · AC
(AM + AN )2
SABC
=
=
≥ 4.
SAM N
AM · AN
AM · AN
9.37. Let us apply Heron’s formula
S 2 = p(p − a)(p − b)(p − c).
Since p−a = (p1 −a1 )+(p2 −a2 ) and (x+y)2 ≥ 4xy, it follows that (p−a)2 ≥ 4(p1 −a1 )(p2 −a2 ).
Similarly,
(p − b)2 ≥ 4(p1 − b1 )(p2 − b2 ), (p − c)2 ≥ 4(p1 − c1 )(p2 − c2 ) and p2 ≥ 4p1 p2 .
Multiplying these inequalities we get the desired statement.
9.38. For definiteness, we may assume that rays BA and CD, BC and AD intersect
(Fig. 102). If we complement triangle ADC to parallelogram ADCK, then point K occurs
inside quadrilateral ABCD. Therefore,
2S ≥ 2SABK + 2SBCK = AB · AK sin α + BC · CK sin β =
AB · CD sin α + BC · AD sin β.
The equality is obtained if point D lies on segment AC.
Figure 102 (Sol. 9.38)
SOLUTIONS
221
Let point D′ be symmetric to point D through the midperpendicular to segment AC.
Then
2S = 2SABCD′ = 2SABD′ + 2SBCD′ ≤
AB · AD′ + BC · CD′ = AB · CD + BC · AD.
9.39. Thanks to the
and the mean arithmetic,
qinequality between the√mean geometric
√
√
b
c
3
a
3 abc
we have α + β + γ ≥ 3 αβγ = 2 because α = 2 bc, β = 2 ca and γ = 2 ab, cf. Problem
1.33.
9.40. The inequalities α < α1 , β < β1 and γ < γ1 cannot hold simultaneously. Therefore,
for instance, α1 ≤ α ≤ 90◦ ; hence, sin α1 ≤ sin α. It follows that 2S1 = a1 b1 sin α1 ≤
k 2 ab sin α = 2k 2 S.
9.41. a) Let chords AE and BD intersect diameter CM at points K and L, respectively.
AC 2
= BC
Then AC 2 = CK · CM and BC 2 = CL · CM . It follows that CK
2 < 4. Moreover,
CL
SACE
AE
AE
AE·CK
= AC < 2. Therefore, SBCD = BD·CL < 8.
BD
b) Let H be the midpoint of segment BC. Since ∠CBD = ∠BCD = ∠ABD, it follows
AB
AD
= BH
> 1.
that D is the intersection point of the bisectors of triangle ABC. Hence, DH
1
Therefore, SM AN > 4 SABC and
BC · AH
SABC
BC · DH
<
=
.
2
4
4
9.42. Let us cut off the obtained polygon rectangles with side h constructed outwards
on the sides of the initial polygon (Fig. 103). Then beside the initial polygon there will
be left several quadrilaterals from which one can compose a polygon circumscribed about a
circle of radius h. The sum of the areas of these quadrilaterals is greater than the area of
the circle of radius h, i.e., greater than πh2 . It is also clear that the sum of areas of the cut
off rectangles is equal to P h.
SBCD =
Figure 103 (Sol. 9.42)
9.43. Let s, s1 , . . . , sn be the areas of the square and the rectangles that constitute it,
respectively; S, S1 , . . . , Sn the areas of the disks circumscribed about the square and the
rectangles, respectively. Let us prove that sk ≤ 2Sπk . Indeed, if the sides of the rectangle are
2
2
equal to a and b, then sk = ab and Sk = πR2 , where R2 = a4 + b4 . Therefore, sk = ab ≤
2
a2 +b2
= 2πR
= 2Sπk . It follows that
2
π
2(S1 + · · · + Sn )
2S
= s = s1 + · · · + sn ≤
.
π
π
222
CHAPTER 9. GEOMETRIC INEQUALITIES
9.44. Let, for definiteness, ABC be the triangle of the least area. Denote the intersection
point of diagonals AD and EC by F . Then SABCDE < SAED + SEDC + SABCF . Since
point F lies on segment EC and SEAB ≥ SCAB , it follows that SEAB ≥ SF AB . Similarly,
SDCB ≥ SF CB . Therefore, SABCF = SF AB +SF CB ≤ SEAB +SDCB . It follows that SABCDE <
SAED + SEDC + SEAB + SDCB and this is even a stronger inequality than the one required.
9.45. a) Denote the intersection points of diagonals AD and CF , CF and BE, BE
and AD by P , Q, R, respectively (Fig. 104). Quadrilaterals ABCP and CDEQ have no
common inner points since sides CP and QC lie on line CF and segments AB and DE lie
on distinct sides of it. Similarly, quadrilaterals ABCP , CDEQ and EF AR have no pairwise
common inner points. Therefore, the sum of their areas does not exceed S.
Figure 104 (Sol. 9.45 a))
It follows that the sum of the areas of triangles ABP , BCP , CDQ, DEQ, EF R, F AR
does not exceed S, i.e., the area of one of them, say ABP , does not exceed 16 S. Point P
lies on segment CF and, therefore, one of the points, C or F , is distant from line AB not
further than point P . Therefore, either SABC ≤ SABP ≤ 61 S or SABF ≤ SABP ≤ 16 S.
b) Let ABCDEF GH be a convex octagon. First, let us prove that quadrilaterals ABEF ,
BCF G, CDGH and DEHA have a common point. Clearly, a convex quadrilateral KLM N
(Fig. 105) is the intersection of ABEF and CDGH. Segments AF and HC lie inside
angles ∠DAH and ∠AHE, respectively; hence, point K lies inside quadrilateral DEHA.
We similarly prove that point M lies inside quadrilateral DEHA, i.e., the whole segment
KM lies inside it. Similarly, segment LN lies inside quadrilateral BCF G. The intersection
point of diagonals KM and LN belongs to all our quadrilaterals; denote it by O.
Figure 105 (Sol. 9.45 b))
Let us divide the 8-gon into triangles by connecting point O with the vertices. The area
of one of these triangles, say ABO, does not exceed 81 S. Segment AO intersects side KL
at a point P , therefore, SABP < SABO ≤ 81 S. Since point P lies on diagonal CH, it follows
that either SABC ≤ SABP ≤ 18 S or SABH ≤ SABP ≤ 18 S.
9.46. Let us draw through all the vertices of the polygon lines parallel to one pair of
sides of the square thus dividing the square into strips. Each such strip cuts off the polygon
SOLUTIONS
223
either a trapezoid or a triangle. It suffices to prove that the length of one of the bases of these
trapezoids is greater than 0.5. Suppose that the length of each base of all the trapezoids does
not exceed 0.5. Then the area of each trapezoid does not exceed a half height of the strip
that confines it. Therefore, the area of the polygon, equal to the sum of areas of trapezoids
and triangles into which it is cut, does not exceed a half sum of heights of the strips, i.e.,
does not exceed 0.5. Contradiction.
9.47. a) Let P1 , . . . , Pn be the given points. Let us connect point P1 with the vertices
of the square. We will thus get four triangles. Next, for k = 2, . . . , n let us perform the
following operation. If point Pk lies strictly inside one of the triangles obtained earlier, then
connect it with the vertices of this triangle.
If point Pk lies on the common side of two triangles, then connect it with the vertices
of these triangles opposite to the common side. Each such operation increases the total
number of triangles by 2. As a result we get 2(n + 1) triangles. The sum of the areas of
1
these triangles is equal to 1, therefore, the area of any of them does not exceed 2(n+1)
.
b) Let us consider the least convex polygon that contains the given points. Let is have
k vertices. If k = n then this k-gon can be divided into n − 2 triangles by the diagonals
that go out of one of its vertices. If k < n, then inside the k-gon there are n − k points and
it can be divided into triangles by the method indicated in heading a). We will thus get
k + 2(n − k − 1) = 2n − k − 2 triangles. Since k < n, it follows that 2n − k − 2 > n − 2.
The sum of the areas of the triangles of the partition is less than 1 and there are not less
1
.
than n − 2 of them; therefore, the area of at least one of them does not exceed n−2
9.48. a) We may assume that the circumscribed n-gon A1 . . . An and the inscribed n-gon
B1 . . . Bn are placed so that lines Ai Bi intersect at the center O of the given circle. Let Ci
and Di be the midpoints of sides Ai Ai+1 and Bi Bi+1 , respectively. Then
SOBi Ci = p · OBi · OCi , SOBi Di = p · OBi · ODi and SOAi Ci = p · OAi · OCi ,
2
=
where p = 21 sin ∠Ai OCi . Since OAi : OCi = OBi : ODi , it follows that SOB
i Ci
SOBi Di SOAi Ci . It remains to notice that the area of the part of the disk confined inside
angle ∠Ai OCi is greater than SOBi Ci and the areas of the parts of the inscribed and circumscribed n-gons confined inside this angle are equal to SOBi Di and SOAi Ci , respectively.
b) Let the radius of the circle be equal to R. Then P1 = 2nR sin πn , P2 = 2nR tan πn and
L = 2πR. We have to prove that sin x tan x > x2 for 0 < x ≤ 31 π. Since
µ
2
sin x
x
4
¶2
≥
µ
x2
1−
6
¶2
x2 x4
+
=1−
3
36
and 0 < cos x ≤ 1 − x2 + x24 (see Supplement to this chapter), it remains to verify that
2
4
2
4
1 − x3 + x36 ≥ 1 − x2 + x24 , i.e., 12x2 > x4 . For x ≤ 13 π this inequality is satisfied.
9.49. Let O be the center of homothety that sends the inscribed circle into the circumscribed one. Let us divide the plane by rays that exit from point O and pass through the
vertices of the polygon and the tangent points of its sides with the inscribed circle (Fig.
106).
It suffices to prove the required inequality for the parts of disks and the polygon confined
inside each of the angles formed by these rays. Let the legs of the angle intersect the
inscribed circle at points P , Q and the circumscribed circle at points R, S so that P is
the tangency point and S is a vertex of the polygon. The areas of the parts of disks are
greater than the areas of triangles OP Q and ORS and, therefore, it suffices to prove that
2SOP S ≤ SOP Q + SORS . Since 2SOP S = 2SOP Q + 2SP QS and SORS = SOP Q + SP QS + SP RS ,
224
CHAPTER 9. GEOMETRIC INEQUALITIES
Figure 106 (Sol. 9.49)
it remains to prove that SP QS ≤ SP RS . This inequality is obvious, because the heights of
triangles P QS and P RS dropped to bases P Q and RS, respectively, are equal and P Q < RS.
9.50. It suffices to prove that both triangles contain the center O of the disk. Let us
prove that if triangle ABC placed in the disk of radius 1 does not contain the center of the
disk, then its area is less than 1. Indeed, for any point outside the triangle there exists a
line that passes through two vertices and separating this point from the third vertex. Let,
for definiteness, line AB separate points C and O. Then hc < 1 and AB < 2, hence,
S = 12 hc · AB < 1.
9.51. a) On the sides of the polygon, construct inwards rectangles whose other side is
equal to R = PS . The rectangles will not cover the whole polygon (these rectangles overlap
and can stick out beyond the limits of the polygon whereas the sum of their areas is equal
to the area of the polygon). An uncovered point is distant from every side of the polygon
further than by R, consequently, the disk of radius R centered at this point entirely lies
inside the polygon.
b) Heading a) implies that in the inner polygon a disk of radius PS22 can be placed. Clearly,
this disk lies inside the outer polygon. It remains to prove that if inside a polygon a disk of
. For this let us connect (with lines) the center O of the disk with
radius R lies, then R ≤ 2S
P
the vertices of the polygon. These lines split the polygon into triangles whose respective
areas are equal to 12 hi ai , where hi is the distance from pointP
O to theP
i-th side and ai is the
length of the i-th side. Since hi ≥ R, we deduce that 2S = hi ai ≥
Rai = RP .
9.52. First, let us consider the case when two sides of a parallelogram lie on lines AB
and AC and the fourth vertex X lies on side BC. If BX : CX = x : (1 − x), then the ratio
of the area of the parallelogram to the area of the triangle is equal to 2x(1 − x) ≤ 21 .
Figure 107 (Sol. 9.52)
In the general case let us draw parallel lines that contain a pair of sides of the given
parallelogram (Fig. 107). The area of the given parallelogram does not exceed the sum
of areas of the shaded parallelograms which fall in the case considered above. If lines that
contain a pair of sides of the given parallelogram only intersect two sides of the triangle,
then we can restrict ourselves to one shaded parallelogram only.
SOLUTIONS
225
9.53. First, let us consider the following case: two vertices A and B of triangle ABC lie
on one side P Q of the parallelogram. Then AB ≤ P Q and the height dropped to side AB is
not longer than the height of the parallelogram. Therefore, the area of triangle ABC does
not exceed a half area of the parallelogram.
Figure 108 (Sol. 9.53)
If the vertices of the triangle lie on distinct sides of the parallelogram, then two of them
lie on opposite sides. Let us draw through the third vertex of the triangle a line parallel to
these sides (Fig. 108). This line cuts the parallelogram into two parallelograms and it cuts
the triangle into two triangles so that two vertices of each of these triangles lie on sides of
the parallelogram. We get the case already considered.
9.54. Let M be the midpoint of the longest side BC of the given acute triangle ABC.
The circle of radius M A centered at M intersects rays M B and M C at points B1 and
C1 , respectively. Since ∠BAC < 90◦ , it follows that M B < M B1 . Let, for definiteness,
∠AM B√≤ ∠AM C, i.e., ∠AM B ≤ 90◦ . Then AM 2 + M B 2 ≤ AB 2 ≤ BC 2 = 4M B 2 , i.e.,
AM ≤ 3BM . If AH is a height of triangle ABC, then AH · BC = 2 and, therefore,
√
√
B1 C1 · AH
SAB1 C1 =
= AM · AH ≤ 3BM · AH = 3.
2
9.55. a) Let AB be the longest of the diagonals and sides of the given polygon M .
Polygon M is confined inside the strip formed by the perpendiculars to segment AB passing
through points A and B. Let us draw two baselines to M parallel to AB. Let them intersect
polygon M at points C and D. As a result we have confined M into a rectangle whose area
is equal to 2SABC + 2SABD ≤ 2S.
Figure 109 (Sol. 9.55)
b) Let M be the initial polygon, l an arbitrary line. Let us consider the polygon M1 one
of whose sides is the projection of M to l and the lengths of the sections of polygons M and
M1 by any line perpendicular to l are equal (Fig. 109). It is easy to verify that M1 is also
a convex polygon and its area is equal to S. Let A be the most distant from l point of M1 .
The line equidistant from point A and line l intersects the sides of M1 at points B and C.
226
CHAPTER 9. GEOMETRIC INEQUALITIES
Let us draw base lines through points B and C. As a result we will circumscribe a
trapezoid about M1 (through point A a base line can also be drawn); the area of this
trapezoid is no less than S. If the height of the trapezoid, i.e., the distance from A to l
is equal to h then its area is equal to h · BC and, therefore, h · BC ≥ S. Let us consider
sections P Q and RS of polygon M by lines perpendicular to l and passing through B and
C. The lengths of these sections are equal to 21 h and, therefore, P QRS is a parallelogram
whose area is equal to 12 BC · h ≥ 21 S.
9.56. a) Let us confine the polygon in the strip formed by parallel lines. Let us shift
these lines parallelly until some vertices A and B of the polygon lie on them. Then let us
perform the same for the strip formed by lines parallel to AB. Let the vertices that lie on
these new lines be C and D (Fig. 110). The initial polygon is confined in a parallelogram
and, therefore, the area of this parallelogram is not less than 1. On the other hand, the
sum of areas of triangles ABC and ADB is equal to a half area of the parallelogram and,
therefore, the area of one of these triangles is not less than 14 .
Figure 110 (Sol. 9.56 a))
b) As in heading a) let us confine the polygon in a strip formed by parallel lines so that
some vertices, A and B, lie on these lines. Let d be the width of this strip. Let us draw
three lines that divide this strip into equal strips of width 14 d. Let the first and the third
lines intersect sides of the polygon at points K, L and M , N , respectively (Fig. 111).
Figure 111 (Sol. 9.56 b))
Let us extend the sides on which points K, L, M and N lie to the intersection with the
sides of the initial strip and with the line that divides it in halves. In this way we form two
trapezoids with the midlines KL and M N and heights of length 21 d each.
Since these trapezoids cover the whole polygon, the sum of their areas is not less than its
area, i.e., 12 (d · KL + d · M N ) ≥ 1. The sum of areas of triangles AM N and BKL contained
in the initial polygon is equal to 18 (3d · M N + 3d · KL) ≥ 43 . Therefore, the area of one of
these triangles is not less than 83 .
9.57. Let us prove that there exists even three last vertices satisfying the required
condition. Let αi be the angle between the i-th and (i + 1)-th sides βi = π − αi ; let ai be
the length of the i-th side.
SOLUTIONS
227
a) The area of the triangle formed by the i-th and (i + 1)-th sides is equal to Si =
Let S be the least of these areas. Then 2S ≤ ai ai+1 sin αi ; hence,
ai ai+1 sin αi
.
2
(2S)n ≤ (a21 . . . a2n )(sin α1 . . . sin αn ) ≤ a21 . . . a2n .
1
By the inequality between the mean arithmetic and the mean geometric we have (a1 . . . an ) n ≤
a1 +···+an
and, therefore,
n
(a1 + · · · + an )2
2S ≤ (a1 . . . an ) ≤
.
n2
Since ai ≤ pi +qi , where pi and qi are projections of the i-th side to a vertical and a horizontal
sides of the square, it follows that
2
n
a1 + · · · + an ≤ (p1 + · · · + pn ) + (q1 + · · · + qn ) ≤ 4.
Hence, 2S ≤ 16n2 , i.e., S ≤ n82 .
b) Let us make use of the inequality
16
(sin α1 . . . sin αn )
n2
proved above. Since sin αi = sin βi and β1 + · · · + βn = 2π, it follows that
1
1
2π
β1 + · · · + βn
(sin α1 . . . sin αn ) n = (sin β1 . . . sin βn ) n ≤
=
.
n
n
Hence, 2S ≤ 32π
, i.e., S ≤ 16π
.
n3
n3
9.58. Let li be the length of the i-th link of the broken line; ai and bi the lengths of its
projections to the sides of the square. Then li ≤ ai + bi . It follows that
2
1
2S ≤ (a1 . . . an ) n (sin α1 . . . sin αn ) n ≤
1000 = l1 + · · · + ln ≤ (a1 + · · · + an ) + (b1 + · · · + bn ),
i.e., either a1 + · · · + an ≥ 500 or b1 + · · · + bn ≥ 500. If the sum of the lengths of the links’
projections on a side of length 1 is not less than 500, then not fewer than 500 distinct lengths
of the broken line are projected into one of the points of this side, i.e., the perpendicular to
the side that passes through this point intersects the broken line at least at 500 points.
9.59. The locus of points distant from the given segment not further than by ε is depicted
on Fig. 112. The area of this figure is equal to πε2 +2εl, where l is the length of the segment.
Figure 112 (Sol. 9.59)
Let us construct such figures for all N links of the given broken lines. Since neighbouring
figures have N − 1 common disks of radius ε centered at vertices of the broken line which
are not its endpoints, it follows that the area covered by these figures does not exceed
N πε2 + 2ε(l1 + · · · + ln ) − (N − 1)πε2 = 2εL + πε2 .
This figure covers the whole square since any point of the square is distant from a point of
1
− πε
.
the broken line by less than ε. Hence, 1 ≤ 2εL + πε2 , i.e., L ≥ 2ε
2
9.60. Let us divide the square into n vertical strips that contain n points each. Inside
each strip let us connect points downwards thus getting n broken lines. These broken lines
can be connected into one broken line in two ways: Fig. 113 a) and b).
228
CHAPTER 9. GEOMETRIC INEQUALITIES
Figure 113 (Sol. 9.60)
Let us consider the segments that connect distinct bands. The union of all such segments
obtained in both ways is a pair of broken lines such that the sum of the lengths of the
horizontal projections of each of them does not exceed 1. Therefore, the sum of the lengths
of horizontal projections of the connecting segments for one of these ways does not exceed
1.
Let us consider such a connection. The sum of the lengths of the horizontal projections
for connecting links does not exceed 1 and for all the other links it does not exceed (n −
1)(h1 + · · · + hn ), where hi is the width of the i-th strip. Clearly, h1 + · · · + hn = 1. The
sum of the vertical projections of all links of the broken line does not exceed n. As a result
we deduce that the sum of the vertical and horizontal projections of all the links does not
exceed 1 + (n − 1) + n = 2n and, therefore, the length of the broken line does not exceed 2n.
9.61. Let M and N be the endpoints of the broken line. Let us traverse along the
broken line from M to N . Let A1 be the first of points of the broken line that we meet
whose distance from a vertex of the square is equal to 0.5. Let us consider the vertices of
the square neighboring to this vertex. Let B1 be the first after A1 point of the broken line
distant from one of these vertices by 0.5. Denote the vertices of the square nearest to points
A1 and B1 by A and B, respectively (Fig. 114).
Figure 114 (Sol. 9.61)
Denote the part of the broken line from M to A1 by L1 and the part from A1 to N by
L2 . Let X and Y be the sets of points that lie on AD and distant not further than by 0.5
from L1 and L2 , respectively. By hypothesis, X and Y cover the whole side AD. Clearly,
A ∈ X and D 6∈ X; hence, D ∈ Y , i.e., both sets, X and Y , are nonempty. But each of
these sets consists of several segments and, therefore, they should have a common point P .
Therefore, on L1 and L2 , there are points F1 and F2 for which P F1 ≤ 0.5 and P F2 ≤ 0.5.
Let us prove that F1 and F2 are the points to be found. Indeed, F1 F2 ≤ F1 P + P F2 ≤ 1.
On the other hand, while traversing from F1 to F2 we should pass through point B; and we
SOLUTIONS
229
have F1 B1 ≥ 99 and F2 B1 ≥ 99 because point B1 is distant from side BC no further than
by 0.5 while F1 and F2 are distant from side AD not further than by 0.5.
9.62. Let ∠A = ∠B. It suffices to prove that if AD < BC; then ∠D > ∠C. On side
BC, take point D1 such that BD1 = AD. Then ABD1 D is an isosceles trapezoid. Hence,
∠D > ∠D1 DA = ∠DD1 B ≥ ∠C.
9.63. Let B1 and C1 be the projections of points B and C on base AD. Since ∠BAB1 <
∠CDC1 and BB1 = CC1 , it follows that AB1 > DC1 and, therefore, B1 D < AC1 . It follows
that
BD2 = B1 D2 + B1 B 2 < AC12 + CC12 = AC 2 .
9.64. Let angles ∠B and ∠D of quadrilateral ABCD be obtuse ones. Then points B
and D lie inside the circle with diameter AC. Since the distance between any two points
that lie inside the circle is less than its diameter, BD < AC.
9.65. In an isosceles trapezoid ABCD diagonals AC and BD are equal. Therefore,
BM + (AM + CM ) ≥ BM + AC = BM + BD ≥ DM.
9.66. Let O be the midpoint of segment BD. Point A lies inside the circle with diameter
BD, hence, OA < 21 BD. Moreover, F O = 12 CD. Therefore, 2F A ≤ 2F O + 2OA <
CD + BD.
1
1
9.67. On rays AB, AC and AD mark segments AB ′ , AC ′ and AD′ of length AB
, AC
1
and AD
. Then AB : AC = AC ′ : AB ′ , i.e., △ABC ∼ △AC ′ B ′ . The similarity coefficient of
1
BC
CD
these triangles is equal to AB·AC
and therefore, B ′ C ′ = AB·AC
. Analogously, C ′ D′ = AC·AD
BD
. Substituting these expressions in the inequality B ′ D′ ≤ B ′ C ′ + C ′ D′
and B ′ D′ = AB·AD
and multiplying both sides by AB · AC · AD, we get the desired statement.
9.68. Clearly,
SABCD = SABC + SACD = 2SAM C + 2SAN C = 2(SAM N + SCM N ).
If segment AM intersects diagonal BD at point A1 , then SCM N = SA1 M N < SAM N . Therefore, SABCD < 4SAM N .
9.69. Diagonals AC and BD intersect at point O. Let, for definiteness, point P lie in
side of AOB. Then AP + BP ≤ AO + BO < AC + BD (cf. the solution of Problem 9.28)
and CP + DP < CB + BA + AD.
9.70. Let ri , Si and pi be the radii of the inscribed circles, the areas and semiperimeters
of the obtained triangles, respectively. Then
X µ Si ¶ 4S
X
X µ Si ¶
=
.
>4
Q≥2
ri = 2
pi
P
P
9.71. Let AC ≤ BD. Let us drop from vertices A and C perpendiculars AA1 and CC1
to diagonal BD. Then AA1 + CC1 ≤ AC ≤ BD and, therefore, either AA1 ≤ 21 BD or
CC1 ≤ 12 BD.
9.72. Let us draw through the endpoints of segment KL lines perpendicular to it and
consider projections to these lines of the vertices of the quadrilateral. Consider also the
intersection points of lines AC and BD with these lines, cf. Fig. 115.
Let, for definiteness, point A lie inside the strip determined by these lines and point B
outside it. Then we may assume that D lies inside the strip, because otherwise BD > KL
and the proof is completed. Since
A1 K
C1 L
CC ′
AA′
≤
=
≤
,
BB ′
B1 K
D1 L
DD′
230
CHAPTER 9. GEOMETRIC INEQUALITIES
Figure 115 (Sol. 9.72)
then either AA′ ≤ CC ′ (and, therefore, AC > KL) or BB ′ ≥ DD′ (and, therefore, BD >
KL).
9.73. Let us introduce the notations as plotted on Fig. 116. All the parallelograms
considered have a common center (thanks to Problem 1.7). The lengths of the sides of
parallelogram P3 are equal to a + a1 and b + b1 and the lengths of the sides of parallelogram
P1 are equal to a + a1 + 2x and b + b1 + 2y, consequently, we have to verify that either
a + a1 + 2x ≤ 2(a + a1 ) or b + b1 + 2y ≤ 2(b + b1 ), i.e., either 2x ≤ a + a1 or 2y ≤ b + b1 .
Figure 116 (Sol. 9.73)
√
√
Suppose that a + a1 < 2x and b + b1 < 2y. Then aa1 ≤ 21 (a + a1 ) < x and bb1 < y.
On the other hand, the equality of the areas of
shaded parallelograms (cf. Problem 4.19)
√ √
shows that ab = xy = a1 b and, therefore, aa1 bb1 = xy. Contradiction.
9.74. Let the angles of the pentagon be equal to α, α + γ, α + 2γ, α + 3γ, α + 4γ,
where α, γ ≥ 0. Since the sum of the angles of the pentagon is equal to 3π, it follows that
5α + 10γ = 3π. Since the pentagon is a convex one, each of its angles is less than π, i.e.,
either α + 4γ < π or −5 21 α − 10γ > − 12 5π. Taking the sum of the latter inequality with
5α + 10γ = 3π we get 5α
> π2 , i.e., α > π5 = 36◦ .
2
9.75. Clearly,
−→ −−→ −−→ −−→
4 = AE 2 = |AB + BC + CD + DE|2 =
−→ −−→
−→ −−→ −−→ −−→
−−→ −−→
|AB + BC|2 + 2(AB + BC, CD + DE) + |CD + DE|2 .
Since ∠ACE = 90◦ , we have
−→ −−→ −−→ −−→
−→ −−→
(AB + BC, CD + DE) = (AC, CE) = 0.
Hence,
−−→ −−→
−→ −−→
4 = |AB + BC|2 + |CD + DE|2 =
−→ −−→
−−→ −−→
AB 2 + BC 2 + CD2 + DE 2 + 2(AB, BC) + 2(CD, DE),
i.e., it suffices to prove that
−→ −−→
−−→ −−→
abc < 2(AB, BC) and bcd < 2(CD, DE).
SOLUTIONS
231
Since
−→ −−→
2(AB, BC) = 2ab cos(180◦ − ∠ABC) = 2ab cos AEC = ab · CE and c < CE,
−→ −−→
it follows that abc < 2(AB, BC).
The second inequality is similarly proved, because in notations A1 = E, B1 = D, C1 =
−−→ −−→
C, a1 = d, b1 = c, c1 = b the inequality bcd < 2(CD, DE) takes the form a1 b1 c1 <
−−−→ −−−→
2(A1 B1 , B1 C1 ).
9.76. Let B be the midpoint of side A1 A2 of the given hexagon A1 . . . A6 and O its
center. We may assume that point P lies inside triangle A1 OB. Then P A3 ≥ 1 because the
distance from point A3 to line BO is equal to 1; since the distances from points A4 and A5
to line A3 A6 are equal to 1, we deduce that P A4 ≥ 1 and P A5 ≥ 1.
9.77. Suppose that the radii of the circumscribed circles of triangles ACE and BDF
are greater than 1. Let O be the center of the circumscribed circle of triangle ACE. Then
∠ABC > ∠AOC, ∠CDE > ∠COE and ∠EF A > ∠EOA and, therefore, ∠B + ∠D + ∠F >
2π. Similarly, ∠A + ∠C + ∠E > 2π, i.e., the sum of the angles of hexagon ABCDEF is
greater than 4π. Contradiction.
Remark. We can similarly prove that the radius of the circumscribed circle of one of
triangles ACE and BDF is not less than 1.
9.78. We may assume that AE ≤ AC ≤ CE. By Problem 9.67
AD · CE ≤ AE · CD + AC · DE < AE + AC ≤ 2CE,
i.e., AD < 2.
9.79. Since ∠A1 = 180◦ −
A6 A4 , it follows that
1
2
⌣ A2 A7 , ∠A3 = 180◦ −
1
2
⌣ A4 A2 and ∠A5 = 180◦ −
1
2
⌣
◦
4 A2 −⌣A6 A4
∠A1 + ∠A3 + ∠A5 = 2 · 180◦ + 360 −⌣A2 A7 −⌣A
=
2
2 · 180◦ + ⌣A27 A6 .
Since the center of the circle lies inside the hexagon, it follows that ⌣ A7 A6 < 180◦ and,
therefore, ∠A1 + ∠A3 + ∠A5 < 360◦ + 90◦ = 450◦ .
9.80. a) We have to prove that if c is the hypothenuse of the right triangle and a and b
√ , i.e., (a + b)2 ≤ 2(a2 + b2 ). Clearly,
are its legs, then c ≥ a+b
2
(a + b)2 = (a2 + b2 ) + 2ab ≤ (a2 + b2 ) + (a2 + b2 ) = 2(a2 + b2 ).
b) Let di be the length of the i-th side of the polygon; xi and yi the lengths of its
projections to coordinate axes. Then x1 + · · · + xn ≥ 2a, y1 + · · · + yn ≥ 2b. By heading a)
i
. Therefore,
di ≥ xi√+y
2
x1 + · · · + x n + y1 + · · · + y n √
√
≥ 2(a + b).
d1 + · · · + dn ≥
2
9.81. Let us take a segment of length P and place the sides of the polygon on the
segment as follows: on one end of the segment place the greatest side, on the other end place
the second long side; place all the other sides between them. Since any side of the polygon
is shorter than 12 P , the midpoint O of the segment cannot lie on these two longest sides.
The length of the side on which point O lies, does not exceed 13 P (otherwise the first two
sides would also have been longer than 13 P and the sum of the three sides would have been
greater than P ) and, therefore, one of its vertices is distant from O not further than by 16 P .
This vertex divides the segment into two segments to be found since the difference of their
lengths does not exceed 26 P = 31 P .
232
CHAPTER 9. GEOMETRIC INEQUALITIES
9.82. Let βk = ∠OAk Ak+1 . Then xk sin βk = dk = xk+1 sin(αk+1 − βk+1 ). Hence,
P
P
2
d
=
xk (sin(α
k
¡ αk k −¢βk ) +Psin βk ) =
P
αk
2 xk sin 2 cos 2 − βk ≤ 2 xk sin α2k .
It is also clear that
Therefore,
Ak Ak+1 = xk cos βk + xk+1 cos(αk+1 − βk+1 ).
P
P
2p =P Ak Ak+1 = ¡ xk (cos(α
k ) + cos βk ) =
¢ k−β
P
2 xk cos α2k cos α2k − βk ≤ 2 xk cos α2k .
In both cases the equality is only attained if αk = 2βk , i.e., O is the center of the inscribed
circle.
9.83. Suppose that the center O of polygon M2 lies outside polygon M1 . Then there
exists a side AB of polygon M1 such that polygon M1 and point O lie on distinct sides of
line AB. Let CD be a side of M1 parallel to AB. The distance between lines AB and CD
is equal to the radius of the inscribed circle S of polygon M2 and, therefore, line CD lies
outside S. On the other hand, segment CD lies inside M2 . Therefore, segment CD is shorter
than a half side of polygon M2 , cf. Problem 10.66. Contradiction.
9.84. Let A1 be the nearest to O vertex of the polygon. Let us divide the polygon into
triangles by the diagonals that pass through vertex A1 . Point O lies inside one of these
triangles, say, in triangle A1 Ak Ak+1 . If point O lies on side A1 Ak , then ∠A1 OAk = π and
the problem is solved.
Therefore, let us assume that point O lies strictly inside triangle A1 Ak Ak+1 . Since A1 O ≤
Ak O and A1 O ≤ Ak+1 O, it follows that ∠A1 Ak O ≤ ∠Ak A1 O and ∠A1 Ak+1 O ≤ ∠Ak+1 A1 O.
Hence,
∠Ak OA1 + ∠Ak+1 OA1 =
(π − ∠OA1 Ak − ∠OAk A1 ) + (π − ∠OA1 Ak+1 − ∠OAk+1 A1 ) ≥
,
2π − 2∠OA1 Ak − 2∠OA1 Ak+1 = 2π − 2∠Ak A1 Ak+1 = 2π − 2π
n
¡
¢
i.e., one of the angles ∠Ak OA1 and ∠Ak+1 OA1 is not less than π 1 − n1 .
9.85. Let d be the length of the longest diagonal (or side) AB of the given n-gon. Then
the perimeter of the n-gon does notPexceed πd (Problem
Let A′i be the projection
P 13.42).
1
1
′
′
of Ai to segment AB. Then either
AAi ≥ 2 nd or
BAi ≥ 2 nd (Problem 9.87); let, for
P
P
definiteness, the first inequality hold. Then
AAi >
AA′i ≥ 21 nd > πd ≥ P because
1
n ≥ 3.5 > π. Any point of the n-gon sufficiently close to vertex A possesses the required
2
property.
9.86. a) First, suppose that ∠Ai > ∠Bi and for all the other considered pairs of angles
an equality takes place. Let us arrange polygons so that vertices A1 , . . . , Ai coincide with B1 ,
. . . , Bi . In triangles A1 Ai An and A1 Ai Bn sides Ai An and Ai Bn are equal and ∠A1 Ai An >
∠A1 Ai Bn ; hence, A1 An > A1 Bn .
If several angles are distinct, then polygons A1 . . . An and B1 . . . Bn can be included in a
chain of polygons whose successive terms are such as in the example considered above.
b) As we completely traverse the polygon we encounter the changes of minus sign by
plus sign as often as the opposite change. Therefore, the number of pairs of neighbouring
vertices with equal signs is an even one. It remains to verify that the number of sign changes
cannot be equal to 2 (the number of sign changes is not equal to zero because the sums of
the angles of each polygon are equal).
Suppose the number of sign changes is equal to 2. Let P and Q, as well as P ′ and Q′ be
the midpoints of sides of polygons A1 . . . An and B1 . . . Bn on which a change of sign occurs.
We can apply the statement of heading a) to pairs of polygons M1 and M1′ , M2 and M2′
SOLUTIONS
233
Figure 117 (Sol. 9.86)
(Fig. 117); we get P Q > P ′ Q′ in the one case, and P Q < P ′ Q′ in the other one, which is
impossible.
9.87. Let A and B be the midpoints
of P
the segment; X1 , . . . , Xn the given
P
P points. 1Since
AXi + BXi = 1, it follows that
AXi + BXi = n. Therefore, either
AXi ≥ 2 n or
P
1
BXi ≥ 2 n.
Figure 118 (Sol. 9.88)
9.88. Let us draw a wire along segment AB circumventing the encountered trees along
the shortest arc as on Fig. 118. It suffices to prove that the way along an arc of the circle
is not more than 1.6 times longer than the way along the line. The ratio of the length of an
arc with the angle value 2ϕ to the chord it subtends is equal to sinϕ ϕ . Since 0 < ϕ ≤ π2 , it
follows that sinϕ ϕ ≤ π2 < 1.6.
9.89. Let the trees of height a1 > a2 > · · · > an grow at points A1 , . . . , An . Then by
the hypothesis
A1 A2 ≤ |a1 − a2 | = a1 − a2 , . . . , An−1 An ≤ an−1 − an .
It follows that the length of the broken line A1 A2 . . . An does not exceed
(a1 − a2 ) + (a2 − a3 ) + · · · + (an−1 − an ) = a1 − an < 100 m.
This broken line can be fenced by a fence, whose length does not exceed 200 m (Fig. 119).
Figure 119 (Sol. 9.89)
9.90. In the obtained pentagon, distinguish the parts that were glued (on Fig. 120
these parts are shaded). All the sides that do not belong to the shaded polygons enter the
234
CHAPTER 9. GEOMETRIC INEQUALITIES
perimeters of the initial and the obtained polygons. The sides of the shaded polygons that
lie on the line along which the folding was performed enter the perimeter of the obtained
polygon whereas all the other sides enter the perimeter of the initial polygon.
Figure 120 (Sol. 9.90)
Since for any polygon the sum of its sides that lie on a line is less than the sum of the
other sides, the perimeter of the initial polygon is always longer than the perimeter of the
obtained one.
9.91. On the broken line, take two points A and B, that divide its perimeter in halves.
Then AB ≤ 12 . Let us prove that all the points of the broken line lie inside the circle of
radius 14 centered at the midpoint O of segment AB. Let M be an arbitrary point of the
broken line and point M1 be symmetric to M through point O. Then
M1 A + AM
BM + AM
1
M1 M
≤
=
≤
MO =
2
2
2
4
because BM + AM does not exceed a half length of the broken line.
9.92. Let acute triangle ABC be placed inside circle S. Let us construct the circumscribed circle S1 of triangle ABC. Since triangle ABC is an acute one, the angle value of
the arc of circle S1 that lies inside S is greater than 180◦ . Therefore, on this arc we can
select diametrically opposite points, i.e., inside circle S a diameter of circle S1 is contained.
It follows that the radius of S is not shorter than the radius of S1 .
A similar statement for an acute triangle is false. An acute triangle lies inside a circle
constructed on the longest side a as on diameter. The radius of this circle is equal to 21 a and
a
a
. Clearly, 21 a < 2 sin
.
the radius of the circle circumscribed about the triangle is equal to 2 sin
α
α
9.93. First solution. Any triangle of perimeter P can be placed in a disk of radius
1
P
and if an acute triangle is placed in a disk of radius R1 , then R1 ≥ R (Problem 9.92).
4
Hence, 14 P = R1 ≥ R.
. Hence,
Second solution. If 0 < x < π2 , then sin x > 2x
π
a + b + c = 2R(sin α + sin β + sin γ) >
2R(2α + 2β + 2γ)
= 4R.
π
Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF
A TRIANGLE
This chapter is in close connection with the preceding one. For background see the
preceding chapter.
§1. Medians
10.1. Prove that if a > b, then ma < mb .
10.2. Medians AA1 and BB1 of triangle ABC intersect at point M . Prove that if
quadrilateral A1 M B1 C is a circumscribed one, then AC = BC.
10.3. Perimeters of triangles ABM , BCM and ACM , where M is the intersection point
of medians of triangle ABC, are equal. Prove that triangle ABC is an equilateral one.
10.4. a) Prove that if a, b, c are the lengths of sides of an arbitrary triangle, then
2
a + b2 ≥ 21 c2 .
b) Prove that m2a + m2b ≥ 82 c2 .
10.5. Prove that m2a + m2b + m2c ≤ 27
R2 .
4
b) Prove that ma + mb + mc ≤ 29 R.
2 −b2 |
2 +b2
10.6. Prove that |a 2c
< mc ≤ a 2c
.
9
< xx1 < 54 .
10.7. Let x = ab + bc + ca, x1 = ma mb + mb mc = mc ma . Prove that 20
See also Problems 9.1, 10.74, 10.76, 17.17.
§2. Heights
10.8. Prove that in any triangle the sum of the lengths of its heights is less than its
semiperimeter.
10.9. Two heights of a triangle are longer than 1. Prove that its area is greater than 12 .
10.10. In triangle ABC, height AM is not shorter than BC and height BH is not
shorter than AC. Find the angles of triangle ABC.
1
10.11. Prove that 2r
< h1a + h1b < 1r .
10.12. Prove that ha + hb + hc ≥ 9r.
10.13. Let a < b. Prove that a + ha ≤ b + hb .
√
10.14. Prove that ha ≤ rb rc .
10.15. Prove that ha ≤ a2 cot α2 .
10.16. Let a ≤ b ≤ c. Prove that
ha + hb + hc ≤
3b(a2 + ac + c2 )
.
4pR
See also Problems 10.28, 10.55, 10.74, 10.79.
§3. The bisectors
p
10.17. Prove that la ≤ p(p − a).
235
236
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.18. Prove that
ha
la
≥
q
2r
.
R
2
lb + lc2
10.19. Prove that a) la2 +
≤√p2 ; b) la + lb + lc ≤
10.20. Prove that la + lb + mc ≤ 3p.
√
3p.
See also Problems 6.38, 10.75, 10.94.
§4. The lengths of sides
9r
≤ a1 + 1b + 1c ≤ 9R
.
10.21. Prove that 2S
4S
2bc cos α
10.22. Prove that b+c < b + c − a < 2bc
.
a
10.23. Prove that if a, b, c are the lengths of sides of a triangle of perimeter 2, then
a2 + b2 + c2 < 2(1 − abc).
10.24. Prove that 20Rr − 4r2 ≤ ab + bc + ca ≤ 4(R + r)2 .
§5. The radii of the circumscribed, inscribed and escribed circles
2
10.25. Prove that rrc ≤ c4 . ¡
¢
10.26. Prove that Rr ≤ 2 sin α2 1 − sin α2 .
10.27. Prove that 6r ≤ a + b.
10.28. Prove that hraa + hrbb + hrcc ≥ 3.
10.29. Prove that 27Rr ≤ 2p2 ≤ 21 27R2 .
10.30. Let O be the centre of √
the inscribed circle of triangle ABC and OA ≥ OB ≥ OC.
Prove that OA ≥ 2r and OB ≥ r 2.
10.31. Prove that the sum of distances from any point inside of a triangle to its vertices
is not less than 6r.
´
³
¢
¡
10.32. Prove that 3 raa + rbb + rcc ≥ 4 raa + rbb + rcc .
10.33. Prove√that:
a) 5R − r ≥ 3p;
i
h√
2
2
.
b) 4R − ra ≥ (p − a) 3 + a +(b−c)
2S
10.34. Prove that 16Rr − 5r2 ≤ p2 ≤ 4R2 + 4Rr + 3r2 .
10.35. Prove that ra2 + rb2 + rc2 ≥ 41 27R2 .
See also Problems 10.11, 10.12, 10.14, 10.18, 10.24, 10.55, 10.79, 10.82, 19.7.
§6. Symmetric inequalities between the angles of a triangle
Let α, β and γ be the angles of triangle ABC. In problems of this section you have to
prove the inequalities indicated.
Remark. If α, β and γ are the angles of a triangle, then there exists a triangle with
angles π−α
, π−β
and π−γ
. Indeed, these numbers are positive and their sum is equal to π. It
2
2
2
follows that if a symmetric inequality holds for sines, cosines, tangents and cotangents of the
angles of any triangle then a similar inequality in which sin x is replaced with cos x2 , cos x
with sin x2 , tan x with cot x2 and cot x with tan x2 is also true.
The converse passage from inequalities for halved angles to inequalities with whole angles
is only possible for acute triangles. Indeed, if α′ = 21 (π − α), then α = π − 2α′ . Therefore,
for an acute triangle with angles α′ , β ′ , γ ′ there exists a triangle with angles π − 2α′ , π − 2β ′
and π − 2γ ′ . Under such a passage sin x2 turns into cos x, etc., but the inequality obtained
can only be true for acute triangles.
§8. INEQUALITIES FOR THE AREA OF A TRIANGLE
237
10.36. a) 1 < cos α + cos β + cos γ ≤ 23 .
b) 1 < sin α2 + sin β2 + sin γ2 ≤ 23 .
√
10.37. a) sin α + sin β + sin γ ≤ 23 3.
√
b) cos α2 + cos β2 + cos γ2 ≤ 23 3.
√
10.38. a) cot α + cot β + cot
γ
≥
3.
√
b) tan α2 + tan β2 + tan γ2 ≥ 3.
√
10.39. cot α2 + cot β2 + cot γ2 ≥ 3 3.
√
b) For an acute triangle tan α + tan β + tan γ ≥ 3 3.
10.40. a) sin α2 sin β2 sin γ2 ≤ 18 .
b) cos α cos β cos γ ≤ 81 .
√
10.41. a) sin α sin β sin γ ≤ 3 8 3 ;
√
b) cos α2 cos β2 cos γ2 ≤ 83 3.
10.42. a) cos2 α + cos2 β + cos2 γ ≥ 43 .
b) For an obtuse triangle
cos2 α + cos2 β + cos2 γ > 1.
10.43. cos α cos β + cos β cos γ + cos γ cos α ≤ 43 .
10.44. For an acute triangle
sin 2α + sin 2β + sin 2γ ≤ sin(α + β) + sin(β + γ) + sin(γ + α).
§7. Inequalities between the angles of a triangle
10.45. Prove that 1 − sin α2 ≤ 2 sin β2 sin γ2 .
c
.
10.46. Prove that sin γ2 ≤ a+b
10.47. Prove that if a + b < 3c, then tan α2 tan β2 < 21 .
10.48. In an acute triangle, if α < β < γ, then sin 2α > sin 2β > sin 2γ.
10.49. Prove that cos 2α + cos 2β − cos 2γ ≤ 32 .
10.50. On median BM of triangle ABC, point X is taken. Prove that if AB < BC,
then ∠XAB < ∠XCB.
10.51. The inscribed circle is tangent to sides of triangle ABC at points A1 , B1 and C1 .
Prove that triangle A1 B1 C1 is an acute one.
10.52. From the medians of a triangle whose angles are α, β and γ a triangle whose
angles are αm , βm and γm is constructed. (Angle αm subtends median AA1 , etc.) Prove that
if α > β > γ, then α > αm , α > βm , γm > β > αm , βm > γ and γm > γ.
See also Problems 10.90, 10.91, 10.93.
§8. Inequalities for the area of a triangle
√
2
2
2
2
√ +c .
10.53. Prove that: a) 3 3r2 ≤ S ≤ 3p√3 ; b) S ≤ a +b
4 3
10.54. Prove that
√
a2 + b2 + c2 − (a − b)2 − (b − c)2 − (c − a)2 ≥ 4 3S.
³ √ ´3
√ √
√
√
10.55. Prove that: a) S 3 ≤ 43 (abc)2 ; b) ha hb hc ≤ 4 3 S ≤ 3 ra rb rc .
238
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
***
10.56. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 , respectively,
S B1 C1
≤ 41 .
are taken so that AA1 , BB1 and CC1 meett at one point. Prove that AS1ABC
10.57. On sides BC, CA and AB of triangle ABC arbitrary points A1 , B1 and C1 are
taken. Let a = SAB1 C1 , b = SA1 BC1 , c = SA1 B1 C and u = SA1 B1 C1 . Prove that
u3 + (a + b + c)u2 ≥ 4abc.
10.58. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 are taken.
Prove that the area of one of the triangles AB1 C1 , A1 BC1 , A1 B1 C does not exceed: a)
1
S
; b) SA1 B1 C1 .
4 ABC
See also Problems 9.33, 9.37, 9.40, 10.9, 20.1, 20.7.
§9. The greater angle subtends the longer side
10.59. In a triangle ABC, prove that ∠ABC < ∠BAC if and only if AC < BC, i.e.,
the longer side subtends the greater angle and the greater angle subtends the longer side.
10.60. Prove that in a triangle ABC angle ∠A is an acute one if and only if mb > 12 a.
10.61. Let ABCD and A1 B1 C1 D1 be two convex quadrilaterals with equal corresponding
sides. Prove that if ∠A > ∠A1 , then ∠B < ∠B1 , ∠C < ∠C1 , ∠D < ∠D1 .
10.62. In an acute triangle ABC the longest height AH is equal to median BM . Prove
that ∠B ≤ 60◦ .
10.63. Prove that a convex pentagon ABCDE with equal sides whose angles satisfy
inequalities ∠A ≥ ∠B ≥ ∠C ≥ ∠D ≥ ∠E is a regular one.
§10. Any segment inside a triangle is shorter than the longest side
10.64. a) Segment M N is placed inside triangle ABC. Prove that the length of M N
does not exceed the length of the longest side of the triangle.
b) Segment M N is placed inside a convex polygon. Prove that the length of M N does
not exceed that of the longest side or of the greatest diagonal of this polygon.
10.65. Segment M N lies inside sector AOB of a disk of radius R = AO = BO. Prove
that either M N ≤ R or M N ≤ AB (we assume that ∠AOB < 180◦ ).
10.66. In an angle with vertex A, a circle tangent to the legs at points B and C is
inscribed. In the domain bounded by segments AB, AC and the shorter arc ⌣ BC a
segment is placed. Prove that the length of the segment does not exceed that of AB.
10.67. A convex pentagon lies inside a circle. Prove that at least one of the sides of the
pentagon is not longer than a side of the regular pentagon inscribed in the circle.
10.68. Given triangle ABC the lengths of whose sides satisfy inequalities a > b > c and
an arbitrary point O inside the triangle. Let lines AO, BO, CO intersect the sides of the
triangle at points P , Q, R, respectively. Prove that OP + OQ + OR < a.
§11. Inequalities for right triangles
In all problems of this section ABC is a right triangle with right angle ∠C.
10.69. Prove that cn > an + bn for n > 2.
10.70. Prove that a + b < c + hc .
10.71. Prove that for a right triangle 0.4 <
the vertex of the right angle.
√
10.72. Prove that rc ≥ 2(1 + 2).
10.73. Prove that m2a + m2b > 29r2 .
r
h
< 0.5, where h is the height dropped from
§13. INEQUALITIES IN TRIANGLES
239
§12. Inequalities for acute triangles
10.74. Prove that for an acute triangle
R
ma mb mc
+
+
≤1+ .
ha
hb
hc
r
10.75. Prove that for an acute triangle
µ
¶
√
1 1 1
1
1
1
.
+ +
+ + ≤ 2
la lb lc
a b c
10.76. Prove that if a triangle is not an obtuse one, then ma + mb + mc ≥ 4R.
10.77. Prove that if in an acute triangle ha = lb = mc , then this triangle is an equilateral
one.
10.78. In an acute triangle ABC heights AA1 , BB1 and CC1 are drawn. Prove that the
perimeter of triangle A1 B1 C1 does not exceed a semiperimeter of triangle ABC.
10.79. Let h be the longest height of a non-obtuse triangle. Prove that r + R ≤ h.
10.80. On sides BC, CA and AB of an acute triangle ABC, points A1 , B1 and C1 ,
respectively, are taken. Prove that
2(B1 C1 cos α + C1 A1 cos β + A1 B1 cos γ) ≥ a cos α + b cos β + c cos γ).
10.81. Prove that a triangle is an acute one if and only if a2 + b2 + c2 > 8R2 .
10.82. Prove that a triangle is an acute one if and only if p > 2R + r.
10.83. Prove that triangle ABC is an acute one if and only if on its sides BC, CA and
AB interior points A1 , B1 and C1 , respectively, can be selected so that AA1 = BB1 = CC1 .
10.84. Prove that triangle ABC is an acute one if and only if the lengths of its projections
onto three distinct directions are equal.
See also Problems 9.93, 10.39, 10.44, 10.48, 10.62.
§13. Inequalities in triangles
10.85. A line is drawn through the intersection point O of the medians of triangle ABC.
The line intersects the triangle at points M and N . Prove that N O ≤ 2M O.
10.86. Prove that if triangle ABC lies inside triangle A′ B ′ C ′ , then rABC < rA′ B ′ C ′ .
10.87. In triangle ABC side c is the longest and a is the shortest. Prove that lc ≤ ha .
10.88. Medians AA1 and BB1 of triangle ABC are perpendicular. Prove that cot ∠A +
cot ∠B ≥ 32 .
10.89. Through vertex A of an isosceles triangle ABC with base AC a circle tangent to
side BC at point M and intersecting side AB at point N is drawn. Prove that AN > CM .
10.90. In an acute triangle ABC bisector AB, median BM and height CH intersect at
one point. What are the limits inside which the value of angle A can vary?
10.91. In triangle ABC, prove that 31 π ≤ πaα + bβ + cγa + b + c < 21 π.
10.92. Inside triangle ABC point O is taken. Prove that
AO sin ∠BOC + BO sin ∠AOC + CO sin ∠AOB ≤ p.
10.93. On the extension of the longest side AC of triangle ABC beyond point C, point
D is taken so that CD = CB. Prove that angle ∠ABD is not an acute one.
10.94. In triangle ABC bisectors AK and CM are drawn. Prove that if AB > BC,
then AM > M K > KC.
10.95. On sides BC, CA, AB of triangle ABC points X, Y , Z are taken so that lines
AX, BY , CZ meet at one point O. Prove that of ratios OA : OX, OB : OY , OC : OZ at
least one is not greater than 2 and one is not less than 2.
240
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.96. Circle S1 is tangent to sides AC and AB of triangle ABC, circle S2 is tangent
to sides BC and AB and, moreover, S1 and S2 are tangent to each other from the outside.
Prove that the sum of radii of these circles is greater than the radius of the inscribed circle
S.
See also Problems 14.24, 17.16, 17.18.
Problems for independent study
10.97. In a triangle ABC, let P = a +b +c, Q = ab +bc +ca. Prove that 3Q < P 2 < 4Q.
10.98. Prove that the product of any two sides of a triangle is greater than 4Rr.
10.99. In triangle ABC bisector AA1 is drawn. Prove that A1 C < AC.
10.100. Prove that if a > b and a + ha ≤ b + hb , then ∠C = 90◦ .
10.101. Let O be the centre of the inscribed circle of triangle ABC. Prove that ab +
bc + ca ≥ (AO + BO + CO)2 .
10.102. On sides of triangle ABC equilateral triangles with centers at D, E and F are
constructed outwards. Prove that SDEF ≥ SABC .
10.103. In plane, triangles ABC and M N K are given so that line M N passes through
the midpoints of sides AB and AC and the intersection of these triangles is a hexagon of
area S with pairwise parallel opposite sides. Prove that 3S < SABC + SM N K .
Solutions
10.1. Let medians AA1 and BB1 meet at point M . Since BC > AC, points A and C
lie on one side of the midperpendicular to segment AB and therefore, both median CC1 and
its point M lie on the same side. It follows that AM < BM , i.e., ma < mb .
10.2. Suppose that, for instance, a > b. Then m < mb (Problem 10.1). Since
quadrilateral A1 M B1 C is a circumscribed one, it follows that 12 a + 13 mb = 12 b + 31 ma , i.e.,
1
(a − b) = 13 (ma − mb ). Contradiction.
2
10.3. Let, for instance, BC > AC. Then M A < M B (cf. Problem 10.1); hence,
BC + M B + M C > AC + M A + M C.
10.4. a) Since c ≤ a + b, it follows that c2 ≤ (a + b)2 = a2 + b2 + 2ab ≤ 2(a2 + b2 ).
b) Let M be the intersection point of medians of triangle ABC. By heading a) M A2 +
M B 2 ≥ 12 AB 2 , i.e., 94 m2a + 49 m2b ≥ 21 c2 .
10.5. a) Let M be the intersection point of medians, O the center of the circumscribed
circle of triangle ABC. Then
AO2 + BO2 + CO2 =
−−→ −−→ 2
−−→ −−→
−−→ −−→
(AM + M O) + (BM + M O)2 + (CM + M O)2 =
−−→ −−→ −−→ −−→
AM 2 + BM 2 + CM 2 + 2(AM + BM + CM , M O) + 3M O2 .
−−→ −−→ −−→ −
→
Since AM + BM + CM = 0 , it follows that
AO2 + BO2 + CO2 = AM 2 + BM 2 + CM 2 + 3M O2 ≥ AM 2 + BM 2 + CM 2 ,
i.e., 3R2 ≥ 94 (m2a + m2b + m2c ).
b) It suffices to notice that (ma + mb + mc )2 ≤ 3(m2a + m2b + m2c ), cf. Supplement to Ch.
9.
10.6. Heron’s formula can be rewritten as
16S 2 = 2a2 b2 + 2a2 c2 + 2b2 c2 − a4 − b4 − c4 .
SOLUTIONS
241
Since m2c = 41 (2a2 + 2b2 − c2 ) (Problem 12.11 a)), it follows that the inequalities
m2c
≤
µ
a2 + b 2
2c
¶2
;
m2c
≥
µ
a2 − b2
2c
¶2
are equivalent to the inequalities 16S 2 ≤ 4a2 b2 and 16S 2 > 0, respectively.
10.7. Let y = a2 + b2 + c2 and y1 = m2a + m2b + m2c . Then 3y = 4y1 (Problem 12.11, b),
y < 2x (Problem 9.7) and 2x1 + y1 < 2x + y because (ma + mb + mc )2 < (a + b + c)2 (cf.
Problem 9.2). By adding 8x1 + 4y1 < 8x + 4y to 3y = 4y1 we get 8x1 < y + 8x < 10x, i.e.,
x1
< 54 .
x
Let M be the intersection point of the medians of triangle ABC. Let us complement
triangle AM B to parallelogram AM BN . Applying the above-proved statement to triangle
(x/4)
< 45 , i.e., xx1 < 20
.
AM N we get (4x
9
1 /9)
10.8. Clearly, ha ≤ b, hb ≤ c, hc ≤ a, where at least one of these inequalities is a strict
one. Hence, ha + hb + hc < a + b + c.
10.9. Let ha > 1 and hb > 1. Then a ≥ hb > 1. Hence, S = 12 aha > 21 .
10.10. By the hypothesis BH ≥ AC and since the perpendicular is shorter than a slanted
line, BH ≥ AC ≥ AM . Similarly, AM ≥ BC ≥ BH. Hence, BH = AM = AC = BC.
Since AC = AM , segments AC and AM coincide, i.e., ∠C = 90◦ ; since AC = BC, the
angles of triangle ABC are equal to 45◦ , 45◦ , 90◦ .
a+b
10.11. Clearly, h1a + h1b = a+b
= (a+b+c)r
and a + b + c < 2(a + b) < 2(a + b + c).
2S
¢
¡
10.12. Since aha = r(a + b + c), it follows that ha = r 1 + ab + ac . Adding these
equalities for ha , hb and hc and taking into account that xy + xy ≥ 2 we get the desired
statement.
¢
¡
and 2S ≤ ab, it follows that ha − hb ≤ b − a.
10.13. Since ha − hb = 2S a1 − 1b = 2S b−a
ab
2
1
1
10.14. By Problem 12.21 ha = rb + rc . Moreover, r1b + r1c ≥ √r2b rc .
10.15. Since
2 sin β sin γ = cos(β − γ) − cos(β + γ) ≤ 1 + cos α,
we have
sin β sin γ
1 + cos α
1
α
ha
=
≤
= cot .
a
sin α
2 sin α
2
2
10.16. Since
b
2R
= sin β, then multiplying by 2p we get
(a + b + c)(ha + hb + hc ) ≤ 3 sin β(a2 + ac + c2 ).
Subtracting 6S from both sides we get
a(hb + hc ) + b(ha + hc ) + c(ha + hb ) ≤ 3 sin β(a2 + c2 ).
2
Since, for instance, ahb = a2 sin γ = a2Rc , we obtain a(b2 + c2 ) − 2b(a2 + c2 ) + c(a2 + b2 ) ≤ 0.
To prove the latter inequality let us consider the quadratic exoression
f (x) = x2 (a + c) − 2x(a2 + c2 ) + ac(a + c).
It is easy to verify that f (a) = −a(a − c)2 ≤ 0 and f (c) = −c(a − c)2 ≤ 0. Since the
coefficient of x is positive and a ≤ b ≤ c, it follows that f (b) ≤ 0.
. Moreover, 4bc ≤ (b + c)2 .
10.17. By Problem 12.35 a) la2 = 4bcp(p−a)
(b+c)2
242
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.18. Clearly,
ha
la
= cos 21 (β − γ). By Problem 12.36 a)
2r
α
β
γ
α
β−γ
β+γ
= 8 sin sin sin = 4 sin [ cos
− cos
] = 4x(q − x),
R
2
2
2
2
2
2
α
β−γ
where x = sin and q = cos
.
2
2
It remains to notice that 4x(q − x) ≤ q 2 .
10.19. a) By Problem 10.17 la2 ≤ p(p − a). Adding three similar inequalities we get the
desired statement.
b) For any numbers la , lb and lc we
lb + lc )2 ≤ 3(la2 + lb2 + lc2 ).
p have (la +p
√
10.20. It suffices to prove that p(p − a) + p(p − b) + mc ≤ 3p. We may assume
that p = 1; let x = 1 − a and y = 1 − b. Then
(x − y)2
2a2 + 2b2 − c2
= 1 − (x + y) +
= m(x, y).
m2c =
4
4
Let us consider the function
p
√
√
f (x, y) = x + y + m(x, y).
√
We have to prove that f (x, y) ≤ 3 for x, y ≥ 0 and x + y ≤ 1. Let
√
√
g(x) = f (x, x) = 2 x + 1 − 2x.
√
1
Since g ′ (x) = √1x − √1−2x
, it follows that as x grows from 0 to 13 and g(x) grows from 1 to 3
√
√
and as x grows from 31 to 12 ; we also see that g(x) diminishes from 3 to 2. Introduce new
√
√
variables: d = x − y and q = x + y. It is easy to verify that (x − y)2 − 2q 2 (x + y) + q 4 = 0,
2
4
i.e., x + y = d 2q+q2 . Hence,
s
q 2 d2 (2 − q 2 )
−
.
f (x, y) = q + 1 −
2
4q 2
√
2
2)
√
Now, observe that q 2 = ( x + y)2 ≤ 2(x + y) ≤ 2, i.e., d (2−q
≥ 0. It follows that for
4q 2
a fixed q the value of function f (x, y) is the maximal one for d = 0, i.e., x = y; the case
x = y(?) is the one considered above.
10.21. Clearly, a1 + 1b + 1c = ha +h2Sb +hc . Moreover, 9r ≤ ha + hb + hc (Problem 10.12) and
ha + hb + hc ≤ ma + mb + mc ≤ 29 R (Problem 10.5 b)).
10.22. First, let us prove that b + c − a < 2bc
. Let 2x = b + c − a, 2y = a + c − b and
a
2z = a + b − c. We have to prove that
2(x + y)(x + z)
, i.e., xy + xz < xy + xz + x2 + yz.
2x <
y+z
The latter inequality is obvious.
Since
2bc cos α = b2 + c2 − a2 = (b + c − a)(b + c + a) − 2bc,
it follows that
¸
·
2bc cos α
2bc
(b + c − a)a
.
=b+c−a+
−
b+c
b+c
b+c
The expression in square brackets is negative because b + c − a < 2bc
.
a
10.23. By Problem 12.30 we have
a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ac) = 4p2 − 2r2 − 2p2 − 8rR = 2p2 − 2r2 − 8rR
SOLUTIONS
243
and abc = 4prR. Thus, we have to prove that
2p2 − 2r2 − 8rR < 2(1 − 4prR), where p = 1.
This inequality is obvious.
10.24. By Problem 12.30, ab + bc + ca = r2 + p2 + 4Rr. Moreover, 16Rr − 5r2 ≤ p2 ≤
4R2 + 4Rr + 3r2 (Problem 10.34).
10.25. Since
r(cot α + cot β) = c = rc (tan α + tan β),
it follows that
µ
¶
tan α tan β
2
c = rrc 2 +
≥ 4rrc .
+
tan β tan β
10.26. It suffices to apply the results of Problems 12.36 a) and 10.45. Notice also that
x(1 − x) ≤ 41 , i.e., Rr ≤ 12 .
10.27. Since hc ≤ a and hc ≤ b, it follows that 4S = 2chc ≤ c(a + b). Hence,
6r(a + b + c) = 12S ≤ 4ab + 4S ≤ (a + b)2 + c(a + b) = (a + b)(a + b + c).
³
´
1
ra
1 ra
1
ra
2
10.28. Since ha = rb + rc (Problem 12.21), it follows that ha = 2 rb + rc . Let us write
similar equalities for hrbb and hrcc and add them. Taking into account that xy + xy ≥ 2 we get
the desired statement.
(cf. Problem 12.1), we obtain 27abc ≤ 8p3 = (a + b + c)3 .
10.29. Since Rr = PpS = abc
4p
Since (a + b + c)2 ≤ 3(a2 + b2 + c2 ) for any numbers a, b and c, we have
3
p2 ≤ (a2 + b2 + c2 ) = m2a + m2b + m2c
4
(cf. Problem 12.11 b)). It remains to notice that m2a + m2b + m2c ≤ 27
R2 (Problem 10.5 a)).
4
10.30. Since OA = sinr A , OB = sinr∠B and OC = sin r∠X and since angles 12 ∠A, 12 ∠B and
2
1
∠C
2
2
2
are acute ones, it follows that ∠A ≤ ∠B ≤ ∠C. Hence, ∠A ≤ 60◦ and ∠B ≤ 90◦ and,
therefore, sin ∠A
≤ 21 and sin ∠B
≤ √12 .
2
2
10.31. If ∠C ≥ 120◦ , then the sum of distances from any point inside the triangle to its
vertices is not less than a + b (Problem 11.21); moreover, a + b ≥ 6r (Problem 10.27).
If each angle of the triangle is less than 120◦ , then at a point the sum of whose distances
from the vertices of the triangle is the least one the square of this sum is equal to 12 (a2 +
√
√
b2 + c2 ) + 2 3S (Problem 18.21 b)). Further, 21 (a2 + b2 + c2 ) ≥ 2 3S (Problem 10.53 b))
√
and 4 3S ≥ 36r2 (Problem 10.53 a)).
α
10.32. Let α = cos ∠A
, β = cos ∠B
and γ = cos ∠C
. By Problem 12.17 b) raa = βγ
,
2
2
2
γ
β
b
= γα and rcc = αβ . Therefore, multiplying by αβγ we express the inequality to be proved
rb
in the form
3(α2 + β 2 + γ 2 ) ≥ 4(β 2 γ 2 + γ 2 α2 + α2 β 2 ).
Since α2 =
1+cos ∠A
,
2
β2 =
1+cos ∠B
2
and γ 2 =
1+cos ∠C
,
2
we obtain the inequality
cos ∠A + cos ∠B + cos ∠C + 2(cos ∠A cos ∠B + cos ∠B cos ∠C + cos ∠C cos ∠A) ≤ 3.
It remains to make use of results of Problems 10.36 and 10.43.
10.33. a) Adding equality 4R+r = ra +rb +rc (Problem 12.24) with inequality R−2r ≥ 0
(Problem 10.26) we get
5R − r ≥ ra + rb + rc = pr((p − a)−1 + (p − b)−1 + (p − c)−1 ) =
2 −b2 −c2 )
p(ab+bc+ca−p2 )
== p(2(ab+bc+ca)−a
.
S
4S
244
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
It remains to observe that
√
2(ab + bc + ca) − a2 − b2 − c2 ≥ 4 3S
(Problem 10.54).
b) It is easy to verify that
4R − ra = rb + rc − r =
pr
pr
(p − a)(p2 − bc)
pr
+
−
=
.
p−b p−c
p
S
It remains to observe that
4(p2 − bc) = a2 + b2 + c2 + 2(ab − bc + ca) =
= 2(ab + bc + ca) =√
2
2
2
2
−a − b − c + 2(a + b2 + c2 − 2bc) ≥ 4 3S + 2(a2 + (b − c)2 ).
10.34. Let a, b and c be the lengths of the sides of the triangle, F = (a−b)(b−c)(c−a) =
A − B, where A = ab2 + bc2 + ca2 and B = a2 b + b2 c + c2 a. Let us prove that the required
inequalities can be obtained by a transformation of an obvious inequality F 2 ≥ 0. Let
σ1 = a + b + c = 2p, σ2 = ab + bc + ca = r2 + p2 + 4rR and σ3 = abc = 4prR, cf. Problem
12.30. It is easy to verify that
F 2 = σ12 σ22 − 4σ23 − 4σ13 σ3 + 18σ1 σ2 σ3 − 27σ32 .
Indeed,
(σ1 σ2 )2 − F 2 = (A + B + 3abc)2 − (A − B)2 = 4AB + 6(A + B)σ3 + 9σ32 =
4(a3 b3 + . . .) + 4(a4 bc + . . .) + 6(A + B)σ3 + 21σ32 .
It is also clear that
4σ23 = 4(a3 b3 + . . .) + 12(A + B)σ3 + 24σ32 ,
4σ13 σ3 = 4(a4 bc + . . .) + 12(A + B)σ3 + 24σ32 ,
18σ1 σ2 σ3 = 18(A + B)σ3 + 54σ32 .
Expressing σ1 , σ2 and σ3 via p, r and R, we obtain
Thus, we obtain
F 2 = −4r2 [(p2 − 2R2 − 10Rr + r2 )2 − 4R(R − 2r)3 ] ≥ 0.
p
2
R(R − 2r) =
p2 ≥ 2R2 +
10Rr
−
r
−
2(R
−
2r)
p
[(R − 2r) − R(R − 2r)]2 + 16Rr − p
5r2 ≥ 16Rr − 5r2
p2 ≤ 2R2 + 10Rr + r2 + 2(R − 2r)
=
p R(R − 2r)
2
2
2
4R + 4Rr + 3r − [(R − 2r) − R(R − 2r)] ≤
4R2 + 4Rr + 3r2 .
10.35. Since ra + rb + rc = 4R + r and ra rb + rb rc + rc ra = p2 (Problems 12.24 and 12.25),
it follows that ra2 + rb2 + rc2 = (4R + r)2 − 2p2 . By Problem 10.34 p2 ≤ 4R2 + 4Rr + 3r2 ;
hence, ra2 + rb2 + rc2 = 8R2 − 5r2 . It remains to notice that r ≤ 12 R (Problem 10.26).
10.36. a) By Problem 12.38 cos α + cos β + cos γ = R+r
. Moreover, r ≤ 21 R (Problem
R
10.26).
b) Follows from heading a), cf. Remark.
√
10.37. a) Clearly, sin α + sin β + sin γ = Rp . Moreover, p ≤ 32 3R (Problem 10.29).
b) Follows from heading a), cf. Remark.
10.38. a) By Problem 12.44 a)
cot α + cot β + cot γ =
a2 + b 2 + c 2
.
4S
SOLUTIONS
245
√
Moreover, a2 + b2 + c2 ≥ 4 3S (Problem 10.53 b)).
b) Follows from heading a), cf. Remark.
10.39. a) By Problem 12.45 a)
α
β
γ
p
cot + cot + cot = .
2
2
2
r
√
Moreover, p ≥ 3 3r (Problem 10.53 a)).
b) Follows from heading a), cf. Remark. For an acute triangle tan α + tan β + tan γ < 0;
cf., for instance, Problem 12.46.
10.40. a) By Problem 12.36 a)
β
γ
r
α
.
sin + sin + sin =
2
2
2
4R
Moreover, r ≤ 21 R (Problem 10.26).
b) For an acute triangle it follows from heading a), cf. Remark. For an obtuse triangle
cos α cos β cos γ < 0.
10.41. a) Since sin x = 2 sin x2 cos x2 , we see that making use of results of Problems 12.36
√
pr
3
3R (Problem 10.29) and
a) and 12.36 c) we obtain sin α sin β sin γ = 2R
2 . Moreover, p ≤ 2
1
r ≤ 2 R (Problem 10.26).
b) Follows from heading a), cf. Remark.
10.42. By Problem 12.39 b)
cos2 α + cos2 β + cos2 γ = 1 − 2 cos α cos β cos γ.
It remains to notice that cos α cos β cos γ ≤
cos α cos β cos γ < 0.
10.43. Clearly,
1
8
(Problem 10.40 b)) and for an obtuse triangle
2(cos α cos β + cos β cos γ + cos γ cos α) =
(cos α + cos β + cos γ)2 − cos2 α − cos2 β − cos2 γ.
It remains to notice that cos α + cos β + cos γ ≤ 32 (Problem 10.36 a)) and cos2 α + cos2 β +
cos2 γ ≥ 43 (Problem 10.42).
10.44. Let the extensions of bisectors of acute triangle ABC with angles α, β and γ
intersect the circumscribed circle at points A1 , B1 and C1 , respectively. Then
SABC =
R2 (sin 2α + sin 2β + sin 2γ)
;
2
S A1 B 1 C 1 =
R2 (sin(α + β) + sin(β + γ) + sin(γ + α))
.
2
It remains to make use of results of Problems 12.72 and 10.26.
10.45. Clearly,
γ
β−γ
β+γ
α
β
− cos
≤ 1 − sin .
2 sin sin = cos
2
2
2
2
2
10.46. Let us drop perpendiculars AA1 and BB1 from vertices A and B to the bisector
of angle ∠ACB. Then AB ≥ AA1 + BB1 = b sin γ2 + a sin γ2 .
a+b−c
. Since a+b < 3c, it follows that a+b−c <
10.47. By Problem 12.32 tan α2 tan β2 = a+b+c
1
(a + b + c).
2
246
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.48. Since π − 2α > 0, π − 2β > 0, π − 2γ > 0 and (π − 2α) + (π − 2β) + (π − 2γ) = π,
it follows that there exists a triangle whose angles are π − 2α, π − 2β, π − 2γ. The lengths
of sides opposite to angles π − 2α, π − 2β, π − 2γ are proportional to sin(π − 2α) = sin 2α,
sin 2β, sin 2γ, respectively. Since π − 2α > π − 2β > π − 2γ and the greater angle subtends
the longer side, sin 2α > sin 2β > sin 2γ.
10.49. First, notice that
cos 2γ = cos 2(π − α − β) = cos 2α cos 2β − sin 2α sin 2β.
Hence,
cos 2α + cos 2β − cos 2γ = cos 2α + cos 2β − cos 2α cos 2β + sin 2α sin 2β.
√
Since a cos ϕ + b sin ϕ ≤ a2 + b2 (cf. Supplement to Ch. 9), it follows that
(1 − cos 2β) cos 2α + sin 2β sin 2α + cos 2β ≤
q
(1 − cos 2β)2 + sin2 2β + cos 2β = 2| sin β| + 1 − 2 sin2 β.
It remains to notice that the greatest value of the quadratic 2t + 1 − 2t2 is attained at point
t = 21 and this value is equal to 32 . The maximal value corresponds to angles α = β = 30◦
and γ = 120◦ .
10.50. Since AB < CB, AX < CX = SABX = SBCX , it follows that sin ∠XAB >
sin ∠XCB. Taking into account that angle ∠XCB is an acute one, we get the desired
statement.
10.51. If the angles of triangle ABC are equal to α, β and γ, then the angles of triangle
A1 B1 C1 are equal to 21 (β + γ), 21 (γ + α) and 12 (α + β).
10.52. Let M be the intersection point of medians AA1 , BB1 and CC1 . Complementing
triangle AM B to parallelogram AM BN we get ∠BM C1 = αm and ∠AM C1 = βm . It is easy
to verify that ∠C1 CB < 12 γ and ∠B1 BC < 21 β. It follows that αm = ∠C1 CB + ∠B1 BC <
1
(β + γ) < β. Similarly, γm = ∠A1 AB + ∠B1 BA > 12 (α + β) > β.
2
First, suppose that triangle ABC is an acute one. Then the heights’ intersection point
H lies inside triangle AM C1 . Hence, ∠AM B < ∠AHB, i.e., π − γm < π − γ and ∠CM B <
∠CHB, i.e., π − αm > π − α. Now, suppose that angle α is an obtuse one. Then angle
CC1 B is also an obtuse one and therefore, angle αm is an acute one, i.e., αm < α. Let us
drop perpendicular M X from point M to BC. Then γm > ∠XM B > 180◦ − ∠HAB > γ.
Since α > αm , it follows that α + (π − αm ) > π, i.e., point M lies inside the circumscribed
circle of triangle AB1 C1 . Therefore, γ = ∠AB1 C1 < ∠AM C1 = βm . Similarly, α =
∠CB1 A1 > ∠CM A1 = βm because γ + (π − γm ) < π.
10.53 a) Clearly,
µ
¶2
p−a+p−b+p−c
p3
S2
= (p − a)(p − b)(p − c) ≤
= .
p
3
27
2
Hence, pr = S ≤ 3p√3 , i.e., r ≤ 3√p 3 . By multiplying the latter inequality by r we get the
desired statement.
b) Since (a + b + c)2 ≤ 3(a2 + b2 + c2 ), it follows that
(a + b + c)2
a2 + b 2 + c 2
p2
√
√
.
≤
S≤ √ =
3 3
12 3
4 3
SOLUTIONS
247
10.54. Let x = p − a, y = p − b, z = p − c. Then
(a2 − (b − c)2 ) + (b2 − (a − c)2 ) + (c2 − (a − b)2 ) =
4(p − b)(p − c) + 4(p − a)(p − c) + 4(p − a)(p − b) =
4(yz + zx + xy)
and
p
p
√
4 3S = 4 3p(p − a)(p − b)(p − c) = 4 3(x + y + z)xyz.
p
Thus, we have to prove that xy + yz + zx ≥ 3(x + y + z)xyz. After squaring and simplification we obtain
x2 y 2 + y 2 z 2 + z 2 x2 ≥ x2 yz + y 2 xz + z 2 xy.
Adding inequalities
x2 (y 2 + z 2 )
y 2 (x2 + z 2 )
z 2 (x2 + y 2 )
x2 yz ≤
, y 2 xz ≤
and z 2 xy ≤
2
2
2
we get the desired statement.
10.55. a) By multiplying three equalities of the form S = 12 ab sin γ we get S 3 =
1
(abc)2 sin γ sin β sin α. It remains to make use of a result of Problem 10.41.
8
³ ´3
√
(2S)6
(2S)6 ( 3/4)3
4
2
3
2
√
and
(abc)
≥
=
b) Since (ha hb hc )2 = (abc)
,
it
follows
that
(h
h
h
)
≤
S
a b c
2
S3
3
√
( 3S)3 .
√
4
Since (ra rb rc )2 = Sr2 (Problem 12.18, c) and r2 ( 3)3 ≤ S (Problem 10.53 a), it follows
√
that (ra rb rc )2 ≥ ( 3S)3 .
BA
1
1
10.56. Let p = BC
, q = CB
and r = AC
. Then
CA
AC
S A1 B 1 C 1
= 1 − p(1 − r) − q(1 − p) − r(1 − q) = 1 − (p + q + r) + (pq + qr + rp).
SABC
By Cheva’s theorem (Problem 5.70) pqr = (1 − p)(1 − q)(1 − r), i.e., 2pqr = 1 − (p + q +
r) + (pq + qr + rp). Moreover,
µ ¶3
1
2
(pqr) = p(1 − p)(1 − q)r(1 − r) ≤
.
4
S
B1 C1
= 2pqr ≤ 14 .
Therefore, AS1ABC
10.57. We can assume that the area of triangle ABC is equal to 1. Then a + b + c = 1
1
1
and, therefore, the given inequality takes the form u2 ≥ 4abc. Let x = BA
, y = CB
and
BC
CA
AC1
z = AB . Then
u = 1 − (x + y + z) + xy + yz + zx and abc = xyz(1 − x)(1 − y)(1 − z) = v(u − v),
where v = xyz. Therefore, we pass to inequality u2 ≥ 4v(u − v), i.e., (u − 2v)2 ≥ 0 which is
obvious.
1
1
1
, y = BC
and z = AC
. We may assume that the area of triangle
10.58. a) Let x = BA
BC
BA
AB
ABC is equal to 1. Then SAB1 C1 = z(1 − y), SA1 BC1 = x(1 − z) and SA1 B1 C = y(1 − x).
Since x(1 − x) ≤ 41 , y(1 − y) ≤ 14 and z(1 − z) ≤ 41 , it follows that the product of numbers
¡ ¢3
SAB1 C1 , SA1 BC1 and SA1 B1 C does not exceed 41 ; hence, one of them does not exceed 14 .
b) Let, for definiteness, x ≥ 12 . If y ≤ 12 , then the homothety with center C and
coefficient 2 sends points A1 and B1 to inner points on sides BC and AC, consequently,
SA1 B1 C ≤ SA1 B1 C1 . Hence, we can assume that y ≥ 12 and, similarly, z ≥ 12 . Let x = 21 (1 + α),
y = 21 (1 + β) and z = 12 (1 + γ). Then SAB1 C1 = 14 (1 + γ − β − βγ), SA1 B1 C1 = 41 (1 + α −
γ − αγ) and SA1 B1 C = 14 (1 + β − α − αβ); hence, SA1 B1 C1 = 14 (1 + αβ + βγ + αγ) ≥ 41 and
SAB1 C1 + SA1 BC1 + SA1 B1 C ≤ 43 .
248
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.59. It suffices to prove that if AC < BC, then ∠ABC < ∠BAC. Since AC < BC, on
side BC point A1 can be selected so that A1 C = AC. Then ∠BAC < ∠AAC = ∠AA1 C >
∠ABC.
10.60. Let A1 be the midpoint of side BC. If AA1 < 21 BC = BA1 = A1 C, then
∠BAA1 > ∠ABA1 and ∠CAA1 > ∠ACA1 ; hence, ∠A = ∠BAA1 + ∠CAA1 < ∠B + ∠C,
i.e., ∠A > 90◦ . Similarly, if AA1 > 21 BC then ∠A > 90◦ .
10.61. If we fix two sides of the triangle, then the greater the angle between these sides
the longer the third side. Therefore, inequality ∠A > ∠A1 implies that BD > B1 D1 , i.e.,
∠C < ∠C1 . Now, suppose that ∠B ≥ ∠B1 . Then AC ≥ A1 C1 , i.e., ∠D ≥ ∠D1 . Hence,
360◦ = ∠A + ∠B + ∠C + ∠D > ∠A1 + ∠B1 + ∠C1 + ∠D1 = 360◦ .
Contradiction; therefore, ∠B < ∠B1 and ∠D < ∠D1 .
10.62. Let point B1 be symmetric to B through point M . Since the height dropped
from point M to side BC is equal to a half of AH, i.e., to a half of BM , it follows that
∠M BC = 30◦ . Since AH is the longest of heights, BC is the shortest of sides. Hence,
AB1 = BC ≤ AB, i.e., ∠ABB1 ≤ ∠AB1 B = ∠M BC = 30◦ . Therefore, ∠ABC =
∠ABB1 + ∠M BC ≤ 30◦ + 30◦ = 60◦ .
10.63. First, let us suppose that ∠A > ∠D. Then BE > EC and ∠EBA < ∠ECD.
Since in triangle EBC side BE is longer than side EC, it follows that ∠EBC < ∠ECB.
Therefore,
∠B = ∠ABE + ∠EBC < ∠ECD + ∠ECB = ∠C
which contradicts the hypothesis. Thus, ∠A = ∠B = ∠C = ∠D. Similarly, the assumption
∠B > ∠E leads to inequality ∠C < ∠D. Hence, ∠B = ∠C = ∠D = ∠E.
10.64. Let us carry out the proof for the general case. Let line M N intersect the sides
of the polygon at points M1 and N1 . Clearly, M N ≤ M1 N1 . Let point M1 lie on side AB
and point N1 lie on P Q. Since ∠AM1 N1 + ∠BM1 N1 = 180◦ , one of these angles is not less
than 90◦ . Let, for definiteness, ∠AM1 N1 ≥ 90◦ . Then AN1 ≥ M1 N1 because the longer side
subtends the greater angle.
We similarly prove that either AN1 ≤ AP or AN1 ≤ AQ. Therefore, the length of
segment M N does not exceed the length of a segment with the endpoints at vertices of the
polygon.
10.65. The segment can be extended to its intersection with the boundary of the sector
because this will only increase its length. Therefore, we may assume that points M and N
lie on the boundary of the disk sector. The following three cases are possible:
1) Points M and N lie on an arc of the circle. Then
M N = 2R sin
∠M ON
∠AOB
≤ 2R sin
= AB
2
2
because 12 ∠M ON ≤ 21 ∠AOB ≤ 90◦ .
2) Points M and N lie on segments AO and BO, respectively. Then M N is not longer
than the longest side of triangle AOB.
3) One of points M and N lies on an arc of the circle, the other one on one of segments
AO or BO. Let, for definiteness, M lie on AO and N on an arc of the circle. Then the
length of M N does not exceed that of the longest side of triangle AN O. It remains to notice
that AO = N O = R and AN ≤ AB.
10.66. If the given segment has no common points with the circle, then a homothety
with center A (and coefficient greater than 1) sends it into a segment that has a common
point X with arc AB and lies in our domain. Let us draw through point X tangent DE to
the circle (points D and E lie on segments AB and AC). Then segments AD and AE are
SOLUTIONS
249
shorter than AB and DE < 21 (DE + AD + AE) = AB, i.e., each side of triangle ADE is
shorter than AB. Since our segment lies inside triangle ADE (or on its side DE), its length
does not exceed that of AB.
10.67. First, suppose that the center O of the circle lies inside the given pentagon
A1 A2 A3 A4 A5 . Consider angles ∠A1 OA2 , ∠A2 OA3 , . . . , ∠A5 OA1 . The sum of these five
angles is equal to 2π; hence, one of them, say, ∠A1 OA2 , does not exceed 52 π. Then segment
A1 A2 can be placed in disk sector OBC, where ∠BOC = 52 π and points B and C lie on the
circle. In triangle OBC, side BC is the longest one; hence, A1 A2 ≤ BC.
If point O does not belong to the given pentagon, then the union of angles ∠A1 OA2 , . . . ,
∠A5 OA1 is less than π and each point of the angle — the union — is covered twice by these
angles. Therefore, the sum of these five angles is less than 2π, i.e., one of them is less than
2
π. The continuation of the proof is similar to the preceding case.
5
If point O lies on a side of the pentagon, then one of the considered angles is not greater
than 14 π and if it is its vertex, then one of them is not greater than 13 π. Clearly, 41 π < 13 π < 52 π.
Figure 121 (Sol. 10.68)
10.68. On sides BC, CA, AB take points A1 and A2 , B1 and B2 , C1 and C2 , respectively,
so that B1 C2 k BC, C1 A2 k CA, A1 B2 k AB (Fig. 121). In triangles A1 A2 O, B1 B2 O, C1 C2 O
sides A1 A2 , B1 O, C2 O, respectively, are the longest ones. Hence, OP < A1 A2 , OQ < BO,
OR ≤ C2 O, i.e.,
OP + OQ + OR < A1 A2 + B1 O + C2 O = A1 A2 + CA2 + BA1 = BC.
10.69. Since c2 = a2 + b2 , it follows that
cn = (a2 + b2 )cn−2 = a2 cn−2 + b2 cn−2 > an + bn .
10.70. The height of any of the triangles considered is longer than 2r. Moreover, in a
right triangle 2r = a + b − c (Problem 5.15).
√
√
2
2
10.71. Since ch = 2S = r(a + b + c) and c = a2 + b2 , it follows that hr = a+b+a√+b
2 +b2 =
a
q
√
2ab
1
2ab
√ a+b
1
+
2.
,
where
x
=
=
2 +b2 . Since 0 < a2 +b2 ≤ 1, it follows that 1 < x ≤
2
2
x+1
a
a +b
Hence,
2
5
<
1√
1+ 2
≤
r
h
< 12 .
√
10.72. Clearly, a + b ≥ 2 ab and c2 + a2 + b2 ≥ 2ab. Hence,
√
√
√ 2
(a + b + c)2 c2
(2 ab + 2ab)2 · 2ab
c2
2) .
=
≥
=
4(1
+
r2
a2 b 2
a2 b 2
10.73. By Problem 12.11 a) m2a + m2b = 41 (4c2 + a2 + b2 ) = 54 c2 . Moreover,
√
√
5c2
≥ 5(1 + 2)2 r2 = (15 + 10 2)r2 > 29r2 ,
4
cf. Problem 10.72.
10.74. Let O be the center of the circumscribed circle, A1 , B1 , C1 the midpoints of
sides BC, CA, AB, respectively. Then ma = AA1 ≤ AO + OA1 = R + OA1 . Similarly,
250
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
mb ≤ R + OB1 and mc ≤ R + OC1 . Hence,
¶
µ
ma mb mc
OA1 OB1 OC1
1
1
1
+
+
+
≤R
+
+
+
+
.
ha
hb
hc
ha hb hc
ha
hb
hc
It remains to make use of the result of Problem √
12.22 and the solution of Problem 4.46.
2 cos(α/2)
1
1
≥ la2 . Adding three analogous inequalities we
10.75. By Problem 4.47 b + c =
la
get the required statement.
10.76. Denote the intersection point of medians by M and the center of the circumscribed
circle by O. If triangle ABC is not an obtuse one, then point O lies inside it (or on its side);
let us assume, for definiteness, that it lies inside triangle AM B. Then AO+BO ≤ AM +BM ,
i.e., 2R ≤ 32 ma + 32 mb or, which is the same, ma + mb ≥ 3R. It remains to notice that since
angle ∠COC1 (where C1 is the midpoint of AB) is obtuse, it follows that CC1 ≥ CO, i.e.,
mc ≥ R.
The equality is attained only for a degenerate triangle.
10.77. In any triangle hb ≤ lb ≤ mb (cf. Problem 2.67); hence, ha = lb ≥ hb and
mc = lb ≤ mb . Therefore, a ≤ b and b ≤ c (cf. Problem 10.1), i.e., c is the length of the
longest side and γ is the greatest angle.
The equality ha = mc yields γ ≤ 60◦ (cf. Problem 10.62). Since the greatest angle γ of
triangle ABC does not exceed 60◦ , all the angles of the triangle are equal to 60◦ .
10.78. By Problem 1.59 the ratio of the perimeters of triangles A1 B1 C1 and ABC is
equal to Rr . Moreover, r ≤ R2 (Problem 10.26).
Remark. Making use of the result of Problem 12.72 it is easy to verify that
r1
≤ 14 .
2R1
SA1 B1 C1
SABC
=
10.79. Let 90◦ ≥ α ≥ β ≥ γ, then CH is the longest height. Denote the centers of the
inscribed and circumscribed circles by I and O, the tangent points of the inscribed circle
with sides BC, CA, AB by K, L, M , respectively (Fig. 122).
Figure 122 (Sol. 10.79)
First, let us prove that point O lies inside triangle KCI. For this it suffices to prove that
CK ≥ KB and ∠BCO ≤ ∠BCI. Clearly, CK = r cot γ2 ≥ r cot β2 = KB and
2∠BCO = 180◦ − ∠BOC = 180◦ − 2α ≤ 180◦ − α − β = γ = 2∠BCI.
Since ∠BCO = 90◦ − α = ∠ACH, the symmetry through CI sends line CO to line CH.
Let O′ be the image of O under this symmetry and P the intersection point of CH and IL.
Then CP ≥ CO′ = CO = R. It remains to prove that P H ≥ IM = r. It follows from the
fact that ∠M IL = 180◦ − α ≥ 90◦ .
10.80. Let B2 C2 be the projection of segment B1 C1 on side BC. Then
BC1 ≥ B2 C2 = BC − BC1 cos β − CB1 cos γ.
SOLUTIONS
251
Similarly,
A1 C1 ≥ AC − AC1 cos α − CA1 cos γ;
A1 B1 ≥ AB − AB1 cos α − BA1 cos β.
Let us multiply these inequalities by cos α, cos β and cos γ, respectively, and add them; we
get
B1 C1 cos α + C1 A1 cos β + AB1 cos γ ≥ a cos α + b cos β + c cos γ−
−(a cos β cos γ + b cos α cos γ + c cos α cos β).
Since c = a cos β + b cos α, it follows that c cos γ = a cos β cos γ + b cos α cos γ. Write three
analogous inequalities and add them; we get
a cos α + b cos β + c cos γ
a cos β cos γ + b cos α cos γ + c cos α cos β =
.
2
10.81. Since
cos2 α + cos2 β + cos2 γ + 2 cos α cos β cos γ = 1
(Problem 12.39 b)), it follows that triangle ABC is an acute one if and only if cos2 α +
cos2 β + cos2 γ < 1, i.e., sin2 α + sin2 β + sin2 γ > 2. Multiplying both sides of the latter
inequality by 4R2 we get the desired statement.
10.82. It suffices to notice that
p2 − (2R + r)2 = 4R2 cos α cos β cos γ
(cf. Problem 12.41 b).
10.83. Let ∠A ≤ ∠B ≤ ∠C. If triangle ABC is not an acute one, then CC1 < AC <
AA1 for any points A1 and C1 on sides BC and AB, respectively. Now, let us prove that for
an acute triangle we can select points A1 , B1 and C1 with the required property. For this it
suffices to verify that there exists a number x satisfying the following inequalities:
ha ≤ x < max(b, c) = c, hb ≤ x < max(a, c) = c and hc ≤ x < max(a, b) = b.
It remains to notice that max(ha , hb , hc ) = ha , min(b, c) = b and ha < h.
10.84. Let ∠A ≤ ∠B ≤ ∠C. First, suppose that triangle ABC is an acute one. As line
l that in its initial position is parallel to AB rotates, the length of the triangle’s projection
on l first varies monotonously from c to hb , then from hb to a, then from a to hc , next from
hc to b, then from b to ha and, finally, from ha to c. Since hb < a, there exists a number x
such that hb < x < a. It is easy to verify that a segment of length x is encountered on any
of the first four intervals of monotonity.
Now, suppose that triangle ABC is not an acute one. As line l that in its initial position
is parallel to AB rotates, the length of the triangle’s projection on l monotonously decreases
first from c to hb , then from hb to hc ; after that it monotonously increases, first, from hc to
ha , then from ha to c. Altogether we have two intervals of monotonity.
10.85. Let points M and N lie on sides AB and AC, respectively. Let us draw through
vertex C the line parallel to side AB. Let N1 be the intersection point of this line with M N .
Then N1 O : M O = 2 but N O ≤ N1 O; hence, N O : M O ≤ 2.
10.86. Sircle S inscribed in triangle ABC lies inside triangle A′ B ′ C ′ . Draw the tangents
to this circle parallel to sides of triangle A′ B ′ C ′ ; we get triangle A′′ B ′′ C ′′ similar to triangle
A′ B ′ C ′ and S is the inscribed circle of triangle A′′ B ′′ C ′′ . Hence, rABC = rA′′ B ′′ C ′′ < rA′ B ′ C ′ .
10.87. The bisector lc divides triangle ABC into two triangles whose doubled areas are
equal to alc sin γ2 and blc sin γ2 . Hence, aha = 2S = lc (a + b) sin γ2 . The conditions of the
a
≤ 21 ≤ sin γ2 .
problem imply that a+b
252
CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE
10.88. Clearly, cot ∠A + cot ∠B = hcc ≥ mcc . Let M be the intersection point of the
medians, N the midpoint of segment AB. Since triangle AM B is a right one, M N = 12 AB.
Therefore, c = 2M N = 32 mc .
10.89. Since BN ·BA = BM 2 and BM < BA, it follows that BN < BM and, therefore,
AN > CN .
10.90. Let us draw through point B the perpendicular to side AB. Let F be the
intersection point of this perpendicular with the extension of side AC (Fig. 123). Let us
prove that bisector AD, median BM and height CH intersect at one point if and only if
AB = CF . Indeed, let L be the intersection point of BM and CH. Bisector AD passes
through point L if and only if BA : AM = BL : LM but BL : LM = F C : CM = F C : AM .
Figure 123 (Sol. 10.90)
If on side AF of right triangle ABF (∠ABF = 90◦ ) segment CF equal to AB is marked,
then angles ∠BAC and ∠ABC are acute ones. It remains to find out when angle ∠ACB is
acute.
Let us drop perpendicular BP from point B to side AF . Angle ACB is an acute one if
F P > F C = AB,
i.e., BF sin ∠A > BF cot ∠A. Therefore, 1−cos2 ∠A = sin2 ∠A > cos ∠A,
√
i.e., cos A < 21 ( 5 − 1). Finally, we see that
√
5−1
◦
90 > ∠A > arccos
≈ 51◦ 50′ .
2
10.91. Since the greater angle subtends the longer side,
(a − b)(α − β) ≥ 0, (b − c)(β − γ) ≥ 0 and (a − c)(α − γ) ≥ 0.
Adding these inequalities we get
i.e., 31 π ≤
i.e.,
aα+bβ+cγ
.
a+b+c
2(aα + bβ + cγ) ≥ a(β + γ) + b(α + γ) + c(α + β) =
(a + b + c)π − aα − bβ − cγ,
The triangle inequality implies that
α(b + c − a) + β(a + c − b) + γ(a + b − c) > 0,
a(β + γ − α) + b(α + γ − β) + c(α + β − γ) > 0.
< 21 π.
Since α +β +γ = π, it follows that a(π −2α)+b(π −2β) = c(π −2γ) > 0, i.e., aα+bβ+cγ
a+b+c
10.92. On rays OB and OC, take points C1 and B1 , respectively, such that OC1 = OC
and OB1 = OB. Let B2 and C2 be the projections of points B1 and C1 , respectively, on a
line perpendicular to AO. Then
BO sin ∠AOC + CO sin ∠AOB = B2 C2 ≤ BC.
Adding three analogous inequalities we get the desired statement. It is also easy to verify
that the conditions B1 C1 ⊥ AO, C1 A1 ⊥ BO and A1 B1 ⊥ CO are equivalent to the fact
that O is the intersection point of the bisectors.
SOLUTIONS
253
10.93. Since ∠CBD = 21 ∠C and ∠B ≤ ∠A, it follows that ∠ABD = ∠B + ∠CBD ≥
1
(∠A + ∠B + ∠C) = 90◦ .
2
10.94. By the bisector’s property, BM : M A = BC : CA and BK : KC = BA : AC.
Hence, BM : M A < BK : KC, i.e.,
BM
BK
CB
AB
=1+
<1+
=
.
AM
MA
KC
CK
Therefore, point M is more distant from line AC than point K, i.e., ∠AKM < ∠KAC =
∠KAM and ∠KM C < ∠M CA = ∠M CK. Hence, AM > M K and M K > KC, cf.
Problem 10.59.
10.95. Suppose that all the given ratios are less than 2. Then
SABO + SAOC < 2SXBO + 2SXOC = 2SOBC ,
SABO + SOBC < 2SAOC , SAOC + SOBC < 2SABO .
Adding these inequalities we come to a contradiction. We similarly prove that one of the
given ratios is not greater than 2.
10.96. Denote the radii of the circles S, S1 and S2 by r, r1 and r2 , respectively. Let
triangles AB1 C1 and A2 BC2 be similar to triangle ABC with similarity coefficients rr1 and
r2
, respectively. Circles S1 and S2 are the inscribed circles of triangles AB1 C1 and A2 BC2 ,
r
respectively. Therefore, these triangles intersect because otherwise circles S1 and S2 would
not have had common points. Hence, AB1 + A2 B > AB, i.e., r1 + r2 > r.
Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM
Background
1) Geometric problems on maximum and minimum are in close connection with geometric
inequalities because in order to solve these problems we always have to prove a corresponding geometric inequality and, moreover, to prove that sometimes it turns into an equality.
Therefore, before solving problems on maximum and minimum we have to skim through
Supplement to Ch. 9 once again with the special emphasis on the conditions under which
strict inequalities become equalities.
2) For elements of a triangle we use the standard notations.
3) Problems on maximum and minimum are sometimes called extremal problems (from
Latin extremum).
Introductory problems
1. Among all triangles ABC with given sides AB and AC find the one with the greatest
area.
2. Inside triangle ABC find the vertex of the smallest angle that subtends side AB.
3. Prove that among all triangles with given side a and height ha . an isosceles triangle
is the one with the greatest value of angle α.
4. Among all triangles with given sides AB and AC (AB < AC), find the one for which
the radius of the circumscribed circle is maximal.
5. The iagonals of a convex quadrilateral are equal to d1 and d2 . What the greatest
value the quadrilateral’s area can attain?
§1. The triangle
11.1. Prove that among all the triangles with fixed angle α and area S, an isosceles
triangle with base BC has the shortest length of side BC.
11.2. Prove that among all triangles with fixed angle α and semiperimeter p, an isosceles
triangle with base BC is of the greatest area.
11.3. Prove that among all the triangles with fixed semiperimeter p, an equilateral
triangle has the greatest area.
11.4. Consider all the acute triangles with given side a and angle α. What is the
maximum of b2 + c2 ?
11.5. Among all the triangles inscribed in a given circle find the one with the maximal
sum of squared lengths of the sides.
11.6. The perimeter of triangle ABC is equal to 2p. On sides AB and AC points M
and N , respectively, are taken so that M N k BC and M N is tangent to the inscribed circle
of triangle ABC. Find the greatest value of the length of segment M N .
11.7. Into a given triangle place a centrally symmetric polygon of greatest area.
11.8. The area of triangle ABC is equal to 1. Let A1 , B1 , C1 be the midpoints of sides
BC, CA, AB, respectively. On segments AB1 , CA1 , BC1 , points K, L, M , respectively, are
taken. What is the least area of the common part of triangles KLM and A1 B1 C1 ?
255
256
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
11.9. What least width From an infinite strip of paper any triangle of area 1 can be cut.
What is the least width of such a strip?
***
11.10. Prove that triangles with the lengths of sides a, b, c and a1 , b1 , c1 , respectively,
are similar if and only if
p
p
√
√
aa1 + bb1 + cc1 = (a + b + c)(a1 + b1 + c1 ).
11.11. Prove that if α, β, γ and α1 , β1 , γ1 are the respective angles of two triangles,
then
cos α1 cos β1 cos γ1
+
+
≤ cot α + cot β + cot γ.
sin α
sin β
sin γ
11.12. Let a, b and c be the lengths of the sides of a triangle of area S; let α1 , β1 and
γ1 be the angles of another triangle. Prove that
a2 cot α1 + b2 cot β1 + c2 cot γ1 ≥ 4S,
where the equality is attained only if the considered triangles are similar.
11.13. In a triangle a ≥ b ≥ c; let x, y and z be the angles of another triangle. Prove
that
a2 + b 2 + c 2
.
bc + ca − ab < bc cos x + ca cos y + ab cos z ≤
2
See also Problem 17.21.
§2. Extremal points of a triangle
11.14. On hypothenuse AB of right triangle ABC point X is taken; M and N are the
projections of X on legs AC and BC, respectively.
a) What is the position of X for which the length of segment M N is the smallest one?
b) What is the position of point X for which the area of quadrilateral CM XN is the
greatest one?
11.15. From point M on side AB of an acute triangle ABC perpendiculars M P and
M Q are dropped to sides BC and AC, respectively. What is the position of point M for
which the length of segment P Q is the minimal one?
11.16. Triangle ABC is given. On line AB find point M for which the sum of the radii
of the circumscribed circles of triangles ACM and ACN is the least possible one.
11.17. From point M of the circumscribed circle of triangle ABC perpendiculars M P
and M Q are dropped on lines AB and AC, respectively. What is the position of point M
for which the length of segment P Q is the maximal one?
11.18. Inside triangle ABC, point O is taken. Let da , db , dc be distances from it to lines
BC, CA, AB, respectively. What is the position of point O for which the product da db dc is
the greatest one?
11.19. Points A1 , B1 and C1 are taken on sides BC, CA and AB, respectively, of triangle
ABC so that segments AA1 , BB1 and CC1 meet at one point M . For what position of point
A1 M B 1 M C 1
M the value of M
· BB1 · CC1 is the maximal one?
AA1
11.20. From point M inside given triangle ABC perpendiculars M A1 , M B1 , M C1 are
dropped to lines BC, CA, AB, respectively. What are points M inside the given triangle
ABC for which the quantity MaA1 + MbB1 + McC1 takes the least possible value?
11.21. Triangle ABC is given. Find a point O inside of it for which the sum of lengths
of segments OA, OB, OC is the minimal one. (Take a special heed to the case when one of
the angles of the triangle is greater than 120◦ .)
§5. POLYGONS
257
11.22. Inside triangle ABC find a point O for which the sum of squares of distances
from it to the sides of the triangle is the minimal one.
See also Problem 18.21 a).
§3. The angle
11.23. On a leg of an acute angle points A and B are given. On the other leg construct
point C the vertex of the greatest angle that subtends segment AB.
11.24. Angle ∠XAY and point O inside it are given. Through point O draw a line that
cuts off the given angle a triangle of the least area.
11.25. Through given point P inside angle ∠AOB draw line M N so that the value
OM + ON is minimal (points M and N lie on legs OA and OB, respectively).
11.26. Angle ∠XAY and a circle inside it are given. On the circle construct a point the
sum of the distances from which to lines AX and AY is the least.
11.27. A point M inside acute angle ∠BAC is given. On legs BA and AC construct
points X and Y , respectively, such that the perimeter of triangle XY M is the least.
11.28. Angle ∠XAY is given. The endpoints B and C of unit segments BO and CO
move along rays AX and AY , respectively. Construct quadrilateral ABOC of the greatest
area.
§4. The quadrilateral
11.29. Inside a convex quadrilateral find a point the sum of distances from which to the
vertices were the least one.
11.30. The diagonals of convex quadrilateral ABCD intersect at point O. What least
area can this quadrilateral have if the area of triangle AOB is equal to 4 and the area of
triangle COD is equal to 9?
11.31. Trapezoid ABCD with base AD is cut by diagonal AC into two triangles. Line
l parallel to the base cuts these triangles into two triangles and two quadrilaterals. What
is the position of line l for which the sum of areas of the obtained triangles is the minimal
one?
11.32. The area of a trapezoid is equal to 1. What is the least value the length of the
longest diagonal of this trapezoid can attain?
11.33. On base AD of trapezoid ABCD point K is given. On base BC find point M
for which the area of the common part of triangles AM D and BKC is maximal.
11.34. Prove that among all quadrilaterals with fixed lengths of sides an inscribed
quadrilateral has the greatest area.
See also Problems 9.35, 15.3 b).
§5. Polygons
11.35. A polygon has a center of symmetry, O. Prove that the sum of the distances
from a point to the vertices attains its minimum at point O.
11.36. Among all the polygons inscribed in a given circle find the one for which the sum
of squared lengths of its sides is minimal.
11.37. A convex polygon A1 . . . An is given. Prove that a point of the polygon for which
the sum of distances from it to all the vertices is maximal is a vertex.
See also Problem 6.69.
258
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
§6. Miscellaneous problems
11.38. Inside a circle centered at O a point A is given. Find point M on the circle for
which angle ∠OM A is maximal.
11.39. In plane, line l and points A and B on distinct sides of l are given. Construct a
circle that passes through points A and B so that line l intercepts on the circle a shortest
chord.
11.40. Line l and points P and Q lying on one side of l are given. On line l, take point
M and in triangle P QM draw heights P P ′ and QQ′ . What is the position of point M for
which segment P ′ Q′ is the shortest?
11.41. Points A, B and O do not lie on one line. Through point O draw line l so that
the sum of distances from it to points A and B were: a) maximal; b) minimal.
11.42. If five points in plane are given, then considering all possible triples of these
points we can form 30 angles. Denote the least of these angles by α. Find the greatest value
of α.
11.43. In a town there are 10 streets parallel to each other and 10 streets that intersect
them at right angles. A closed bus route passes all the road intersections. What is the least
number of turns such a bus route can have?
11.44. What is the greatest number of cells on a 8 × 8 chessboard that one straight line
can intersect? (An intersection should have a common inner point.)
11.45. What is the greatest number of points that can be placed on a segment of length
1 so that on any segment of length d contained in this segment not more than 1 + 1000d2
points lie?
See also Problems 15.1, 17.20.
§7. The extremal properties of regular polygons
11.46. a) Prove that among all n-gons circumscribed about a given circle a regular n-gon
has the least area.
b) Prove that among all the n-gons circumscribed about a given circle a regular n-gon
has the least perimeter.
11.47. Triangles ABC1 and ABC2 have common base AB and ∠AC1 B = ∠AC2 B.
Prove that if |AC1 − C1 B| < |AC2 − C2 B|, then:
a) the area of triangle ABC1 is greater than the area of triangle ABC2 ;
b) the perimeter of triangle ABC1 is greater than the perimeter of triangle ABC2 .
11.48. a) Prove that among all the n-gons inscribed in a given circle a regular n-gon has
the greatest area.
b) Prove that among all n-gons inscribed in a given circle a regular n-gon has the greatest
perimeter.
Problems for independent study
11.49. On a leg of an acute angle with vertex A point B is given. On the other leg
construct point X such that the radius of the circumscribed circle of triangle ABX is the
least possible.
11.50. Through a given point inside a (given?) circle draw a chord of the least length.
11.51. Among all triangles with a given sum of lengths of their bisectors find a triangle
with the greatest sum of lengths of its heights.
11.52. Inside a convex quadrilateral find a point the sum of squared distances from
which to the vertices is the least possible.
SOLUTIONS
259
11.53. Among all triangles inscribed in a given circle find the one for which the value
+ 1b + 1c is the least possible.
11.54. On a chessboard with the usual coloring draw a circle of the greatest radius so
that it does not intersect any white field.
11.55. Inside a square, point O is given. Any line that passes through O cuts the square
into two parts. Through point O draw a line so that the difference of areas of these parts
were the greatest possible.
11.56. What is the greatest length that the shortest side of a triangle inscribed in a
given square can have?
11.57. What greatest area can an equilateral triangle inscribed in a given square can
have?
1
a
Solutions
11.1. By the law of cosines
a2 = b2 + c2 − 2bc cos α = (b − c)2 + 2bc(1 − cos α) =
(b − c)2 +
4S(1 − cos α)
.
sin α
Since the last summand is constant, a is minimal if b = c.
11.2. Let an escribed circle be tangent to sides AB and AC at points K and L, respectively. Since AK = AL = p, the escribed circle Sa is fixed. The radius r of the inscribed
circle is maximal if it is tangent to circle Sa , i.e., triangle ABC is an isosceles one. It is also
clear that S = pr.
2
11.3. By Problem 10.53 a) we have S ≤ 3p√3 , where the equality is only attained for an
equilateral triangle.
11.4. By the law of cosines b2 + c2 = a2 + 2bc cos α. Since 2bc ≤ b2 + c2 and cos α > 0,
a2
it follows that b2 + c2 ≤ a2 + (b2 + c2 ) cos α, i.e., b2 + c2 ≤ 1−cos
. The equality is attained if
α
b = c.
11.5. Let R be the radius of the given circle, O its center; let A, B and C be the vertices
−→
−−→
−→
of the triangle; a = OA, b = OB, c = OC. Then
Since
it follows that
AB 2 + BC 2 + CA2 = |a − b|2 + |b − c|2 + |c − a|2 =
2(|a|2 + |b|2 + |c|2 ) − −2(a, b) − 2(b, c) − 2(c, a).
|a + b + c|2 = |a|2 + |b|2 + |c|2 + 2(a, b) + 2(b, c) + 2(c, a),
AB 2 + BC 2 + CA2 = 3(|a|2 + |b|2 + |c|2 ) − |a + b + c|2 ≤
3(|a|2 + |b|2 + |c|2 ) = 9R2 ,
where the equality is only attained if a + b + c = 0. This equality means that triangle ABC
is an equilateral one.
11.6. Denote the length of the height dropped
△AM N ∼
¢ BC by h. S Since
¡ on 2rside
h−2r
ah
N
.
Since
r
=
=
,
i.e.
M
N
=
a
1
−
=
,
we deduce
△ABC, it follows that M
h
h
´ BC
³ p ´2p
³
that M N = a 1 − ap . The maximum of the quadratic expression a 1 − ap = a(p−a)
in a
p
260
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
is attained for a = 12 p. This maximum is equal to p4 . It remains to notice that there exists a
triangle of perimeter 2p with side a = 21 p (set b = c = 34 p).
11.7. Let O be the center of symmetry of polygon M lying inside triangle T , let S(T )
be the image of triangle T under the symmetry through point O. Then M lies both in T
and in S(T ). Therefore, among all centrally symmetric polygons with the given center of
symmetry lying in T the one with the greatest area is the intersection of T and S(T ). Point
O lies inside triangle T because the intersection of T and S(T ) is a convex polygon and a
convex polygon always contains its center of symmetry.
Figure 124 (Sol. 11.7)
Let A1 , B1 and C1 be the midpoints of sides BC, CA and AB, respectively, of triangle
T = ABC. First, let us suppose that point O lies inside triangle A1 B1 C1 . Then the
intersection of T and S(T ) is a hexagon (Fig. 124). Let side AB be divided by the sides of
triangle S(T ) in the ratio of x : y : z, where x + y + z = 1. Then the ratio of the sum of
areas of the shaded triangles to the area of triangle ABC is equal to x2 + y 2 + z 2 and we
have to minimize this expression. Since
1 = (x + y + z)2 = 3(x2 + y 2 + z 2 ) − (x − y)2 − (y − z)2 − (z − x)2 ,
it follows that x2 + y 2 + z 2 ≥ 31 , where the equality is only attained for x = y = z; the latter
equality means that O is the intersection point of the medians of triangle ABC.
Now, consider another case: point O lies inside one of the triangles AB1 C1 , A1 BC1 ,
A1 B1 C; for instance, inside AB1 C1 . In this case the intersection of T and S(T ) is a parallelogram and if we replace point O with the intersection point of lines AO and B1 C1 , then
the area of this parallelogram can only increase. If point O lies on side B1 C1 , then this is
actually the case that we have already considered (set x = 0).
The polygon to be found is a hexagon with vertices at the points that divide the sides of
the triangles into three equal parts. Its area is equal to 32 of the area of the triangle.
11.8. Denote the intersection point of lines KM and BC by T and the intersection
points of the sides of triangles A1 B1 C1 and KLM as shown on Fig. 125.
Figure 125 (Sol. 11.8)
SOLUTIONS
261
Then T L : RZ = KL : KZ = LC : ZB1 . Since T L ≥ BA1 = A1 C ≥ LC, it follows that
RZ ≥ ZB1 , i.e., SRZQ ≥ SZB1 Q . Similarly, SQY P ≥ SY A1 P and SP XR ≥ SXC1 R . Adding all
these inequalities and the inequality SP QR > 0 we see that the area of hexagon P XRZQY
is not less than the area of the remaining part of triangle A1 B1 C1 , i.e., its area is not less
S
than A1 B2 1 C1 = 18 . The equality is attained, for instance, if point K coincides with B1 and
point M with B.
√
2
11.9. Since the area of an equilateral triangle with side a is equal to a 4 3 , the side of an
√
4
2
and
its
height
is
equal
to
3. Let us prove that
equilateral triangle of area 1 is equal to √
4
3
√
it is impossible to cut an equilateral triangle of area 1 off a strip of√width less than 4 3.
Let equilateral triangle ABC lie inside a strip of width less than 4 3. Let, for definiteness,
the projection of vertex B on the boundary of the strip lie between the projections of vertices
A and C. Then the line drawn through point B perpendicularly to the boundary of the strip
intersects segment AC at a point M . The length of a height of triangle ABC does not exceed
BM and BM is not greater
than the width of the strip and, therefore, a height of triangle
√
4
ABC is shorter than 3, i.e., its area is less than 1.
√
It remains to prove that any triangle of area 1 can be cut off a strip
of width 4 3. Let us
√
prove that any triangle of area 1 has a height that does not exceed 4 3. For this it suffices
2
to prove that it has a side not shorter than √
4 . Suppose that all sides of triangle ABC are
3
√
shorter than 2 4 3. Let α be the smallest angle of this triangle. Then α ≤ 60◦ and
¶2 Ã √ !
µ
3
2
AB · AC sin α
< √
= 1.
SABC =
4
2
4
3
√
We have obtained a contradiction.
A triangle that has a height not exceeding 4 3 can be
√
placed in a strip of width 4 3: place the side to which this height is dropped on a boundary
of the strip.
11.10. Squaring both sides of the given equality we easily reduce the equality to the
form
p
p
p
p
√
√
( ab1 − a1 b)2 + ( ca1 − c1 a)2 + ( bc1 − cb1 )2 = 0,
i.e., aa1 = bb1 = cc1 .
11.11. Fix angles α, β and γ. Let A1 B1 C1 be a triangle with angles α1 , β1 and γ1 .
−−−→ −−−→
−−−→
Consider vectors a, b and c codirected with vectors B1 C1 , C1 A1 and A1 B1 and of length
sin α, sin β and sin γ, respectively. Then
cos α1 cos β1 cos γ1
[(a, b) + (b, c) + (c, a)]
+
+
=−
.
sin α
sin β
sin γ
sin α sin β sin γ
Since
2[(a, b) + (b, c) + (c, a)] = |a + b + c|2 − |a|2 − |b|2 − |c|2 ,
the quantity (a, b) + (b, c) + (c, a) attains its minimum when a + b + c = 0, i.e., α1 = α,
β1 = β and γ1 = γ.
11.12. Let x = cot α1 and y = cot β1 . Then x + y > 0 (since α1 + β1 < π) and
cot γ1 =
x2 + 1
1 − xy
=
− x.
x+y
x+y
Therefore,
2
2
2
2
2
2
2
a cot α1 + b cot β1 + c cot γ1 = (a − b − c )x + b (x + y) + c
2x
2
+1
.
x+y
262
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
2
+1
, i.e.,
For a fixed x this expression is minimal for a y such that b2 (x + y) = c2 xx+y
sin γ1
c
x+y
= sin α1 (cot α1 + cot β1 ) =
=√
.
2
b
sin β1
1+x
Similar arguments show that if a : b : c = sin α1 : sin β1 : sin γ1 , then the considered
expression is minimal. In this case triangles are similar and a2 cot α+b2 cot β +c2 cot γ = 4S,
cf. Problem 12.44 b).
11.13. Let f = bc cos x + ca cos y + ab cos z. Since cos x = − cos y cos z + sin y sin z, it
follows that
f = c(a − b cos z) cos y + bc sin y sin z + ab cos z.
Consider a triangle the lengths of whose two sides are equal to a and b and the angle between
them is equal to z; let ξ and η be the angles subtending sides a and b; let t be the length of
the side that subtends angle z. Then
t2 + a2 − b2
a2 + b2 − t2
and cos η =
;
2ab
2at
η
. Therefore, f = ct cos(η − y) + 21 (a2 + b2 − t2 ).
hence, a−btcos z = cos η. Moreover, bt = sin
sin z
Since cos(η − y) ≤ 1, it follows that f ≤ 21 (a2 + b2 + c2 ) − 21 ((c − t)2 ) ≤ 21 (a2 + b2 + c2 ).
Since a ≥ b, it follows that ξ ≥ η, consequently, −ξ ≤ −η < y − η < π − z − η = ξ, i.e.,
cos(y − ψ) > cos ξ. Hence,
cos z =
c − b 2 c(b2 − a2 ) a2 + b2
a2 + b2 − t2
=
t +
+
= g(t).
2
2b
2b
2
The coefficient of t2 is either negative or equal to zero; moreover, t < a + b. Hence, g(t) ≥
g(a + b) = bc + ca − ab.
11.14. a) Since CM XN is a rectangle, M N = CX. Therefore, the length of segment
M N is the least possible if CX is a height.
2 ·S
2 ·S
b) Let SABC = S. Then SAM X = AX
and SBN X = BX
. Since AX 2 + BX 2 ≥ 12 AB 2
AB 2
AB 2
(where the equality is only attained if X is the midpoint of segment AB), it follows that
SCM XN = S − SAM X − SBN X ≤ 12 S. The area of quadrilateral CM XN is the greatest if X
is the midpoint of side AB.
11.15. Points P and Q lie on the circle constructed on segment CM as on the diameter.
In this circle the constant angle C intercepts chord P Q, therefore, the length of chord P Q
is minimal if the diameter CM of the circle is minimal, i.e., if CM is a height of triangle
ABC.
11.16. By the law of sines the radii of the circumscribed circles of triangles ACM and
BCM are equal to 2 sinAC
and 2 sinBC
, respectively. It is easy to verify that sin AM C =
AM C
BM C
sin BM C. Therefore,
BC
AC + BC
AC
+
=
.
2 sin AM C 2 sin BM C
2 sin BM C
The latter expression is minimal if sin BM C = 1, i.e., CM ⊥ AB.
11.17. Points P and Q lie on the circle with diameter AM , hence, P Q = AM sin P AQ =
AM sin A. It follows that the length of segment P Q is maximal if AM is a diameter of the
circumscribed circle.
11.18. Clearly, 2SABC = ada + bdb + cdc . Therefore, the product (ada )(bdb )(cdc ) takes
its greatest value if ada = bdb = cdc (cf. Supplement to Ch. 9, the inequality between the
mean arithmetic and the mean geometric). Since the value abc is a constant, the product
(ada )(bdb )(cdc ) attains its greatest value if and only if the product da db dc takes its greatest
value.
f > ct cos ξ +
SOLUTIONS
263
Let us show that equality ada = bdb = cdc means that O is the intersection point of the
medians of triangle ABC. Denote the intersection point of lines AO and BC by A1 . Then
BA1 : A1 C = SABA1 : SACA1 = SABO : SACO = (cdc ) : (bdb ) = 1,
i.e., AA1 is a median. We similarly prove that point O lies on medians BB1 and CC1 .
A1
B1
C1
11.19. Let α = M
,β=M
and γ = M
. Since α + β + γ = 1 (cf. Problem 4.48 a)),
AA1
BB1
CC1
√
1
1
we have αβγ ≤ 3 (α + β + γ) = 3 , where the equality is attained when α = β = γ = 31 ,
i.e., M is the intersection point of the medians.
11.20. Let x = M A1 , y = M B1 and z = M C1 . Then
ax + by + cz = 2SBM C + 2SAM C + 2SAM B = 2SABC .
Hence,
³
a
x
+
b
y
+
c
z
´
³
a
x
b
y
c
z
´
· 2SABC =
³
´
³
´
¢
¡
y
y
x
z
2
2
2
(ax + by + cz) = a + b + c + ab y + x + bc z + y + ac xz + xz ≥
a2 + b2 + c2 + 2ab + 2bc + 2ac,
+
+
where the equality is only attained if x = y = z, i.e., M is the center of the inscribed circle
of triangle ABC.
11.21. First, suppose that all the angles of triangle ABC are less than 120◦ . Then inside
triangle ABC there exists a point O — the vertex of angles of 120◦ that subtend each side.
Let us draw through vertices A, B and C lines perpendicular to segments OA, OB and OC,
respectively. These lines form an equilateral triangle A1 B1 C1 (Fig. 126).
Figure 126 (Sol. 11.21)
Let O′ be any point that lies inside triangle ABC and is distinct from O. Let us prove
then that O′ A + O′ B + O′ C > OA + OB + OC, i.e., O is the desired point. Let A′ , B ′ and
C ′ be the bases of the perpendiculars dropped from point O′ on sides B1 C1 , C1 A1 and A1 B1 ,
respectively, a the length of the side of equilateral triangle A1 B1 C1 . Then
O′ A′ + O′ B ′ + O′ C ′ =
2(SO′ B1 C1 + SO′ A1 C1 )
=
a
2SA1 B1 C1
= OA + OB + OC.
a
Since a slanted line is longer than the perpendicular,
O′ A + O′ B + O′ C > O′ A′ + O′ B ′ + O′ C ′ = OA + OB + OC.
264
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
Figure 127 (Sol. 11.21)
Now, let one of the angles of triangle ABC, say ∠C, be greater than 120◦ . Let us
draw through points A and B perpendiculars B1 C1 and C1 A1 to segments CA and CB and
through point C line A1 B1 perpendicular to the bisector of angle ∠ACB (Fig. 127).
Since ∠AC1 B = 180◦ − ∠ACB < 60◦ , it follows that B1 C1 > A1 B1 . Let O′ be any point
that lies inside triangle A1 B1 C1 . Since
it follows that
B1 C1 · O′ A′ + C1 A1 · O′ B ′ + A1 C1 · O′ C ′ = 2SA1 B1 C1 ,
(O′ A′ + O′ B ′ + O′ C ′ ) · B1 C1 = 2SA1 B1 C1 + (B1 C1 − A1 B1 ) · O′ C ′ .
Since B1 C1 > A1 B1 , the sum O′ A′ + O′ B ′ + O′ C ′ is minimal for points that lie on side B1 A1 .
It is also clear that
O′ A + O′ B + O′ C ≥ O′ A′ + O′ B ′ + O′ C ′ .
Therefore, vertex C is the point to be found.
11.22. Let the distances from point O to sides BC, CA and AB be equal to x, y and z,
respectively. Then
ax + by + cz = 2(SBOC + SCOA + SAOB ) = 2SABC .
It is also clear that
µ
¶ µ
¶ µ
¶
SBOC
SCOA
SAOB
x:y:z=
:
:
.
a
b
c
Equation ax + by + cz = 2S determines a plane in 3-dimensional space with coordinates
x, y, z; vector (a, b, c) is perpendicular to this plane because if ax1 + by1 + cz1 = 2S and
ax2 + by2 + cz2 = 2S, then a(x1 − x2 ) + b(y1 − y2 ) + c(z1 − z2 ) = 0.
We have to find a point (x0 , y0 , z0 ) on this plane at which the minimum of expression
2
x + y 2 + z 2 is attained and verify that an inner point of the triangle corresponds to this
point. Since x2 + y 2 + z 2 is the squared distance from the origin to point (x, y, z), it follows
that the base of the perpendicular dropped from the origin to the plane is the desired point,
i.e., x : y : z = a : b : c. It remains to verify that inside the triangle there exists point O for
which x : y : z = a : b : c. This equality is equivalent to the condition
µ
¶ µ
¶ µ
¶
SBOC
SCOA
SAOB
:
:
= a : b : c,
a
b
c
i.e., SBOC : SCOA : SAOB = a2 : b2 : c2 . Since the equality SBOC : SAOB = a2 : c2 follows
from equalities SBOC : SCOA = a2 : b2 and SCOA : SAOB = b2 : c2 , the desired point is the
intersection point of lines CC1 and AA1 that divide sides AB and BC, respectively, in the
ratios of BC1 : C1 A = a2 : b2 and CA1 : A1 B = b2 : c2 , respectively.
SOLUTIONS
265
11.23. Let O be the vertex of the given angle. Point C is the tangent point of a leg with
the circle that passes through points A and B, i.e., OC 2 = OA · OB. To find the length of
segment OC, it suffices to draw the tangent to any circle that passes through points A and
B.
11.24. Let us consider angle ∠X ′ A′ Y ′ symmetric to angle ∠XAY through point O. Let
B and C be the intersection points of the legs of these angles. Denote the intersection points
of the line that passes through point O with the legs of angles ∠XAY and ∠X ′ A′ Y ′ by B1 ,
C1 and B1′ , C1′ , respectively (Fig. 128).
Figure 128 (Sol. 11.24)
Since SAB1 C1 = SA′ B1′ C1′ , it follows that SAB1 C1 = 21 (SABA′ C + SBB1 C1′ + SCC1 B1′ ). The area
of triangle AB1 C1 is the least if B1 = B and C1 = C, i.e., line BC is the one to be found.
11.25. On legs OA and OB, take points K and L so that KP k OB and LP k OA.
Then KM : KP = P L : LN and, therefore,
√
√
√
KM + LN ≥ 2 KM · LN = 2 KP · P L = 2 OK · OL
√
where the equality is attained when KM = LN = OK · OL. It is also clear that OM +
ON = (OK + OL) + (KM + LN ).
11.26. On rays AX and AY , mark equal segments AB an AC. If point M lies on segment
+SACM )
ABC
BC, then the sum of distances from it to lines AB and AC is equal to 2(SABMAB
= 2SAB
.
Therefore, the sum of distances from a point to lines AX and AY is the lesser, the lesser is
the distance between point A and the point’s projection on the bisector of angle ∠XAY .
11.27. Let points M1 and M2 be symmetric to M through lines AB and AC, respectively.
Since ∠BAM1 = ∠BAM and ∠CAM2 = ∠CAM , it follows that ∠M1 AM2 = 2∠BAC <
180◦ . Hence, segment M1 M2 intersects rays AB and AC at certain points X and Y (Fig.
129). Let us prove that X and Y are the points to be found.
Figure 129 (Sol. 11.27)
Indeed, if points X1 and Y1 lie on rays AB and AC, respectively, then M X1 = M1 X1
and M Y1 = M2 Y1 , i.e., the perimeter of triangle M X1 Y1 is equal to the length of the broken
266
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
line M X1 Y1 M2 . Of all the broken lines with the endpoints at M1 and M2 segment M1 M2 is
the shortest one.
11.28. Quadrilateral ABOC of the greatest area is a convex one. Among all the triangles
ABC with the fixed angle ∠A and side BC an isosceles triangle with base BC has the greatest
area. Therefore, among all the considered quadrilaterals ABOC with fixed diagonal BC the
quadrilateral with AB = AC, i.e., for which point O lies on the bisector of angle ∠A, is of
greatest area.
Further, let us consider triangle ABO in which angle ∠BAO equal to 12 ∠A and side BO
are fixed. The area of this triangle is maximal when AB = AO.
11.29. Let O be the intersection point of the diagonals of convex quadrilateral ABCD
and O1 any other point. Then AO1 + CO1 ≥ AC = AO + CO and BO1 + DO1 ≥ BD =
BO + DO, where at least one of the inequalities is a strict one. Therefore, O is the point to
be found.
11.30. Since SAOB : SBOC = AO : OC = SAOD
√ : SDOC , it follows that SBOC · SAOD =
SAOB · SDOC = 36. Therefore, SBOC + SAOD ≥ 2 SBOC · SAOD = 12, where the equality
takes place if SBOC = SAOD , i.e., SABC = SABD . This implies that AB k CD. In this case
the area of the triangle is equal to 4+9+12=25.
11.31. Let S0 and S be the considered sums of areas of triangles for line l0 that passes
through the intersection point of the diagonals of the trapezoid and for another line l. It is
easy to verify that S = S0 + s, where s is the area of the triangle formed by diagonals AC
and BD and line l. Hence, l0 is the line to be found.
11.32. Denote the lengths of the diagonals of the trapezoid by d1 and d2 and the lengths
of their projections on the bottom base by p1 and p2 , respectively; denote the lengths of the
bases by a and b and that of the height by h. Let, for definiteness, d1 ≥ d2 . Then p1 ≥ p2 .
Clearly, p1 + p2 ≥ a + b. Hence, p1 ≥ a+b
= Sh = h1 . Therefore, d21 = p21 + h2 ≥ h12 + h2 ≥ 2,
2
√
where the equality is attained only if p1 = p2 = h = 1. In this case d1 = 2.
11.33. Let us prove that point M that divides side BC in the ratio of BM : N C =
AK : KD is the desired one. Denote the intersection points of segments AM and BK, DM
and CK by P and Q, respectively. Then KQ : QC = KD : M C = KA : M B = KP : P B,
i.e., line P Q is parallel to the basis of the trapezoid.
Let M1 be any other point on side BC. For definiteness, we may assume that M1 lies
on segment BM . Denote the intersection points of AM1 and BK, DM1 and CK, AM1 and
P Q, DM1 and P Q, AM and DM1 by P1 , Q1 , P2 , Q2 , O, respectively (Fig. 130).
Figure 130 (Sol. 11.33)
We have to prove that SM P KQ > SM1 P1 KQ1 , i.e., SM OQ1 Q > SM1 OP P1 . Clearly, SM OQ1 Q >
SM OQ2 Q = SM1 OP P2 > SM1 OP P1 .
11.34. By Problem 4.45 a) we have
∠B + ∠D
.
S 2 = (p − a)(p − b)(p − c)(p − d) − abcd cos2
2
SOLUTIONS
267
= 0, i.e., ∠B + ∠D = 180◦ .
This quantity takes its maximal value when cos ∠B+∠D
2
′
11.35. If A and A are vertices of the polygon symmetric through point O, then the sum
of distances from any point of segment AA′ to points A and A′ is the same whereas for any
other point it is greater. Point O belongs to all such segments.
11.36. If in triangle ABC, angle ∠B is either obtuse or right, then by the law of sines
AC 2 ≥ AB 2 + BC 2 . Therefore, if in a polygon the angle at vertex B is not acute, then
deleting vertex B we obtain a polygon with the sum of squared lengths of the sides not less
than that of the initial polygon. Since for n ≥ 3 any n-gon has a nonacute angle, it follows
that by repeating such an operation we eventually get a triangle. Among all the triangles
inscribed in the given circle an equilateral triangle has the greatest sum of squared lengths
of the sides, cf. Problem 11.5.
−−→
11.37. If point X divides segment P Q in the ratio of λ : (1 − λ), then Ai X = (1 −
−−→
−−→
λ)Ai P + λAi Q; hence, Ai X ≤ (1 − λ)Ai P + λAi Q. Therefore,
X
X
X
f (X) =
Ai X ≤ (1 − λ)
Ai P + λ
Ai Q = (1 − λ)f (P ) + λf (Q).
Let, for instance, f (P ) ≤ f (Q), then f (X) ≤ f (Q); hence, on segment P Q the function f
attains its maximal value at one of the endpoints; more precisely, inside the segment there
can be no point of strict maximum of f . Hence, if X is any point of the polygon, then
f (X) ≤ f (Y ), where Y is a point on a side of the polygon and f (Y ) ≤ f (Z), where Z is a
vertex.
11.38. The locus of points X for which angle ∠OXA is a constant consists of two arcs
of circles S1 and S2 symmetric through line OA.
Consider the case when the diameter of circles S1 and S2 is equal to the radius of the
initial circle, i.e., when these circles are tangent to the initial circle at points M1 and M2
for which ∠OAM1 = ∠OAM2 = 90◦ . Points M1 and M2 are the desired ones because if
∠OXA > ∠OM1 A = ∠OM2 A, then point X lies strictly inside the figure formed by circles
S1 and S2 , i.e., cannot lie on the initial circle.
11.39. Let us denote the intersection point of line l and segment AB by O. Let us
consider an arbitrary circle S that passes through points A and B. It intersects l at certain
points M and N . Since M O · N O = AO · BO is a constant,
√
√
M N = M O + N O ≥ 2 M O · N O = 2 AO · BO,
where the equality is only attained if M O = N O. In the latter case the center of S is the
intersection point of the midperpendicular to AB and the perpendicular to l that passes
through point O.
11.40. Let us construct the circle with diameter P Q. If this circle intersects with l,
then any of the intersection points is the desired one because in this case P ′ = Q′ . If the
circle does not intersect with l, then for any point M on l angle ∠P M Q is an acute one
and ∠P ′ P Q′ = 90◦ ± ∠P M Q. Now it is easy to establish that the length of chord P ′ Q′ is
minimal if angle ∠P M Q is maximal.
To find point M it remains to draw through points P and Q circles tangent to l (cf.
Problem 8.56 a)) and select the needed point among the tangent points.
11.41. Let the sum of distances from points A and B to line l be equal to 2h. If l
intersects segment AB at point X, then SAOB = h · OX and, therefore, the value of h is
extremal when the value of OX is extremal, i.e., when line OX corresponds to a side or a
height of triangle AOB.
If line l does not intersect segment AB, then the value of h is equal to the length of the
midline of the trapezoid confined between the perpendiculars dropped from points A and B
268
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
on line l. This quantity is an extremal one when l is either perpendicular to median OM of
triangle AOB or corresponds to a side of triangle AOB. Now it only remains to select two
of the obtained four straight lines.
11.42. First, suppose that the points are the vertices of a convex pentagon. The sum
◦
of angles of the pentagon is equal to 540◦ ; hence, one of its angles does not exceed 540
=
5
108◦ . The diagonals divide this angle into three angles, hence, one of them does not exceed
108◦
= 36◦ . In this case α ≤ 36◦ .
3
If the points are not the vertices of a convex pentagon, then one of them lies inside the
triangle formed by some other three points. One of the angles of this triangle does not exceed
60◦ . The segment that connects the corresponding vertex with an inner point divides this
angle into two angles, hence, one of them does not exceed 30◦ . In this case α ≤ 30◦ . In all
the cases α ≤ 36◦ . Clearly, for a regular pentagon α = 36◦ .
11.43. A closed route that passes through all the road crossings can have 20 turns (Fig.
131). It remains to prove that such a route cannot have less than 20 turns. After each turn
a passage from a horizontal street to a vertical one or from a vertical street to a horizontal
one occurs.
Figure 131 (Sol. 11.43)
Hence, the number of horizontal links of a closed route is equal to the number of vertical
links and is equal to half the number of turns. Suppose that a closed route has less than 20
turns. Then there are streets directed horizontally, as well as streets directed vertically, along
which the route does not pass. Therefore, the route does not pass through the intersection
point of these streets.
11.44. A line can intersect 15 cells (Fig. 132). Let us prove now that a line cannot
intersect more than 15 cells. The number of cells that the line intersects is by 1 less than
the number of intersection points of the line with the segments that determine the sides of
the cells. Inside a square there are 14 such segments.
Hence, inside a square there are not more than 14 intersection points of the line with
sides of cells. No line can intersect the boundary of the chessboard at more than 2 points;
hence, the number of intersection points of the line with the segments does not exceed 16.
Hence, the maximal number of cells on the chessboard of size 8 × 8 that can be intersected
by one line is equal to 15.
11.45. First, let us prove that 33 points are impossible to place in the required way.
Indeed, if on a segment of length 1 there are 33 points, then the distance between some two
1
of them does not exceed 32
. The segment with the endpoints at these points contains two
points, i.e., not less than two points.
points and it should contain not more than 1 + 1000
322
Now, let us prove that it is possible to place 32 points. Let us take 32 points that divide
the segment into equal parts (the endpoints of the given segment should be among these 32
SOLUTIONS
269
Figure 132 (Sol. 11.44)
points). Then a segment of length d contains either [31d] or [31d] + 1 points. (Recall that
[x] denotes the integer part of the number x, i.e., the greatest integer that does not exceed
x.) We have to prove that [31d] ≤ 1000d2 . If 31d < 1, then [31d] = 0 < 1000d2 . If 31d ≥ 1,
then [31d] ≤ 31d ≤ (31d)2 = 961d2 < 1000d2 .
11.46. a) Let a non-regular n-gon be circumscribed about circle S. Let us circumscribe
a regular n-gon about this circle and let us circumscribe circle S1 about this regular n-gon
(Fig. 133). Let us prove that the area of the part of the non-regular n-gon confined inside
S1 is greater than the area of the regular n-gon.
Figure 133 (Sol. 11.46)
All the tangents to S cut off S1 equal segments. Hence, the sum of areas of the segments
cut off S1 by the sides of the regular n-gon is equal to the sum of segments cut off S1 by the
sides of the non-regular n-gon or by their extensions.
But for the regular n-gon these segments do not intersect (more exactly, they do not have
common interior points) and for the non-regular n-gon some of them must overlap, hence,
the area of the union of these segments for a regular-gon is greater than for a non-regular
one. Therefore, the area of the part of the non-regular n-gon confined inside S1 is greater
than the area of the regular n-gon and the area of the whole non-regular n-gon is still greater
than the area of the regular one.
b) This heading follows from heading a) because the perimeter of the polygon circum, where S is the area of the polygon.
scribed about a circle of radius R is equal to 2S
R
11.47. The sides of triangle ABC are proportional to sin α, sin β and sin γ. If angle γ is
fixed, then the value of
¯
¯
¯
¯ α−β
γ
sin ¯¯
| sin α − sin β| = 2 ¯¯sin
2
2
270
CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM
is the greater the greater is ϕ = |α − β|. It remains to observe that quantities
S = 2R2 sin α sin β sin γ = R2 sin γ(cos α − β + cos γ) =
R2 sin γ(cos ϕ + cos γ)
and sin α + sin β = 2 cos γ2 cos ϕ2
monotonously decrease as ϕ increases.
11.48. a) Denote the length of the side of a regular n-gon inscribed in the given circle by
an . Consider an arbitrary non-regular n-gon inscribed in the same circle. It will necessarily
have a side shorter than an .
On the other hand, it can have no side longer than an and in such a case such a polygon
can be confined in a segment cut off a side of the regular n-gon. Since the symmetry through
a side of a regular n-gon sends the segment cut off this side inside the n-gon, the area of
the n-gon is greater than the area of the segment(??). Therefore, we may assume that the
considered n-gon has a side shorter than an and a side longer than an .
We can replace neighbouring sides of the n-gon, i.e., replace A1 A2 A3 . . . An with polygon
A1 A′2 A3 . . . An , where point A′2 is symmetric to A2 through the midperpendicular to segment
A1 A3 (Fig. 134). Clearly, both polygons are inscribed in the same circle and their areas
are equal. It is also clear that with the help of this operation we can make any two sides
of the polygon neighbouring ones. Therefore, let us assume that for the n-gon considered,
A1 A2 > an and A2 A3 < an .
Figure 134 (Sol. 11.48)
Let A′2 be the point symmetric to A2 through the midperpendicular to segment A1 A3 . If
point A′′2 lies on arc ⌣ A2 A′2 , then the difference of the angles at the base A1 A3 of triangle
A1 A′′2 A3 is less than that of triangle A1 A2 A3 because the values of angles ∠A1 A2 A′′2 and
∠A3 A1 A′′2 are confined between the values of angles ∠A1 A3 A2 and ∠A3 A1 A2 .
Since A1 A′2 < an and A1 A2 > an , on arc ⌣ A2 A′2 there exists a point A′′2 for which
A1 A′′2 = an . The area of triangle A1 A′′2 A3 is greater than the area of triangle A1 A2 A3 , cf.
Problem 11.47 a). The area of polygon A1 A′′2 A3 . . . An is greater than the area of the initial
polygon and it has at least by 1 more sides equal to an .
After finitely many steps we get a regular n-gon and at each step the area increases.
Therefore, the area of any non-regular n-gon inscribed in a circle is less than the area of a
regular n-gon inscribed in the same circle.
b) Proof is similar to the proof of heading a); one only has to make use of the result of
Problem 11.47 b) instead of that of Problem 11.47 a).
Chapter 12. CALCULATIONS AND METRIC RELATIONS
Introductory problems
1. Prove the law of cosines:
BC 2 = AB 2 + AC 2 − 2AB · AC cos ∠A.
2. Prove the law of sines:
b
c
a
=
=
= 2R.
sin α
sin β
sin γ
p
3. Prove that the area of a triangle is equal to p(p − a)(p − b)(p − c), where p is semiperimeter (Heron’s formula.)
4. The sides of a parallelogram are equal to a and b and its diagonals are equal to d and
e. Prove that 2(a2 + b2 ) = d2 + e2 .
5. Prove that for convex quadrilateral ABCD with the angle ϕ between the diagonals
we have SABCD = 21 AC · BD sin ϕ.
§1. The law of sines
.
12.1. Prove that the area S of triangle ABC is equal to abc
4R
12.2. Point D lies on base AC of equilateral triangle ABC. Prove that the radii of the
circumscribed circles of triangles ABD and CBD are equal.
12.3. Express the area of triangle ABC in terms of the length of side BC and the value
of angles ∠B and ∠C.
cos
α−β
sin
α−β
12.4. Prove that a+b
= sin γ2 and a−b
= cos 2γ .
c
c
2
2
12.5. In an acute triangle ABC heights AA1 and CC1 are drawn. Points A2 and C2 are
symmetric to A1 and C1 through the midpoints of sides BC and AB, respectively. Prove
that the line that connects vertex B with the center O of the circumscribed circle divides
segment A2 C2 in halves.
12.6. Through point S lines a, b, c and d are drawn; line l intersects them at points A,
does not depend on the choice of line l.
B, C and D. Prove that the quantity AC·BD
BC·AD
12.7. Given lines a and b that intersect at point O and an arbitrary point P . Line l that
passes through point P intersects lines a and b at points A and B. Prove that the value of
OA
OB
P A does not depend on the choice of line l.
PB
12.8. Denote the vertices and the intersection points of links of a (non-regular) fiveangled star as shown on Fig. 135. Prove that
A1 C · B1 D · C1 E · D1 A · E1 B = A1 D · B1 E · C1 A · D1 B · E1 C.
12.9. Two similar isosceles triangles have a common vertex. Prove that the projections
of their bases on the line that connects the midpoints of the bases are equal.
12.10. On the circle with diameter AB, points C and D are taken. Line CD and the
tangent to the circle at point B intersect at point X. Express BX in terms of the radius R
of the circle and angles ϕ = ∠BAC and ψ = ∠BAD.
271
272
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
Figure 135 (12.8)
§2. The law of cosines
12.11. Prove that:
2
2
2
a) m2a = 2b +2c4 −a ;
2
2
2
b) m2a + m2b + m2c = 3(a +b4 +c ) .
12.12. Prove that 4S = (a2 − (b − c)2 ) cot α2 .
12.13. Prove that
p(p − a)
α
(p − b)(p − c)
α
and sin2 =
.
cos2 =
2
bc
2
bc
12.14. The lengths of sides of a parallelogram are equal to a and b; the lengths of the
diagonals are equal to m and n. Prove that a4 + b4 = m2 n2 if and only if the acute angle of
the parallelogram is equal to 45◦ .
12.15. Prove that medians AA1 and BB1 of triangle ABC are perpendicular if and only
2
if a + b2 = 5c2 .
12.16. Let O be the center of the circumscribed circle of scalane triangle ABC, let M
be the intersection point of the medians. Prove that line OM is perpendicular to median
CC1 if and only if a2 + b2 = 2c2 .
§3. The inscribed, the circumscribed and escribed circles; their radii
12.17. Prove
that: ¢
¡
r cos α
a) a = r cot β2 + cot γ2 = sin β sin2 γ ;
2
2
¡
¢
r cos α
b) a = ra tan β2 + tan γ2 = cosa β cos2 γ ;
2
2
c) p − b = r cot β2 = ra tan γ2 ;
d) p = ra cot α2 .
12.18. Prove that:
a) rp = ra (p − a), rra = (p − b)(p − c) and rb rc = p(p − a);
b) S 2 = p(p − a)(p − b)(p − c); (Heron’s formula.)
c) S 2 = rra rb rc .
12.19. Prove that S = rc2 tan α2 tan β2 cot γ2 .
a rb
12.20. Prove that S = rcra +r
.
b
1
2
12.21. Prove that ha = rb + r1c .
12.22. Prove that h1a + h1b + h1c = r1a + r1b + r1c = 1r.
12.23. Prove that
1
1
1
1
+
+
= 2.
(p − a)(p − b) (p − b)(p − c) (p − c)(p − a)
r
§5. THE SINES AND COSINES OF A TRIANGLE’S ANGLES
12.24.
12.25.
12.26.
12.27.
Prove
Prove
Prove
Prove
273
that ra + rb + rc = 4R + r.
that ra rb + rb rc + rc ra = p2 .
that r13 − r13 − r13 − r12 = 12R
.
S2
a
c
b
that
a(b + c) = (r + ra )(4R + r − ra ) and a(b − c) = (rb − rc )(4R − rb − rc ).
2
12.28. Let O be the center of the inscribed circle of triangle ABC. Prove that OA
+
bc
OC 2
OB 2
+ ab = 1.
ac
12.29. a) Prove that if for a triangle we have p = 2R + r, then this triangle is a right
one.
b) Prove that if p = 2R sin ϕ + r cot ϕ2 , then ϕ is one of the angles of the triangle (we
assume here that 0 < ϕ < π).
§4. The lengths of the sides, heights, bisectors
12.30.
12.31.
12.32.
12.33.
12.34.
that abc = 4prR and ab + bc + ca = r2 + p2 + 4rR.
1
1
1
+ bc1 + ca
= 2Rr
.
that ab
β
a+b−c
α
that a+b+c = tan 2 tan 2 .
bc
that ha = 2R
.
that
Prove
Prove
Prove
Prove
Prove
2(p − a) cos β2 cos γ2
2(p − b) sin β2 cos γ2
ha =
=
.
cos α2
sin α2
12.35. q
Prove that the length of bisector la can be computed from the following formulas:
;
a) la = 4p(p−a)bc
(b+c)2
2bc cos
α
b) la = b+c 2 ;
sin β sin γ
;
c) la = 2Rcos
β−γ
2
d) la =
4p sin β2 sin γ2
sin β+sin γ
.
§5. The sines and cosines of a triangle’s angles
Let α, β and γ be the angles of triangle ABC. In the problems of this section one should
prove the relations indicated.
r
12.36. a) sin α2 sin β2 sin γ2 = 4R
;
β
γ
r
α
b) tan 2 tan 2 tan 2 = p ;
p
c) cos α2 cos β2 cos γ2 = 4R
.
β
α
;
12.37. a) cos 2 sin 2 sin γ2 = p−a
4R
β
γ
ra
α
b) sin 2 cos 2 cos 2 = 4R .
.
12.38. cos α + cos β + cos γ = R+r
R
12.39. a) cos 2α + cos 2β + cos 2γ + 4 cos α cos β cos γ + 1 = 0;
b) cos2 α + cos2 β + cos2 γ + 2 cos α cos β cos γ = 1.
12.40. sin 2α + cos 2β + cos 2γ = 4 sin α sin β sin γ.
2
2
12.41. a) sin2 α + sin2 β + sin2 γ = p −r2R−4rR
.
2
b) 4R2 cos α cos β cos γ = p2 − (2R + r)2 .
2
2
2
12.42. ab cos γ + bc cos α + ca cos β = a +b2 +c .
12.43.
cos2
a
α
2
+
cos2
b
β
2
+
cos2
c
γ
2
=
p
.
4Rr
274
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
§6. The tangents and cotangents of a triangle’s angles
In problems of this section one has to prove the relations indicated between the values
α, β and γ of the angles of triangle ABC.
2
2
2
12.44. a) cot α + cot β + cot γ = a +b4S+c ;
b) a2 cot α + b2 cot β + c2 cot γ = 4S.
12.45. a) cot α2 + cot β2 + cot³γ2 = pr ;
´
b) tan α2 + tan β2 + tan γ2 = 21 raa + rbb + rcc .
12.46. tan α + tan β + tan γ = tan σ tan β tan γ.
12.47. tan α2 tan β2 + tan β2 tan γ2 + tan γ2 tan α2 = 1.
12.48. a) cot α cot β + cot β cot γ + cot α cot γ = 1;
b) cot α + cot β + cot γ − cot α cot β cot γ = sin α sin1 β sin γ .
12.49. For a non-right triangle we have
tan σ + tan β + tan γ =
a2
+
b2
4S
.
+ c2 − 8R2
§7. Calculation of angles
12.50. Two intersecting circles, each of radius R with the distance between their centers
greater than R are given. Prove that β = 3α (Fig. 136).
Figure 136 (12.50)
12.51. Prove that if 1b + 1c = l1a , then ∠A = 120◦ .
12.52. In triangle ABC height AH is equal to median BM . Find angle ∠M BC.
12.53. In triangle ABC bisectors AD and BE are drawn. Find the value of angle ∠C
if it is given that AD · BC = BE · AC and AC 6= BC.
12.54. Find angle ∠B of triangle ABC if the length of height CH is equal to a half
length of side AB and ∠BAC = 75◦ .
12.55. In right triangle ABC with right angle ∠A the circle is constructed with height
AD of the triangle as a diameter; the circle intersects leg AB at point K and leg AC at
point M . Segments AD and KM intersect at point L. Find the acute angles of triangle
ABC if AK : AL = AL : AM .
12.56. In triangle ABC, angle ∠C = 2∠A and b = 2a. Find the angles of triangle ABC.
12.57. In triangle ABC bisector BE is drawn and on side BC point K is taken so that
∠AKB = 2∠AEB. Find the value of angle ∠AKE if ∠AEB = α.
***
12.58. In an isosceles triangle ABC with base BC angle at vertex A is equal to 80◦ .
Inside triangle ABC point M is taken so that ∠M BC = 30◦ and ∠M CB = 10◦ . Find the
value of angle ∠AM C.
§9. MISCELLANEOUS PROBLEMS
275
12.59. In an isosceles triangle ABC with base AC the angle at vertex B is equal to 20◦ .
On sides BC and AB points D and E, respectively, are taken so that ∠DAC = 60◦ and
∠ECA = 50◦ . Find angle ∠ADE.
12.60. In an acute triangle ABC segments BO and CO, where O is the center of the
circumscribed circle, are extended to their intersection at points D and E with sides AC
and AB, respectively. It turned out that ∠BDE = 50◦ and ∠CED = 30◦ . Find the value
of the angles of triangle ABC.
§8. The circles
12.61. Circle S with center O on base BC of isosceles triangle ABC is tangent to equal
sides AB and AC. On sides AB and AC, points P and Q, respectively, are taken so that
segment P Q is tangent to S. Prove that 4P B · CQ = BC 2 .
12.62. Let E be the midpoint of side AB of square ABCD and points F and G are
taken on sides BC and CD, respectively, so that AG k EF . Prove that segment F G is
tangent to the circle inscribed in square ABCD.
12.63. A chord of a circle is distant from the center by h. A square is inscribed in each
of the disk segments subtended by the chord so that two neighbouring vertices of the square
lie on an arc and two other vertices lie either on the chord or on its extension (Fig. 137).
What is the difference of lengths of sides of these squares?
Figure 137 (12.63)
12.64. Find the ratio of sides of a triangle one of whose medians is divided by the
inscribed circle into three equal parts.
***
12.65. In a circle, a square is inscribed; in the disk segment cut off the disk by one of
the sides of this square another square is inscribed. Find the ratio of the lengths of the sides
of these squares.
12.66. On segment AB, point C is taken and on segments AC, BC and AB as on
diameters semicircles are constructed lying on one side of line AB. Through point C the
line perpendicular to AB is drawn and in the obtained curvilinear triangles ACD and BCD
circles S1 and S2 are inscribed (Fig. 138). Prove that the radii of these circles are equal.
12.67. The centers of circles with radii 1, 3 and 4 are positioned on sides AD and BC of
rectangle ABCD. These circles are tangent to each other and lines AB and CD as shown
on Fig. 139. Prove that there exists a circle tangent to all these circles and line AB.
§9. Miscellaneous problems
12.68. Find all the triangles whose angles form an arithmetic projection and sides form
a) an arithmetic progression; b) a geometric progression.
276
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
Figure 138 (12.66)
Figure 139 (12.67)
12.69. Find the height of a trapezoid the lengths of whose bases AB and CD are equal
to a and b (a < b), the angle between the diagonals is equal to 90◦ , and the angle between
the extensions of the lateral sides is equal to 45◦ .
12.70. An inscribed circle is tangent to side BC of triangle ABC at point K. Prove
that the area of the triangle is equal to BK · KC cot α2 .
12.71. Prove that if cot α2 = a+b
, then the triangle is a right one.
a
12.72. The extensions of the bisectors of triangle ABC intersect the circumscribed circle
at points A1 , B1 and C1 . Prove that SASABC
, where r and R are the radii of the inscribed
= 2r
R
1 B1 C1
and circumscribed circles, respectively, of triangle ABC.
12.73. Prove that the sum of cotangents of the angles of triangle ABC is equal to the
sum of cotangents of the angles of the triangle formed by the medians of triangle ABC.
12.74. Let A4 be the orthocenter of triangle A1 A2 A3 . Prove that there exist
P 1numbers
= 0.
λ1 , . . . , λ4 such that Ai A2j = λi + λj and if the triangle is not a right one, then
λi
§10. The method of coordinates
12.75. Coordinates of the vertices of a triangle are rational numbers. Prove that then
the coordinates of the center of the circumscribed circle are also rational.
12.76. Diameters AB and CD of circle S are perpendicular. Chord EA intersects
diameter CD at point K, chord EC intersects diameter AB at point L. Prove that if
CK : KD = 2 : 1, then AL : LB = 3 : 1.
12.77. In triangle ABC angle ∠C is a right one. Prove that under the homothety with
center C and coefficient 2 the inscribed circle turns into a circle tangent to the circumscribed
circle.
SOLUTIONS
277
12.78. A line l is fixed. Square ABCD is rotated about its center. Find the locus of the
midpoints of segments P Q, where P is the base of the perpendicular dropped from point D
on l and Q is the midpoint of side AB.
See also Problems 7.6, 7.14, 7.47, 22.15.
Problems for independent study
12.79. Each of two circles is tangent to both sides of the given right angle. Find the
ratio of the circles’ radii if it is known that one of the circles passes through the center of
the other one.
12.80. Let the extensions of sides AB and CD, BC and AD of convex quadrilateral ABCD intersect at points K and M , respectively. Prove that the radii of the circles circumscribed about triangles ACM , BDK, ACK, BDM are related by the formula
RACM · RBDK = RACK · RBDM .
12.81. Three circles of radii 1, 2, 3 are tangent to each other from the outside. Find the
radius of the circle that passes through the tangent points of these circles.
12.82. Let point K lie on side BC of triangle ABC. Prove that
AC 2 · BK + AB 2 · CK = BC(AK 2 + BK · KC).
12.83. Prove
that the length of the bisector of an outer angle ∠A of triangle ABC is
2bc sin α
2
equal to |b−c| .
12.84. Two circles of radii R and r are placed so that their common inner tangents are
perpendicular. Find the area of the triangle formed by these tangents and their common
outer tangent.
12.85. Prove that the sum of angles at rays of any (nonregular) five-angled star is equal
to 180◦ .
12.86. Prove that in any triangle S = (p − a)2 tan α2 cot β2 cot γ2 .
12.87. Let a < b < c be the lengths of sides of a triangle; la , lb , lc and la′ , lb′ , lc′ the lengths
of its bisectors and the bisectors of its outer angles, respectively. Prove that ala1l′ + cl1c l′ = bl1b l′ .
a
c
b
12.88. In every angle of a triangle a circle tangent to the inscribed circle of the triangle
is inscribed. Find the radius of the inscribed circle if the radii of these smaller circles are
known.
12.89. The inscribed circle is tangent to sides AB, BC, CA at points K, L, M , respectively. Prove³that:
´
K2
KL2
LM 2
;
+
+
a) S = 21 M
sin α
sin β
sin γ
b) S 2 = 41 (bcM K 2 + caKL2 + abLM 2 );
2
2
K2
+ hKL
+ LM
= 1.
c) M
hb hc
ha h b
c ha
Solutions
12.1. By the law of sines sin γ = c2R; hence, S = 21 ab sin γ = abc
.
4R
12.2. The radii of the circumscribed circles of triangles ABD and CBD are equal to
AB
and 2 sinBC
. It remains to notice that AB = BC and sin ∠ADB = sin ∠BDC.
2 sin ∠ADB
∠BDC
2 sin β sin γ
a sin β
sin β
= sin(β+y)
and, therefore, S = 21 ab sin γ = a2 sin(β+γ)
.
12.3. By the law of sines b = asin
α
sin α+sin β
1
12.4. By the law of sines 2 (a + b) = sin γ . Moreover,
α−β
γ
α−β
α+β
cos
= 2 cos cos
2
2
2
2
γ
γ
and sin γ = 2 sin 2 cos 2 . The second equality is similarly proved.
sin α + sin β = 2 sin
278
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
12.5. In triangle A2 BC2 , the lengths of sides A2 B and BC2 are equal to b cos γ and
b cos α; line BO divides angle ∠A2 BC2 into angles of 90◦ − γ and 90◦ − α. Let line BO
intersect segment A2 C2 at point M . By the law of sines
A2 M =
b cos γ cos α
A2 B sin ∠A2 BM
=
= C2 M.
sin ∠A2 M B
sin ∠C2 M B
12.6. Let α = ∠(a, c), β = ∠(c, d) and γ = ∠(d, b). Then
AC
AS
BC
BS
Hence
12.7. Since
OA
PA
=
=
sin α
and
sin(β + γ)
BD
BS
AD
AS
=
sin γ
.
sin(α + β)
(AC · BD)
sin α sin γ
=
.
(BC · AD)
sin(α + β) sin(β + γ)
sin ∠OP A
sin ∠P OA
and
OB
PB
=
sin ∠OP B
,
sin ∠P OB
it follows that
(OA : OB)
sin ∠P OB
=
.
(P A : P B)
sin ∠P OA
D1 A
∠B
12.8. It suffices to multiply five equalities of the form D
= sin
.
sin ∠A
1B
12.9. Let O be the common vertex of the given triangles, M and N the midpoints of
the bases, k the ratio of the lengths of the bases to that of heights. The projections of the
bases of given triangles on line M N are equal to k · OM sin ∠OM N and k · ON sin ∠ON M .
ON
OM
= sin ∠OM
.
It remains to notice that sin ∠ON
M
N
12.10. By the law of sines
BX
BD
2R sin ψ
=
=
.
sin ∠BDX
sin ∠BXD
sin ∠BXD
Moreover, sin ∠BDX = sin ∠BDC = sin ϕ and the value of angle ∠BXD is easy to calculate: if points C and D lie on one side of AB, then ∠BXD = π − ϕ − ψ and if they lie on
ϕ sin ψ
distinct sides, then ∠BXD = |ϕ − ψ|. Hence, BX = 2Rsinsin|ϕ±ψ|
.
12.11. a) Let A1 be the midpoint of segment BC. Adding equalities
AB 2 = AA21 + A1 B 2 − 2AA1 · BA1 cos ∠BA1 A
and
AC 2 = AA21 + A1 C 2 − 2AA1 · A1 C cos ∠CA1 A
and taking into account that cos ∠BA1 A = − cos ∠CA1 A we get the statement desired.
b) Follows in an obvious way from heading a).
12.12. By the law of cosines
a2 − (b − c)2 = 2bc(1 − cos α) =
2
2
4S(1 − cos α)
α
= 4S tan .
sin α
2
2
12.13. By the law of cosines cos α = b +c2bc−a . It remains to make use of the formulas
cos2 α2 = 12 (1 + cos α) and sin2 α2 = 12 (1 − cos α).
12.14. Let α be the angle at a vertex of the parallelogram. By the law of cosines
m2 = a2 + b2 + 2ab cos α and n2 = a2 + b2 − 2ab cos α.
Hence,
m2 n2 = (a2 + b2 )2 − (2ab cos α)2 = a4 + b4 + 2a2 b2 (1 − 2 cos2 α).
Therefore, m2 n2 = a4 + b4 if and only if cos2 α = 21 .
SOLUTIONS
279
12.15. Let M be the intersection point of medians AA1 and BB1 . Angle ∠AM B is
a right one if and only if AM 2 + BM 2 = AB 2 , i.e. 49 (m2a + m2b ) = c2 . By Problem 12.11
2
2
2
m2a + m2b = 4c +a4 +b .
12.16. Let m = C1 M and ϕ = ∠C1 M O. Then
and
OC12 = C1 M 2 + OM 2 − 2OM · C1 M cos ϕ
BO2 = CO2 = OM 2 + M C 2 + 2OM · CM cos ϕ = OM 2 + 4C1 M 2 + 4OM · C1 M cos ϕ.
Hence,
i.e.,
BC12 = BO2 − OC12 = 3C1 M 2 + 6OM · C1 M cos ϕ,
c2 = 4BC12 = 12m2 + 24OM · C1 M cos ϕ.
2
It is also clear that 18m2 = 2m2c = a2 + b2 − c2 , cf. Problem 12.11. Therefore, equality
2
a2 + b2 = 2c2 is equivalent to the fact that 18m2 = 3c2 , i.e., c2 = 12m2 . Since c2 = 12m2 +
24OM · C1 M cos ϕ, equality a2 + b2 = 2c2 is equivalent to the fact that ∠C1 M O = ϕ = 90◦ ,
i.e., CC1 ⊥ OM .
12.17. Let the inscribed circle be tangent to side BC at point K and the escribed one
at point L. Then
β
γ
BC = BK + KC = t cot + r cot
2
2
and
γ
β
BC = BL + LC = ra cot LBOa + ra cot LCOa = ra tan + ra tan .
2
2
¡β γ ¢
α
Moreover, cos 2 = sin 2 + 2 .
By Problem 3.2, p − b = BK = r cot β2 and p − b = CL = ra tan γ2 .
If the inscribed circle is tangent to the extensions of sides AB and AC at points P and
Q, respectively, then p = AP = AQ = ra cot α2 .
12.18. a) By Problem 12.17,
p = ra cot α2 and r cot α2 = p − a;
r cot β2 = p − b and ra tan β2 = p − c;
rc tan β2 = p − a and rb cot β2 = p.
By multiplying these pairs of equalities we get the desired statement.
b) By multiplying equalities rp = ra (p − a) and rra = (p − b)(p − c) we get r2 p =
(p − a)(p − b)(p − c). It is also clear that S 2 = p(r2 p).
c) It suffices to multiply rra = (p − b)(p − c) and rb rc = p(p − a) and make use of Heron’s
formula.
12.19. By Problem 12.17, r = rc tan α2 tan β2 and p = rc cot γ2 .
rp
rp
12.20. By Problem 12.18 a), ra = p−a
and rb = p−b
. Hence,
cra rb =
rpc
cr2 p2
and ra + rb =
(p − a)(p − b)
(p − a)(p − b)
a rb
and, therefore, rcra +r
= rp = S.
b
a
and r1c = p−c
, hence, r1b + r1c = pr
= Sa = h2a .
12.21. By Problem 12.18 a), r1b = p−b
pr
pr
a
and r1a = p−a
. Adding similar equalities we get
12.22. It is easy to verify that h1a = 2pr
pr
the desired statement.
280
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
1
12.23. By Problem 12.18 a) (p−b)(p−c)
= rr1a . It remains to add similar equalities and
make use of the result of Problem 12.22.
12.24. By Problem 12.1, 4SR = abc. It is also clear that
abc = p(p − b)(p − c) + p(p − c)(p − a) + p(p − a)(p − b) − (p − a)(p − b)(p − c) =
S2
S2
S2
+ p−b
+ p−c
− S 2 p = S(ra + rb + rc − r).
p−a
12.25. By Problem 12.18 a)
ra rb = p(p − c), rb rc = p(p − a) and rc ra = p(p − b).
Adding these equalities we get the desired statement.
12.26. Since
S = rp = ra (p − a) = rb (p − b) = rc (p − c),
the right-most expression is equal to
3abc
p3 − (p − a)3 − (p − b)3 − (p − c)3
= 3 .
3
S
S
abc
It remains to observe that S = 4R (Problem 12.1).
12.27. Let the angles of triangle ABC be equal to 2α, 2β and 2γ. Thanks to Problems
12.36 a) and 12.37 b) we have r = 4R sin α sin β sin γ and ra = 4R sin α cos β cos γ. Therefore,
(r + ra )(4R + r − ra ) =
16R2 sin α · (sin β sin γ + cos β cos γ)(1 + sin α(sin β sin γ − cos β cos γ)) =
16R2 sin α cos(β − γ)(1 − sin α cos(β + γ)) =
16R2 sin α cos(β − γ) cos2 α.
It remains to notice that 4R sin α cos α = a and
4R sin(β + γ) cos(β − γ) = 2R(sin 2β + sin 2γ) = b + c.
The second equality is similarly proved.
2S
12.28. Since OA = sinr α and bc = sin
, it follows that
α
2
r2 cot α2
r(p − a)
OA2
=
=
,
bc
S
S
cf. Problem 12.17 c). It remains to notice that r(p − a + p − b + p − c) = rp = S.
sin ϕ
12.29. Let us solve heading b); heading a) is its particular case. Since cot ϕ2 = 1−cos
, it
ϕ
follows that
p2 (1 − x)2 = (1 − x2 )(2R(1 − x) + r)2 , where x = cos ϕ.
The root x0 = 1 of this equation is of no interest to us because in this case cot ϕ2 is undefined;
therefore, by dividing both parts of the equation by 1 − x we get a cubic equation. Making
use of results of Problems 12.38, 12.41 b) and 12.39 b) we can verify that this equation
coincides with the equation
(x − cos α)(x − cos β)(x − cos γ) = 0,
where α, β and γ are the angles of the triangle. Therefore the cosine of ϕ is equal to the
cosine of one of the angles of the triangle; moreover, the cosine is monotonous on the interval
[0, π].
, i.e., 4prR = abc. To prove the second
12.30. It is clear that 2pr = 2S = ab sin γ = abc
2R
2
equality make use of Heron’s formula: S = p(p − a)(p − b)(p − c), i.e.,
pr2 = (p − a)(p − b)(p − c) = p3 − p2 (a + b + c) + p(ab + bc + ca) − abc =
= −p3 + p(ab + bc + ca) − 4prR.
SOLUTIONS
281
By dividing by p we get the desired equality.
12.31. Since abc = 4RS (Problem 12.1), the expression in the left-hand side is equal to
2p
c+a+b
1
= 4Rpr
= 2Rr
.
4RS
= rrc (Problem 12.18 a)),
12.32. It suffices to observe that p−c
p
c cos α2 cos β2
c sin α2 sin β2
and rc =
r=
cos γ2
cos γ2
(Problem 12.17).
12.33. By Problem 12.1, S =
abc
.
4R
On the other hand, S =
12.34. Since aha = 2S = 2(p − a)ra and
ha =
ra
a
=
cos β2 cos
cos α
2
γ
2
aha
.
2
Hence, ha =
bc
.
2R
(Problem 12.17 b)), we have
2(p − a) cos β2 cos γ2
.
cos α2
Taking into account that (p − a) cot β2 = rc = (p − b) cot α2 (Problem 12.17 c)), we get
2(p−b) sin
β
cos
γ
2
2
.
ha =
sin α
2
12.35. a) Let the extension of bisector AD intersect the circumscribed circle of triangle
ABC at point M . Then AD · DM = BD · DC and since △ABC ∼ △AM C, it follows that
AB · AC = AD · AM = AD(AD + DM ) = AD2 + BD · DC.
Moreover, BD =
ac
b+c
and DC =
ab
.
b+c
Hence,
AD2 = bc −
bca2
4p(p − a)bc
=
.
(b + c)2
(b + c)2
b) See the solution of Problem 4.47.
c) Let AD be a bisector, AH a height of triangle ABC. Then AH = c sin β = 2R sin β sin γ.
On the other hand,
³
π+β−γ
β−γ
α´
= la sin
= la cos
.
AH = AD sin ∠ADH = la sin β +
2
2
2
d) Taking into account that p = 4R cos α2 cos β2 cos γ2 (Problem 12.36 c)) and
β−γ
α
β−γ
β+γ
cos
= 2 cos cos
2
2
2
2
we arrive at the formula of heading c).
12.36. a) Let O be the center of the inscribed circle, K the tangent point of the inscribed
circle with side AB. Then
µ
¶
α
β
α
β
α+β
2R sin γ = AB = AK + KB = r cot + cot
sin sin .
= r sin
2
2
2
2
2
sin β + sin γ = 2 sin
= cos γ2 we get the desired stateTaking into account that sin γ = 2 sin γ2 cos γ2 and sin α+β
2
ment.
b) By Problem 3.2, p−a = AK = r cot α2 . Similarly, p−b = r cot β2 and p−c = r cot γ2 . By
multiplying these equalities and taking into account that p(p − a)(p − b)(p − c) = S 2 = (pr)2
we get the desired statement.
c) Obviously follows from headings a) and b).
12.37. a) By multiplying equalities r cos α2 sin α2 = p − a and
sin
β
γ
r
α
sin sin =
2
2
2
4R
282
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
(cf. Problems 12.17 c) and 12.36 a)) we get the desired statement.
b) By Problem 12.17 c), ra tan γ2 = p − b = r cot β2 . By multiplying this equality by
r
= sin α2 sin β2 sin γ2 we get the desired statement.
4R
12.38. By adding equalities
cos α + cos β = 2 cos
α+β
α−β
cos
2
2
cos γ = − cos(α + β) = −2 cos2
α+β
+1
2
and taking into account that
cos
α+β
α
β
α−β
− cos
= 2 sin sin
2
2
2
2
we get
cos α + cos β + cos γ = 4 sin
α
β
γ
r
sin sin + 1 = + 1,
2
2
2
R
cf. Problem 12.36 a).
12.39. a) Adding equalities
cos 2α + cos 2β = 2 cos(α + β) cos(α − β) = −2 cos γ cos(α − β);
cos 2γ = 2 cos2 γ − 1 = −2 cos γ cos(α + β) − 1
and taking into account that
cos(α + β) + cos(α − β) = 2 cos α cos β
we get the desired statement.
b) It suffices to substitute expressions of the form cos 2α = 2 cos2 α − 1 in the equality
obtained in heading a).
12.40. Adding equalities
sin 2α + sin 2β = 2 sin(α + β) cos(α − β) = 2 sin γ cos(α − β);
sin 2γ = 2 sin γ cos γ = −2 sin γ cos(α + β)
and taking into account that
cos(α − β) − cos(α + β) = 2 sin α sin β
we get the desired statement.
12.41. a) Clearly,
sin2 α + sin2 β + sin2 γ =
a2 + b 2 + c 2
4R
and
a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca) = 4p2 − 2(r2 + p2 + 4rR),
cf. Problem 12.30.
b) By Problem 12.39 b)
2 cos α cos β cos γ = sin2 α + sin2 β + sin2 γ − 2.
It remains to make use of a result of heading a).
2
2
2
12.42. The law of cosines can be expressed as ab cos γ = a +b2 −c . By adding three
similar equalities we get the desired statement.
cos2 α
. It remains to notice that p(p − a) + p(p − b) +
12.43. By Problem 12.13 a 2 = p(p−a)
abc
p(p − c) = p2 and abc = 4SR = 4prR.
SOLUTIONS
283
12.44. a) Since bc cos α = 2S cot α, it follows that a2 = b2 + c2 − 4S cot α. By adding
three similar equalities we get the desired statement.
b) For an acute triangle a2 cot α = 2R2 sin 2α = 4SBOC , where O is the center of the
circumscribed circle. It remains to add three analogous equalities. For a triangle with an
obtuse angle α the quality SBOC should be taken with the minus sign.
12.45. By Problem 12.17 cot α2 + cot β2 = rc and tan α2 + tan β2 = rcc . It remains to add
such equalities for all pairs of angles of the triangle.
12.46. Clearly,
tan α + tan β
.
tan γ = − tan(α + β) = −
1 − tan α tan β
By multiplying both sides of equality by 1 − tan α tan β we get the desired statement.
12.47.
µ
¶ ·
¸·
¸
α
β
α
β
α β
γ
+
= 1 − tan tan
tan + tan
.
tan = cot
2
2
2
2
2
2
2
It remains to multiply both sides of the equality by tan α2 + tan β2 .
12.48. a) Let us multiply both sides of the equality by sin α sin β sin γ. Further on:
cos γ(sin α cos β + sin β cos α) + sin γ(cos α cos β − sin α sin β) =
cos γ sin(α + β) + sin γ cos(α + β) =
cos γ sin γ − sin γ cos γ = 0.
b) Let us multiply both sides of the equality by sin α sin β sin γ. Further on:
cos α(sin β sin γ − cos β cos γ) + sin α(cos β sin γ + cos γ sin β) = cos2 α + sin2 α = 1.
12.49. Since
sin2 α + sin2 β + sin2 γ − 2 = 2 cos α cos β cos γ
(see Problem 12.39 b) and S = 2R2 sin α sin β sin γ, it remains to verify that
(tan α + tan β + tan γ) cos α cos β cos γ = sin γ sin β sin α.
The latter equality is proved in the solution of Problem 12.48 a).
12.50. Let A and B be the vertices of angles α and β, let P be the intersection point
of non-coinciding legs of these angles, Q the common point of the given circles that lies
on segment P A. Triangle AQB is an isosceles one, hence, ∠P QB = 2α. Since ∠P QB +
∠QP B = β + ∠QBA, it follows that β = α3α.
2 cos
12.51. By Problem 4.47, 1b + 1c = la 2 , hence, cos α2 = 21 , i.e., α = 120◦ .
12.52. Let us drop perpendicular M D from point M to line BC. Then M D = 12 AH =
1
BM . In right triangle BDM , leg M D is equal to a half hypothenuse BM . Hence,
2
∠M BC = ∠M BD = 30◦ .
12.53. The quantities AD · BC sin ADB and BE · AC sin AEB are equal because each
of them is equal to the doubled area of triangle ABC. Hence, sin ADB = sin AEB. Two
cases are possible:
1) ∠ADB = ∠AEB. In this case points A, E, D, B lie on one circle; hence, ∠EAD =
∠EBD, i.e., ∠A = ∠B which contradicts the hypothesis.
2) ∠ADB + ∠AEB = 180◦ . In this case ∠ECD + ∠EOD = 180◦ , where O is the
intersection point of bisectors. Since ∠EOD = 90◦ + ∠C
(Problem 5.3), it follows that
2
◦
∠C = 60 .
12.54. Let B ′ be the intersection point of the midperpendicular to segment AC with line
AB. Then AB ′ = CB ′ and ∠AB ′ C = 180◦ −2·75◦ = 30◦ . Hence, AB ′ = CB ′ = 2CH = AB,
i.e., B ′ = B and ∠B = 30◦ .
284
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
12.55. Clearly, AKDM is a rectangle and L the intersection point of its diagonals. Since
AD ⊥ BC and AM ⊥ BA, it follows that ∠DAM = ∠ABC. Similarly, ∠KAD = ∠ACB.
Let us drop perpendicular AP from point A to line KM . Let, for definiteness, ∠B < ∠C.
Then point P lies on segment KL. Since △AKP ∼ △M KA, it follows that AK : AP =
M K : M A. Hence, AK · AM = AP · M K = AP · AD = 2AP · AL. By the hypothesis
AL2 = AK · AM ; hence, AL = 2AP , i.e., ∠ALP = 30◦ . Clearly, ∠KM A = ∠ALP
= 15◦ .
2
◦
◦
Therefore, the acute angles of triangle ABC are equal to 15 and 75 .
ac
. On the other hand, △BDC ∼ △BCA,
12.56. Let CD be a bisector. Then BD = a+b
a2
consequently, BD : BC = BC : BA, i.e., BD = c . Hence c2 = a(a + b) = 3a2 . The lengths
√
of the sides of triangle ABC are equal to a, 2a and 3a; hence, its angles are equal to 30◦ ,
90◦ and 60◦ , respectively.
12.57. Let ∠ABC = 2x. Then the outer angle ∠A of triangle ABE is equal to ∠ABE +
∠AEB = x + α. Further,
∠KAE = ∠BAE − ∠BAK = (180◦ − x − α) − (180◦ − 2x − 2α) = x + α.
Therefore, AE is the bisector of the outer angle ∠A of triangle ABK. Since BE is the
bisector of the inner angle ∠B of triangle ABK, it follows that E is the center of its escribed
circle tangent to side AK. Hence, ∠AKE = 21 ∠AKC = 90◦ − α.
12.58. Let A1 . . . A18 be a regular 18-gon. For triangle ABC we can take triangle
A14 A1 A9 . By Problem 6.59 b) the diagonals A1 A12 , A2 A14 and A9 A18 meet at one point,
hence, ∠AM C = 21 (⌣ A18 A2 + ⌣ A9 A14 ) = 70◦ .
12.59. Let A1 . . . A18 be a regular 18-gon, O its center. For triangle ABC we can take
triangle A1 OA18 . The diagonals A2 A14 and A18 A6 are symmetric through diameter A1 A10 ;
diagonal A2 A14 passes through the intersection point of diagonals A1 A12 and A9 A18 (cf. the
solution of Problem 12.58), therefore, ∠ADE = 21 (⌣ A1 A2 + ⌣ A12 A14 ) = 30◦ .
12.60. Since ∠BDE = 50◦ and ∠CDE = 30◦ , it follows that ∠BOC = ∠EOD =
180◦ − 50◦ − 30◦ = 100◦ . Let us assume that diameters BB ′ and CC ′ of the circle are fixed,
∠BOC = 100◦ and point A moves along arc ⌣ B ′ C ′ . Let D be the intersection point of
BB ′ and AC, E the intersection point of CC ′ and AB (Fig. 140). As point A moves from
B ′ to C ′ , segment OE increases while OD decreases, consequently, angle ∠OED decreases
and angle ∠ODE increases. Therefore, there exists a unique position of point A for which
∠CED = ∠OED = 30◦ and ∠BDE = ∠ODE = 50◦ .
Figure 140 (Sol. 12.60)
Now, let us prove that triangle ABC with angles ∠A = 50◦ , ∠B = 70◦ , ∠C = 60◦
possesses the required property. Let A1 . . . A18 be a regular 18-gon. For triangle ABC we
can take triangle A2 A14 A9 . Diagonal A1 A12 passes through point E (cf. solution of Problem
12.58). Let F be the intersection point of lines A1 A12 and A5 A14 ; line A9 A16 is symmetric to
line A1 A12 through line A5 A14 and, therefore, it passes through point F . In triangle CDF ,
SOLUTIONS
285
ray CE is the bisector of angle ∠C and line F E is the bisector of the outer angle at vertex
F . Hence, DE is the bisector of angle ∠ADB, i.e., ∠ODE = 14 (⌣ A2 A14 + ⌣ A5 A9 ) = 50◦ .
12.61. Let D, E and F be the tangent points of the circle with BP, P Q and QC,
respectively; ∠BOD = 90◦ − ∠B = 90◦ − ∠C = ∠COF = α, ∠DOP = ∠P OE = β and
∠EOQ = ∠QOF = γ. Then 180◦ = ∠BOC = 2α + 2β + 2γ, i.e., α + β + γ = 90◦ . Since
∠BP O = 21 ∠DP E = 21 (180◦ − ∠DOE) = 90◦ − β and ∠QOC = γ + α = 90◦ − β, it follows
that ∠BP O = ∠COQ. It is also clear that ∠P BO = ∠OCQ. Hence, △BP O ∼ △COQ,
i.e., P B · CQ = BO · CO = 41 BC 2 .
12.62. Let P and Q be the midpoints of sides BC and CD, respectively. Points P and
Q are the tangent points of the inscribed circle with sides BC and CD. Therefore, it suffices
to verify that P F + GQ = F G. Indeed, if F ′ G′ is the segment parallel to F G and tangent
to the inscribed circle, then P F ′ + G′ Q = F ′ G′ ; hence, F ′ = F and G′ = G.
We may assume that the side of the square is equal to 2. Let GD = x. Since BF : EB =
AD : GD, then BF = x2 . Therefore, CG = 2 − x, GQ = x − 1, CF = 2 − x2 , F P = x2 − 1,
i.e., P F + GQ = x + x2 − 2 and
¶2
µ
2
F G = CG + CF = (2 − x) + 2 −
=
x
2
2
2
2
8
4
4 − 4x + x + 4 − + 2 =
x x
2
µ
¶2
2
x + − 2 = (P F + GQ)2 .
x
12.63. Denote the vertices of the squares as shown on Fig. 141. Let O be the center of
the circle, H the midpoint of the given chord, K the midpoint of segment AA1 .
Figure 141 (Sol. 12.63)
Since tan AHB = 2 = tan A1 HD1 , point H lies on line AA1 . Let α = ∠AHB =
∠A1 HD1 , then
AB − A1 D1 = (AH − A1 H) · sin α = 2KH sin α = 2OH sin2 α.
Since tan α = 2 and 1 + cot2 α = sin12 α , it follows that sin2 α = 54 . Therefore, the difference
of the lengths of the squares’ sides is equal to 58 h.
12.64. Let median BM of triangle ABC intersect the inscribed circle at points K and L,
where BK = KL = LM = x. Let, for definiteness, the tangent point of the inscribed circle
with side AC lie on segment M C. Then since the symmetry through the midperpendicular
to segment BM interchanges points B and M and fixes the inscribed circle, tangent M C
turns into tangent BC. Therefore, BC = M C = 12 AC, i.e., b = 2a.
2
2
2
2
2
2
2 −b2
by Problem 12.11 a), we have 9x2 = 2a +2c4 −4a = c −a
. Let
Since BM 2 = 2a +2c
4
2
c−a
a+c−b
P be the tangent point of the inscribed circle with side BC. Then BP = 2 = 2 . On
the other hand, by a property of the tangent, BP 2 = BK · BL, i.e., BP 2 = 2x2 . Hence,
286
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
¡ ¢2
¡ ¢2
2
2
2x2 = c−a
. Multiplying inequalities 9x2 = c −a
= 2x2 we get c+a
and c−a
= 49 , i.e.,
2
2
2
c−a
c : a = 13 : 5. As a result we get a : b : c = 5 : 10 : 13.
12.65. Let 2a and 2b be the length of the side of the first and second squares, respectively.
Then the distance from the center
p of the circle to any of the vertices of the second square
(a + 2b)2 + b2 . On the √
other hand, this distance is equal
that
lie
on
the
circle
is
equal
to
√
2
2
2
to 2a. Therefore, (a + 2b) + b = 2a , i.e., a = 2b ± 4b2 + 5b2 = (2 ± 3)b. Only the
solution a = 5b is positive.
12.66. Let P and Q be the midpoints of segments AC and AB, respectively, R the
center of circle S1 ; a = 21 AC, b = 21 BC, x the radius of circle S1 . It is easy to verify that
P R = a + x, QR = a + b − x and P Q = b. In triangle P QR, draw height RH. The distance
from point R to line CD is equal to x, hence, P H = a − x, consequently, QH = |b − a + x|.
It follows that
(a + x)2 − (a − x)2 = RH 2 = (a + b − x)2 − (b − a + x)2 ,
ab
i.e., ax = b(a − x). As a result we get x = a+b
.
For the radius of circle S2 we get precisely the same expression.
12.67. Let x be the radius of circle S tangent to circles S1 and S2 and ray AB, let y be
the radius of circle S ′ tangent to circles S2 and S3 and ray BA. The position of the circle
tangent to circle S1 and ray AB (resp. S3 and BA) is uniquely determined by its radius,
consequently, it suffices to verify that x = y.
By equating two expressions for the squared distance from the center of circle S to line
AD we get
4
(x + 1)2 − (x − 1)2 = (3 + x)2 − (5 − x)2 , i.e., x = .
3
Considering circles S2 and S3 it is easy to verify that AB 2 = (3 + 4)2 − 12 = 48. On the
other hand, the squared distances from the center of circle S ′ to lines AD and BC are equal
to (y + 3)2 − (5 − √
y)2 = 16(y − 1) and (4 + y)2 − (4 − y)2 = 16y, respectively. Therefore,
√
√
4 y − 1 + 4 y = 48, i.e., y = 34 .
12.68. If the angles of a triangle form an arithmetic progression, then they are equal to
α − γ, α, α + γ, where γ ≥ 0. Since the sum of the angles of a triangle is equal to 180◦ ,
we deduce that α = 60◦ . The sides of this triangle are equal to 2R sin(α − γ), 2R sin α,
2R sin(α + γ). Since the greater side subtends the greater angle, sin(α − γ) ≤ sin α ≤
sin(α + γ).
a) If the numbers sin(α − γ) ≤ sin α ≤ sin(α + γ) form an arithmetic progression, then
sin α = 21 (sin(α + γ) + sin(α − γ)) = sin α cos γ, i.e., either cos γ = 1 or γ = 0. Therefore,
each of the triangle’s angles is equal to 60◦ .
b) If the numbers sin(α − γ) ≤ sin α ≤ sin(α + γ) form a geometric progression, then
sin2 α = sin(α − γ) sin(α + γ) = sin2 α cos2 γ − sin2 γ cos2 α ≤ sin2 α cos2 γ.
Hence, cos γ = 1, i.e., each of the triangle’s angles is equal to 60◦ .
12.69. Let us complement triangle ABC to parallelogram ABCE (Fig. 142). Let
BC = x and AD = y. Then (b − a)h = 2SAED = xy sin 45◦ and
(b − a)2 = x2 + y 2 − 2xy cos 45◦ = x2 + y 2 − 2xy sin 45◦ .
By Pythagoras theorem
a2 + b2 = (AO2 + BO2 ) + (CO2 + DO2 ) = (BO2 + CO2 ) + (DO2 + AO2 ) = x2 + y 2 .
Therefore,
(b − a)2 = x2 + y 2 − 2xy sin 45◦ = a2 + b2 − 2(b − a)h,
SOLUTIONS
287
Figure 142 (Sol. 12.69)
ab
.
i.e., h = b−a
12.70. Since BK = 12 (a + c − b) and KC = 21 (a + b − c) (cf. Problem 3.2), it follows
2
2
= S tan α2 , cf. Problem 12.12.
that BK · KC = a −(b−c)
4
cos
β−γ
12.71. Since b+c
= sin α2 (Problem 12.4), it follows that cos β−γ
= cos α2 , i.e., β − γ =
a
2
2
±α. If β = γ + α, then β = 90◦ and if β + α = γ, then γ = 90◦ .
12.72. It is easy to verify that SABC = 2R2 sin α sin β sin γ. Analogously,
β+γ
α+γ
α+β
α
β
γ
SA1 B1 C1 = 2R2 sin
sin
sin
= 2R2 cos cos cos .
2
2
2
2
2
2
Hence,
SABC
α
β
γ
2r
= 8 sin sin sin = ,
S A1 B 1 C 1
2
2
2
R
cf. Problem 12.36 a).
2
2
2
12.73. The sum of cotangents of the angles of a triangle is equal to a +b4S+c , cf. Problem
2
2
2
12.44 a). Moreover, m2a + m2b + m2c = 3(a +b4 +c ) (by Problem 12.11 b)) and the area of the
triangle formed by the medians of triangle ABC is equal to 43 SABC (Problem 1.36).
12.74. One of the points Ai lies inside the triangle formed by the other three points;
hence, we can assume that triangle A1 A2 A3 is an acute one (or a right one). Numbers λ1 ,
λ2 and λ3 are easy to obtain from the corresponding system of equations; as a result we get
a2 + c 2 − b 2
a2 + b 2 − c 2
b 2 + c 2 − a2
, λ2 =
and λ3 =
,
λ1 =
2
2
2
where a = A2 A3 , b = A1 A3 and c = A1 A2 . By Problem 5.45 b) A1 A24 = 4R2 − a2 , where R
is the radius of the circumscribed circle of triangle A1 A2 A3 . Hence,
a2 + b 2 + c 2
λ4 = A1 A24 − λ1 = 4R2 −
= A2 A24 − λ2 = A3 A24 − λ3 .
2
P 1
2
2
2
Now, let us verify that
= 0. Since b +c2 −a = bc cos α = 2S cot α, it follows that
λi
α
1
= tan 2S
. It remains to observe that
λ1
a2
+
b2
tan α + tan β + tan γ
2
=
2
2
+ c − 8R
2S
?Problem 12.49.
12.75. Let (a1 , b1 ), (a2 , b2 ) and (a3 , b3 ) be the coordinates of the triangle’s vertices. The
coordinates of the center of the circumscribed circle of the triangle are given by the system
of equations
(x − a1 )2 + (y − b1 )2 = (x − a2 )2 + (y − b2 )2 ,
(x − a1 )2 + (y − b1 )2 = (x − a3 )2 + (y − b3 )2 .
288
CHAPTER 12. CALCULATIONS AND METRIC RELATIONS
It is easy to verify that these equations are actually linear ones and, therefore, the solution
of the considered system is a rational one.
12.76. On segments AB and CD, take points K and L that divide the segments in
the ratios indicated. It suffices to prove that the intersection point of lines AK and CL
lies on circle S. Let us take the coordinate system with the origin at the center O of circle
S and axes Ox and Oy directed along rays OB and OD. The radius of circle S can be
assumed to be equal to 1. Lines AK and CL are given by equations y = x+1
and y = 2x − 1,
3
respectively. Therefore, the coordinates of their intersection point are x0 = 45 and y0 = 53 .
Clearly, x20 + y02 = 1.
12.77. Let d be the distance between the center of the circumscribed circle and the
image of the center of the inscribed circle under the considered homothety. It suffices to
verify that R = d + 2r. Let (0, 0), (2a, 0) and (0, 2b) be the coordinates of the vertices of
the given triangle. Then (a, b) are the coordinates of the center of the circumscribed circle,
(r, r) the coordinates of the center of the inscribed circle, where r = a + b − R. Therefore,
d2 = (2r − a)2 + (2r − b)2 = a2 + b2 − 4r(a + b − r) + 4r2 = (R − 2r)2
because a2 + b2 = R2 .
12.78. Let us consider the coordinate system with the origin at the center of the square
and the Ox-axis parallel to line l. Let the coordinates of the vertices of the square be (A(x, y),
B(y, −x), C(−x, −y) and ¡D(−y, x);¢ let line l be given by the equation y = a. Then the
x+y y−x
coordinates of point Q are
, 2 ¢ and those of P are (−y, a). Therefore, the locus to be
2
¡
. It remains to observe that the quantity
found consists of points t, −t + 21 a , where t = x−y
4
p
2
2
x − y varies from − 2(x + y ) = −AB to AB.
Chapter 13. VECTORS
Background
1. We will make use of the following notations:
−→
a) AB and a denote vectors;
b) AB and |a| denote the lengths of these vectors; sometimes the length of vector a will
be denoted by a; a unit vector is a vector of unit length;
−→ −−→
−→
c) (AB, CD), (a, b) and (AB, a) denote the inner products of the vectors;
d) (x, y) is the vector with coordinates x, y;
−
→
e) 0 or 0 denotes the zero vector.
2. The oriented angle between the nonzero vectors a and b (notation ∠(a, b)) is the
angle through which one should rotate the vector a counterclockwise to make it directed as
b is. The angles that differ by 360 degrees are assumed to be equal. It is easy to verify the
following properties of oriented angles between vectors:
a) ∠(a, b) = −∠(b, a);
b) ∠(a, b) + ∠(b, c) = ∠(a, c);
c) ∠(−a, b) = ∠(a, b) + 180◦ .
3. The inner product of vectors a and b is the number
(a, b) = |a| · |b| cos ∠(a, b)
(if one of these vectors is the zero one, then by definition (a, b) = 0). The following properties
of the inner product are easily verified:
a) (a, b) = (b, a);
b) |(a, b)| ≤ |a| · |b|;
c) (λa + µb, c) = λ(a, c) + µ(b, c);
d) if a, b 6= 0 then (a, b) = 0 if and only if a ⊥ b.
4. Many of vector inequalities can be proved with the help of the following fact.
Given two sets of vectors such that the sum of lengths of projections of the vectors of
the first set to any straight line does not exceed the sum of the lengths of projections of the
vectors from the second set to the same line, the sum of the lengths of the vectors from the
first set does not exceed the sum of the lengths of the vectors of the second set, cf. Problem
13.39.
In this way a problem on a plane reduces to a problem on a straight line which is usually
easier.
Introductory problems
−−→
−→ −→
1. Let AA1 be the median of triangle ABC. Prove that AA1 = 21 (AB + AC).
2. Prove that |a + b|2 + |a − b|2 = 2(|a|2 + |b|2 ).
3. Prove that if vectors a + b and a − b are perpendicular, then |a| = |b|.
−→ −−→ −→ −
→
4. Let OA + OB + OC = 0 and OA = OB = OC. Prove that ABC is an equilateral
triangle.
289
290
CHAPTER 13. VECTORS
5. Let M and N be the midpoints of segments AB and CD, respcetively. Prove that
−−→ 1 −→ −−→
M N = 2 (AC + BD).
§1. Vectors formed by polygons’ (?) sides
13.1. a) Prove that from the medians of a triangle one can construct a triangle.
b) From the medians of triangle ABC one constructed triangle A1 B1 C1 and from the
medians of triangle A1 B1 C1 one constructed triangle A2 B2 C2 . Prove that triangles ABC
and A2 B2 C2 are similar with simlarity coefficient 34 .
13.2. The sides of triangle T are parallel to the respective medians of triangle T1 . Prove
that the medians of T are parallel to the corresponding sides of T1 .
13.3. Let M1 , M2 , . . . , M6 be the midpoints of a convex hexagon A1 A2 . . . A6 . Prove
that there exists a triangle whose sides are equal and parallel to the segments M1 M2 , M3 M4 ,
M5 M 6 .
13.4. From a point inside a convex n-gon, the rays are drawn perpendicular to the sides
and intersecting the sides (or their continuations). On these rays the vectors a1 , . . . , an
whose lengths are equal to the lengths of the corresponding sides are drawn. Prove that
a1 + · · · + an = 0.
13.5. The sum of four unit vectors is equal to zero. Prove that the vectors can be divided
into two pairs of opposite vectors.
13.6. Let E and F be the midpoints of sides AB and CD of quadrilateral ABCD and
K, L, M and N are the midpoints of segments AF , CE, BF and DE, respectively. Prove
that KLM N is a parallelogram.
13.7. Consider n pairwise noncodirected vectors (n ≥ 3) whose sum is equal to zero.
Prove that there exists a convex n-gon such that the set of vectors formed by its sides
coincides with the given set of vectors.
13.8. Given four pairwise nonparallel vectors whose sum is equal to zero. Prove that we
can construct from them:
a) a nonconvex quadrilateral;
b) a self-intersecting broken line of four links.
13.9. Given four pairwise nonparallel vectors a, b, c and d whose sum is equal to zero,
prove that
|a| + |b| + |c| + |d| > |a + b| + |a + c| + |a + d|.
13.10. In a convex pentagon ABCDE side BC is parallel to diagonal AD, in addition
we have CD k BE, DE k AC and AE k BD. Prove that AB k CE.
§2. Inner product. Relations
13.11. Prove that if the diagonals of quadrilateral ABCD are perpendicular to each
other, then the diagonals of any other quadrilateral with the same lengths of its sides are
perpendicular to each other.
13.12. a) Let A, B, C and D be arbitrary points on a plane. Prove that
−→ −−→
−−→ −−→
−→ −−→
(AB, CD) + (BC, AD) + (CA, BD) = 0.
b) Prove that the hights of a triangle intersect at one point.
13.13. Let O be the center of the circle inscribed in triangle ABC and let point H
satisfy OH = OA + OB + OC. Prove that H is the intersection point of heights of triangle
ABC.
§3. INEQUALITIES
291
13.14. Let a1 , . . . , an be vectors formed by the sides of an n-gon, ϕij = ∠(ai , aj ). Prove
that
X
a21 = a22 + · · · + a2n + 2
ai aj cos ϕij , where ai = |ai |.
i>j>1
13.15. Given quadrilateral ABCD and the numbers
u = AD2 , v = BD2 , w = CD2 , U = BD2 + CD2 − BC 2 ,
V = AD2 + CD2 − AC 2 , W = AD2 + BD2 − AB 2 .
Prove that
((Gauss).)
uU 2 + vV 2 + wW 2 = U V W + 4uvw.
−−→ −−→
13.16. Points A, B, C and D are such that for any point M the numbers (M A, M B)
−−→ −−→
−→ −−→
and (M C, M D) are distinct. Prove that AC = DB.
13.17. Prove that in a convex k-gon the sum of distances from any inner point to the
sides of the k-gon is constant if and only if the sum of vectors of unit exterior normals to
the sides is equal to zero.
13.18. In a convex quadrilateral the sum of distances from a vertex to the sides is the
same for all vertices. Prove that this quadrilateral is a parallelogram.
§3. Inequalities
13.19. Given points A, B, C and D. Prove that
AB 2 + BC 2 + CD2 + DA2 ≥ AC 2 + BD2 ,
where the equality is attained only if ADCD is a parallelogram.
13.20. Prove that from any five vectors one can always select two so that the length of
their sum does not exceed the length of the sum of the remaining three vectors.
13.21. Ten vectors are such that the length of the sum of any nine of them is smaller
than the length of the sum of all the ten vectors. Prove that there exists an axis such that
the projection of every of the ten vectors to the axis is positive.
−−→
−−→ −
→
13.22. Points A1 , . . . , An lie on a circle with center O and OA1 + · · · + OAn = 0 . Prove
that for any point X we have
XA1 + · · · + XAn ≥ nR,
where R is the radius of the circle.
13.23. Given eight real numbers a, b, c, d, e, f , g, h. Prove that at least one of the six
numbers
ac + bd, ae + bf, ag + bh, ce + df, cg + dh, eg + f h
is nonnegative.
13.24. On the circle of radius 1 with center O there are given 2n + 1 points P1 , . . . , P2n+1
which lie on one side of a diameter. Prove that
−−→
−−−−→
|OP1 + · · · + OP2n+1 | ≥ 1.
13.25. Let a1 , a2 , . . . , an be vectors whose length does not exceed 1. Prove that in the
sum
c = ±a1 ± a2 ± · · · ± an
√
we can select signs so that |c| ≤ 2.
292
CHAPTER 13. VECTORS
13.26. Point O is the beginning point of n unit vectors such that in any half plane
bounded by a straight line through O there are contained not less than k vectors (we assume
that the boundary line belongs to the half-plane). Prove that the length of the sum of these
vectors does not exceed n − 2k.
§4. Sums of vectors
13.27. Prove that point X belongs to line AB if and only if
−−→
−→
−−→
OX = tOA + (1 − t)OB
for some t and any point O.
13.28. We are given several points and for several pairs (A, B) of these points the vectors
AB are taken in such a way that as many vectors exit from every point as terminate in it.
Prove that the sum of all the selected vectors is equal to 0.
13.29. Inside triangle ABC, point O is taken. Prove that
−→
−−→
−→ −
→
SBOC · OA + SAOC · OB + SAOB · OC = 0 .
p
q
13.30. Points A and B move along two fixed rays with common origin O so that OA
+ OB
is a constant. Prove that line AB passes through a fixed point.
13.31. Through the intersection point M of medians of triangle ABC a straight line is
drawn intersecting BC, CA and AB at points A1 , B1 and C1 , respectively. Prove that
1
1
1
)+(
)+(
) = 0.
(
M A1
M B1
M C1
(Segments M A1 , M B1 and M C1 are assumed to be oriented.)
13.32. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 , respectively,
are taken. Segments BB1 and CC1 , CC1 and AA1 , AA1 and BB1 intersect at points A1 , B2
−−→ −−→ −−→ −
→
and C2 , respectively. Prove that if AA2 + BB2 + CC2 = 0 , then
AB1 : B1 C = CA1 : A1 B = BC1 : C1 A.
13.33. Quadrilateral ABCD is an inscribed one. Let Ha be the orthocenter of BCD,
let Ma be the midpoint of AHa ; let points Mb , Mc and Md be similarly defined. Prove that
points Ma , Mb , Mc and Md coincide.
13.34. Quadrilateral ABCD is inscribed in a circle of radius R.
a) Let Sa be the circle of radius R with center at the orthocenter of triangle BCD; let
circles Sb , Sc and Sd be similarly defined. Prove that these four circles intersect at one point.
b) Prove that the circles of nine points of triangles ABC, BCD, CDA and DAB intersect
at one point.
§5. Auxiliary projections
13.35. Point X belongs to the interior of triangle ABC; let α = SBXC , β = SCXA
and γ = SAXB . Let A1 , B1 and C1 be the projections of points A, B and C, respectively,
−−→
−−→
−−→
on an arbitrary line l. Prove that the length of vector αAA1 + β BB1 + γ CC1 is equal to
(α + β + γ)d, where d is the distance from X to l.
13.36. A convex 2n-gon A1 A2 . . . A2n is inscribed into a unit circle. Prove that
−−−→ −−−→
−−−−−−→
|A1 A2 + A3 A4 + · · · + A2n−1 A2n | ≤ 2.
13.37. Let a, b and c be the lengths of the sides of triangle ABC; let na , nb and nc be
unit vectors perpendicular to the corresponding sides and directed outwards. Prove that
−−→
a3 na + b2 nb + c2 nc = 12S · M O,
§7. PSEUDOINNER PRODUCT
293
where S is the area, M the intersection point of the medians, O the center of the circle
inscribed into triangle ABC.
13.38. Let O and R be the center and the radius, respectively, of an escribed circle
of triangle ABC; let Z and r be the center and the radius of the inscribed circle, K the
intersection point of the medians of the triangle with vertices at the tangent points of the
inscribed circle of triangle ABC with the sides of triangle ABC. Prove that Z belongs to
segment OK and
OZ : ZK = 3R : r.
§6. The method of averaging
13.39. Given two sets of vectors a1 , . . . , an and b1 , . . . , bm such that the sum of the
lengths of the projections of the vectors from the first set to any straight line does not
exceed the sum of the lengths of the projections of the vectors from the second set to the
same straight line. Prove that the sum of the lengths of the vectors from the first set does
not exceed the sum of the lengths of the vectors from the second set.
13.40. Prove that if one convex polygon lies inside another one, then the perimeter of
the inner polygon does not exceed the perimeter of the outer one.
13.41. The sum of the length of several vectors on a plane is equal to L. Prove that
from these vectors one can select several vectors (perhaps, just one) so that the length of
their sum is not less than Lπ .
13.42. Prove that if the lengths of any side and diagonal of a convex polygon are shorter
than d, then its perimeter is shorter than πd.
13.43. On the plane, there are given four vectors a, b, c and d whose sum is equal to
zero. Prove that
|a| + |b| + |c| + |d| ≥ |a + d| + |b + d| + |b + d|.
−−→
−−→
13.44. Inside a convex n-gon A1 A2 . . . An a point O is selected so that OA1 +· · ·+ OAn =
−
→
0 . Let d = OA1 + · · · + OAn . Prove that the perimeter of the polygon is not shorter than
4d
for n even and not shorter than n4dn
2 −1 for n odd.
n
13.45. The length of the projection of a closed convex curve to any line is equal to 1.
Prove that its length is equal to π.
13.46. Given several convex polygons so that it is impossible to draw a line which does
not intersect any of the polygons and at least one polygon would lie on both sides of it.
Prove that all the polygons are inside a polygon whose perimeter does not exceed the sum
of the perimeters of the given polygons.
§7. Pseudoinner product
The pseudoinner product of nonzero vectors a and b is the number
c = |a| · |b| sin ∠(a, b);
the pseudoinner product is equal to 0 if at least one of the vectors a or b is zero. The
pseudoinner product is denoted by c = a ∨ b. Clearly, a ∨ b = −b ∨ a.
The absolute value of the pseudoinner product of a and b is equal to the area of the
parallelogram spanned by these vectors. In this connection the oriented area of the triple of
points A, B and C is the number
1 −→ −→
S(A, B, C) = (AB ∨ AC).
2
The absolute value of S(A, B, C) is equal to the area of triangle ABC.
13.47. Prove that:
294
CHAPTER 13. VECTORS
a) (λa) ∨ b = λ(a, b);
b) a ∨ (b + c) = a ∨ b + a ∨ c.
13.48. Let a = (a1 , a2 ) and b = (b1 , b2 ). Prove that
13.49. a) Prove that
a ∨ b = a 1 b 2 − a2 b 1 .
S(A, B, C) = −S(B, A, C) = S(B, C, A).
b) Prove that for any points A, B, C and D we have
S(A, B, C) = S(D, A, B) + S(D, B, C) + S(D, C, A).
13.50. Three runners A, B and C run along the parallel lanes with constant speeds. At
the initial moment the area of triangle ABC is equal to 2 in 5 seconds it is equal to 3. What
might be its value after 5 more seconds?
13.51. Three pedestrians walk at constant speeds along three straight roads. At the
initial moment the pedestrians were not on one straight line. Prove that the pedestrians can
occure on one straight line not more than twice.
13.52. Prove Problem 4.29 b) with the help of a pseudoinner product.
13.53. Points P1 , P2 and P3 not on one line are inside a convex 2n-gon A1 . . . A2n . Prove
that if the sum of the areas of triangles A1 A2 Pi , A3 A4 Pi , . . . , A2n−1 A2n Pi is equal to the
same number c for i = 1, 2, 3, then for any inner point P the sum of the areas of these
triangles is equal to c.
13.54. Given triangle ABC and point P . Let point Q be such that CQ k AP and point
R be such that AR k BQ and CR k BP . Prove that SABC = SP QR .
13.55. Let H1 , H2 and H3 be the orthocenters of triangles A2 A3 A4 , A1 A3 A4 and A1 A2 A4 .
Prove that the areas of triangles A1 A2 A3 and H1 H2 H3 are equal.
13.56. In a convex 5-gon ABCDE whose area is equal to S the areas of triangles ABC,
BCD, CDE, DEA and EAB are equal to a, b, c, d and e, respectively. Prove that
S 2 − S(a + b + c + d + e) + ab + bc + cd + de + ea = 0.
Problems for independent study
13.57. Let M and N be the midpoints of segments AB and AC, respectively, P the
−→ −−→ −→
−→
midpoint of M N and O an arbitrary point. Prove that 2OA + OB + OC = 4OP .
13.58. Points A, B and C move uniformly with the same angle velocities along the three
circles in the same direction. Prove that the intersection point of the medians of triangle
ABC moves along a circle.
13.59. Let A, B, C, D and E be arbitrary points. Is there a point O such that
−→ −−→ −→ −−→ −−→
OA + OB + OC = OD + OE? Find all such points, if any.
13.60. Let P and Q be the midpoints of the diagonals of a convex quadrilateral ABCD.
Prove that
AB 2 + BC 2 + CD2 + DA2 = AC 2 + BD2 + 4P Q2 .
13.61. The midpoints of segments AB and CD are connected by a segment; so are
the midpoints of segments BC and DE. The midpoints of the segments obtained are also
connected by a segment. Prove that the last segment is parallel to segment AE and its
length is equal to 41 AE.
13.62. The inscribed circle is tangent to sides BC, CA and AB of triangle ABC at
−−→ −−→ −−→
−
→
points A1 , B1 and C1 , respectively. Prove that if AA1 + BB1 + CC1 = 0 , then triangle
ABC is an equilateral one.
SOLUTIONS
295
13.63. Quadrilaterals ABCD, AEF G, ADF H, F IJE and BIJC are parallelograms.
Prove that quadrilateral AF HG is also a parallelogram.
Solutions
−−→
−→
−→
13.1. a) Let a = BC, b = CA and c = AB; let AA′ , BB ′ and CC ′ be medians of
−−→
−−→
−−→
triangle ABC. Then AA′ = 21 (c − b), BB ′ = 21 (a − c) and CC ′ = 21 (b − c). Therefore,
−−→′ −−→′ −−→′ −
→
AA + BB + CC = 0 .
−−→
−−→
−−→
b) Let a1 = AA′ , b1 = BB ′ and c = CC ′ . Then 12 (c1 − b1 ) = 41 (b − a − a + c) = − 34 a is
the vector of one of the sides of triangle A2 B2 C2 .
13.2. Let a, b and c be the vectors of the sides of T . Then 21 (b − a), 21 (a − c) and
1
(c − b) are the vectors of its medians. We may assume that a, b and c are the vectors
2
directed from the intersection point of the medians of triangle T1 to its vertices. Then b − a,
a − c and c − a are the vectors of its sides.
−−−−→ −−−→ −−−→ −−−→ −−−−→ −−−→
−−−−→
13.3. It is clear that 2M1 M2 = A1 A2 + A2 A3 = A1 A3 , 2M3 M4 = A3 A5 and 2M5 M6 =
−−−→
−−−−→ −−−−→ −−−−→ −
→
A5 A1 . Therefore, M1 M2 + M3 M4 + M5 M6 = 0 .
13.4. After rotation through 90◦ the vectors a1 , . . . , an turn into the vectors of sides of
the n-gon.
13.5. From given vectors one can construct a convex quadrilateral. The lengths of all
the sides of this quadrilateral are equal to 1, therefore, this quadrilateral is a rhombus; the
pairs of its opposite sides provide us with the division desired.
−→
−−→
−−→
−−→
−→
13.6. Let a = AE, b = DF and v = AD. Then 2AK = b + v and 2AL = a + v + 2b
−−→
−−→ −→ −−→
and, therefore, KL = AL − AK = 21 (a + b). Similarly, N M = 21 (a + b).
13.7. Let us draw the given vectors from one point and index them clockwise: a1 , . . . ,
−−−−→
an . Consider a closed broken line A1 . . . An , where Ai Ai+1 = ai . Let us prove that A1 . . . An
is a convex polygon. Introduce a coordinate system and direct the Ox-axis along a1 . Let the
vectors a2 , . . . , ak lie on one side of Ox-axis and the vectors ak+1 , . . . , an lie on the other side
(if there is a vector directed opposite to a1 it can be referred to either of these two groups).
The projections of the vectors from the first group on the Oy-axis are of one sign and
the projections of the vectors of the other group are of the opposite sign. Therefore, the
second coordinate of the points A2 , A3 , . . . , Ak+1 and the points Ak+1 , . . . , An , A1 vary
monotonously: for the first group from 0 to a quantity d, for the second group they decrease from d to 0. Since there are two intervals of monotonity, all the vertices of the
polygon lie on one side of the line A1 A2 .
For the other lines passing through the sides of the polygon the proof is similar.
13.8. Thanks to Problem 13.7 the given vectors form a convex quadrilateral. The rest
is clear from Fig. 143.
Figure 143 (Sol. 13.8)
296
CHAPTER 13. VECTORS
13.9. By Problem 13.8 b) from the given vectors we can construct a self-intersecting
broken line of four links; this broken line can be viewed as the two diagonals and two opposite
sides of a convex quadrilateral. Two cases are possible: the vector a can be either a side or
a diagonal of this quadrilateral.
But in both cases the sum in the left-hand side of the inequality is the sum of lengths of
two opposite sides and two diagonals of the quadrilateral and the sum in the right-hand side
is constituted by the length of the sum of vectors of the same opposite sides and the lengths
of the two other opposite sides. It only remains to observe that the sum of lengths of two
vectors is not shorter than the length of their sum and the sum of the length of diagonals
of a convex quadrilateral is longer than the sum of lengths of the two opposite sides: cf.
Problem 19.14.
13.10. Let diagonal BE intersect diagonals AD and AC at points F and G, respectively.
The respective sides of triangles AF E and BCD are parallel; hence, the triangles are similar
and AF : F E = BC : CD. Therefore,
AD : BE = (AF + BC) : (EF + CD) = BC : CD.
Similarly, AE : BD = DE : AC. From the similarity of BED and EGA we deduce that
AE : DB = EG : BE = CD : BE. Thus,
CD
AE
DE
BC
=
=
=
= λ.
AD
BE
BD
AC
Clearly,
−−→ −−→ −−→ −→ −→ −
→
BC + CD + DE + EA + AB = 0 ,
−−→ −−→ −→ −−→ −−→ −
→
AD + BE + CA + DB + EC = 0
and
−−→
−−→ −−→
−−→ −−→
−→ −→
−−→
BC = λAD, CD = λBE, DE = λCA, EA = λDB.
It follows that
−−→ −−→ −→ −−→
−→
−−→ −→
−
→
0 = λ(AD + BE + CA + DB) + AB = −λEC + AB,
−→
−−→
i.e., AB = λEC. Hence, AB k EC.
−→
−−→
−−→
−−→
13.11. Let a = AB, b = BC, c = CD and d = DA. It suffices to verify that AC ⊥ BD
if and only if a2 + c2 = b2 + d2 . Clearly,
d2 = |a + b + c|2 = a2 + b2 + c2 + 2[(a, b) + (b, c) + (c, a)].
Therefore, the condition AC ⊥ BD, i.e.,
0 = (a + b, b + c) = b2 + (b, c) + (a, c) + (a, b)
is equivalent to the fact that
d2 = a2 + b2 + c2 − 2b2 .
−→ −−→
13.12. a) Let us express all the vectors that enter the formula through AB, BC and
−−→
−−→
−→ −−→ −−→ −→
−→ −−→
−−→
−−→ −−→
CD, i.e., let us write AD = AB + BC + CD, CA = −AB − BC and BD = BC + CD.
After simplification we get the statement desired.
b) Let D be the intersection point of heights drawn from vertices A and C of triangle
ABC. Then in the formula proved in heading a) the first two summands are zero and,
therefore, the last summand is also zero, i.e., BD ⊥ AC.
SOLUTIONS
297
−−→ −→ −−→ −→ −→ −−→ −→
13.13. Let us prove that AH ⊥ BC. Indeed, AH = AO + OH = AO + OA+ OB + OC =
−−→ −→
−−→ −−→ −→
−−→ −→
OB + OC and BC = BO + OC = −OB + OC and, therefore,
−−→ −−→
(AH, BC) = OC 2 − OB 2 = R2 − R2 = 0
because O is the center of the circumscribed circle. We similarly prove that BH ⊥ AC and
CH ⊥ AB.
13.14. Let αi = ∠(ai , a1 ). Considering the projections
parallel to a1
P to the straight lineP
and the straight line perpendicular to a1 we get a1 = i>1 ai cos αi and 0 = i>1 ai sin αi ,
respectively. Squaring these equalities and summing we get
P
P
a21 = i>1 a2i (cos2 αi + sin2 αi ) + 2 Pi>j>1 ai aj (cos αi cos αj + sin αi sin αj ) =
a22 + · · · + a2n + 2 i>j>1 ai aj cos(αi − αj ).
It remains to notice that αi − αj = ∠(ai , a1 ) − ∠(aj , a1 ) = ∠(ai , aj ) = ϕij .
−−→
−−→
−−→
13.15. Let a = AD, b = BD and c = CD. Since BC 2 = |b−c|2 = BD2 +CD2 −2(b, c),
it follows that U = 2(b, c). Similarly, V = 2(a, c) and W = 2(a, b). Let α = ∠(a, b) and
β = ∠(b, c). Multiplying the equality
cos2 α + cos2 β + cos2 (α + β) = 2 cos α cos β cos(α + β) + 1
(cf. Problem 12.39 b)) by 4uvw = 4|a|2 |b|2 |c|2 we get the statement desired.
−−→
−→
−−→
13.16. Fix an arbitrary point O. Let m = OM , a = OA, . . . , d = OD. Then
−−→ −−→
−−→ −−→
(M A, M B) − (M C, M D) = (a − m, b − m) − (c − m, d − m) =
(c + d − a − b, m) + (a, b) − (c, d).
If v = c + d − a − b 6= 0, then as the point M runs over the plane the value (v, m)
attains all the real values, in particular, it takes the value (c, d) − (a, b). Hence, v = 0, i.e.,
−→ −−→ −→ −−→
−→ −−→
OC + OD = OA + OB and, therefore, AC = DB.
13.17. Let n1 , . . . , nk be the unit exterior normals to the sides and let M1 , . . . , Mk be
arbitrary points on these sides. For any point X inside the polygon the distance from X to
−−−→
the i-th side is equal to (XMi , ni ). Therefore, the sums of distances from the inner points
A and B to the sides of the polygon are equal if and only if
k
k
X
X
−−→
−−→
(AMi , ni ) =
(BMi , ni ) =
i=1
i=1
k
k
X
X
−→
−−→
(BA, ni ) +
(AMi , ni ),
i=1
i=1
−→ P
i.e., (BA, ki=1 ni ) = 0. Hence, theP
sum of distances from any inner point of the polygon to
the sides is constant if and only if
ni = 0.
13.18. Let l be an arbitrary line, n the unit vector perpendicular to l. If points A
and B belong to the same half-plane given by the line l the vector n belongs to, then
−→
ρ(B, l) − ρ(A, l) = (AB, n), where ρ(X, l) is the distance from X to l.
Let n1 , n2 , n3 and n4 be unit vectors perpendicular to the consecutive sides of quadrilateral ABCD and directed inwards. Denote the sum of distances from point X to the sides
298
CHAPTER 13. VECTORS
P
of quadrilateral ABCD by (X). Then
X
X
−→
0=
(B) −
(A) = (AB, n1 + n2 + n3 + n4 ).
Similarly,
−−→
(BC, n1 + n2 + n3 + n4 ) = 0.
Since points A, B and C do not belong to the same line, n1 + n2 + n3 + n4 = 0. It remains
to make use of the result
−→ of Problem
−−→ 13.5. −−→
−−→
−→
13.19. Let a = AB, b = BC and c = CD. Then AD = a + b + c, AC = a + b and
−−→
BD = b + c. It is also clear that
|a|2 + |b|2 + |c|2 + |a + b + c|2 − |a + b|2 − |b + c|2 = |a|2 + 2(a,
c) + |c|2 = |a + c|2 ≥ 0.
The equality is only attained if a = −c, i.e., ABCD is a parallelogram.
13.20. Consider five vectors a1 , a2 , a3 , a4 , a5 and suppose that the length of the sum
of any two of them is longer than the length of the sum of the three remaining ones. Since
|a1 + a2 | > |a3 + a4 + a5 |, it follows that
|a1 |2 + 2(a1 , a2 ) + |a2 |2 > |a3 |2 + |a4 |2 + |a5 |2 + 2(a3 , a4 ) + 2(a4 , a5 ) + 2(a3 , a5 ).
Adding such inequalities for all ten pairs of vectors we get
4(|a1 |2 + . . . ) + 2((a1 , a2 ) + . . . ) > 6(|a1 |2 + . . . ) + 6((a1 , a2 ) + . . . )
i.e., |a1 + a2 + a3 + a4 + a5 |2 < 0. Contradiction.
−→
13.21. Denote the given vectors by e1 , . . . , e10 . Let AB = e1 + · · · + e10 . Let us prove
−→
−→
that the ray AB determines the required axis. Clearly, |AB − ei |2 = AB 2 − 2(AB, ei ) + |ei |2 ,
−→
−→
−→
i.e., (AB, ei ) = 21 (AB 2 +|ei |2 −|AB −ei |2 ). By the hypothesis AB > |AB −ei | and, therefore,
−→
(AB, ei ) > 0, i.e., the projection of ei to AB is positive.
−−→
−−→
−−→
13.22. Let ai = OAi and x = OX. Then |ai | = R and XAi = ai − x. Therefore,
X
XAi =
X
|ai − x| =
X |ai − x| · |ai |
R
≥
X ai − x, ai
R
=
X ai , ai
R
−
(x,
P
R
ai )
.
P
It remains to observe that (ai , ai ) = R2 and
ai = 0.
13.23. On the plane, consider four vectors (a, b), (c, d), (e, f ) and (g, h). One of the
◦
= 90◦ . If the angle between the vectors
angles between these vectors does not exceed 360
4
does not exceed 90◦ , then the inner product is nonnegative.
The given six numbers are inner products of all the pairs of our four vectors and, therefore,
at least one of them is nonnegative.
13.24. Let us prove this statement by induction. For n = 0 the statement is obviously
true. Let us assume that the statement is proved for 2n + 1 vectors. In a system of 2n + 3
vectors consider two extreme vectors (i.e., the vectors the angle between which is maximal).
−−→
−−−−→
For definiteness sake, suppose that these are vectors OP1 and OP2n+3 . By the inductive
−→ −−→
−→
−→
hypothesis the length of OR = OP2 + · · · + OP 2n+2 is not shorter than 1. The vector OR
belongs to the interior of angle ∠P1 OP2n+3 and, therefore, it forms an acute angle with the
−→ −→
−→
−→ −→
vector OS = OP 1 + OP 2n+3 . Hence, |OS + OR| ≥ OR ≥ 1.
13.25. First, let us prove that if a, b and c are vectors whose length does not exceed 1,
then at least one of the vectors a ± b, a ± c, b ± c is not longer than 1.
SOLUTIONS
299
Indeed, two of the vectors ±a, ±b, ±c form an angle not greater than 60◦ and, therefore,
the difference of these two vectors is not longer than 1 (if (??) in triangle ABC we have
AB ≤ 1, BC ≤ 1 and ∠ABC ≤ 60◦ , then AC is not the greatest side and AC ≤ 1).
Thus, we can reduce the discussion to two vectors a and b. Then either the angle between
√
vectors a and√b or between vectors a and −b does not exceed 90◦ ; hence, either |a − b| ≤ 2
or |a + b| ≤ 2.
13.26. We can assume that the sum a of the given vectors is nonzero because otherwise
the statement of the problem is obvious.
Figure 144 (Sol. 13.26)
Let us introduce a coordinate system directing Oy-axis along a. Let us enumerate the
vectors of the lower half-plane clockwise: e1 , e2 , . . . as on Fig. 144. By the hypothesis there
are not less than k of these vectors. Let us prove that among the given vectors there are
also vectors v1 , . . . , vk such that the second coordinate of the vector vi + ei is nonpositive
for any i = 1, . . . , k. This will prove the required statement.
Indeed, the length of the sum of the given vectors is equal to the sum of the second
coordinates (the coordinate system was introduced just like this). The second coordinate of
the sum of the vectors e1 , v1 , . . . , ek , vk is nonpositive and the second coordinate of any of
the remaining vectors does not exceed 1. Therefore, the second coordinate of the sum of all
the given vectors does not exceed n − 2k.
Let vectors e1 , . . . , ep belong to the fourth quadrant. Let us start assigning to them the
vectors v1 , . . . , vp . Let us rotate the lower half plane that consists of points with nonpositive
second coordinate by rotating the Ox-axis clockwise through an angle between 0◦ and 90◦ .
If one of the two vectors that belongs to the half plane rotated this way lies in the fourth
quadrant, then their sum has a nonpositive second coordinate. As the Ox-axis rotates beyond
vector e1 , at least one vector that belongs to the half plane should be added to the vectors
e2 , . . . , ek ; hence, the vector which follows ek should be taken for v1 .
Similarly, while the Ox-axis is rotated beyond e2 we get vector v2 , and so on. These
arguments remain valid until the Ox-axis remains in the fourth quadrant. For the vectors
ep+1 , . . . , ek which belong to the third quadrant the proof is given similarly (if the first
coordinate of the vector ep+1 is zero, then we should first disregard it; then take any of the
remaining vectors for its(whose?) partner).
−−→
−→
13.27. Point X belongs to line AB if and only if AX = λAB, i.e.,
−−→ −→ −−→
−→
−−→
OX = OA + AX = (1 − λ)OA + λOB.
13.28. Let us take an arbitrary point O and express all the selected vectors in the form
−−−→ −→
−→
−→
Ai Aj = OAj − OAi . By the hypothesis every vector OAi enters the sum of all the chosen
vectors with the “plus” sign as many times as with the “minus” sign.
300
CHAPTER 13. VECTORS
−→ −−→
−→
13.29. Let e1 , e2 and e3 be unit vectors codirected with vectors OA, OB and OC,
respectively; let α = ∠BOC, β = ∠COA and γ = ∠AOB. We have to prove that
−
→
e1 sin α + e2 sin β + e3 sin γ = 0 .
Consider triangle A1 B1 C1 whose sides are parallel to lines OC, OA and OB. Then
−
→ −−−→ −−−→ −−−→
0 = A1 B1 + B1 C1 + C1 A1 = ±2R(e1 sin α + e2 sin β + e3 sin γ),
where R is the radius of the circumscribed circle of triangle ABC.
13.30. Let a and b be unit vectors codirected with rays OA and OB, let λ = OA and
µ = OB. Line AB consists of all points X such that
−−→
−→
−−→
OX = tOA + (1 − t)OB = tλa + (1 − t)µb.
We have to find numbers x0 and y0 such that xλ0 = t = 1 − yµ0 for all the considered values
p
q
of λ and µ. It remains to set x0 = pc and y0 = qc . As a result we see that if OA
+ OB
= c,
−−→ pa+qb
then line AB passes through a point X such that OX = c .
−−→
−−→
−−→
−−→
13.31. Let a = M A, b = M B and c = M C. Then e = M C 1 = pa + (1 − p)b and
−−→
M A1 = qc + (1 − q)b = −qa + (1 − 2q)b.
−−→
On the other hand, M A1 = αe. Similarly,
−−→
βe = M B 1 = −rb + (1 − 2r)a.
We have to show that 1 + α1 + β1 = 0. Since αpa + α(1 − p)b = αe = −qa + (1 − 2q)b,
it follows that αp = 1 − 2r and α(1 − p) = 1 − 2q and, therefore, α1 = 1 − 3p. Similarly,
βp = 1 − 2r and β(1 − p) = −r and, therefore, β1 = 3p − 2.
−→
−−→
−→
−−→
−−→
−−→
−
→
13.32. Summing up the equalities AA2 + BB 2 + CC 2 = 0 and A2 B 2 + B2 C 2 + C2 A2 =
−→
−−→
−→
−→
−−→ −−→
−−→
−
→
−
→
0 we get AB 2 + BC 2 + CA2 = 0 . It follows that AB 2 = λC2 B 2 , BC 2 = λA2 C 2 and
−→
−−→
−→
−→
CA2 = λB2 A2 . Let E be a point on line BC such that A2 E k AA1 . Then BA1 = λEA1 and
−
−→
−→
−−→ −−→ −→
−→
A1 C
= λ−1
EC = λEA1 ; hence, A1 C = EC − EA1 = (λ − 1)EA1 . Therefore, BA
. Similarly,
λ
1
AB1
B1 C
=
BC1
C1 A
=
λ−1
.
λ
−→
13.33. Let O be the center of the inscribed circle of the given quadrilateral, a = OA,
−−→
−−→
−→
−−→
b = OB, c = OC and d = OD. If Ha is the orthocenter of triangle BCD, then OH a =
b + c + d (cf. Problem 13.13). Therefore,
−−→
−−→
−−→
−−→
1
OM a = (a + b + c + d) = OM b = OM c = OM d .
2
13.34. Let O be the center of the circumscribed circle of the given quadrilateral; a =
−→
−−→
−→
−−→
OA, b = OB, c = OC and d = OD. If Hd is the orthocenter of triangle ABC, then
−−→
OH d = a + b + c (Problem 13.13).
−−→
a) Take a point K such that OK = a + b + c + d. Then
−−→ −−→
KHd = |OK − OH d | = |d| = R,
i.e., K belongs to circle Sd . We similarly prove that K belongs to circles Sa , Sb and Sc .
b) Let Od be the center of the circle of nine points of triangle ABC, i.e., the midpoint of
−→
−−→
−−→
OHd . Then OOd = OH d /2 = (a+b+c)/2. Take point X such that OX = (a+b+c+d)/2.
Then XOd = 12 |d| = 21 R, i.e., X belongs to the circle of nine points of triangle ABC. We
SOLUTIONS
301
similarly prove that X belongs to the circles of nine points of triangles BCD, CDA and
DAB.
−→
−−→
−→
13.35. Let X1 be the projection of X on l. Vector αAA1 +β BB 1 +γ CC 1 is the projection
−−→
−−→
−−→
of vector αAX 1 + β BX 1 + γ CX 1 to a line perpendicular to l. Since
−−→
−−→
−−→
−−→
−−→
−−→
−−→
αAX 1 + β BX 1 + γ CX 1 = αAX + β BX + γ CX + (α + β + γ)XX 1
−−→
−−→
−−→ −
→
and αAX + β BX + γ CX = 0 (by Problem 13.29), we get the statement required.
−−→
−−→
−−−−−−→
(?)13.36. Let a = A1 A2 + A3 A4 + · · · + A2n−1 A2n and a 6= 0. Introduce the coordinate
−−−→
system directing the Ox-axis along vector a. Since the sum of projections of vectors A1 A2 ,
−−−→
−−−−−−→
A3 A4 , . . . , A2n−1 A2n on Oy is zero, it follows that the length of a is equal to the absolute
value of the difference between the sum of the lengths of positive projections of these vectors
to the Ox-axis and the sum of lengths of their negative projections.
Therefore, the length of a does not exceed either the sum of the lengths of the positive
projections or the sum of the lengths of the negative projections.
It is easy to verify that the sum of the lengths of positive projections as well as the sum
of the lengths of negative projections of the given vectors on any axis does not exceed the
diameter of the circle, i.e., does not exceed 2.
13.37. In the proof of the equality of vectors it suffices to verify the equality of their
projections (minding the sign) on lines BC, CA and AB. Let us carry out the proof, for
example, for the projections on line BC, where the direction of ray BC will be assumed to
be the positive one. Let P be the projection of point A on line BC and N the midpoint of
BC. Then
−−→ −→ −−→ b2 + a2 − c2 a
b2 − c 2
P N = P C + CN =
− =
2a
2
2a
2
2
2
2
(P C is found from the equation AB − BP = AC − CP ). Since N M : N A = 1 : 3,
−−→
−−→
2 −c2
the projection of M O on line BC is equal to 31 P N = b 6a
. It remains to notice that the
projection of vector a3 na + b3 nb + c3 nc on BC is equal to
abc b2 − c2
b2 − c 2
b3 c − c 3 b
=
·
= 2S
.
b sin γ − c sin β =
2R
2R
a
a
13.38. Let the inscribed circle be tangent to sides AB, BC and CA at points U , V and
−−→
−→
−→ −→ −−→
−→
ZK, i.e., OZ = Rr (ZU + ZV + ZW ). Let
W , respectively. We have to prove that OZ = 3R
r
us prove, for example, that the (oriented) projections of these vectors on line BC are equal;
the direction of ray BC will be assumed to be the positive one.
Let N be the projection of point O on line BC. Then the projection of vector OZ on
line BC is equal to
3
3
−−→ −−→ −−→
(a + b − c)
(c − b)
a
=
.
N V = N C + CV = ( ) −
2
2
2
−→ −→ −−→
The projection of vector ZU + ZV + ZW on this line is equal to the projection of vector
−→ −−→
ZU + ZW , i.e., it is equal to
r(c − b)
.
2R
13.39. Introduce the coordinate system Oxy. Let lϕ be the straight line through O and
constituting an angle of ϕ (0 < ϕ < π) with the Ox-axis, i.e., if point A belongs to lϕ and
the second coordinate of A is positive, then ∠AOX = ϕ; in particular, l0 = lπ = Ox.
−r sin V ZU + r sin V ZW = −r sin B + r sin C =
302
CHAPTER 13. VECTORS
If vector a forms an angle of α with the Ox-axis (the angle is counted counterclockwise
from the Ox-axis to the vectorRa), then the length of the projection of a on lϕ is equal to
π
|a| · | cos(ϕ − α)|. The integral o |a| · | cos(ϕ − α)|dϕ = 2|a| does not depend on α.
Let vectors a1 , . . . , an ; b1 , . . . , bm constitute angles of α1 , . . . , αn ; β1 , . . . , βn , respectively, with the Ox-axis. Then by the hypothesis
|a1 | · | cos(ϕ − α1 )| + · · · + |an | · | cos(ϕ − αn )| ≤
|b1 | · | cos(ϕ − β1 )| + · · · + |bm | · | cos(ϕ − βm )|
for any ϕ. Integrating these inequalities over ϕ from 0 to π we get
|a1 | + · · · + |an | ≤ |b1 | + · · · + |bm |.
Rb
1
f (x)dx is called the mean value of the function f on the
Remark. The value b−a
a
segment [a, b]. The equality
Z π
|a| · | cos(ϕ − α)|dϕ = 2|a|
0
means that the mean value of the length of the projection of vector a is equal to π2 |a|; more
precisely, the mean value of the function f (ϕ) equal to the length of the projection of a to
lϕ on the segment [0, π] is equal to π2 |a|.
13.40. The sum of the lengths of the projections of a convex polygon on any line is equal
to twice the length of the projection of the polygon on this line. Therefore, the sum of the
lengths of the projections of vectors formed by edges on any line is not longer for the inner
polygon than for the outer one. Hence, by Problem 13.39 the sum of the lengths of vectors
formed by the sides, i.e., the perimeter of the inner polygon, is not longer than that of the
outer one.
13.41. If the sum of the lengths of vectors is equal to L, then by Remark to Problem
13.39 the mean value of the sum of the lengths of projections of these vectors is equal to
2L/π.
The value of function f on segment [a, b] cannot be always less than its mean value c
because otherwise
Z b
1
(b − a)c
c=
= c.
f (x)dx <
a−b a
b−a
Therefore, there exists a line l such that the sum of the lengths of the projections of the
initial vectors on l is not shorter than 2L/π.
On l, select a direction. Then either the sum of the lengths of the positive projections
to this directed line or the sum of the lengths of the negative projections is not shorter than
L/π. Therefore, either the length of the sum of vectors with positive projections or the
length of the sum of vectors with negative porjections is not shorter than L/π.
13.42. Let AB denote the projection of the polygon on line l. Clearly, points A and B
are projections of certain vertices A1 and B1 of the polygon. Therefore, A1 B1 ≥ AB, i.e.,
the length of the projection of the polygon is not longer than A1 B1 and A1 B1 < d by the
hypothesis. Since the sum of the lengths of the projections of the sides of the polygon on l
is equal to 2AB, it does not exceed 2d.
The mean value of the sum of the lengths of the projections of sides is equal to π2 P ,
where P is a perimeter (see Problem 13.39). The mean value does not exceed the maximal
one; hence, π2 P < 2d, i.e., P < πd.
SOLUTIONS
303
13.43. By Problem 13.39 it suffices to prove the inequality
|a| + |b| + |c| + |d| ≥ |a + d| + |b + d| + |c + d|
for the projections of the vectors on a line, i.e., we may assume that a, b, c and d are vectors
parallel to one line, i.e., they are just numbers such that a + b + c + d = 0. Let us assume
that d ≥ 0 because otherwise we can change the sign of all the numbers.
We can assume that a ≤ b ≤ c. We have to consider three cases:
1) a, b, c ≤ 0;
2) a ≤ 0 and b, c ≥ 0;
3) a, b ≤ 0, c ≥ 0.
All arising inequalities are quite easy to verify. In the third case we have to consider
separately the subcases |d| ≤ |b|, |b| ≤ |d| ≤ |a| and |a| ≤ |d| (in the last subcase we have to
take into account that |d| = |a| + |b| − |c| ≤ |a| + |b|).
13.44. By Problem 13.39 it suffices to prove the inequality for the projections of vectors
−→
−→
on any line. Let the projections of OA1 , . . . , OAn on a line l be equal (up to a sign) to
a1 , . . . , an . Let us divide the numbers a1 , . . . , an into two groups: x1 ≥ x2 ≥ · · · ≥ xk > 0
′
and y1′ ≤ y2′ ≤ · · · ≤ yn−k
≤ 0. Let yi = −yi′ . Then x1 + · · · + xk = y1 + · · · + yn−k = a and,
a
a
therefore, x1 ≥ k and y1 ≥ n−k
. To the perimeter the number 2(x1 + y1 ) in the projection
−→
corresponds. To the sum of the vectors OAi the number x1 + · · · + xk + y1 + · · · + yn−k = 2a
in the projection corresponds. And since
2(x1 + y1 )
n
2((a/k) + (a/(n − k)))
=
,
≥
x1 + · · · + yn−k
2a
k(n − k)
it remains to notice that the quantity k(n − k) is maximal for k = n/2 if n is even and for
k = (n ± 1)/2 if n is odd.
13.45. By definition the length of a curve is the limit of perimeters of the polygons
inscribed in it. [Vo vvedenie]
Consider an inscribed polygon with perimeter P and let the length of the projection on
line l be equal to di . Let 1 − ε < di < 1 for all lines l. The polygon can be selected so
that ε is however small. Since the polygon is a convex one, the sum of the lengths of the
projections of its sides on l is equal to 2di .
By Problem 13.39 the mean value of the quantity 2di is equal to π2 P (cf. Problem 13.39)
and, therefore, 2 − 2ε < π2 P < 2, i.e., π − πε < P < π. Tending ε to zero we see that the
length of the curve is equal to π.
13.46. Let us prove that the perimeter of the convex hull of all the vertices of given
polygons does not exceed the sum of their perimeters. To this end it suffices to notice that
by the hypothesis the projections of given polygons to any line cover the projection of the
convex hull.
13.47. a) If λ < 0, then
(λa) ∨ b = −λ|a| · |b| sin ∠(−a, b) = λ|a| · |a| sin ∠(a, b) = λ(a ∨ b).
For λ > 0 the proof is obvious.
−→
−−→
−→
b) Let a = OA, b = OB and c = OC. Introduce the coordinate system directing the
Oy-axis along ray OA. Let A = (0, y1 ), B = (x2 , y2 ) and C = (x3 , y3 ). Then
a ∨ b = x 2 y1 , a ∨ c = x 3 y1 ;
a ∨ (b + c) = (x2 + x3 )y1 = a ∨ b + a ∨ c.
13.48. Let e1 and e2 be unit vectors directed along the axes Ox and Oy. Then e1 ∨ e2 =
−e2 ∨ e1 = 1 and e1 ∨ e1 = e2 ∨ e2 = 0; hence,
a ∨ b = (a1 e1 + a2 e2 ) ∨ (b1 e1 + b2 e2 ) = a1 b2 − a2 b1 .
304
CHAPTER 13. VECTORS
13.49. a) Clearly,
−→ −→ −→ −→ −−→
−→ −−→ −−→ −→
AB ∨ AC = AB ∨ (AB + BC) = −BA ∨ BC = BC ∨ BA.
b) In the proof it suffices to make use of the chain of inequalities
−→ −→
−−→ −−→
−−→ −−→
AB ∨ AC = (AD + DB) ∨ (AD + DC) =
−−→ −−→ −−→ −−→ −−→ −−→
AD ∨ DC + DB ∨ AD + DB ∨ DC =
−−→ −−→ −−→ −−→ −−→ −−→
= DC ∨ DA + DA ∨ DB + DB ∨ DC.
−→
−→
13.50. Let at the initial moment, i.e., at t = 0 we have AB = v and AC = w. Then at
−→
−→
the moment t we get AB = v + t(a − b) and AC = w + t(c − a), where a, b and c are the
velocity vectors of the runners A, B and C, respectively. Since vectors a, b and c are parallel,
−→ −→
it follows that (b − a) ∨ (c − a) = 0 and, therefore, |S(A, B, C)| = 21 |AB ∨ AC| = |x + ty|,
where x and y are some constants.
Solving the system |x| = 2, |x + 5y| = 3 we get two solutions with the help of which we
express the dependence of the area of triangle ABC of time t as |2 + 5t | or |2 − t|. Therefore,
at t = 10 the value of the area can be either 4 or 8.
13.51. Let v(t) and w(t) be the vectors directed from the first pedestrian to the second
and the third ones, respectively, at time t. Clearly, v(t) = ta + b and w(t) = tc + d. The
pedestrians are on the same line if and only if v(t) k w(t), i.e., v(t) ∨ w(t) = 0. The function
f (t) = v(t) ∨ w(t) = t2 a ∨ c + t(a ∨ d + b ∨ c) + b ∨ d
is a quadratic and f (0) 6= 0. We know that a quadratic not identically equal to zero has not
more than 2 roots.
−→
−−→
−−→
−→
13.52. Let OC = a, OB = λa, OD = b and OA = µb. Then
−→ −→ a + µb λa + b
1 − λµ
±2SOP Q = OP ∨ OQ =
∨
=
(a ∨ b)
2
2
4
and
±SABCD = ±2(SCOD − SAOB ) = ±(a ∨ b − λa ∨ µb) = ±(1 − λµ)a ∨ b.
−−→
13.53. Let aj = P1 Aj . Then the doubled sum of the areas of the given triangles is equal
for any inner point P to
(x + a1 ) ∨ (x + a2 ) + (x + a3 ) ∨ (x + a4 ) + · · · + (x + a2n−1 ) ∨ (x + a2n ),
−→
where x = P P 1 and it differs from the doubled sum of the areas of these triangles for point
P1 by
x ∨ (a1 − a2 + a3 − a4 + · · · + a2n−1 − a2n ) = x ∨ a.
−−→
−−→
By the hypothesis x ∨ a = 0 for x = P1 P 1 and x = P3 P 1 and these vectors are not
parallel. Hence, a = 0, i.e., x ∨ a = 0 for any x.
−→
−−→
−→
−→
−→
−−→
13.54. Let a = AP , b = BQ and c = CR. Then QC = αa, RA = βb and P B = γc;
we additionally have
(1 + α)a + (1 + β)b + (1 + γ)c = 0.
−→ −→ −→ −→
It suffices to verify that AB ∨ CA = P Q ∨ RP . The difference between these quantities is
equal to
(a + γc) ∨ (c + βb) − (γc + b) ∨ (a + βb) = a ∨ c + βa ∨ b + a ∨ b + γa ∨ c =
= a ∨ [(1 + γ)c + (1 + β)b] = −a ∨ (1 + α)a = 0.
SOLUTIONS
305
−−→
−−→
13.55. Let ai = A4 Ai and wi = A4 H i . By Problem 13.49 b) it suffices to verify that
a1 ∨ a2 + a2 ∨ a3 + a3 ∨ a1 = w1 ∨ w2 + w2 ∨ w3 + w3 ∨ w1 .
Vectors a1 − w2 and a2 − w1 are perpendicular to vector a3 and, therefore, they are parallel
to each other, i.e., (a1 − w2 ) ∨ (a2 − w1 ) = 0. Adding this equality to the equalities
(a2 − w3 ) ∨ (a3 − w2 ) = 0 and (a3 − w1 ) ∨ (a1 − w3 ) = 0 we get the statement required.
13.56. Let x = x1 e1 + x2 e2 . Then e1 ∨ x = x2 (e1 ∨ e2 ) and x ∨ e2 = x1 (e1 ∨ e2 ), i.e.,
(x ∨ e2 )e1 + (e1 ∨ x)e2
.
x=
e1 ∨ e2
Multiplying this expression by (e1 ∨ e2 )y from the right we get
(1)
(x ∨ e2 )(e1 ∨ y) + (e1 ∨ x)(e2 ∨ y) + (e2 ∨ e1 )(x ∨ y) = 0.
−→
−→
−−→
−→
Let e1 = AB, e2 = AC, x = AD and y = AE. Then
i.e.,
S = a + x ∨ e2 + d = c + y ∨ e2 + a = d + x ∨ e1 + b,
x ∨ e2 = S − a − d, y ∨ e2 = S − c − a
and x ∨ e1 = S − d − b. Substituting these expressions into (1) we get the statement required.
Chapter 14. THE CENTER OF MASS
Background
1. Consider a system of mass points on a plane, i.e., there is a set of pairs (Xi , mi ), where
Xi is a point on the plane and mi a positive number. The center of mass of the system of
points X1 , . . . , Xn with masses m1 , . . . , mn , respectively, is a point, O, which satisfies
−−→
−−→ −
→
m1 OX1 + · · · + mn OXn = 0 .
The center of mass of any system of points exists and is unique (Problem 14.1).
2. A careful study of the solution of Problem 14.1 reveals that the positivity of the
numbers mi is not actually used; it is only important that their sum is nonzero. Sometimes
it is convenient to consider systems of points for which certain masses are positive and certain
are negative (but the sum of masses is nonzero).
3. The most important property of the center of mass which lies in the base of almost
all its applications is the following
Theorem on mass regroupping. The center of mass of a system of points does not
change if part of the points are replaced by one point situated in their center of mass and
whose mass is equal to the sum of their masses (Problem 14.2).
4. The moment of inertia of a system of points X1 , . . . , Xn with masses m1 , . . . , mn
with respect to point M is the number
IM = m1 M X12 + · · · + mn M Xn2 .
The applications of this notion in geometry are based on the relation IM = IO + mOM 2 ,
where O is the center of mass of a system and m = m1 + · · · + mn (Problem 14.17).
§1. Main properties of the center of mass
14.1. a) Prove that the center of mass exists and is unique for any system of points.
b) Prove that if X is an arbitrary point and O the center of mass of points X1 , . . . , Xn
with masses m1 , . . . , mn , then
−−→
XO =
−−→
−−−→
1
(m1 XX1 + · · · + mn XXn ).
m1 + · · · + m n
14.2. Prove that the center of mass of the system of points X1 , . . . , Xn , Y1 , . . . , Ym with
masses a1 , . . . , an , b1 , . . . , bm coincides with the center of mass of two points — the center
of mass X of the first system with mass a1 + · · · + an and the center of mass Y of the second
system with mass b1 + · · · + bm .
14.3. Prove that the center of mass of points A and B with masses a and b belongs to
segment AB and divides it in the ratio of b : a.
307
308
CHAPTER 14. THE CENTER OF MASS
§2. A theorem on mass regroupping
14.4. Prove that the medians of triangle ABC intersect at one point and are divided by
it in the ratio of 2 : 1 counting from the vertices.
14.5. Let ABCD be a convex quadrilateral; let K, L, M and N be the midpoints of
sides AB, BC, CD and DA, respectively. Prove that the intersection point of segments
KM and LN is the midpoint of these segments and also the midpoint of the segment that
connects the midpoints of the diagonals.
14.6. Let A1 , B1 , . . . , F1 be the midpoints of sides AB, BC, . . . , F A, respectively, of a
hexagon. Prove that the intersection points of the medians of triangles A1 C1 E1 and B1 D1 F1
coincide.
14.7. Prove Ceva’s theorem (Problem 4.48 b)) with the help of mass regrouping.
14.8. On sides AB, BC, CD and DA of convex quadrilateral ABCD points K, L, M
and N , respectively, are taken so that AK : KB = DM : M C = α and BL : LC = AN :
N D = β. Let P be the intersection point of segments KL and LN . Prove that N P : P L = α
and KP : P M = β.
14.9. Inside triangle ABC find point O such that for any straight line through O,
AK
CL
intersecting AB at K and intersecting BC at L the equality p KB
+ q LB
= 1 holds, where p
and q are given positive numbers.
14.10. Three flies of equal mass crawl along the sides of triangle ABC so that the center
of their mass is fixed. Prove that the center of their mass coincides with the intersection
point of medians of ABC if it is known that one fly had crawled along the whole boundary
of the triangle.
14.11. On sides AB, BC and CA of triangle ABC, points C1 , A1 and B1 , respectively,
are taken so that straight lines CC1 , AA1 and BB1 intersect at point O. Prove that
CA1
CO
1
=A
+ CB
;
a) OC
B1 A
1
1B
BO
CO
AO
BO
CO
AO
+ OB
+ OC
+ 2 ≥ 8.
b) OA1 · OB1 · OC1 = OA
1
1
1
14.12. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 , respectively,
1
1
1
= CB
= CAC
. Prove that the centers of mass of triangles ABC and
are taken so that BA
A1 C
B1 A
1B
A1 B1 C1 coincide.
14.13. On a circle, n points are given. Through the center of mass of n − 2 points a
straight line is drawn perpendicularly to the chord that connects the two remaining points.
Prove that all such straight lines intersect at one point.
14.14. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 , respectively,
are taken so that segments AA1 , BB1 and CC1 intersect at point P . Let la , lb , lc be the
lines that connect the midpoints of segments BC and B1 C1 , CA and C1 A1 , AB and A1 B1 ,
respectively. Prove that lines la , lb and lc intersect at one point and this point belongs to
segment P M , where M is the center of mass of triangle ABC.
14.15. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 , respectively,
are taken; straight lines B1 C1 , BB1 and CC1 intersect straight line AA1 at points M , P and
Q, respectively. Prove that:
1Q
a) AM1 M
= AP1AP + AQA
;
A
1
1
: CB
.
b) if P = Q, then M C1 : M B1 = BC
AB
AC
14.16. On line AB points P and P1 are taken and on line AC points Q and Q1 are
taken. The line that connects point A with the intersection point of lines P Q and P1 Q1
§3. THE MOMENT OF INERTIA
309
intersects line BC at point D. Prove that
BD
=
CD
BP
PA
CQ
QA
−
−
BP1
P1 A
CQ1
Q1 A
.
§3. The moment of inertia
For point M and a system of mass points X1 , . . . , Xn with masses m1 , . . . , mn the
quantity IM = m1 M X12 + · · · + mn M Xn2 is called the moment of inertia with respect to M .
14.17. Let O be the center of mass of a system of points whose sum of masses is equal
to m. Prove that the moments of inertia of this system with respect to O and with respect
to an arbitrary point X are related as follows: IX = IO + mXO2 .
14.18. a) Prove that the moment of
Pinertia with respect to the center of mass of a system
of points of unit masses is equal to n1 i<j a2ij , where n is the number of points and aij the
distance between points whose indices are i and j.
b) Prove that the moment of inertia with respect
of mass of a system of
P to the center
1
2
points whose masses are m1 , . . . , mn is equal to m i<j mi mj aij , where m = m1 + · · · + mn
and aij is the distance between the points whose indices are i and j.
14.19. a) Triangle ABC is an equilateral one. Find the locus of points X such that
AX 2 = BX 2 + CX 2 .
b) Prove that for the points of the locus described in heading a) the pedal triangle with
respect to the triangle ABC is a right one.
14.20. Let O be the center of the circumscribed circle of triangle ABC and H the
intersection point of the heights of triangle ABC. Prove that a2 + b2 + c2 = 9R2 − OH 2 .
14.21. Chords AA1 , BB1 and CC1 in a disc with center O intersect at point X. Prove
that
BX
CX
AX
+
+
=3
XA1 XB1 XC1
if and only if point X belongs to the circle with diameter OM , where M is the center of
mass of triangle ABC.
14.22. On sides AB, BC, CA of triangle ABC pairs of points A1 and B2 , B1 and C2 ,
C1 and A2 , respectively, are taken so that segments A1 A2 , B1 B2 and C1 C2 are parallel to
the sides of triangle ABC and intersect at point P . Prove that
P A1 · P A2 + P B1 · P B2 + P C1 · P C2 = R2 − OP 2 ,
where O is the center of the circumscribed circle.
14.23. Inside a circle of radius R, consider n points. Prove that the sum of squares of
the pairwise distances between the points does not exceed n2 R2 .
14.24. Inside triangle ABC point P is taken. Let da , db and dc be the distances from P
to the sides of the triangle; Ra , Rb and Rc the distances from P to the vertices. Prove that
3(d2a + d2b + d2c ) ≥ (Ra sin A)2 + (Rb sin B)2 + (Rc sin C)2 .
14.25. Points A1 , . . . , An belong to the same circle and M is their center of mass. Lines
M A1 , . . . , M An intersect this circle at points B1 , . . . , Bn (distinct from A1 , . . . , An ). Prove
that
M A 1 + · · · + M A n ≤ M B1 + · · · + M B n .
310
CHAPTER 14. THE CENTER OF MASS
§4. Miscellaneous problems
14.26. Prove that if a polygon has several axes of symmetry, then all of them intersect
at one point.
14.27. A centrally symmetric figure on a graph paper consists of n “corners” and k
rectangles of size 1 × 4 depicted on Fig. 145. Prove that n is even.
Figure 145 (14.27)
14.28. Solve Problem 13.44 making use the properties of the center of mass.
14.29. On sides BC and CD of parallelogram ABCD points K and L, respectively, are
taken so that BK : KC = CL : LD. Prove that the center of mass of triangle AKL belongs
to diagonal BD.
§5. The barycentric coordinates
Consider triangle A1 A2 A3 whose vertices are mass points with masses m1 , m2 and m3 ,
respectively. If point X is the center of mass of the triangle’s vertices, then the triple
(m1 : m2 : m3 ) is called the barycentric coordinates of point X with respect to triangle
A1 A2 A3 .
14.30. Consider triangle A1 A2 A3 . Prove that
a) any point X has some barycentric coordinates with respect to △A1 A2 A3 ;
b) provided m1 + m2 + m3 = 1 the barycentric coordinates of X are uniquely defined.
14.31. Prove that the barycentric coordinates with respect to △ABC of point X which
belongs to the interior of ABC are equal to (SBCX : SCAX : SABX ).
14.32. Point X belongs to the interior of triangle ABC. The straight lines through
X parallel to AC and BC intersect AB at points K and L, respectively. Prove that the
barycentric coordinates of X with respect to △ABC are equal to (BL : AK : LK).
14.33. Consider △ABC. Find the barycentric coordinates with respect to △ABC of
a) the center of the circumscribed circle;
b) the center of the inscribed circle;
c) the orthocenter of the triangle.
14.34. The baricentric coordinates of point X with respect to △ABC are (α : β : γ),
−−→
−→
−→
where α + β + γ = 1. Prove that XA = β BA + γ CA.
14.35. Let (α : β : γ) be the barycentric coordinates of point X with respect to △ABC
and α + β + γ = 1 and let M be the center of mass of triangle ABC. Prove that
−−→
−→
−−→
−→
3XM = (α − β)AB + (β − γ)BC + (γ − α)CA.
14.36. Let M be the center of mass of triangle ABC and X an arbitrary point. On
lines BC, CA and AB points A1 , B1 and C1 , respectively, are taken so that A1 X k AM ,
B1 X k BM and C1 X k CM . Prove that the center of mass M1 of triangle A1 B1 C1 coincides
with the midpoint of segment M X.
14.37. Find an equation of the circumscribed circle of triangle A1 A2 A3 (kto sut’ indexy?
iz 14.36?) in the barycentric coordinates.
SOLUTIONS
311
14.38. a) Prove that the points whose barycentric coordinates with respect to △ABC
are (α : β : γ) and (α−1 : β −1 : γ −1 ) are isotomically conjugate with respect to triangle ABC.
b) The lengths of the sides of triangle ABC are equal to a, b and c. Prove that the points
2
2
2
whose barycentric coordinates with respect to △ABC are (α : β : γ) and ( aα : bβ : cγ ) are
isogonally conjugate with respect to ABC.
Solutions
14.1. Let X and O be arbitrary points. Then
−−→
−−→
m1 OX1 + · · · + mn OXn =
−−→
−−→
−−−→
(m1 + · · · + mn )OX + m1 XX1 + · · · + mn XXn
and, therefore, O is the center of mass of the given system of points if and only if
−−→
−−→
−−−→ −
→
(m1 + · · · + mn )OX + m1 XX1 + · · · + Mn XXn = 0 ,
−−→
−−−→
−−→
1
(m
XX
+
·
·
·
+
m
XXn ).
i.e., OX = m1 +···+m
1
1
n
n
This argument gives a solution to the problems of both headings.
14.2. Let Z be an arbitrary point; a = a1 + · · · + an and b = b1 + · · · + bm . Then
−
−−
→
−−→
−
−−
→
−−→
−−→
−→
b1 ZY1 +···+bm ZYm
n ZXn
ZX = a1 ZX1 +···+a
and
ZY
=
. If O is the center of mass of point X
a
b
whose mass is a and of point Y whose mass is b, then
−−→
−→
−→ aZX + bZY
=
ZO =
a+b
−−→
−−→
−−→
−−→
a1 ZX1 + · · · + an ZXn + b1 ZY1 + · · · + bm ZYm
,
a+b
i.e., O is the center of mass of the system of points X1 , . . . , Xn and Y1 , . . . , Ym with masses
a1 , . . . , a n , b 1 , . . . , b m .
−→
−−→
−
→
14.3. Let O be the center of mass of the given system. Then aOA + bOB = 0 and,
therefore, O belongs to segment AB and aOA = bOB, i.e., AO : OB = b : a.
14.4. Let us place unit masses at points A, B and C. Let O be the center of mass
of this system of points. Point O is also the center of mass of points A of mass 1 and A1
of mass 2, where A1 is the center of mass of points B and C of unit mass, i.e., A1 is the
midpoint of segment BC. Therefore, O belongs to median AA1 and divides it in the ratio
AO : OA1 = 2 : 1. We similarly prove that the remaining medians pass through O and are
divided by it in the ratio of 2 : 1.
14.5. Let us place unit masses in the vertices of quadrilateral ABCD. Let O be the
center of mass of this system of points. It suffices to prove that O is the midpoint of segments
KM and LN and the midpoint of the segment connecting the midpoints of the diagonals.
Clearly, K is the center of mass of points A and B while M is the center of mass of points
C and D. Therefore, O is the center of mass of points K and M of mass 2, i.e., O is the
center of mass of segment KM .
Similarly, O is the midpoint of segment LN . Considering centers of mass of pairs of
points (A, C) and (B, D) (i.e., the midpoints of diagonals) we see that O is the midpoint of
the segment connecting the midpoints of diagonals.
312
CHAPTER 14. THE CENTER OF MASS
14.6. Let us place unit masses in the vertices of the hexagon; let O be the center of
mass of the obtained system of points. Since points A1 , C1 and E1 are the centers of mass of
pairs of points (A, B), (C, D) and (E, F ), respectively, point O is the center of mass of the
system of points A1 , C1 and E1 of mass 2, i.e., O is the intersection point of the medians of
triangle A1 C1 E1 (cf. the solution of Problem 14.4).
We similarly prove that O is the intersection point of medians of triangle B1 D1 F1 .
14.7. Let lines AA1 and CC1 intersect at O and let AC1 : C1 B = p and BA1 : A1 C = q.
We have to prove that line BB1 passes through O if and only if CB1 : B1 A = 1 : pq.
Place masses 1, p and pq at points A, B and C, respectively. Then point C1 is the center
of mass of points A and B and point A1 is the center of mass of points B and C. Therefore,
the center of mass of points A, B and C with given masses is the intersection point O of
lines CC1 and AA1 .
On the other hand, O belongs to the segment which connects B with the center of mass
of points A and C. If B1 is the center of mass of points A and C of masses 1 and pq,
respectively, then AB1 : B1 C = pq : 1. It remains to notice that there is one point on
segment AC which divides it in the given ratio AB1 : B1 C.
14.8. Let us place masses 1, α, αβ and β at points A, B, C and D, respectively. Then
points K, L, M and N are the centers of mass of the pairs of points (A, B), (B, C), (C, D)
and (D, A), respectively. Let O be the center of mass of points A, B, C and D of indicated
mass. Then O belongs to segment N L and N O : OL = (αβ + α) : (1 + β) = α. Point O
belongs to the segment KM and KO : OM = (β + αβ) : (1 + α) = β. Therefore, O is the
intersection point of segments KM and LN , i.e., O = P and N P : P L = N O : OL = α,
KP : P M = β.
14.9. Let us place masses p, 1 and q in vertices A, B and C, respectively. Let O be the
center of mass of this system of points. Let us consider a point of mass 1 as two coinciding
points of mass xa and xc , where xa + xc = 1. Let K be the center of mass of points A and B
of mass p and xa and L the center of mass of points C and B of mass q and xc , respectively.
Then AK : KB = xa : p and CL : LB = xc : q, whereas point O which is the center of mass
of points K and L of mass p + xa and q + xc , respectively, belongs to line KL. By varying
xa from 0 to 1 we get two straight lines passing through O and intersecting sides AB and
BC. Therefore, for all these lines we have
pAK qCL
+
= xa + xc = 1.
KB
LB
14.10. Denote the center of mass of the flies by O. Let one fly be sited in vertex A
and let A1 be the center of mass of the two other flies. Clearly, point A1 lies inside triangle
ABC and point O belongs to segment AA1 and divides it in the ratio of AO : OA1 = 2 : 1.
Therefore, point O belongs to the interior of the triangle obtained from triangle ABC by a
homothety with coefficient 23 and center A.
Considering such triangles for all the three vertices of triangle ABC we see that their
unique common point is the intersection point of the medians of triangle ABC. Since one
fly visited all the three vertices of the triangle ABC and point O was fixed during this, O
should belong to all these three small triangles, i.e., O coincides with the intersection point
of the medians of triangle ABC.
14.11. a) Let AB1 : B1 C = 1 : p and BA1 : A1 C = 1 : q. Let us place masses p, q, 1 at
points A, B, C, respectively. Then points A1 and B1 are the centers of mass of the pairs of
points (B, C) and (A, C), respectively. Therefore, the center of mass of the system of points
A, B and C belongs both to segment AA1 and to segment BB1 , i.e., coincides with O. It
SOLUTIONS
313
follows that C1 is the center of mass of points A and B. Therefore,
CO
CB1 CA1
+
.
=p+q =
OC1
B1 A A1 B
b) By heading a) we have
1+q 1+p p+q
AO BO CO
·
·
=
·
·
=
OA1 OB1 OC1
p
q
1
p q 1 1
AO
BO
CO
p+q+ + + + +2=
+
+
+ 2.
q p p q
OA1 OB1 OC1
It is also clear that
1
1
p q
≥ 2, q + ≥ 2 and + ≥ 2.
p
q
q p
14.12. Let M be the center of mass of triangle ABC. Then
−−→ −−→ −−→ −
→
MA + MB + MC = 0 .
p+
Moreover,
−−→ −−→ −−→
−→ −→ −−→
−
→
AB1 + BC1 + CA1 = k(AC + BA + CB) = 0 .
−−−→ −−−→ −−−→ −
→
Adding these identities we get M B1 + M C1 + M A1 = 0 , i.e., M is the center of mass of
triangle A1 B1 C1 .
Remark. We similarly prove a similar statement for an arbitrary n-gon.
14.13. Let M1 be the center of mass of n − 2 points; K the midpoint of the chord
connecting the two remaining points, O the center of the circle, and M the center of mass
of all the given points. If line OM intersects a(?) line drawn through M1 at point P , then
n−2
OM
KM
=
=
2
MP
M M1
and, therefore, the position of point P is uniquely determined by the position of points O
and M (if M = O, then P = O).
14.14. Let P be the center of mass of points A, B and C of masses a, b and c, respectively,
M the center of mass of points A, B and C (the mass of M is a + b + c) and Q the center of
mass of the union of these two systems of points. The midpoint of segment AB is the center
of mass of points A, B and C of mass a + b + c − abc , a + b + c − abc and 0, respectively, and
the midpoint of segment A1 B1 is the center of mass of points A, B and C of mass a(b+c)
,
c
b(a+c)
and (b + c) + (a + c), respectively. Point O is the center of mass of the union of these
c
systems of points.
14.15. a) Place masses β, γ and b + c in points B, C and A so that CA1 : BA1 = β : γ,
BC1 : AC1 = b : β and AB1 : CB1 = γ : c. Then M is the center of mass of this system
b+c
1M
and, therefore, AAM
= β+γ
. Point P is the center of mass of points A, B and C of masses c,
A1 P
c
b
1Q
β and γ and, therefore, P A = β+γ
. Similarly, AAQ
= b+γ
.
c+γ BC1
M C1
b
AC
b) As in heading a), we get M B1 = b+β , AB = b+β and CB
= c+γ
. Moreover, b = c
c
1
because straight lines AA1 , BB1 and CC1 intersect at one point (cf. Problem 14.7).
14.16. The intersection point of lines P Q and P1 Q1 is the center of mass of points A,
B and C of masses a, b and c and P is the center of mass of points A and B of masses a − x
314
CHAPTER 14. THE CENTER OF MASS
and b while Q is the center of mass of points A and C of masses x and c. Let p =
= xc . Then pb + qc = a. Similarly, p1 b + q1 c = a. It follows that
and q = CQ
QA
BP
PA
=
a−x
b
c
BD
(p − p1 )
=− =
.
b
(q − q1 )
CD
14.17. Let us enumerate the points of the given system. Let xi be the vector
P with the
beginning at O and the end at the point of index i and of mass mi . Then
mi xi = 0.
−−→
Further, let a = OX. Then
P
IO = Pm2ii ,
P
P
P
IM = mi (xi + a)2 = mi x2i + 2( mi xi , a) + mi a2 = IO + ma2 .
14.18. a) Let xi be the vector with the beginning at the center of mass O and the end
at the point of index i. Then
X
X
X
(xi − xj )2 =
(x2i + x2j ) − 2
(xi , xj ),
i,j
i,j
i,j
where the sum runs over all the possible pairs of indices. Clearly,
X
X
X
X
X
x2i = 2nIO ;
(xi , xj ) =
(xi ,
xj ) = 0.
(x2i + x2j ) = 2n
i
i,j
i,j
i
j
Therefore, 2nIO = i,j (xi − xj )2 = 2 i<j a2ij .
b) Let xi be the vector with the beginning at the center of mass O and the end at the
point with index i. Then
X
X
X
mi mj (xi − xj )2 =
mi mj (x2i + x2j ) − 2
mi mj (xi , xj ).
P
P
i,j
i,j
i,j
It is clear that
X
X X
X
mi mj (x2i + x2j ) =
mi
(mj x2i + mj x2j ) =
mi (mx2i + IO ) = 2mIO
i,j
i
j
i
and
X
mi mj (xi , xj ) =
i,j
Therefore,
2mIO =
X
i
X
i,j
mi (xi ,
X
mj xj ) = 0.
j
mi mj (xi − xj )2 = 2
X
mi mj a2ij .
i<j
14.19. a) Let M be the point symmetric to A through line BC. Then M is the center
of mass of points A, B and C whose masses are −1, 1 and 1, respectively, and, therefore,
−AX 2 + BX 2 + CX 2 = IX = IM + (−1 + 1 + 1)M X 2 = (−3 + 1 + 1)a2 + M X 2 ,
where a is the length of the side of triangle ABC. As a result we see that the locus to be
found is the circle of radius a with the center at M .
b) Let A′ , B ′ and C ′ be the projections of point X to lines BC, CA and AB, respectively. Points B ′ and
C ′ belong to the circle √
with diameter AX and,
therefore, B ′ C ′ =
√
√
3
3
3
′ ′
′ ′
′
′
AX sin B AC = 2 AX. Similarly, C A = 2 BX and A B = 2 CX. Therefore, if
AX 2 = BX 2 + CX 2 , then ∠B ′ A′ C ′ = 90◦ .
SOLUTIONS
315
14.20. Let M be the center of mass of the vertices of triangle ABC with unit masses in
them. Then
1
IO = IM + 3M O2 = (a2 + b2 + c2 ) + 3M O2
3
(cf. Problems 14.17 and 14.18 a)). Since OA = OB = OC = R, it follows that IO = 3R2 . It
remains to notice that OH = 3OM (Problem 5.105).
14.21. It is clear that
AX
AX 2
AX 2
=
= 2
.
XA1
AX · XA1
R − OX 2
Therefore, we have to verify that AX 2 + BX 2 + CX 2 = 3(R2 − OX 2 ) if and only if OM 2 =
OX 2 + M X 2 . To this end it suffices to notice that
AX 2 + BX 2 + CX 2 = IX = IM + 3M X 2 =
IO − 3M O2 + 3M X 2 = 3(R2 − M O2 + M X 2 ).
14.22. Let P be the center of mass of points A, B and C whose masses are α, β and γ,
respectively. We may assume that α + β + γ = 1. If K is the intersection point of lines CP
and AB, then
CP + P K
CP
α+β
1
CK
BC
=
=1+
=1+
= .
=
P A1
PK
PK
PK
γ
γ
Similar arguments show that the considered quantity is equal to βγa2 +γαb2 +αβc2 = IP (cf.
Problem 14.18 b)). Since IO = αR2 + βR2 + γR2 = R2 , we have IP = IO − OP 2 = R2 − OP 2 .
14.23. Let us place unit masses in the given points. As follows from the result of Problem
14.18 a) the sum of squared distances between the given points is equal to nI, where I is the
moment of inertia of the system of points with respect to its center of mass. Now, consider
the moment of inertia of the system with respect to the center O of the circle. On the one
hand, I ≤ IO (see Problem 14.17). On the other hand, since the distance from O to any of
the given points does not exceed R, it follows that IO ≤ nR2 . Therefore, nI ≤ n2 R2 and the
equality is attained only if I = IO (i.e., when the center of mass coincides with the center of
the circle) and IO = nR2 (i.e., all the points lie on the given circle).
14.24. Let A1 , B1 and C1 be projections of point P to sides BC, CA and AB, respectively; let M be the center of mass of triangle A1 B1 C1 . Then
3(d2a + d2b + d2c ) = 3IP ≥
3IM = A1 B12 + B1 C12 + C1 A21 = (Rc sin C)2 + (Ra sin A)2 + (Rb sin B)2
because, for example, segment A1 B1 is a chord of the circle with diameter CP .
14.25. Let O be the center of the given circle. If chord AB passes through M , then
AM · BM = R2 − d2 , where d = M O. Denote by IX the moment of inertia of the system
of points A1 , . . . , An with respect to X. Then IO = IM + nd2 (see Problem 14.17). On the
other hand, since OAi = R, we deduce that IO = nR2 . Therefore,
1
Ai M · Bi M = R2 − d2 = (A1 M 2 + · · · + An M 2 ).
n
Set ai = Ai M . Then the inequality to be proved takes the form
1
1
1
a1 + · · · + an ≤ (a21 + · · · + a2n )( + · · · + ).
n
a1
an
316
CHAPTER 14. THE CENTER OF MASS
To prove this inequality we have to make use of the inequality
x+y ≤(
y2
x2
)+( )
y
x
which is obtained from the inequality xy ≤ x2 − xy + y 2 by multiplying both of its sides by
x+y
.
xy
14.26. Let us place unit masses in the vertices of the polygon. Under the symmetry
through a line this system of points turns into itself and, therefore, its center of mass also
turns into itself. It follows that all the axes of symmetry pass through the center of mass of
the vertices.
14.27. Let us place unit masses in the centers of the cells which form “corners” and
rectangles. Let us split each initial small cell of the graph paper into four smaller cells
getting as a result a new graph paper. It is easy to verify that now the center of mass of a
corner belongs to the center of a new small cell and the center of mass of a rectangle is a
vertex of a new small cell, cf. Fig. 146.
Figure 146 (Sol. 14.27)
It is clear that the center of mass of a figure coincides with its center of symmetry and
the center of symmetry of the figure consisting of the initial cells can only be situated in a
vertex of a new cell. Since the masses of corners and bars (rectangles) are equal, the sum
of vectors with the source in the center of mass of a figure and the targets in the centers of
mass of all the corners and bars is equal to zero. If the number of corners had been odd,
then the sum of the vectors would have had half integer coordinates and would have been
nonzero. Therefore, the number of corners is an even one.
14.28. Let us place unit masses in the vertices of the polygon A1 . . . An . Then O is the
−−−→
−−−→
−−→
center of mass of the given system of points. Therefore, Ai O = n1 (Ai A1 + · · · + Ai An ) and
Ai O ≤ n1 (Ai A1 + · · · + Ai An ); it follows that
n
1 X
d = A1 O + · · · + A n O ≤
Ai Aj .
n i,j=1
We can express the number n either in the form n = 2m or in the form n = 2m + 1. Let P
be the perimeter of the polygon. It is clear that
A1 A2 + · · · + An A1 = P,
A1 A3 + A2 A4 + · · · + An A2 ≤ 2P,
...........................
A1 Am+1 + A2 Am+2 + · · · + An Am ≤ mP
SOLUTIONS
317
and in the left-hand sidesP
of these inequalities all the sides and diagonals are encountered.
Since they enter the sum ni,j=1 Ai Aj twice, it is clear that
n
1 X
2
m(m + 1)
d≤
Ai Aj ≤ (P + 2P + · · · + mP ) =
P.
n i,j=1
n
n
For n even this inequality can be strengthened due to the fact that in this case every diagonal
occuring in the sum A1 Am+1 + · · · + An Am+n is counted twice, i.e., instead of mP we can
take m2 P . This means that for n even we have
d≤
m
m2
2
(P + 2P + · · · + (m − 1)P + P ) =
P.
n
2
n
Thus, we have
d≤
½
m2
P = n4 P
n
2
m(m+1)
P = n4n−1 P
n
if n is even
if n is odd.
DL
14.29. Let k = BK
= 1 − DC
. Under the projection to a line perpendicular to diagonal
BC
BD points A, B, K and L pass into points A′ , B ′ , K ′ and L′ , respectively, such that
B ′ K ′ + B ′ L′ = kA′ B ′ + (1 − k)A′ B ′ = A′ B ′ .
It follows that the center of mass of points A′ , K ′ and L′ coincides with B ′ . It remains to
notice that under the projection a center of mass turns into a center of mass.
−−−→
−−−→
−−→
14.30. Introduce the following notations: e1 = A3 A1 , e2 = A3 A2 and x = XA3 . Point
X is the center of mass of the vertices of triangle A1 A2 A3 with masses m1 , m2 , m3 attached
to them if and only if
m1 (x + e1 ) + m2 (x + e2 ) + m3 x = 0,
i.e., mx = −(m1 e1 +m2 e2 ), where m = m1 +m2 +m3 . Let us assume that m = 1. Any vector
x on the plane can be represented in the form x = −m1 e1 − m2 e2 , where the numbers m1
and m2 are uniquely defined. The number m3 is found from the relation m3 = 1 − m1 − m2 .
14.31. This problem is a reformulation of Problem 13.29.
Remark. If we assume that the areas of triangles BCX, CAX and ABX are oriented,
then the statement of the problem remains true for all the points situated outside the triangle
as well.
−−→
−−→
14.32. Under the projection to line AB parallel to line BC vector u = XA · BL + XB ·
−−→
−→
−→
−→
AK + XC · LK turns into vector LA · BL + LB · AK + LB · LK. The latter vector is the
−→ −−→ −−→
zero one since LA = LK + KA. Considering the projection to line AB parallel to line AC
we get u = 0.
14.33. Making use of the result of Problem 14.31 it is easy to verify that the answer is
as follows: a) (sin 2α : sin 2β : sin 2γ); b) (a : b : c); c) (tan α : tan β : tan γ).
−−→ −
−−→
−−→
−−→
→
14.34. Adding vector (β + γ)XA to both sides of the equality αXA + β XB + γ XC = 0
we get
−−→
−−→
−−→
−−→
−→
−→
XA = (β + γ)XA + β BX + γ CX = β BA + γ CA.
−−→
−−→ −−→ −−→
−−→
14.35. By Problem 14.1 b) we have 3XM = XA + XB + XC. Moreover, XA =
−→
−→ −−→
−→
−−→
−−→
−→
−−→
β BA + γ CA, XB = αAB + γ CB and XC = αAC + β BC (see Problem 14.34).
14.36. Let the lines through point X parallel to AC and BC intersect the line AB
at points K and L, respectively. If (α : β : γ) are the barycentric coordinates of X and
α + β + γ = 1, then
−−→ −−→ −−→
−→
−−→
2XC1 = XK + XL = γ CA + γ CB
318
CHAPTER 14. THE CENTER OF MASS
(see the solution of Problem 14.42). Therefore,
−−−→ −−→ −−→ −−→
3XM1 = XA1 + XB1 + XC1 =
−→ −→
−→ −−→
−→ −−→
−−→
1
(α(AB + AC) + β(BA + BC) + γ CA + CB) = 23 XM
2
(see Problem 14.35).
14.37. Let X be an arbitrary point, O the center of the circumscribed circle of the given
−−→
−−→
triangle, ei = OAi and a = XO. If the barycentric coordinates of X are (x1 : x2 : x3 ), then
P
P −−→
xi (a + ei ) =
xi XAi = 0P
because X P
is the center of mass of points A1 , A2 , A3 with
masses x1 , x2 , x3 . Therefore, ( xi )a = − xi ei .
Point X belongs to the circumscribed circle of the triangle if and only if |a| = XO = R,
where R is the radius of this circle. Thus, the circumscribed circle of the triangle is given in
the barycentric coordinates by the equation
X
X
R2 (
xi )2 = (
xi ei )2 ,
i.e.,
R
2
X
x2i + 2R2
X
xi xj = R 2
i<j
X
x2i + 2
X
xi xj (ei , ej )
i<j
because |ei | = R. This equation can be rewritten in the form
X
xi xj (R2 − (ei , ej )) = 0.
i<j
2
Now notice that 2(R − (ei , ej )) = a2ij , where aij is the length of side Ai Aj . Indeed,
a2ij = |ei − ej |2 = |ei |2 + |ej |2 − 2(ei , ej ) = 2(R2 − (ei , ej )).
As a result we see that the circumscribed
circle of triangle A1 A2 A3 is given in the barycentric
P
coordinates by the equation i<j xi xj aij = 0, where aij is the length of side Ai Aj .
14.38. a) Let X and Y be the points with barycentric coordinates (α : β : γ) and
−1
(α : β −1 : γ −1 ) and let lines CX and CY intersect line AB at points X1 and Y1 , respectively.
Then
AX1 : BX1 = β : α = α−1 : β −1 = BY1 : AY1 .
Similar arguments for lines AX and BX show that points X and Y are isotomically conjugate
with respect to triangle ABC.
b) Let X be the point with barycentric coordinates (α : β : γ). We may assume that
α + β + γ = 1. Then by Problem 14.34 we have
−→
−→
−−→
−→
−→
AB
AC
AX = β AB + γ AC = βc(
) + γb(
).
c
b
Let Y be the point symmetric to X through the bisector of angle ∠A and (α′ : β ′ : γ ′ ) the
2
2
barycentric coordinates of Y . It suffices to verify that β ′ : γ ′ = bβ : cγ . The symmetry
−→
−→
−→
through the bisector of angle ∠A interchanges unit vectors AB
and AC
, consequently, AY =
c
b
−→
−→
βc AC
+ γb AB
. It follows that
b
c
β′ : γ′ =
b2 c 2
γb
: βcb =
: .
c
β γ
Chapter 15. PARALLEL TRANSLATIONS
Background
−→
1. The parallel translation by vector AB is the transformation which sends point X into
−−→ −→
point X ′ such that XX ′ = AB.
2. The composition (i.e., the consecutive execution) of two parallel translations is, clearly,
a parallel translation.
Introductory problems
1. Prove that every parallel translation turns any circle into a circle.
2. Two circles of radius R are tangent at point K. On one of them we take point A,n
on the other one we take point B such that ∠AKB = 90◦ . Prove that AB = 2R.
3. Two circles of radius R intersect at points M and N . Let A and B be the intersection
points of these circles with the perpendicular erected at the midpoint of segment M N . It so
happens that the circles lie on one side of line M N . Prove that M N 2 + AB 2 = 4R2 .
4. Inside rectangle ABCD, point M is taken. Prove that there exists a convex quadrilateral with perpendicular diagonals of the same length as AB and BC whose sides are equal
to AM , BM , CM , DM .
§1. Solving problems with the aid of parallel translations
15.1. Where should we construct bridge M N through the river that separates villages
A and B so that the path AM N B from A to B was the shortest one? (The banks of the
river are assumed to be parallel lines and the bridge perpendicular to the banks.)
15.2. Consider triangle ABC. Point M inside the triangle moves parallel to side BC to
its intersection with side CA, then parallel to AB to its intersection with BC, then parallel to
AC to its intersection with AB, and so on. Prove that after a number of steps the trajectory
of the point becomes a closed one.
15.3. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively,
of convex quadrilateral ABCD.
a) Prove that KM ≤ 12 (BC + AD) and the equality is attained only if BC k AD.
b) For given lengths of the sides of quadrilateral ABCD find the maximal value of the
lengths of segments KM and LN .
15.4. In trapezoid ABCD, sides BC and AD are parallel, M the intersection point of
the bisectors of angles ∠A and ∠B, and N the intersection point of the bisectors of angles
∠C and ∠D. Prove that 2M N = |AB + CD − BC − AD|.
15.5. From vertex B of parallelogram ABCD heights BK and BH are drawn. It is
known that KH = a and BD = b. Find the distance from B to the intersection point of the
heights of triangle BKH.
15.6. In the unit square a figure is placed such that the distance between any two of its
points is not equal to 0.001. Prove that the area of this figure does not exceed a) 0.34; b)
0.287.
319
320
CHAPTER 15. PARALLEL TRANSLATIONS
§2. Problems on construction and loci
15.7. Consider angle ∠ABC and straight line l. Construct a line parallel to l on which
the legs of angle ∠ABC intercept a segment of given length a.
15.8. Consider two circles S1 , S2 and line l. Draw line l1 parallel to l so that:
a) the distance between the intersection points of l1 with circles S1 and S2 is of a given
value a;
b) S1 and S2 intercept on l1 equal chords;
c) S1 and S2 intercept on l1 chords the sum (or difference) of whose lengths is equal to
a given value.
15.9. Consider nonintersecting chords AB and CD on a circle. Construct a point X on
the circle so that chords AX and BX would intercept on chord CD a segment, EF , of a
given length a.
15.10. Construct quadrilateral ABCD given the quadrilateral’s angles and the lengths
of sides AB = a and CD = b.
15.11. Given point A and circles S1 and S2 . Through A draw line l so that S1 and S2
intercept on l equal chords.
15.12. a) Given circles S1 and S2 intersect at points A and B. Through point A draw
line l so that the intercept of this line between circles S1 and S2 were of a given length.
b) Consider triangle ABC and triangle P QR. In triangle ABC inscribe a triangle equal
to P QR.
15.13. Construct a quadrilateral given its angles and diagonals.
***
15.14. Find the loci of the points for which the following value is given: a) the sum, b)
the difference of the distances from these points to the two given straight lines.
15.15. An angle made of a transparent material moves so that two nonintersecting circles
are tangent to its legs from the inside. Prove that on the angle a point circumscribing an
arc of a circle can be marked.
Problems for independent study
15.16. Consider two pairs of parallel lines and point P . Through P draw a line on which
both pairs of parallel lines intercept equal segments.
15.17. Construct a parallelogram given its sides and an angle between the diagonals.
15.18. In convex quadrilateral ABCD, sides AB and CD are equal. Prove that
a) lines AB and CD form equal angles with the line that connects the midpoints of sides
AC and BD;
b) lines AB and CD form equal angles with the line that connects the midpoints of
diagonals BC and AD.
15.19. Among all the quadrilaterals with given lengths of the diagonals and an angle
between them find the one of the least perimeter.
15.20. Given a circle and two neighbouring vertices of a parallelogram. Construct the
parallelogram if it is known that its other two (not given) vertices belong to the given circle.
Solutions
−−→
15.1. Let A′ be the image of point A under the parallel translation by M N . Then
A′ N = AM and, therefore, the length of path AM N B is equal to A′ N + N B + M N .
Since the length of segment M N is a constant, we have to find point N for which the sum
SOLUTIONS
321
A′ N + N B is the least one. It is clear that the sum is minimal if N belongs to segment A′ B,
i.e., N is the closest to B intersection point of the bank and segment A′ B.
Figure 147 (Sol. 15.2)
15.2. Denote the consecutive points of the trajectory on the sides of the triangle as on
Fig. 147:
A1 , B1 , B2 , C2 , C3 , A3 , A4 , B4 , . . .
Since A1 B1 k AB2 , B1 B2 k CA1 and B1 C k B2 C2 , it is clear that triangle AB2 C2 is the
image of triangle A1 B1 C under a parallel translation. Similarly, triangle A3 BC3 is the image
of triangle AB2 C2 under a parallel translation and A4 B4 C is obtained in the same way from
A3 BC3 . But triangle A1 B1 C is also the image of triangle A3 BC3 under a parallel translation,
hence, A1 = A4 , i.e., after seven steps the trajectory becomes closed. (It is possible for the
trajectory to become closed sooner. Under what conditions?)
15.3. a) Let us complement triangle CBD to parallelogram CBDE. Then 2KM =
AE ≤ AD + DE = AD + BC and the equality is attained only if AD k BC.
b) Let a = AB, b = BC, c = CD and d = DA. If |a − c| = |b − d| =
6 0 then by heading
a) the maximum is attained in the degenerate case when all points A, B, C and D belong
to one line. Now suppose that, for example, |a − c| < |b − d|. Let us complement triangles
ABL and LCD to parallelograms ABLP and LCDQ, respectively; then P Q ≥ |b − d| and,
therefore,
1
1
LN 2 = (2LP 2 + 2LQ2 − P Q2 ) ≤ (2(a2 + c2 ) − (b − d)2 ).
4
4
1
Moreover, by heading a) KM ≤ 2 (b + d). Both equalities are attained when ABCD is a
trapezoid with bases AD and BC.
15.4. Let us construct circle S tangent to side AB and rays BC and AD; translate
triangle CN D parallelly (in the direction of bases BC and AD) until N ′ coincides with
point M , i.e., side C ′ D′ becomes tangent to circle S (Fig. 148).
Figure 148 (Sol. 15.4)
For the circumscribed trapezoid ABC ′ D′ the equality 2M N ′ = |AB+C ′ D′ −BC ′ −AD′ | is
obvious because N ′ = M . Under the passage from trapezoid ABC ′ D′ to trapezoid ABCD
the left-hand side of this equality accrues by 2N ′ N and the right-hand side accrues by
CC ′ + DD′ = 2N N ′ . Hence, the equality is preserved.
322
CHAPTER 15. PARALLEL TRANSLATIONS
15.5. Denote the intersection point of heights of triangle BKH by H1 . Since HH1 ⊥ BK
and KH1 ⊥ BH, it follows that HH1 k AD and KH1 k DC, i.e., H1 HDK is a parallelogram.
−−→
Therefore, under the parallel translation by vector H1 H point K passes to point D and point
B passes to point P (Fig. 149). Since P D k BK, it follows that BP DK is a rectangle and
P K = BD = b. Since BH1 ⊥ KH, it follows that P H ⊥ KH. It is also clear that
P H = BH1 .
Figure 149 (Sol. 15.5)
In right triangle
√ P KH, hypothenuse KP = b and the leg KH = a are known; therefore,
BH1 = P H = b2 − a2 .
15.6. a) Denote by F the figure that lies inside the unit square ABCD; let S be its
−−→
−−→
area. Let us consider two vectors AA1 and AA2 , where point A1 belongs to side AD and
AA1 = 0.001 and where point A2 belongs to the interior of angle ∠BAD, ∠A2 AA1 = 60◦
and AA2 = 0.001 (Fig. 150).
Figure 150 (Sol. 15.6 a))
−−→
Let F1 and F2 be the images of F under the parallel translations by vectors AA1 and
−−→
AA2 , respectively. The figures F , F1 and F2 have no common points and belong to the
interior of the square with side 1.001. Therefore, 2S < 1.0012 , i.e., S < 0.335 < 0.34.
−−→
−−→ −−→
−−→
b) Consider vector AA3 = AA1 + AA2 . Let us rotate AA3 about point A through an
acute angle counterclockwise so that point A3 turns into point A4 such that A3 A4 = 0.001.
−−→
−−→
Let us also consider vectors AA5 and AA6 of length 0.001 each constituting an angle of 30◦
−−→
with vector AA4 and situated on both sides of it (Fig. 151).
−−→
Denote by Fi the image of figure F under the parallel translation by the vector AAi .
Denote the area of the union of figures A and B by S(A ∪ B) and by S(A ∩ B) the area of
their intersection.
For definiteness, let us assume that S(F4 ∩ F ) ≤ S(F3 ∩ F ). Then S(F4 ∩ F ) ≤ 21 S
and, therefore, S(F4 ∪ F ) ≥ 23 S. The figures F5 and F6 do not intersect either each other
SOLUTIONS
323
Figure 151 (Sol. 15.6 b))
or figures F or F4 and, therefore, S(F ∪ F4 ∪ F5 ∪ F6 ) ≥ 27 S. (If it would have been that
S(F3 ∩ F ) ≤ S(F4 ∩ F ), then instead of figures F5 and F6 we should have taken F1 and F2 .)
√
−−→
considered lie inside
Since the lengths of vectors √
AAi do not exceed 0.001 3, all the
√ figures
2
a square with side 1 + 0.002 3. Therefore, 7S/2 ≤ (1 + 0.002 3) and S < 0.287.
15.7. Given two vectors ±a parallel to l and of given length a. Consider the images of
ray BC under the parallel translations by these vectors. Their intersection point with ray
BA belongs to the line to be constructed (if they do not intersect, then the problem has no
solutions).
15.8. a) Let S1′ be the image of circle S1 under the parallel translation by a vector
of length a parallel to l (there are two such vectors). The desired line passes through the
intersection point of circles S1′ and S2 .
b) Let O1 and O2 be the projections of the centers of circles S1 and S2 to line l; let S1′
−−−→
be the image of the circle S1 under the parallel translation by vector O1 O2 . The desired line
passes through the intersection point of circles S1′ and S2 .
c) Let S1′ be the image of circle S1 under the parallel translation by a vector parallel to l.
Then the lengths of chords cut by the line l1 on circles S1 and S1′ are equal. If the distance
between the projections of the centers of circles S1′ and S2 to line l is equal to 21 a, then the
sum of difference of the lengths of chords cut by the line parallel to l and passing through
the intersection point of circles S1′ and S2 is equal to a. Now it is easy to construct circle S1′ .
−→
15.9. Suppose that point X is constructed. Let us translate point A by vector EF , i.e.,
−→ −−→
let us construct point A′ such that EF = AA′ . This construction can be performed since we
−→
know vector EF : its length is equal to a and it is parallel to CD.
Figure 152 (Sol. 15.9)
324
CHAPTER 15. PARALLEL TRANSLATIONS
Since AX k A′ F , it follows that ∠A′ F B = ∠AXB and, therefore, angle ∠A′ F B is
known. Thus, point F belongs to the intersection of two figures: segment CD and an arc
of the circle whose points are vertices of the angles equal to ∠AXB that subtend segment
A′ B, see Fig. 152.
15.10. Suppose that quadrilateral ABCD is constructed. Denote by D1 the image of
−−→
point D under the parallel translation by vector CB. In triangle ABD1 , sides AB, BD1 and
angle ∠ABD1 are known. Hence, the following construction.
Let us arbitrarily construct ray BC ′ and then draw rays BD1′ and BA′ so that ∠D1′ BC ′ =
180◦ − ∠C, ∠A′ BC ′ = ∠B and these rays lie in the half plane on one side of ray BC ′ .
On rays BA′ and BD1′ , draw segments BA = a and BD1 = b, respectively. Let us draw
ray AD′ so that ∠BAD′ = ∠A and rays BC ′ , AD′ lie on one side of line AB. Vertex D is
the intersection point of ray AD′ and the ray drawn from D1 parallel to ray BC ′ . Vertex C
is the intersection point of BC ′ and the ray drawn from D parallel to ray D1 B.
15.11. Suppose that points M and N at which line l intersects circle S2 are constructed.
Let O1 and O2 be the centers of circles S1 and S2 ; let O1′ be the image of point O1 under the
parallel translation along l such that O1′ O2 ⊥ M N ; let S1′ be the image of circle S1 under
the same translation.
Let us draw tangents AP and AQ to circles S1′ and S2 , respectively. Then AQ2 =
AM · AN = AP 2 and, therefore, O1′ A2 = AP 2 + R2 , where R is the radius of circle S1′ . Since
segment AP can be constructed, we can also construct segment AO1′ . It remains to notice
that point O1′ belongs to both the circle of radius AO1′ with the center at A and to the circle
with diameter O1 O2 .
15.12. a) Let us draw through point A line P Q, where P belongs to circle S and Q
belongs to circle S2 . From the centers O1 and O2 of circles S1 and S2 , respectively, draw
perpendiculars O1 M and O2 N to line P Q. Let us parallelly translate segment M N by a
−−−→
vector M O1 . Let C be the image of point N under this translation.
Triangle O1 CO2 is a right one and O1 C = M N = 12 P Q. It follows that in order to construct line P Q for which P Q = a we have to construct triangle O1 CO2 of given hypothenuse
O1 O2 and leg O1 C = 12 a and then draw through A the line parallel to O1 C.
b) It suffices to solve the converse problem: around the given triangle P QR circumscribe
a triangle equal (?) to the given triangle ABC. Suppose that we have constructed triangle
ABC whose sides pass through given points P , Q and R. Let us construct the arcs of circles
whose points serve as vertices for angles ∠A and ∠B that subtend segments RP and QP ,
respectively. Points A and B belong to these arcs and the length of segment AB is known.
By heading a) we can construct line AP through P whose intercept between circles S1
and S2 is of given length. Draw lines AR and BQ; we get triangle ABC equal to the given
triangle since these triangles have by construction equal sides and the angles adjacent to it.
15.13. Suppose that the desired quadrilateral ABCD is constructed. Let D1 and D2 be
−→
−→
the images of point D under the translations by vectors AC and CA, respectively. Let us
circumscribe circles S1 and S2 around triangles DCD1 and DAD2 , respectively. Denote the
intersection points of lines BC and BA with circles S1 and S2 by M and N , respectively,
see Fig. 153. It is clear that ∠DCD1 = ∠DAD2 = ∠D, ∠DCM = 180◦ − ∠C and
∠DAN = 180◦ − ∠A.
This implies the following construction. On an arbitrary line l, take a point, D, and
construct points D1 and D2 on l so that DD1 = DD2 = AC. Fix one of the half planes Π
determined by line l and assume that point B belongs to this half plane. Let us construct
a circle S1 whose points belonging to Π serve as vertices of the angles equal to ∠D that
subtend segment DD1 .
SOLUTIONS
325
Figure 153 (Sol. 15.13)
We similarly construct circle S2 . Let us construct point M on S1 so that all the points
of the part of the circle that belongs to Π serve as vertices of the angles equal to 180◦ − ∠C
that subtend segment DM .
Point N is similarly constructed. Then segment M N subtends angle ∠B, i.e., B is the
intersection point of the circle with center D of radius DB and the arc of the circle serve
as vertices of the angles equal to ∠B that subtend segment M N (it also belongs to the half
plane Π). Points C and A are the intersection points of lines BM and BN with circles S1
and S2 , respectively.
15.14. From a point X draw perpendiculars XA1 and XA2 to given lines l1 and l2 ,
respectively. On ray A1 X, take point B so that A1 B = a. Then if XA1 ± XA2 = a, we
−−→
have XB = XA2 . Let l1′ be the image of line l1 under the parallel translation by vector A1 B
and M the intersection point of lines l1′ and l2 . Then in the indicated cases ray M X is the
bisector of angle ∠A2 M B. As a result we get the following answer.
Let the intersection points of lines l1 and l2 with the lines parallel to lines l1 and l2 and
distant from them by a form rectangle M1 M2 M3 M4 . The locus to be found is either a) the
sides of this rectangle; or b) the extensions of these sides.
15.15. Let leg AB of angle ∠BAC be tangent to the circle of radius r1 with center O1
and leg AC be tangent to the circle of radius r2 with center O2 . Let us parallelly translate
line AB inside angle ∠BAC by distance r1 and let us parallelly translate line AC inside
angle ∠BAC by distance r2 . Let A1 be the intersection point of the translated lines (Fig.
154).
Figure 154 (Sol. 15.15)
Then ∠O1 A1 O2 = ∠BAC. The constant(?) angle O1 A1 O2 subtends fixed segment O1 O2
and, therefore, point A1 traverses an arc of a(?) circle.
Chapter 16. CENTRAL SYMMETRY
Background
1. The symmetry through point A is the transformation of the plane which sends point
X into point X ′ such that A is the midpoint of segment XX ′ . The other names of such a
transformation: the central symmetry with center A or just the symmetry with center A.
Notice that the symmetry with center A is a particular case of two other transformations:
it is the rotation through an angle of 180◦ with center A and also the homothety with center
A and coefficient −1.
2. If a figure turns into itself under the symmetry through point A, then A is called the
center of symmetry of this figure.
3. The following notations for transformations are used in this chapter:
SA — the symmetry with center A;
Ta — the translation by vector a.
4. We will denote the composition of symmetries through points A and B by SB ◦SA ; here
we assume that we first perform symmetry SA and then symmetry SB . This notation might
look unnatural at first glance, but it is, however, justified by the identity (SB ◦ SA )(X) =
SB (SA (X)).
The composition of maps is associative: F ◦ (G ◦ H) = (F ◦ G) ◦ H. Therefore, the order
of the compositions is inessential and we may simply write F ◦ G ◦ H.
5. The compositions of two central symmetries or of a symmetry with a parallel translation are calculated according to the following formulas (Problem 16.9):
→;
a) SB ◦ SA = T2−
AB
−→
b) Ta ◦ SA = SB and SB ◦ Ta = SA , where a = 2AB.
Introductory problems
1. Prove that under any central symmetry any circle turns into a circle.
2. Prove that a quadrilateral with a center of symmetry is a parallelogram.
3. The opposite sides of a convex hexagon are equal and parallel. Prove that the hexagon
has a center of symmetry.
4. Consider parallelogram ABCD and point M . The lines parallel to lines M C, M D,
M A and M B are drawn through points A, B, C and D, respectively. Prove that the lines
drawn intersect at one point.
5. Prove that the opposite sides of a hexagon formed by the sides of a triangle and the
tangents to its circumscribed circle parallel to the sides of the triangle are equal.
§1. Solving problems with the help of a symmetry
16.1. Prove that if in a triangle a median and a bisector coincide, then the triangle is
an isosceles one.
327
328
CHAPTER 16. CENTRAL SYMMETRY
16.2. Two players lay out nickels on a rectangular table taking turns. It is only allowed
to place a coin onto an unoccupied place. The loser is the one who can not make any move.
Prove that the first player can always win in finitely many moves.
16.3. A circle intersects sides BC, CA, AB of triangle ABC at points A1 and A2 , B1
and B2 , C1 and C2 , respectively. Prove that if the perpendiculars to the sides of the triangle
drawn through points A1 , B1 and C1 intersect at one point, then the perpendiculars to the
sides drawn through A2 , B2 and C2 also intersect at one point.
16.4. Prove that the lines drawn through the midpoints of the circumscribed quadrilateral perpendicularly to the opposite sides intersect at one point.
16.5. Let P be the midpoint of side AB of convex quadrilateral ABCD. Prove that if
the area of triangle P CD is equal to a half area of quadrilateral ABCD, then BC k AD.
16.6. Unit circles S1 and S2 are tangent at point A; the center O of circle S of radius 2
belongs to S1 . Circle S1 is tangent to circle S at point B. Prove that line AB passes through
the intersection point of circles S2 and S.
16.7. In triangle ABC medians AF and CE are drawn. Prove that if ∠BAF = ∠BCE =
◦
30 , then triangle ABC is an equilateral one.
16.8. Consider a convex n-gon with pairwise nonparallel sides and point O inside it.
Prove that it is impossible to draw more than n lines through O so that each line divides
the area of the n-gon in halves.
§2. Properties of the symmetry
16.9. a) Prove that the composition of two central symmetries is a parallel translation.
b) Prove that the composition of a parallel translation with a central symmetry (in either
order) is a central symmetry.
16.10. Prove that if a point is reflected symmetrically through points O1 , O2 and O3
and then reflected symmetrically once again through the same points, then it assumes the
initial position.
16.11. a) Prove that a bounded figure cannot have more than one center of symmetry.
b) Prove that no figure can have precisely two centers of symmetry.
c) Let M be a finite set of points on a plane. Point O will be called an “almost center of
symmetry” of the set M if we can delete a point from M so that O becomes the center of
symmetry of the remaining set. How many “almost centers of symmetry” can a set have?
16.12. On segment AB, consider n pairs of points symmetric through the midpoint; n
of these 2n points are painted blue and the remaining are painted red. Prove that the sum
of distances from A to the blue points is equal to the sum of distances from B to the red
points.
§3. Solving problems with the help of a symmetry. Constructions
16.13. Through a common point A of circles S1 and S2 draw a straight line so that these
circles would intercept on it equal chords.
16.14. Given point A, a line and a circle. Through A draw a line so that A divides the
segment between the intersection points of the line drawn with the given line and the given
circle in halves.
16.15. Given angle ABC and point D inside it. Construct a segment with the endpoints
on the legs of the given angle and with the midpoint at D.
16.16. Consider an angle and points A and B inside it. Construct a parallelogram for
which points A and B are opposite vertices and the two other vertices belong to the legs of
the angle.
SOLUTIONS
329
16.17. Given four pairwise nonparallel straight lines and point O not belonging to these
lines. Construct a parallelogram whose center is O and the vertices lie on the given lines,
one on each.
16.18. Consider two concentric circles S1 and S2 . Draw a line on which these circles
intercept three equal segments.
16.19. Consider nonintersecting chords AB and CD of a circle and point J on chord
CD. Construct point X on the circle so that chords AX and BX would intercept on chord
CD segment EF which J divides in halves.
16.20. Through a common point A of circles S1 and S2 draw line l so that the difference
of the lengths of the chords intercepted by circles S1 and S2 on l were of given value a.
16.21. Given m = 2n + 1 points — the midpoints of the sides of an m-gon — construct
the vertices of the m-gon.
Problems for independent study
16.22. Construct triangle ABC given medians ma , mb and angle ∠C.
16.23. a) Given a point inside a parallelogram; the point does not belong to the segments
that connect the midpoints of the opposite sides. How many segments divided in halves by
the given point are there such that their endpoints are on the sides of the parallelogram?
b) A point inside the triangle formed by the midlines of a given triangle is given. How
many segments divided in halves by the given point and with the endpoints on the sides of
the given triangle are there?
16.24. a) Find the locus of vertices of convex quadrilaterals the midpoints of whose sides
are the vertices of a given square.
b) Three points are given on a plane. Find the locus of vertices of convex quadrilaterals
the midpoints of three sides of each of which are the given points.
16.25. Points A, B, C, D lie in the indicated order on a line and AB = CD. Prove that
for any point P on the plane we have AP + DP ≥ BP + CP .
Solutions
16.1. Let median BD of triangle ABC be a bisector as well. Let us consider point B1
symmetric to B through point D. Since D is the midpoint of segment AC, the quadrilateral
ABCB1 is a parallelogram. Since ∠ABB1 = ∠B1 BC = ∠AB1 B, it follows that triangle
B1 AB is an isosceles one and AB = AB1 = BC.
16.2. The first player places a nickel in the center of the table and then places nickels
symmetrically to the nickels of the second player with respect to the center of the table.
Using this strategy the first player has always a possibility to make the next move. It is also
clear that the play will be terminated in a finite number of moves.
16.3. Let the perpendiculars to the sides drawn through points A1 , B1 and C1 intersect
at point M . Denote the center of the circle by O. The perpendicular to side BC drawn
through point A1 is symmetric through point O to the perpendicular to side BC drawn
through A2 . It follows that the perpendiculars to the sides drawn through points A2 , B2 and
C2 intersect at the point symmetric to M through point O.
16.4. Let P , Q, R and S be the midpoints of sides AB, BC, CD and DA, respectively,
and M the intersection point of segments P R and QS (i.e., the midpoint of both of these
segments, see Problem 14.5); O the center of the circumscribed circle and O′ the point
symmetric to O through M . Let us prove that the lines mentioned in the formulation of
the problem pass through O′ . Indeed, O′ P OR is a parallelogram and, therefore, O′ P k OR.
Since R is the midpoint of chord CD, it follows that OR ⊥ CD, i.e., O′ P ⊥ CD.
330
CHAPTER 16. CENTRAL SYMMETRY
For lines O′ Q, O′ R and O′ S the proof is similar.
16.5. Let point D′ be symmetric to D through P . If the area of triangle P CD is equal
to a half area of quadrilateral ABCD, then it is equal toSP BC + SP AD , i.e., it is equal to
SP BC + SP BD′ . Since P is the midpoint of segment DD′ , it follows that SP CD′ = SP CD =
SP BC + SP BD′ and, therefore, point B belongs to segment D′ C. It remains to notice that
D′ B k AD.
16.6. Circles S1 and S2 are symmetric through point A. Since OB is the diameter of
circle S1 , it follows that ∠BAO = 90◦ and, therefore, under the symmetry through A point
B becomes on the circle S again. It follows that under the symmetry through A point B
turns into the intersection point of circles S2 and S.
16.7. Since ∠EAF = ∠ECF = 30◦ , we see that points A, E, F and C belong to one
circle S and if O is its center, then ∠EOF = 60◦ . Point B is symmetric to A through E and,
therefore, B belongs to circle S1 symmetric to circle S through E. Similarly, point B belongs
to circle S2 symmetric to circle S through point F . Since triangle EOF is an equilateral
one, the centers of circles S, S1 and S2 form an equilateral triangle with side 2R, where R is
the radius of these circles. Therefore, circles S1 and S2 have a unique common point — B
— and triangle BEF is an equilateral one. Thus, triangle ABC is also an equilateral one.
16.8. Consider a polygon symmetric to the initial one through point O. Since the sides
of the polygons are pairwise nonparallel, the contours of these polygons cannot have common
segments but could only have common points. Since the polygons are convex ones, each side
has not more than two intersection points; therefore, there are not more than 2n intersection
points of the contours (more precisely, not more than n pairs of points symmetric through
O).
Let l1 and l2 be the lines passing through O and dividing the area of the initial polygon
in halves. Let us prove that inside each of the four parts into which these lines divide the
plane there is an intersection point of the contours.
Suppose that one of the parts has no such points between lines l1 and l2 . Denote the
intersection points of lines l1 and l2 with the sides of the polygon as indicated on Fig. 12.
Figure 154 (Sol. 16.8)
Let points A′ , B ′ , C ′ and D′ be symmetric trough O to points A, B, C and D, respectively.
For definiteness sake, assume that point A is closer to O than C ′ . Since segments AB and
C ′ D′ do not intersect, point B is closer to O than D′ . It follows that SABO < SC ′ D′ O = SCDO ,
where ABO is a convex figure bounded by segments AO and BO and the part of the boundary
of the n-gon between points A and B.
On the other hand, SABO = SCDO because lines l1 and l2 divide the area of the polygon
in halves. Contradiction.
Therefore, between every pair of lines which divide the area of the polygon in halves
there is a pair of symmetric intersection points of contours; in other words, there are not
more than n such lines.
16.9. a) Let the central symmetry through O1 send point A into A1 ; let the central
symmetry through O2 send point A1 into A2 . Then O1 O2 is the midline of triangle AA1 A2
−−→
−−−→
and, therefore, AA2 = 2O1 O2 .
SOLUTIONS
331
b) Let O2 be the image of point O1 under the translation by vector 21 a. By heading a) we
have SO1 ◦ SO2 = Ta . Multiplying this equality by SO1 from the right or by SO2 from the left
and taking into account that SX ◦ SX is the identity transformation we get SO1 = SO2 ◦ Ta
and SO2 = Ta ◦ SO1 .
→ ; therefore,
16.10. By the preceding problem SB ◦ SA = T2−
AB
−−→ −−−→ −−−→
SO3 ◦ SO2 ◦ SO1 ◦ SO3 ◦ SO2 ◦ SO1 = T2(−
O2 O3 + O3 O1 + O1 O2 )
is the identity transformation.
16.11. a) Suppose that a bounded figure has two centers of symmetry: O1 and O2 .
Let us introduce a coordinate system whose absciss axis is directed along ray O1 O2 . Since
−−−→
−−→ , the figure turns into itself under the translation by vector 2O1 O2 . A
SO2 ◦ SO1 = T2−
O1 O2
bounded figure cannot possess such a property since the image of the point with the largest
absciss does not belong to the figure.
b) Let O3 = SO2 (O1 ). It is easy to verify that SO3 = SO2 ◦ SO1 ◦ SO2 and, therefore, if
O1 and O2 are the centers of symmetry of a figure, then O3 is also a center of symmetry,
moreover, O3 6= O1 and O3 6= O2 .
c) Let us demonstrate that a finite set can only have 0, 1, 2 or 3 “almost centers of
symmetry”. The corresponding examples are given on Fig. 13. It only remains to prove
that a finite set cannot have more than three “almost centers of symmetry”.
Figure 155 (Sol. 16.11)
There are finitely many “almost centers of symmetry” since they are the midpoints of
the segments that connect the points of the set. Therefore, we can select a line such that the
projections of “almost centers of symmetry” to the line are distinct. Therefore, it suffices to
carry out the proof for the points which belong to one line.
Let n points on a line be given and x1 < x2 < · · · < xn−1 < xn be their coordinates.
If we discard the point x1 , then only point 21 (x2 + xn ) can serve as the center of symmetry
of the remaining set; if we discard xn , then only point 12 (x1 + xn−1 ) can be the center of
symmetry of the remaining set and if we discard any other point, then only point 21 (x1 + xn )
can be the center of symmetry of the remaining set. Therefore, there can not be more than
3 centers of symmetry.
16.12. A pair of symmetric points is painted different colours, therefore, it can be
discarded from the consideration; let us discard all such pairs. In the remaining set of points
the number of blue pairs is equal to the number of red pairs. Moreover, the sum of the
distances from either of points A or B to any pair of symmetric points is equal to the length
of segment AB.
332
CHAPTER 16. CENTRAL SYMMETRY
16.13. Consider circle S1′ symmetric to circle S1 through point A. The line to be found
passes through the intersection points of S1′ and S2 .
16.14. Let l′ be the image of line l under the symmetry through point A. The desired
line passes through point A and an intersection point of line l′ with the circle S.
Figure 156 (Sol. 16.15)
16.15. Let us construct the intersection points A′ and C ′ of the lines symmetric to the
lines BC and AB through the point D with lines AB and BC, respectively, see Fig. 14. It is
clear that point D is the midpoint of segment A′ C ′ because points A′ and C ′ are symmetric
through D.
16.16. Let O be the midpoint of segment AB. We have to construct points C and D
that belong to the legs of the angle so that point O is the midpoint of segment CD. This
construction is described in the solution of the preceding problem.
16.17. Let us first separate the lines into pairs. This can be done in three ways. Let the
opposite vertices A and C of parallelogram ABCD belong to one pair of lines, B and D to
the other pair. Consider the angle formed by the first pair of lines and construct points A
and C as described in the solution of Problem 16.15. Construct points B and D in a similar
way.
16.18. On the smaller circle, S1 , take an arbitrary point, X. Let S1′ be the image of S1
under the symmetry with respect to X, let Y be the intersection point of circles S1′ and S2 .
Then XY is the line to be found.
Figure 157 (Sol. 16.19)
16.19. Suppose X is constructed. Denote the images of points A, B and X under the
symmetry through point J by A′ , B ′ and X ′ , respectively, see Fig. 15. Angle ∠A′ F B =
180◦ − ∠AXB is known and, therefore, point F is the intersection point of segment CD
with the arc of the circle whose points serve as vertices of angles of value 180◦ − ∠AXB that
subtend segment BA′ . Point X is the intersection point of line BF with the given circle.
16.20. Suppose that line l is constructed. Let us consider circle S1′ symmetric to circle
S1 through point A. Let O1 , O1′ and O2 be the centers of circles S1 , S1′ and S2 , as shown on
Fig. 16.
SOLUTIONS
333
Figure 158 (Sol. 16.20)
Let us draw lines l1′ and l2 through O1′ and O2 perpendicularly to line l. The distance
between lines l1′ and l2 is equal to a half difference of the lengths of chords intercepted by l
on circles S1 and S2 . Therefore, in order to construct l, we have to construct the circle of
radius 12 a with center O1′ ; line l2 is tangent to this circle. Having constructed l2 , drop the
perpendicular from point A to l2 ; this perpendicular is line l.
16.21. Let B1 , B2 , . . . , Bm be the midpoints of sides A1 A2 , A2 A3 , . . . , Am A1 of polygon
A1 A2 . . . Am . Then SB1 (A1 ) = A2 , SB2 (A2 ) = A3 , . . . , SBm (Am ) = A1 . It follows that
SBm ◦ · · · ◦ SB1 (A1 ) = A1 , i.e., A1 is a fixed point of the composition of symmetries SBm ◦
SBm−1 ◦ · · · ◦ SB1 . By Problem 16.9 the composition of an odd number of central symmetries
is a central symmetry, i.e., has a unique fixed point. This point can be constructed as the
midpoint of the segment that connects points X and SBm ◦ SBm−1 ◦ · · · ◦ SB1 (X), where X
is an arbitrary point.
Chapter 17. THE SYMMETRY THROUGH A LINE
Background
1. The symmetry through a line l (notation: Sl ) is a transformation of the plane which
sends point X into point X ′ such that l is the midperpendicular to segment XX ′ . Such a
transformation is also called the axial symmetry and l is called the axis of the symmetry.
2. If a figure turns into itself under the symmetry through line l, then l is called the axis
of symmetry of this figure.
3. The composition of two symmetries through axes is a parallel translation, if the axes
are parallel, and a rotation, if they are not parallel, cf. Problem 17.22.
Axial symmetries are a sort of “bricks” all the other motions of the plane are constructed
from: any motion is a composition of not more than three axial symmetries (Problem 17.35).
Therefore, the composition of axial symmetries give much more powerful method for solving
problems than compositions of central symmetries. Moreover, it is often convenient to decompose a rotation into a composition of two symmetries with one of the axes of symmetry
being a line passing through the center of the rotation.
Introductory problems
1. Prove that any axial symmetry sends any circle into a circle.
2. A quadrilateral has an axis of symmetry. Prove that this quadrilateral is either an
equilateral trapezoid or is symmetric through a diagonal.
3. An axis of symmetry of a polygon intersects its sides at points A and B. Prove that
either point A is a vertex of the polygon or the midpoint of a side perpendicular to the axis
of symmetry.
4. Prove that if a figure has two perpendicular axes of symmetry, it has a center of
symmetry.
§1. Solving problems with the help of a symmetry
17.1. Point M belongs to a diameter AB of a circle. Chord CD passes through M and
intersects AB at an angle of 45◦ . Prove that the sum CM 2 + DM 2 does not depend on the
choice of point M .
17.2. Equal circles S1 and S2 are tangent to circle S from the inside at points A1 and
A2 , respectively. An arbitrary point C of circle S is connected by segments with points A1
and A2 . These segments intersect S1 and S2 at points B1 and B2 , respectively. Prove that
A1 A2 k B1 B2 .
17.3. Through point M on base AB of an isosceles triangle ABC a line is drawn. It
intersects sides CA and CB (or their extensions) at points A1 and B1 . Prove that A1 A :
A1 M = B1 B : B1 M .
335
336
CHAPTER 17. THE SYMMETRY THROUGH A LINE
§2. Constructions
17.4. Construct quadrilateral ABCD whose diagonal AC is the bisector of angle ∠A
knowing the lengths of its sides.
17.5. Construct quadrilateral ABCD in which a circle can be inscribed knowing the
lengths of two neighbouring sides AB and AD and the angles at vertices B and D.
17.6. Construct triangle ABC knowing a, b and the difference of angles ∠A − ∠B.
17.7. Construct triangle ABC given its side c, height hc and the difference of angles
∠A − ∠B.
17.8. Construct triangle ABC given a) c, a − b (a > b) and angle ∠C; b) c, a + b and
angle ∠C.
17.9. Given line l and points A and B on one side of it. Construct point X on l such
that AX + XB = a, where a is given.
17.10. Given acute angle ∠M ON and points A and B inside it. Find point X on leg
OM such that triangle XY Z, where Y and Z are the intersection points of lines XA and
XB with ON , were isosceles, i.e., XY = XZ.
17.11. Given line M N and two points A and B on one side of it. Construct point X on
M N such that ∠AXM = 2∠BXN .
***
17.12. Given three lines l1 , l2 and l3 intersecting at one point and point A1 on l1 .
Construct triangle ABC so that A1 is the midpoint of its side BC and lines l1 , l2 and l3 are
the midperpendiculars to the sides.
17.13. Construct triangle ABC given points A, B and the line on which the bisector of
angle ∠C lies.
17.14. Given three lines l1 , l2 and l3 intersecting at one point and point A on line l1 .
Construct triangle ABC so that A is its vertex and the bisectors of the triangle lie on lines
l1 , l2 and l3 .
17.15. Construct a triangle given the midpoints of two of its sides and the line that
contains the bisector drawn to one of these sides.
§3. Inequalities and extremals
17.16. On the bisector of the exterior angle ∠C of triangle ABC point M distinct from
C is taken. Prove that M A + M B > CA + CB.
17.17. In triangle ABC median AM is drawn. Prove that 2AM ≥ (b + c) cos( 21 α).
17.18. The inscribed circle of triangle ABC is tangent to sides AC and BC at points
B1 and A1 . Prove that if AC > BC, then AA1 > BB1 .
17.19. Prove that the area of any convex quadrilateral does not exceed a half-sum of
the products of opposite sides.
17.20. Given line l and two points A and B on one side of it, find point X on line l such
that the length of segment AXB of the broken line was minimal.
17.21. Inscribe a triangle of the least perimeter in a given acute triangle.
§4. Compositions of symmetries
17.22. a) Lines l1 and l2 are parallel. Prove that Sl1 ◦ Sl2 = T2a , where Ta is the parallel
translation that sends l1 to l2 and such that a ⊥ l1 .
2α
α
b) Lines l1 and l2 intersect at point O. Prove that Sl2 ◦ Sl1 = RO
, where RO
is the
rotation about O through the angle of α that sends l1 to l2 .
§6. CHASLES’S THEOREM
337
17.23. On the plane, there are given three lines a, b, c. Let T = Sa ◦ Sb ◦ Sc . Prove that
T ◦ T is a parallel translation (or the identity map).
17.24. Let l3 = Sl1 (l2 ). Prove that Sl3 = Sl1 ◦ Sl2 ◦ Sl1 .
17.25. The inscribed circle is tangent to the sides of triangle ABC at points A1 , B1
and C1 . Points A2 , B2 and C2 are symmetric to these points through the bisectors of the
corresponding angles of the triangle. Prove that A2 B2 k AB and lines AA2 , BB2 and CC2
intersect at one point.
17.26. Two lines intersect at an angle of γ. A grasshopper hops from one line to another
one; the length of each jump is equal to 1 m and the grasshopper does not jump backwards
whenever possible. Prove that the sequence of jumps is periodic if and only if γ/π is a
rational number.
17.27. a) Given a circle and n lines. Inscribe into the circle an n-gon whose sides are
parallel to given lines.
b) n lines go through the center O of a circle. Construct an n-gon circumscribed about
this circle such that the vertices of the n-gon belong to these lines.
17.28. Given n lines, construct an n-gon for which these lines are a) the midperpendiculars to the sides; b) the bisectors of the inner or outer angles at the vertices.
17.29. Given a circle, a point and n lines. Into the circle inscribe an n-gon one of whose
sides passes through the given point and the other sides are parallel to the given lines.
§5. Properties of symmetries and axes of symmetries
17.30. Point A lies at the distance of 50 cm from the center of the disk of radius 1 cm.
It is allowed to reflect point A symmetrically through any line intersecting the disk. Prove
that a) after 25 reflexions point A can be driven inside the given circle; b) it is impossible
to perform this in 24 reflexions.
17.31. On a circle with center O points A1 , . . . , An which divide the circle into equal
archs and a point X are given. Prove that the points symmetric to X through lines OA1 ,
. . . , OAn constitute a regular polygon.
17.32. Prove that if a planar figure has exactly two axes of symmetry, then these axes
are perpendicular to each other.
17.33. Prove that if a polygon has several (more than 2) axes of symmetry, then all of
them intersect at one point.
17.34. Prove that if a polygon has an even number of axes of symmetry, then it has a
center of symmetry.
§6. Chasles’s theorem
A transformation which preserves distances between points (i.e., such that if A′ and B ′
are the images of points A and B, respectively, then A′ B ′ = AB) is called a movement. A
movement of the plane that preserves 3 points which do not belong to one line preserves all
the other points.
17.35. Prove that any movement of the plane is a composition of not more than three
symmetries through lines.
A movement which is the composition of an even number of symmetries through lines is
called a first type movement or a movement that preserves the orientation of the plane.
A movement which is the composition of an odd number of symmetries through lines is
called a second type movement or a movement inversing the orientation of the plane.
338
CHAPTER 17. THE SYMMETRY THROUGH A LINE
We will not prove that the composition of an odd number of symmetries through lines
is impossible to represent in the form of the composition of an odd number of symmetries
through lines and the other way round because this fact, though true, is beyond the scope
of our book.
17.36. Prove that any first type movement is either a rotation or a parallel translation.
The composition of a symmetry through line l and the translation by a vector parallel
to l (this vector might be the zero one) is called a transvection.
17.37. Prove that any second type movement is a transvection.
Problems for independent study
17.38. Given a nonconvex quadrilateral of perimeter P . Prove that there exists a convex
quadrilateral of the same perimeter but of greater area.
17.39. Can a bounded figure have a center of symmetry and exactly one axis of symmetry?
17.40. Point M belongs to the circumscribed circle of triangle ABC. Prove that the
lines symmetric to the lines AM , BM and CM through the bisectors of angles ∠A, ∠B and
∠C are parallel to each other.
17.41. The vertices of a convex quadrilateral belong to different
sides of a square. Prove
√
that the perimeter of this quadrilateral is not shorter than 2 2a, where a is the length of
the square’s side.
17.42. A ball lies on a rectangular billiard table. Construct a trajectory traversing along
which the ball would return to the initial position after one reflexion from each side of the
table.
Solutions
17.1. Denote the points symmetric to points C and D through line AB by C ′ and D′ ,
respectively. Since ∠C ′ M D = 90◦ , it follows that CM 2 + M D2 = C ′ M 2 + M D2 = C ′ D2 .
Since ∠C ′ CD = 45◦ , chord C ′ D is of constant length.
17.2. In circle S, draw the diameter which is at the same time the axis of symmetry
of circles S1 and S2 . Let points C ′ and B2′ be symmetric to points C and B2 through this
diameter: see Fig. 17.
Figure 159 (Sol. 17.2)
Circles S1 and S are homothetic with the center of homothety at point A1 ; let this
homothety send line B1 B2′ into line CC ′ . Therefore, these lines are parallel to each other. It
is also clear that B2 B2′ k CC ′ . Therefore, points B1 , B2′ and B2 belong to one line and this
line is parallel to line CC ′ .
SOLUTIONS
339
17.3. Let the line symmetric to line A1 B1 through line AB intersect sides CA and CB
(or their extensions) at points A2 and B2 , respectively. Since ∠A1 AM = ∠B2 BM and
∠A1 M A = ∠B2 M B, it follows that A1 AM ∼ B2 BM , i.e., A1 A : A1 M = B2 B : B2 M .
Moreover, since M B is a bisector in triangle B1 M B2 , it follows that B2 B : B2 M = B1 B :
B1 M .
17.4. Suppose that quadrilateral ABCD is constructed. Let, for definiteness sake,
AD > AB. Denote by B ′ the point symmetric to B through diagonal AC. Point B ′ belongs
to side AD and B ′ D = AD − AB. In triangle B ′ CD, the lengths of all the sides are known:
B ′ D = AD − AB and B ′ C = BC. Constructing triangle B ′ CD on the extension of side
B ′ D beyond B ′ let us construct point A.
Further construction is obvious.
17.5. Suppose that quadrilateral ABCD is constructed. For definiteness sake, assume
that AD > AB. Let O be the center of the circumscribed circle; let point D′ be symmetric
to D through line AO; let A′ be the intersection point of lines AO and DC; let C ′ be the
intersection point of lines BC and A′ D′ (Fig. 18).
Figure 160 (Sol. 17.5)
In triangle BC ′ D′ , side BD′ and adjacent angles are known: ∠D′ BC ′ = 180◦ − ∠B and
∠BD′ C ′ = ∠D. Let us construct triangle BC ′ D′ given these elements. Since AD′ = AD,
we can construct point A. Further, let us construct O — the intersection point of bisectors
of angles ABC ′ and BD′ C ′ . Knowing the position of O we can construct point D and the
inscribed circle. Point C is the intersection point of line BC ′ and the tangent to the circle
drawn from D.
17.6. Suppose that triangle ABC is constructed. Let C ′ be the point symmetric to C
through the midperpendicular to segment AB. In triangle ACC ′ there are known AC = b,
AC ′ = a and ∠CAC ′ = ∠A − ∠B. Therefore, the triangle can be constructed. Point B is
symmetric to A through the midperpendicular to segment CC ′ .
17.7. Suppose that triangle ABC is constructed. Denote by C ′ the point symmetric to
C through the midperpendicular to side AB and by B ′ the point symmetric to B through
line CC ′ . For definiteness, let us assume that AC < BC. Then
∠ACB ′ = ∠ACC ′ + ∠C ′ CB = 180◦ − ∠A + ∠C ′ CB = 180◦ − (∠A − ∠B)
i.e., angle ∠ACB ′ is known.
Triangle ABB ′ can be constructed because AB = c, BB ′ = 2hc and ∠ABB ′ = 90◦ .
Point C is the intersection point of the midperpendicular to segment BB ′ and the arc of
the circle whose points serve as vertices of angles of value 180◦ − (∠A − ∠B) that subtend
segment AB ′ .
340
CHAPTER 17. THE SYMMETRY THROUGH A LINE
17.8. a) Suppose triangle ABC is constructed. Let C ′ be the point symmetric to A
through the bisector of angle ∠C. Then
1
1
∠BC ′ A = 180◦ − ∠AC ′ C = 180◦ − (180◦ − ∠C) = 90◦ + ∠C
2
2
′
and BC = a − b.
In triangle ABC ′ , there are known AB = c, BC ′ = a − b and ∠C ′ = 90◦ + 21 ∠C. Since
∠C ′ > 90◦ , triangle ABC ′ is uniquely constructed from these elements. Point C is the
intersection point of the midperpendicular to segment AC ′ with line BC ′ .
b) The solution is similar to that of heading a). For C ′ we should take the point symmetric
to A through the bisector of the outer angle ∠C in triangle ABC.
Since ∠AC ′ B = 21 ∠C < 90◦ , the problem can have two solutions.
17.9. Let S be the circle of radius a centered at B, let S ′ be the circle of radius AX
with center X and A′ the point symmetric to A through line l. Then circle S ′ is tangent
to circle S and point A′ belongs to circle S ′ . It remains to draw circle S ′ through the given
points A and A′ tangent to the given circle S and find its center X, cf. Problem 8.56 b).
Figure 161 (Sol. 17.10)
17.10. Let the projection of point A to line ON be closer to point O than the projection
of point B. Suppose that the isosceles triangle XY Z is constructed. Let us consider point
A′ symmetric to point A through line OM . Let us drop perpendicular XH from point X to
line ON (Fig. 19). Since
∠A′ XB = ∠A′ XO + ∠OXA + ∠Y XH + ∠HXZ =
2∠OXY + 2∠Y XH = 2∠OXH = 180◦ − 2∠M ON,
angle ∠A′ XB is known. Point X is the intersection point of line OM and the arc whose
points serve as vertices of angles of 180◦ − 2∠M ON that subtend A′ B. In addition, the
projection of X onto ON must lie between the projections of A and B.
Conversely, if ∠A′ XB = 180◦ − ∠M ON and the projection of X to line ON lies between
the projections of A and B, then triangle XY Z is an isosceles one.
17.11. Suppose that point X is constructed. Let B ′ be the point symmetric to point B
through line M N ; the circle of radius AB ′ with center B ′ intersects line M N at point A′ .
Then ray B ′ X is the bisector of angle ∠AB ′ A′ . It follows that X is the intersection point of
lines B ′ O and M N , where O is the midpoint of segment AA′ .
17.12. Through point A1 draw line BC perpendicular to line l1 . Vertex A of triangle
ABC to be found is the intersection point of lines symmetric to line BC through lines l2
and l3 .
17.13. Let point A′ be symmetric to A through the bisector of angle ∠C. Then C is the
intersection point of line A′ B and the line on which the bisector of angle ∠C lies.
SOLUTIONS
341
17.14. Let A2 and A3 be points symmetric to A through lines l2 and l3 , respectively.
Then points A2 and A3 belong to line BC. Therefore, points B and C are the intersection
points of line A2 A3 with lines l2 and l3 , respectively.
17.15. Suppose that triangle ABC is constructed and N is the midpoint of AC, M the
midpoint of BC and the bisector of angle ∠A lies on the given line, l. Let us construct
point N ′ symmetric to N through line l. Line BA passes through point N ′ and is parallel
to M N . In this way we find vertex A and line BA. Having drawn line AN , we get line AC.
It remains to construct a segment whose endpoints belong to the legs of angle ∠BAC and
whose midpoint is M , cf. the solution of Problem 16.15.
17.16. Let points A′ and B ′ be symmetric to A and B, respectively, through line CM .
Then AM + M B = A′ M + M B > A′ B = A′ C + CB = AC + CB.
17.17. Let points B ′ , C ′ and M ′ be symmetric to points B, C and M through the
bisector of the outer angle at vertex A. Then
1
1
1
AM + AM ′ = M M ′ = (BB ′ + CC ′ ) = (b + c) sin(90◦ − α) = (b + c) cos( α).
2
2
2
′
17.18. Let point B be symmetric to B through the bisector of angle ∠ACB. Then
′
B A1 = BB1 , i.e., it remains to verify that B ′ A1 < AA1 . To this end it suffices to notice
that ∠AB ′ A1 > ∠AB ′ B > 90◦ .
17.19. Let D′ be the point symmetric to D through the midperpendicular to segment
AC. Then
1
1
SABCD = SABCD′ = SBAD′ + SBCD′ ≤ AB · AD′ + BC · CD′ =
2
2
1
(AB · CD + BC · AD).
2
17.20. Let point A′ be symmetric to A through line l. Let X be a point on line l. Then
AX + XB = A′ X + XB ≥ A′ B and the equality is attained only if X belongs to segment
A′ B. Therefore, the point to be found is the intersection point of line l with segment A′ B.
17.21. Let P QR be the triangle determined by the bases of the heights of triangle ABC
and let P ′ Q′ R′ be any other triangle inscribed in triangle ABC. Further, let points P1 and
P2 (respectively P1′ and P2′ ) be symmetric to point P (resp. P ′ ) through lines AB and AC,
respectively (Fig. 20).
Figure 162 (Sol. 17.21)
Points Q and R belong to segment P1 P2 (see Problem 1.57) and, therefore, the perimeter
of triangle P QR is equal to the length of segment P1 P2 . The perimeter of triangle P ′ Q′ R′
is, however, equal to the length of the broken segment P1′ R′ Q′ P2′ , i.e., it is not shorter than
342
CHAPTER 17. THE SYMMETRY THROUGH A LINE
the length of segment P1′ P2′ . It remains to notice that (P1′ P2′ )2 = P1 P22 + 4d2 , where d is the
distance from point P1′ to line P1 P2 .
17.22. Let X be an arbitrary point, X1 = Sl1 (X) and X2 = Sl2 (X1 ).
a) On line l1 , select an arbitrary point O and consider a coordinate system with O as the
origin and the absciss axis directed along line l1 . Line l2 is given in this coordinate system
by the equation y = a. Let y, y1 and y2 be ordinates of points X, X1 and X2 , respectively.
It is clear that y1 = −y and y2 = (a − y1 ) + a = y + 2a. Since points X, X1 and X2 have
identical abscisses, it follows that X2 = T2a (X), where Ta is the translation that sends l1 to
l2 , and a ⊥ l1 .
b) Consider a coordinate system with O as the origin and the absciss axis directed along
line l1 . Let the angle of rotation from line l1 to l2 in this coordinate system be equal to
α and the angles of rotation from the absciss axes to rays OX, OX1 and OX2 be equal to
ϕ, ϕ1 and ϕ2 , respectively. Clearly, ϕ1 = −ϕ and ϕ2 = (α − ϕ1 ) + α = ϕ + 2α. Since
2α
α
OX = OX1 = OX2 , it follows that X2 = RO
(X), where RO
is the translation that sends l1
to l2 .
17.23. Let us represent T ◦ T as the composition of three transformations:
T ◦ T = (Sa ◦ Sb ◦ Sc ) ◦ (Sa ◦ Sb ◦ Sc ) = (Sa ◦ Sb ) ◦ (Sc ◦ Sa ) ◦ (Sb ◦ Sc ).
Here Sa ◦ Sb , Sc ◦ Sa and Sb ◦ Sc are rotations through the angles of 2∠(b, a), 2∠(a, c) and
2∠(c, b), respectively. The sum of the angles of the rotations is equal to
2(∠(b, a) + ∠(a, c) + ∠(c, b)) = 2∠(b, b) = 0◦
and this value is determined up to 2 · 180◦ = 360◦ . It follows that this composition of
rotations is a parallel translation, cf. Problem 18.33.
17.24. If points X and Y are symmetric through line l3 , then points Sl1 (X) and Sl1 (Y )
are symmetric through line l2 , i.e., Sl1 (X) = Sl2 ◦ Sl1 (Y ). It follows that Sl1 ◦ Sl3 = Sl2 ◦ Sl1
and Sl3 = Sl1 ◦ Sl2 ◦ Sl1 .
17.25. Let O be the center of the inscribed circle; let a and b be lines OA and OB.
Then Sa ◦ Sb (C1 ) = Sa (A1 ) = A2 and Sb ◦ Sa (C1 ) = Sb (B1 ) = B2 . Points A2 and B2 are
obtained from point C1 by rotations with center O through opposite angles and, therefore,
A2 B2 k AB.
Similar arguments show that the sides of triangles ABC and A2 B2 C2 are parallel and,
therefore, these triangles are homothetic. Lines AA2 , BB2 and CC2 pass through the center
of homothety which sends triangle ABC to A2 B2 C2 . Notice that this homothety sends the
circumscribed circle of triangle ABC into the inscribed circle, i.e., the center of homothety
belongs to the line that connects the centers of these circles.
17.26. For every jump vector there are precisely two positions of a grasshopper for which
the jump is given by this vector. Therefore, a sequence of jumps is periodic if and only if
there exists but a finite number of distinct jump vectors.
Let a1 be the jump vector of the grasshopper from line l2 to line l1 ; let a2 , a3 , a4 ,. . . be
vectors of the successive jumps. Then a2 = Sl2 (a1 ), a3 = Sl1 (a2 ), a4 = Sl2 (a3 ) , . . . Since the
composition Sl1 ◦ Sl2 is a rotation through an angle of 2γ (or 2π − 2γ), it follows that vectors
a3 , a5 , a7 , . . . are obtained from a1 by rotations through angles of 2γ, 4γ, 6γ, . . . (or through
angles of 2(π − γ), 4(π − γ), 6(π − γ), . . . ). Therefore, the set a1 , a3 , a5 , . . . contains a finite
number of distinct vectors if and only if γ/π is a rational number. The set a2 , a4 , a6 , . . . is
similarly considered.
17.27. a) Suppose polygon A1 A2 . . . An is constructed. Let us draw through the center
O of the circle the midperpendiculars l1 , l2 , . . . , ln to chords A1 A2 , A2 A3 , . . . , An A1 ,
respectively. Lines l1 , . . . , ln are known since they pass through O and are perpendicular to
SOLUTIONS
343
the given lines. Moreover, A2 = Sl1 (A1 ), A3 = Sl2 (A2 ), . . . , A1 = Sln (An ), i.e., point A1 is
a fixed point of the composition of symmetries Sln ◦ · · · ◦ Sl1 . For n odd there are precisely
two fixed points on the circle; for n even there are either no fixed points or all the points are
fixed.
b) Suppose the desired polygon A1 . . . An is constructed. Consider polygon B1 . . . Bn
formed by the tangent points of the circumscribed polygon with the circle. The sides of
polygon B1 . . . Bn are perpendicular to the given lines, i.e., they have prescribed directions
and, therefore, the polygon can be constructed (see heading a)); it remains to draw the
tangents to the circle at points B1 , . . . , Bn .
17.28. Consider the composition of consecutive symmetries through given lines l1 , . . . ,
ln . In heading a) for vertex A1 of the desired n-gon we have to take a fixed point of this
composition, and in heading b) for line A1 An we have to take the(a) fixed line.
17.29. The consecutive symmetries through lines l1 , . . . , ln−1 perpendicular to given
lines and passing through the center of the circle send vertex A1 of the desired polygon to
vertex An .
If n is odd, then the composition of these symmetries is a rotation through a known angle
and, therefore, we have to draw through point M chord A1 An of known length.
If n is even, then the considered composition is a symmetry through a line and, therefore,
from M we have to drop perpendicular to this line.
17.30. Let O be the center of the given disk, DR the disk of radius R with center O. Let
us prove that the symmetries through the lines passing through D1 send the set of images of
points of DR into disk DR+2 . Indeed, the images of point O under the indicated symmetries
fill in disk D2 and the disks of radius R with centers in D2 fill in disk DR+2 .
It follows that after n reflexions we can obtain from points of D1 any point of D2n+1 and
only them. It remains to notice that point A can be “herded” inside DR after n reflexions if
and only if we can transform any point of DR into A after n reflexions.
17.31. Denote symmetries through lines OA1 , . . . , OAn by S1 , . . . , Sn , respectively. Let
Xk = Sk (X) for k = 1, . . . , n. We have to prove that under a rotation through point O the
system of points X1 , . . . , Xn turns into itself. Clearly,
Sk+1 ◦ Sk (Xk ) = Sk+1 ◦ Sk ◦ Sk (X) = Xk+1 .
Transformations Sk+1 ◦ Sk are rotations about O through an angle of
b).
4π
,
n
see Problem 17.22
Remark. For n even we get an n2 -gon.
17.32. Let lines l1 and l2 be axes of symmetry of a plane figure. This means that if point
X belongs to the figure, then points Sl1 (X) and Sl2 (X) also belong to the figure. Consider
line l3 = Sl1 (l2 ). Thanks to Problem 17.24 Sl3 (X) = Sl1 ◦ Sl2 ◦ Sl1 (X) and, therefore, l3 is
also an axis of symmetry.
If the figure has precisely two axes of symmetry, then either l3 = l1 or l3 = l2 . Clearly,
l3 6= l1 and, therefore, l3 = l2 i.e., line l2 is perpendicular to line l1 .
17.33. Suppose that the polygon has three axes of symmetry which do not intersect at
one point, i.e., they form a triangle. Let X be the point of the polygon most distant from
an inner point M of this triangle. Points X and M lie on one side of one of the considered
axes of symmetry, l. If X ′ is the point symmetric to X through l, then M ′ X > M X and
point X ′ is distant from M further than X. The obtained contradiction implies that all the
axes of symmetry of a polygon intersect at one point.
17.34. All the axes of symmetry pass through one point O (Problem 17.33). If l1 and
l2 are axes of symmetry, then l3 = Sl1 (l2 ) is also an axis of symmetry, see Problem 17.24.
344
CHAPTER 17. THE SYMMETRY THROUGH A LINE
Select one of the axes of symmetry l of our polygon. The odd axes of symmetry are divided
into pairs of lines symmetric through l. If line l1 perpendicular to l and passing through O
is not an axis of symmetry, then there is an odd number of axes of symmetry. Therefore, l1
180◦
is a central symmetry i.e., O is the center
is an axis of symmetry. Clearly, Sl1 ◦ Sl = RO
of symmetry.
17.35. Let F be a movement sending point A into A′ and such that A and A′ are distinct;
S the symmetry through the midperpendicular l to segment AA′ . Then S ◦ F (A) = A, i.e.,
A is a fixed point of S ◦ F . Moreover, if X is a fixed point of transformation F , then
AX = A′ X, i.e., point X belongs to line l; hence, X is a fixed point of S ◦ F . Thus, point
A and all the fixed points of F are fixed points of the transformation S ◦ F .
Take points A, B and C not on one line and consider their images under the given
movement G. We can construct transformations S1 , S2 and S3 which are either symmetries
through lines or identity transformations such that S3 ◦ S2 ◦ S1 ◦ G preserves points A, B
and C, i.e., is the identity transformation E. Multiplying the equality S3 ◦ S2 ◦ S1 ◦ G = E
from the left consecutively by S3 , S2 and S1 and taking into account that Si ◦ Si = E we get
G = S 1 ◦ S2 ◦ S3 .
17.36. Thanks to Problem 17.35 any first type movement is a composition of two
symmetries through lines. It remains to make use of the result of Problem 17.22.
17.37. By Problem 17.35 any second type movement can be represented in the form
S3 ◦ S2 ◦ S1 , where S1 , S2 and S3 are symmetries through lines l1 , l2 and l3 , respectively.
First, suppose that the lines l2 and l3 are not parallel. Then under the rotation of the lines
l2 and l3 about their intersection point through any angle the composition S3 ◦ S2 does not
change (see Problem 17.22 b)), consequently, we can assume that l2 ⊥ l1 . It remains to
rotate lines l1 and l2 about their intersection point so that line l2 became parallel to line l3 .
Now, suppose that l2 k l3 . If line l1 is not parallel to these lines, then it is possible to
rotate l1 and l2 about their intersection point so that lines l2 and l3 become nonparallel. If
l1 k l2 , then it is possible to perform a parallel transport of l1 and l2 so that lines l2 and l3
coincide.
Chapter 18. ROTATIONS
Background
1. We will not give a rigorous definition of a rotation. To solve the problems it suffices
to have the following idea on the notion of the rotation: a rotation with center O (or about
the point O) through an angle of ϕ is the transformation of the plane which sends point X
into point X ′ such that:
a) OX ′ = OX;
−−→
−−→
b) the angle from vector OX to vector OX ′ is equal to ϕ.
2. In this chapter we make use of the following notations for the transformations and
their compositions:
Ta is a translation by vector a;
SO is the symmetry through point O;
Sl is the symmetry through line l;
ϕ
RO
is the rotation with center O through an angle of ϕ;
F ◦ G is the composition of transformations F and G defined as (F ◦ G)(X) = F (G(X)).
3. The problems solvable with the help of rotations can be divided into two big classes:
problems which do not use the properties of compositions of rotations and properties which
make use of these properties. To solve the problems which make use of the properties of the
β
α
compositions of rotations the following result of Problem 18.33 is handy: RB
= RCγ ,
◦ RA
1
1
where γ = α + β and ∠BAC = 2 α, ∠ABC = 2 β.
Introductory problems
1. Prove that any rotation sends any circle into a circle.
2. Prove that a convex n-gon is a regular one if and only if it turns into itself under the
◦
about a point.
rotation through an angle of 360
n
3. Prove that triangle ABC is an equilateral one if and only if under the rotation through
60◦ (either clockwise or counterclockwise) about point A vertex B turns into vertex C.
4. Prove that the midpoints of the sides of a regular polygon determine a regular polygon.
5. Through the center of a square two perpendicular lines are drawn. Prove that their
intersection points with the sides of the square determine a square.
§1. Rotation by 90◦
18.1. On sides BC and CD of square ABCD points M and K, respectively, are taken
so that ∠BAM = ∠M AK. Prove that BM + KD = AK.
18.2. In triangle ABC median CM and height CH are drawn. Through an arbitrary
point P of the plane in which ABC lies the lines are drawn perpendicularly to CA, CM and
CB. They intersect CH at points A1 , M1 and B1 , respectively. Prove that A1 M1 = B1 M1 .
18.3. Two squares BCDA and BKM N have a common vertex B. Prove that median
BE of triangle ABK and height BF of triangle CBH belong to one line.
The vertices of each square are counted clockwise.
345
346
CHAPTER 18. ROTATIONS
18.4. Inside square A1 A2 A3 A4 point P is taken. From vertex A1 we drop the perpendicular on A2 P ; from A2 on A3 P ; from A3 on A4 P and from A4 on A1 P . Prove that all four
perpendiculars (or their extensions) intersect at one point.
18.5. On sides CB and CD of square ABCD points M and K are taken so that the
perimeter of triangle CM K is equal to the doubled length of the square’s side. Find the
value of angle ∠M AK.
18.6. On the plane three squares (with same orientation) are given: ABCD, AB1 C1 D1
and A2 B2 CD2 ; the first square has common vertices A and C with the two other squares.
Prove that median BM of triangle BB1 B2 is perpendicular to segment D1 D2 .
18.7. Triangle ABC is given. On its sides AB and BC squares ABM N and BCP Q are
constructed outwards. Prove that the centers of these squares and the midpoints of segments
M Q and AC form a square.
18.8. A parallelogram is circumscribed about a square. Prove that the perpendiculars
dropped from the vertices of the parallelograms to the sides of the square form a square.
§2. Rotation by 60◦
18.9. On segment AE, on one side of it, equilateral triangles ABC and CDE are
constructed; M and P are the midpoints of segments AD and BE. Prove that triangle
CP M is an equilateral one.
18.10. Given three parallel lines. Construct an equilateral triangle so that its vertices
belong to the given lines.
18.11. Geven a square, consider all possible equilateral triangles P KM with fixed vertex
P and vertex K belonging to the square. Find the locus of vertices M .
18.12. On sides BC and CD of parallelogram ABCD, equilateral triangles BCP and
CDQ are constructed outwards. Prove that triangle AP Q is an equilateral one.
18.13. Point M belongs to arc ⌣ AB of the circle circumscribed about an equilateral
triangle ABC. Prove that M C = M A + M B.
18.14. Find the locus of points M that lie inside equilateral triangle ABC and such that
M A2 = M B 2 + M C 2 .
18.15. Hexagon ABCDEF is a regular one, K and M are the midpoints of segments
BD and EF , respectively. Prove that triangle AM K is an equilateral one.
18.16. Let M and N be the midpoints of sides CD and DE, respectively, of regular
hexagon ABCDEF , let P be the intersection point of segments AM and BN .
a) Find the value of the angle between lines AM and BN .
b) Prove that SABP = SM DN P .
18.17. On sides AB and BC of an equilateral triangle ABC points M and N are taken
so that M N k AC; let E be the midpoint of segment AN and D the center of mass of
triangle BM N . Find the values of the angles of triangle CDE.
18.18. On the sides of triangle ABC equilateral triangles ABC1 , AB1 C and A1 BC are
constructed outwards. Let P and Q be the midpoints of segments A1 B1 and A1 C1 . Prove
that triangle AP Q is an equilateral one.
18.19. On sides AB and AC of triangle ABC equilateral triangles ABC ′ and AB ′ C are
constructed outwards. Point M divides side BC in the ratio of BM : M C = 3 : 1; points
K and L are the midpoints of sides AC ′ and B ′ C, respectively. Prove that the angles of
triangle KLM are equal to 30◦ , 60◦ and 90◦ .
18.20. Equilateral triangles ABC, CDE, EHK (vertices are circumvent counterclock−−→
−−→
wise) are placed on the plane so that AD = DK. Prove that triangle BHD is also an
equilateral one.
§4. COMPOSITIONS OF ROTATIONS
347
18.21. a) Inside an acute triangle find a point the sum of distances from which to the
vertices is the least one.
b) Inside triangle ABC all the angles of which are smaller than 120◦ a point O is taken; it
serves as vertex of the angles of 120◦ that subtend the√sides. Prove that the sum of distances
from O to the vertices is equal to 12 (a2 + b2 + c2 ) + 2 3S.
18.22. Hexagon ABCDEF is inscribed in a circle of radius R and AB = CD = EF = R.
Prove that the midpoints of sides BC, DE and F A determine an equilateral triangle.
18.23. On sides of a convex centrally symmetric hexagon ABCDEF equilateral triangles
are constructed outwards. Prove that the midpoints of the segments connecting the vertices
of neighbouring triangles determine a regular hexagon.
§3. Rotations through arbitrary angles
18.24. Given points A and B and circle S construct points C and D on S so that
AC k BD and the value of arc ⌣ CD is a given quantity α.
18.25. A rotation with center O transforms line l1 into line l2 and point A1 on l1 into point
A2 . Prove that the intersection point of lines l1 and l2 belongs to the circle circumscribed
about triangle A1 OA2 .
18.26. Two equal letters Γ lie on the plane. Denote by A and A′ the endpoints of
the shorter segments of these letters. Points A1 , . . . , An−1 and A′1 , . . . , A′n−1 divide the
longer segments into n equal parts (the division points are numbered starting from the outer
endpoints of longer segments). Lines AAi and A′ A′i intersect at point Xi . Prove that points
X1 , . . . , Xn−1 determine a convex polygon.
18.27. Along two lines that intersect at point P two points are moving with the same
speed: point A along one line and point B along the other one. They pass P not simultaneously. Prove that at all times the circle circumscribed about triangle ABP passes through
a fixed point distinct from P .
18.28. Triangle A1 B1 C1 is obtained from triangle ABC by a rotation through an angle
of α (α < 180◦ ) about the center of its circumscribed circle. Prove that the intersection
points of sides AB and A1 B1 , BC and B1 C1 , CA and C1 A1 (or their extensions) are the
vertices of a triangle similar to triangle ABC.
18.29. Given triangle ABC construct a line which divides the area and perimeter of
triangle ABC in halves.
−−→
18.30. On vectors Ai Bi , where i = 1, . . . , k similarly oriented regular n-gons Ai Bi Ci Di . . .
(n ≥ 4) are constructed (a given vector serving as a side). Prove that k-gons C1 . . . Ck and
D1 . . . Dk are regular and similarly oriented ones if and only if the k-gons A1 . . . Ak and
B1 . . . Bk are regular and similarly oriented ones.
18.31. Consider a triangle. Consider three lines symmetric through the triangles sides
to an arbitrary line passing through the intersection point of the triangle’s heights. Prove
that the three lines intersect at one point.
18.32. A lion runs over the arena of a circus which is a disk of radius 10 m. Moving
along a broken line the lion covered 30 km. Prove that the sum of all the angles of his turns
is not less than 2998 radian.
§4. Compositions of rotations
18.33. Prove that the composition of two rotations through angles whose sum is not
proportional to 360◦ is a rotation. In which point is its center and what is the angle of the
rotation equal to? Investigate also the case when the sum of the angles of rotations is a
multiple of 360◦ .
348
CHAPTER 18. ROTATIONS
***
18.34. On the sides of an arbitrary convex quadrilateral squares are constructed outwards. Prove that the segments that connect the centers of opposite squares have equal
lengths and are perpendicular to each other.
18.35. On the sides of a parallelogram squares are constructed outwards. Prove that
their centers form a square.
18.36. On sides of triangle ABC squares with centers P , Q and R are constructed
outwards. On the sides of triangle P QR squares are constructed inwards. Prove that their
centers are the midpoints of the sides of triangle ABC.
18.37. Inside a convex quadrilateral ABCD isosceles right triangles ABO1 , BCO2 ,
CDO3 and DAO4 are constructed. Prove that if O1 = O3 , then O2 = O4 .
***
18.38. a) On the sides of an arbitrary triangle equilateral triangles are constructed
outwards. Prove that their centers form an equilateral triangle.
b) Prove a similar statement for triangles constructed inwards.
c) Prove that the difference of the areas of equilateral triangles obtained in headings a)
and b) is equal to the area of the initial triangle.
18.39. On sides of triangle ABC equilateral triangles A′ BC and B ′ AC are constructed
outwards and C ′ AB inwards; M is the center of mass of triangle C ′ AB. Prove that A′ B ′ M
is an isosceles triangle such that ∠A′ M B ′ = 120◦ .
18.40. Let angles α, β, γ be such that 0 < α, β, γ < π and α + β + γ = π. Prove that if
2β
2α
the composition of rotations RC2γ ◦ RB
is the identity transformation, then the angles
◦ RA
of triangle ABC are equal to α, β, γ.
18.41. Construct an n-gon given n points which are the vertices of isosceles triangles
constructed on the sides of this n-gon and such that the angles of these triangles at the
vertices are equal to α1 , . . . , αn .
18.42. On the sides of an arbitrary triangle ABC isosceles triangles A′ BC, AB ′ C and
ABC ′ are constructed outwards with angles α, β and γ at vertices A′ , B ′ and C ′ , respectively,
such that α + β + γ = 2π. Prove that the angles of triangle A′ B ′ C ′ are equal to 21 α, 12 β and
1
γ.
2
18.43. Let AKL and AM N be similar isosceles triangles with vertex A and angle α at
the vertex; GN K and G′ LM similar isosceles triangles with angle π − α at the vertex. Prove
that G = G′ . (All the triangles are oriented ones.)
18.44. On sides AB, BC and CA of triangle ABC points P , Q and R, respectively, are
taken. Prove that the centers of the circles circumscribed about triangles AP R, BP Q and
CQR constitute a triangle similar to triangle ABC.
Problems for independent study
18.45. On the plane, the unit circle with center at O is drawn. Two neighbouring
vertices of a square belong to this circle. What is the maximal distance from point O that
the two other of the square’s vertices can have?
18.46. On the sides of convex quadrilateral ABCD, equilateral triangles ABM , CDP
are constructed outwards and BCN , ADK inwards. Prove that M N = AC.
18.47. On the sides of a convex quadrilateral ABCD, squares with centers M , N , P ,
Q are constructed outwards. Prove that the midpoints of the diagonals of quadrilaterals
ABCD and M N P Q form a square.
SOLUTIONS
349
18.48. Inside an equilateral triangle ABC lies point O. It is known that ∠AOB = 113◦ ,
∠BOC = 123◦ . Find the angles of the triangle whose sides are equal to segments OA, OB,
OC.
18.49. On the plane, there are drawn n lines (n > 2) so that no two of them are parallel
and no three intersect at one point. It is known that it is possible to rotate the plane about
a point O through an angle of α (α < 180◦ ) so that each of the drawn lines coincides with
some other of the drawn lines. Indicate all n for which this is possible.
18.50. Ten gears of distinct shapes are placed so that the first gear is meshed with the
second one, the second one with the third one, etc., the tenth is meshed with the first one.
Is it possible for such a system to rotate? Can a similar system of 11 gears rotate?
18.51. Given a circle and a point. a) Construct an equilateral triangle whose heights
intersect at the given point and two vertices belong to the given circle.
b) Construct a square two vertices of which belong to the given circle and the diagonals
intersect at the given point.
Solutions
18.1. Let us rotate square ABCD about point A through 90◦ so that B turns into
D. This rotation sends point M into point M ′ and point K into point K ′ . It is clear
that ∠BM A = ∠DM ′ A. Since ∠M AK = ∠M AB = ∠M ′ AD, it follows that ∠M AD =
∠M ′ AK. Therefore,
∠M A′ K = ∠M AD = ∠BM A = ∠DM ′ A.
Hence, AK = KM ′ = KD + DM ′ = KD + BM .
18.2. Under the rotation through 90◦ about point P lines P A1 , P B1 , P M1 and CH
turn into lines parallel to CA, CB, CM and AB, respectively. It follows that under such a
rotation of triangle P A1 B1 segment P M1 turns into a median of the (rotated) triangle.
18.3. Consider a rotation through 90◦ about point B which sends vertex K into vertex
N and vertex C into A. This rotation sends point A into point A′ and point E into point
E ′ . Since E ′ and B are the midpoints of sides A′ N and A′ C of triangle A′ N C, it follows
that BE ′ k N C. But ∠EBE ′ = 90◦ and, therefore, BE ⊥ N C.
18.4. A rotation through an angle of 90◦ about the center of the square sends point A1
to point A2 . This rotation sends the perpendiculars dropped from points A1 , A2 , A3 and
A4 into lines A2 P , A3 P , A4 P and A1 P , respectively. Therefore, the intersection point is the
image of point P under the inverse rotation.
18.5. Let us turn the given square through an angle of 90◦ about point A so that vertex
B would coincide with D. Let M ′ be the image of M under this rotation. Since by the
hypothesis
M K + M C + CK = (BM + M C) + (KD + CK),
it follows that M K = BM + KD = DM ′ + KD = KM ′ . Moreover, AM = AM ′ ; hence,
△AM K = △AM ′ K, consequently, ∠M AK = ∠M ′ AK = 21 ∠M AM ′ = 45◦ .
−−→
−→
18.6. Let R be the rotation through an angle of 90◦ that sends BC to BA. Further,
−−→
−−→
−−→
−→
−−→
−−→
let BC = a, CB2 = b and AB1 = c. Then BA = Ra, D2 C = Rb and AD1 = Rc. Hence,
−−−→
−−→
−−→
−−−→
D2 D1 = Rb − a + Ra + Rc and 2BM = a + b + Ra + c. Therefore, R(2BM ) = D2 D1
because R(Ra) = −a.
−−→
−−→
18.7. Let us introduce the following notations: a = BM , b = BC; let Ra and Rb be
the vectors obtained from vectors a and b under a rotation through an angle of 90◦ , i.e.,
−→
−−→
Ra = BA, Rb = BQ. Let O1 , O2 , O3 and O4 be the midpoints of segments AM , M Q, QC
350
CHAPTER 18. ROTATIONS
and CA, respectively. Then
−−→ (a + Ra) −−→ (a + Rb)
, BO2 =
,
BO1 =
2
2
−−→ (b − Rb) −−→ (b + Ra)
BO3 =
, BO4 =
.
2
2
−−−→
−−−→
−−−→
Therefore, O1 O2 = 21 (Rb − Ra) = −O3 O4 and O2 O3 = 12 (b − a) = −O4 O1 . Moreover,
−−−→
−−−→
O1 O2 = R(O2 O3 ).
18.8. Parallelogram A1 B1 C1 D1 is circumscribed around square ABCD so that point A
belongs to side A1 B1 , B to side B1 C1 , etc. Let us drop perpendiculars l1 , l2 , l3 and l4 from
vertices A1 , B1 , C1 and D1 , respectively to the sides of the square. To prove that these
perpendiculars form a square, it suffices to verify that under a rotation through an angle of
90◦ about the center O of square ABCD lines l1 , l2 , l3 and l4 turn into each other. Under
the rotation about O through an angle of 90◦ points A1 , B1 , C1 and D1 turn into points A2 ,
B2 , C2 and D2 (Fig. 21).
Figure 163 (Sol. 18.8)
Since AA2 ⊥ B1 B and BA2 ⊥ B1 A, it follows that B1 A2 ⊥ AB. This means that line
l1 turns under the rotation through an angle of 90◦ about O into l2 . For the other lines the
proof is similar.
18.9. Let us consider a rotation through an angle of 60◦ about point C that turns E
into D. Under this rotation B turns into A, i.e., segment BE turns into AD. Therefore, the
midpoint P of segment BE turns into the midpoint M of segment AD, i.e., triangle CP M
is an equilateral one.
18.10. Suppose that we have constructed triangle ABC so that its vertices A, B and C
lie on lines l1 , l2 and l3 , respectively. Under the rotation through an angle of 60◦ with center
A point B turns into point C and, therefore, C is the intersection point of l3 and the image
of l2 under the rotation through an angle of 60◦ about A.
18.11. The locus to be found consists of two squares obtained from the given one by
rotations through angles of ±60◦ about P .
SOLUTIONS
351
−→
−→
−−→
18.12. Under the rotation through an angle of 60◦ vectors QC and CP turn into QD
−−→ −−→
−→ −→ −→
and CB = DA, respectively. Therefore, under this rotation vector QP = QC + CP turns
−−→ −−→ −→
into vector QD + DA = QA.
18.13. Let M ′ be the image of M under the rotation through an angle of 60◦ about B
that turns A into C. Then ∠CM ′ B = ∠AM B = 120◦ . Triangle M M ′ B is an equilateral
one and, therefore, ∠BM ′ M = 60◦ . Since ∠CM ′ B + ∠BM ′ M = 180◦ , point M ′ belongs to
segment M C. Therefore, M C = M M ′ + M ′ C = M B + M A.
18.14. Under the rotation through an angle of 60◦ about A sending B to C point M turns
into point M ′ and point C into point D. The equality M A2 = M B 2 + M C 2 is equivalent to
the equality M ′ M 2 = M ′ C 2 + M C 2 , i.e., ∠M CM ′ = 90◦ and, therefore,
∠M CB + ∠M BC = ∠M CB + ∠M ′ CD = 120◦ − 90◦ = 30◦
that is ∠BM C = 150◦ . The locus to be found is the arc of the circle situated inside the
triangle and such that the pionts of the arc serve as vertices of angles of 150◦ subtending
segment BC.
18.15. Let O be the center of a hexagon. Consider a rotation about A through an angle
of 60◦ sending B to O. This rotation sends segment OC into segment F E. Point K is the
midpoint of diagonal BD of parallelogram BCDO because it is the midpoint of diagonal
CO. Therefore, point K turns into M under our rotation; in other words, triangle AM K is
an equilateral one.
18.16. There is a rotation through an angle of 60◦ about the center of the given hexagon
that sends A into B. It sends segment CD into DE and, therefore, sends M into N .
Therefore, this rotation sends AM into BN , that is to say, the angle between these segments
is equal to 60◦ . Moreover, this rotation turns pentagon AM DEF into BN EF A; hence, the
areas of the pentagons are equal. Cutting from these congruent pentagons the common part,
pentagon AP N EF , we get two figures of the same area: triangle ABP and quadrilateral
M DN P .
18.17. Consider the rotation through an angle of 60◦ about C sending B to A. It sends
points M , N and D into M ′ , N ′ and D′ , respectively. Since AM N N ′ is a parallelogram,
the midpoint E of diagonal AN is its center of symmetry. Therefore, under the symmetry
through point E triangle BM N turns into M ′ AN ′ and, therefore, D turns into D′ . Hence,
E is the midpoint of segment DD′ . Since triangle CDD′ is an equilateral one, the angles of
triangle CDE are equal to 30◦ , 60◦ and 90◦ .
18.18. Consider a rotation about A sending point C1 into B. Under this rotation
equilateral triangle A1 BC turns into triangle A2 F B1 and segment A1 C1 into segment A2 B.
It remains to notice that BA1 A2 B1 is a parallelogram, i.e., the midpoint of segment A2 B
coincides with the midpoint of segment A1 B1 .
−→
−→
−→
18.19. Let AB = 4a, CA = 4b. Further, let R be the rotation sending vector AB
−−→
−−→
−→
−−→
−−→
into AC ′ (and, therefore, sending CA into CB ′ ). Then LM = (a + b) − 2Rb and LK =
−−→
−−→
−2Rb + 4b + 2Ra. It is easy to verify that b + R2 b = Rb. Hence, 2R(LM ) = LK which
implies the required statement.
18.20. Under the rotation about point C through an angle of 60◦ counterclockwise point
−−→ −−→
−−→
A turns into B and D into E and, therefore, vector DK = AD turns into BE. Since the
−−→
rotation about point H through an angle of 60◦ counterclockwise sends K into E and DK
−−→
into BE, it sends D into B which means that triangle BHD is an equilateral one.
18.21. a) Let O be a point inside triangle ABC. The rotation through an angle of 60◦
about A sends B, C and O into some points B ′ , C ′ and O′ , respectively, see Fig. 22. Since
352
CHAPTER 18. ROTATIONS
Figure 164 (Sol. 18.21)
AO = OO′ and OC = O′ C ′ , we have:
BO + AO + CO = BO + OO′ + O′ C ′ .
The length of the broken line BOO′ C ′ is minimal if and only if this broken line is a segment,
i.e., if ∠AOB = ∠AO′ C ′ = ∠AOC = 120◦ . To construct the desired point, we can make
use of the result of Problem 2.8.
b) The sum of distances from O to the vertices is equal to the length of segment BC ′
obtained in heading a). It is also clear that
◦
(BC ′ )2 = b2 + c2 − 2bc cos(α
√ + 60 ) =
2
2
3 sin α =
b + c − bc cos α + bc √
1 2
2
2
(a + b + c ) + 2 3S.
2
18.22. Let P , Q and R be the midpoints of sides BC, DE and F A; let O be the center
of the circumscribed circle. Suppose that triangle P QR is an equilateral one. Let us prove
then that the midpoints of sides BC, DE ′ and F ′ A of hexagon ABCDE ′ F ′ in which vertices
E ′ and F ′ are obtained from vertices E and F after a rotation through an angle about point
O also form an equilateral triangle.
This will complete the proof since for a regular hexagon the midpoints of sides BC,
DE and F A constitute an equilateral triangle and any of the considered hexagons can be
obtained from a regular one with the help of rotations of triangles OCD and OEF .
Figure 165 (Sol. 18.22)
Let Q′ and R′ be the midpoints of sides DE ′ and AF ′ , see Fig. 23. Under the rotation
−−→
−−→
−−→
−−→
−−→
−−→
through an angle of 60◦ vector EE ′ turns into F F ′ . Since QQ′ = 21 EE ′ and RR′ = 12 F F ′ ,
−−→
−−→
this rotation sends QQ′ into RR′ . By hypothesis, triangle P QR is an equilateral one, i.e.,
−→
−→
under the rotation through an angle of 60◦ vector P Q turns into P R. Therefore, vector
SOLUTIONS
353
−−→′ −→ −−→′
−−→ −→ −−→
P Q = P Q + QQ turns into vector P R′ = P R + RR′ under a rotation through an angle of
60◦ . This means that triangle P Q′ R′ is an equilateral one.
18.23. Let K, L, M and N be vertices of equilateral triangles constructed (wherewards?)
on sides BC, AB, AF and F E, respectively; let also B1 , A1 and F1 be the midpoints of
segments KL, LM and M N (see Fig. 24).
Figure 166 (Sol. 18.23)
−−→
−→
−→
−→
Further, let a = BC = F E, b = AB and c = AF ; let R be the rotation through an
−−→
−−→
−−→
−−→
angle of 60◦ that sends BC into BK. Then AM = −R2 c and F N = −R2 a. Therefore,
−−−→
−−−→
−−−→
−−−→
2A1 B1 = R2 c + Ra + b and 2F1 A1 = R2 a − c + Rb, i.e., F1 A1 = R(A1 B1 ).
18.24. Suppose a rotation through an angle of α about the center of circle S sends C
into D. This rotation sends point A into point A′ . Then ∠(BD, DA′ ) = α, i.e., point D
belongs to the arc of the circle whose points serve as vertices of the angles of α that subtend
segment A′ B.
18.25. Let P be the intersection point of lines l1 and l2 . Then
∠(OA1 , A1 P ) = ∠(OA1 , l1 ) = ∠(OA2 , l2 ) = ∠(OA2 , A2 P ).
Therefore, points O, A1 , A2 and P belong to one circle.
18.26. It is possible to identify similar letters Γ after a rotation about O (unless they
can be identified by a parallel translation in which case AAi k A′ A′i ). Thanks to Problem
18.25 point Xi belongs to the circle circumscribed about triangle A′ OA. It is clear that the
points that belong to one circle constitute a convex polygon.
18.27. Let O be the center of rotation R that sends segment A(t1 )A(t2 ) into segment
B(t1 )B(t2 ), where t1 and t2 are certain time moments. Then this rotation sends A(t) into
B(t) at any moment t. Therefore, by Problem 18.25 point O belongs to the circle circumscribed about triangle AP B.
18.28. Let A and B be points on the circle with center O; let A1 and B1 be the images
of these points under the rotation through an angle of α about O. Let P and P1 be the
midpoints of segments AB and A1 B1 ; let M be the intersection point of lines AB and
A1 B1 . The right triangles P OM and P1 OM have a common hypothenuse and equal legs
P O = P1 O, therefore, these triangles are equal and ∠M OP = ∠M OP1 = 12 α. Point M is
obtained from point P under a rotation through an angle of 21 α and a subsequent homothety
with coefficient cos(11 α) and center O.
2
The intersection points of lines AB and A1 B1 , AC and A1 C1 , BC and B1 C1 are the
vertices of a triangle which is homothetic with coefficient cos(11 α) to the triangle determined
2
354
CHAPTER 18. ROTATIONS
by the midpoints of the sides of triangle ABC. It is clear that the triangle determined by
the midpoints of the sides of triangle ABC is similar to triangle ABC.
18.29. By Problem 5.50 the line which divides in halves both the area and the perimeter
of a triangle passes through the center of its inscribed circle. It is also clear that if the line
passes through the center of the inscribed circle of a triangle and divides its perimeter in
halves, then it divides in halves its area as well. Therefore, we have to draw a line passing
through the center of the inscribed circle of the triangle and dividing its perimeter in halves.
Suppose we have constructed points M and N on sides AB and AC of triangle ABC so
that line M N passes through the center O of the inscribed circle and divides the perimeter
of the triangle in halves. On ray AC construct point D so that AD = p, where p is a
semiperimeter of triangle ABC. Then AM = N D. Let Q be the center of rotation R
that sends segment AM into segment DN (so that A goes to D and M to N ). Since the
angle between lines AM and CN is known, it is possible to construct Q: it is the vertex
of isosceles triangle AQD, where ∠AQD = 180◦ − ∠A and points B and Q lie on one side
of line AD. The rotation R sends segment OM into segment O′ N . We can now construct
point O′ . Clearly, ∠ON O′ = ∠A because the angle between lines OM and O′ N is equal to
∠A. Therefore, point N is the intersection point of line AC and the arc of the circle whose
points serve as vertices for the angles equal to ∠A that subtend segment OO′ . Constructing
point N , draw line ON and find point M .
It is easy to verify that if the constructed points M and N belong to sides AB and AC,
then M N is the desired line. The main point of the proof is the proof of the fact that the
rotation about Q through an angle of 180◦ − ∠A sends M into N . To prove this fact, one
has to make use of the fact that ∠ON O′ = ∠A, i.e., this rotation sends line OM into line
O′ N .
18.30. Suppose that the k-gons C1 . . . Ck and D1 . . . Dk are regular and similarly oriented.
−−→
−−→
Let C and D be the centers of these k-gons; let ci = CCi and di = DDi . Then
−−→ −−→ −−→ −−→
−−→
Ci Di = Ci C + CD + DDi = −ci + CD + di .
−−→
−−→
The rotation Rϕ , where ϕ is the angle at a vertex of a regular n-gon, sends Ci Di into Ci Bi .
Therefore,
−−→ −−→
−−→ −−→
−−→
XBi = XC + ci + Ci Bi = XC + ci + Rϕ (−ci + CD + di ).
−−→
−−→
−−→
−
→
Let us select point X so that XC + Rϕ (CD) = 0 . Then XBi = ci + Rϕ (di − ci ) = Riψ u,
where u = ck + Rϕ (dk − ck ) and Rψ is the rotation sending ck to c1 . Hence, B1 . . . Bk is a
regular k-gon with center X.
We similarly prove that A1 . . . Ak is a regular k-gon.
The converse statement is similarly proved.
18.31. Let H be the intersection point of heights of triangle ABC; let H1 , H2 and H3 be
points symmetric to H through sides BC, CA and AB, respectively. Points H1 , H2 and H3
belong to the circle circumscribed about triangle ABC (Problem 5.9). Let l be a line passing
through H. The line symmetric to l through BC (resp. through CA and AB) intersects the
circumscribed circle at point H1 (resp. H2 and H3 ) and at a point P1 (resp. P2 and P3 ).
Consider another line l′ passing through H. Let ϕ be the angle between l and l′ . Let
us construct points P1′ , P2′ and P3′ for line l′ in the same way as points P1 , P2 and P3 were
constructed for line l. Then ∠Pi Hi Pi′ = ϕ, i.e., the value of arc ⌣ Pi Pi′ is equal to 2ϕ
(the direction of the rotation from Pi to Pi′ is opposite to that of the rotation from l to
l′ ). Therefore, points P1′ , P2′ and P3′ are the images of points P1 , P2 and P3 under a certain
rotation. It is clear that if for l′ we take the height of the triangle dropped from vertex A,
then P1′ = P2′ = P3′ = A, and, therefore, P1 = P2 = P3 .
SOLUTIONS
355
18.32. Suppose that the lion ran along the broken line A1 A2 . . . An . Let us rectify the
lion’s trajectory as follows. Let us rotate the arena of the circus and all(?) the further
trajectory about point A2 so that point A3 would lie on ray A1 A2 . Then let us rotate the
arena and the further trajectory about point A3 so that point A4 were on ray A1 A2 , and so
on. The center O of the arena turns consecutively into points O1 = O, O2 , . . . , On−1 ; and
points A1 , . . . , An into points A′1 , . . . , A′n all on one line (Fig. 25).
Figure 167 (Sol. 18.32)
Let αi−1 be the angle of through which the lion turned at point A′i . Then ∠Oi−1 A′i Oi =
αi−1 and A′i Oi−1 = A′i Oi ≤ 10; hence, Oi Oi−1 ≤ 10αi−1 . Hence,
30000 = A′1 A′n ≤ A′1 O1 + O1 O2 + · · · + On−2 On−1 + On−1 A′n ≤
10 + 10(α1 + · · · + αn−2 ) + 10
i.e., α1 + · · · + αn−2 ≥ 2998.
β
α
18.33. Consider the composition of the rotations RB
◦ RA
. If A = B, then the statement
of the problem is obvious and, therefore, let us assume that A 6= B. Let l = AB; let lines
a and b pass through points A and B, respectively, so that ∠(a, l) = 12 α and ∠(l, b) = 12 β.
Then
β
α
RB
◦ RA
= S b ◦ Sl ◦ Sl ◦ Sa = S b ◦ Sa .
If a k b, then Sa ◦ Sb = T2u , where Tu is a parallel translation sending a into b and such
that u ⊥ a. If lines a and b are not parallel and O is their intersection point, then Sa ◦ Sb is
the rotation through an angle of α + β with center O. It is also clear that a k b if and only
if 12 α + 12 β = kπ, i.e., α + β = 2kπ.
18.34. Let P , Q, R and S be the centers of squares constructed outwards on sides
AB, BC, CD and DA, respectively. On segments QR and SP , construct inwards isosceles
180◦
90◦
90◦
(B) and B =
(B) = RO
◦ RQ
right triangles with vertices O1 and O2 . Then D = RR
1
180◦
90◦
90◦
RP ◦ RS (D) = RO2 (D), i.e., O1 = O2 is the midpoint of segment BD.
The rotation through an angle of 90◦ about point O = O1 = O2 that sends Q into R
sends point S into P , i.e., it sends segment QS into RP and, therefore, these segments are
equal and perpendicular to each other.
18.35. Let P , Q, R and S be the centers of squares constructed outwards on the sides
AB, BC, CD and DA of parallelogram ABCD. By the previous problem P R = QS and
P R ⊥ QS. Moreover, the center of symmetry of parallelogram ABCD is the center of
symmetry of quadrilateral P QRS. This means that P QRS is a parallelogram with equal
and perpendicular diagonals, hence, a square.
18.36. Let P , Q and R be the centers of squares constructed outwards on sides AB, BC
and CA. Let us consider a rotation through an angle of 90◦ with center R that sends C to
A. Under the rotation about P through an angle of 90◦ in the same direction point A turns
into B. The composition of these two rotations is a rotation through an angle of 180◦ and,
356
CHAPTER 18. ROTATIONS
therefore, the center of this rotation is the midpoint of segment BC. On the other hand,
the center of this rotation is a vertex of an isoscceles right triangle with base P R, i.e., it is
the center of a square constructed on P R. This square is constructed inwards on a side of
triangle P QR.
◦
180◦
180◦
90◦
90◦
90◦
= E. Therefore,
◦ RO
= RO
◦ RA
◦ RC90 ◦ RB
18.37. If O1 = O3 , then RD
1
3
◦
◦
◦
◦
◦
◦
◦
◦
−90
90
90
90
90
180
180
= RA
E = RA
◦ RD
◦ E ◦ RA
RC90 ◦ RB
= RO
◦ RO
,
4
2
where E is the identity transformation, i.e., O4 = O2 .
18.38. a) See solution of a more general Problem 18.42 (it suffices to set α = β = γ =
120◦ ). In case b) proof is analogous.
b) Let Q and R (resp. Q1 and R1 ) be the centers of equilateral triangles constructed
outwards (resp. inwards) on sides AC and AB. Since AQ = √13 b, AR = √13 c and ∠QAR =
60◦ + α, it follows that 3QR2 = b2 + c2 − 2bc cos(α + 60◦ ). Similarly, 3Q1 R12 = b2 + c2 −
2bc cos(α − 60◦ ). Therefore, the difference of areas of the obtained equilateral triangles is
equal to
√
bc sin α sin 60◦
(QR2 − Q1 R12 ) 3
√
= SABC .
=
4
3
18.39. The combination of a rotation through an angle of 60◦ about A′ that sends B to
C, a rotation through an angle of 60◦ about B ′ that sends C to A and a rotation through
an angle of 120◦ about M that sends A to B has B as a fixed point. Since the first two
rotations are performed in the direction opposite to the direction of the last rotation, it
follows that the composition of these rotations is a parallel translation with a fixed point,
i.e., the identity transformation:
◦
◦
◦
◦
◦
◦
−120
60
60
◦ RB
RM
′ ◦ RA′ = E.
◦
◦
60
60
120
60
60
, i.e., M is the center of the rotation RB
= RM
Therefore, RB
′ ◦ RA′ . It follows
′ ◦ R A′
that ∠M A′ B ′ = ∠M B ′ A′ = 30◦ , i.e., A′ B ′ M is an isosceles triangle and ∠A′ M B ′ = 120◦ .
2β
2α
◦ RA
, i.e., point C is
18.40. The conditions of the problem imply that RC−2γ = RB
2β
2α
the center of the composition of rotations RB ◦ RA . This means that ∠BAC = α and
∠ABC = β (see Problem 18.33). Therefore, ∠ACB = π − α − β = γ.
18.41. Denote the given points by M1 , . . . , Mn . Suppose that we have constructed
polygon A1 A2 . . . An so that triangles A1 M1 A2 , A2 M2 A3 , . . . , An Mn A1 are isosceles, where
∠Ai Mi Ai+1 = αi and the sides of the polygon are bases of these isosceles triangles. Clearly,
α1
αn
RM
◦ · · · ◦ RM
(A1 ) = A1 . If α1 + · · · + αn 6= k · 360◦ , then point A1 is the center of the
n
1
α1
αn
.
rotation RMn ◦ · · · ◦ RM
1
We can construct the center of the composition of rotations. The construction of the
other vertices of the polygon is done in an obvious way. If α1 + · · · + αn = k · 360◦ , then the
problem is ill-posed: either an arbitrary point A1 determines a polygon with the required
property or there are no solutions.
β
γ
β
γ
α
18.42. Since RCγ ′ ◦ RB
′ ◦ RA′ (B) = RC ′ ◦ RB ′ (C) = RC ′ (A) = B, it follows that B is a
β
α
fixed point of the composition RCγ ′ ◦ RB
′ ◦ RA′ . Since α + β + γ = 2π, it follows that this
composition is a parallel translation with a fixed point, i.e., the identity transformation. It
remains to make use of the result of Problem 18.40.
π−α
π−α
α
α
◦ RA
18.43. Since RG
(N ) = L and RG
◦ RA
(L) = N , it follows that the transfor′
π−α
π−α
α
α
mations RG′ ◦ RA and RG ◦ RA are central symmetries with respect to the midpoint of
π−α
π−α
π−α
π−α
α
α
= RG
and G′ = G.
. Therefore, RG
= RG
◦ RA
◦ RA
segment LN , i.e., RG
′
′
18.44. Let A1 , B1 and C1 be the centers of the circumscribed circles of triangles AP R,
BP Q and CQR. Under the successive rotations with centers A1 , B1 and C1 through angles
SOLUTIONS
357
2α, 2β and 2γ point R turns first into P , then into Q, and then returns home. Since
2α+2β+2γ = 360◦ , the composition of the indicated rotations is the identity transformation.
It follows that the angles of triangle A1 B1 C1 are equal to α, β and γ (see Problem 18.40).
Chapter 19. HOMOTHETY AND ROTATIONAL
HOMOTHETY
Background
1. A homothety is a transformation of the plane sending point X into point X ′ such that
−−→′
−−→
OX = k OX, where point O and the number k are fixed. Point O is called the center of
homothety and the number k the coefficient of homothety.
We will denote the homothety with center O and coefficient k by HOk .
2. Two figures are called homothetic if one of them turns into the other one under a
homothety.
3. A rotational homothety is the composition of a homothety and a rotation with a
ϕ
ϕ
common center. The order of the composition is inessential since RO
◦ HOk = HOk ◦ RO
.
180◦
◦
We may assume that the coefficient of a rotational homothety is positive since RO
−k
k
HO = HO .
4. The composition of two homotheties with coefficients k1 and k2 , where k1 k2 6= 1, is a
homothety with coefficient k1 k2 and its center belongs to the line that connects the centers
of these homotheties (see Problem 19.23).
5. The center of a rotational homothety that sends segment AB into segment CD is the
intersection point of the circles circumscribed about triangles ACP and BDP , where P is
the intersection point of lines AB and CD (see Problem 19.41).
Introductory problems
1. Prove that a homothety sends a circle into a circle.
2. Two circles are tangent at point K. A line passing through K intersects these circles
at points A and B. Prove that the tangents to the circles through A and B are parallel to
each other.
3. Two circles are tangent at point K. Through K two lines are drawn that intersect the
first circle at points A and B and the second one at points C and D. Prove that AB k CD.
4. Prove that points symmetric to an arbitrary point with respect to the midpoints of a
square’s sides are vertices of a square.
5. Two points A and B and a line l on the plane are given. What is the trajectory of
movement of the intersection point of medians of triangle ABC when C moves along l?
§1. Homothetic polygons
19.1. A quadrilateral is cut by diagonals into four triangles. Prove that the intersection
points of their medians form a parallelogram.
19.2. The extensions of the lateral sides AB and CD of trapezoid ABCD intersect at
point K and its diagonals intersect at point L. Prove that points K, L, M and N , where
M and N are the midpoints of bases BC and AD, respectively, belong to one line.
19.3. The intersection point of diagonals of a trapezoid is equidistant from the lines to
which the sides of the trapezoid belong. Prove that the trapezoid is an isosceles one.
359
360
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
19.4. Medians AA1 , BB1 and CC1 of triangle ABC meet at point M ; let P be an
arbitrary point. Line la passes through point A parallel to line P A1 ; lines lb and lc are
similarly defined. Prove that:
a) lines la , lb and lc meet at one point, Q;
b) point M belongs to segment P Q and P M : M Q = 1 : 2.
19.5. Circle S is tangent to equal sides AB and BC of an isosceles triangle ABC at
points P and K, respectively, and is also tangent from the inside to the circle circumscribed
about triangle ABC. Prove that the midpoint of segment P K is the center of the circle
inscribed into triangle ABC.
19.6. A convex polygon possesses the following property: if all its sides are pushed by
distance 1 outwards and extended, then the obtained lines form a polygon similar to the
initial one. Prove that this polygon is a circumscribed one.
19.7. Let R and r be the radii of the circumscribed and inscribed circles of a triangle.
Prove that R ≥ 2r and the equality is only attained for an equilateral triangle.
19.8. Let M be the center of mass of an n-gon A1 . . . An ; let M1 , . . . , Mn be the centers
of mass of the (n − 1)-gons obtained from the given n-gon by discarding vertices A1 , . . . , An ,
respectively. Prove that polygons A1 . . . An and M1 . . . Mn are homothetic to each other.
19.9. Prove that any convex polygon Φ contains two nonintersecting polygons Φ1 and
Φ2 similar to Φ with coefficient 21 .
See also Problem 5.87.
§2. Homothetic circles
19.10. On a circle, points A and B are fixed and point C moves along this circle. Find
the locus of the intersection points of the medians of triangles ABC.
19.11. a) A circle inscribed into triangle ABC is tangent to side AC at point D, and
DM is its diameter. Line BM intersects side AC at point K. Prove that AK = DC.
b) In the circle, perpendicular diameters AB and CD are drawn. From point M outside
the circle there are drawn tangents to the circle that intersect AB at points E and H and
also lines M C and M D that intersect AB at points F and K, respectively. Prove that
EF = KH.
19.12. Let O be the center of the circle inscribed into triangle ABC, let D be the point
where the circle is tangent to side AC and B1 the midpoint of AC. Prove that line B1 O
divides segment BD in halves.
19.13. The circles α, β and γ are of the same radius and are tangent to the sides of
angles A, B and C of triangle ABC, respectively. Circle δ is tangent from the outside to all
the three circles α, β and γ. Prove that the center of δ belongs to the line passing through
the centers of the circles inscribed into and circumscribed about triangle ABC.
19.14. Consider triangle ABC. Four circles of the same radius ρ are constructed so that
one of them is tangent to the three other ones and each of those three is tangent to two sides
of the triangle. Find ρ given the radii r and R of the circles inscribed into and circumscribed
about the triangle.
§3. Costructions and loci
19.15. Consider angle ∠ABC and point M inside it. Construct a circle tangent to the
legs of the angle and passing through M .
19.16. Inscribe two equal circles in a triangle so that each of the circles were tangent to
two sides of the triangle and the other circle.
§5. ROTATIONAL HOMOTHETY
361
19.17. Consider acute triangle ABC. Construct points X and Y on sides AB and BC,
respectively, so that a) AX = XY = Y C; b) BX = XY = Y C.
19.18. Construct triangle ABC given sides AB and AC and bisector AD.
19.19. Solve Problem 16.18 with the help of homothety.
19.20. On side BC of given triangle ABC, construct a point such that the line that
connects the bases of perpendiculars dropped from this point to sides AB and AC is parallel
to BC.
***
19.21. Right triangle ABC is modified so that vertex A of the right angle is fixed
whereas vertices B and C slide along fixed circles S1 and S2 tangent to each other at A from
the outside. Find the locus of bases D of heights AD of triangles ABC.
See also problems 7.26–7.29, 8.15, 8.16, 8.70.
§4. Composition of homotheties
19.22. A transformation f has the following property: if A′ and B ′ are the images of
−−→
−→
points A and B, then A′ B ′ = k AB, where k is a constant. Prove that:
a) if k = 1, then f is a parallel translation;
b) if k 6= 1, then f is a homothety.
19.23. Prove that the composition of two homotheties with coefficients k1 and k2 , where
k1 k2 6= 1, is a homothety with coefficient k1 k2 and its center belongs to the line that connects
the centers of these homotheties. Investigate the case k1 k2 = 1.
19.24. Common outer tangents to the pairs of circles S1 and S2 , S2 and S3 , S3 and S1
intersect at points A, B and C, respectively. Prove that points A, B and C belong to one
line.
19.25. Trapezoids ABCD and AP QD have a common base AD and the length of all
their bases are distinct. Prove that the intersections points of the following pairs of lines
belong to one line:
a) AB and CD, AP and DQ, BP and CQ;
b) AB and CD, AQ and DP , BQ and CP .
§5. Rotational homothety
19.26. Circles S1 and S2 intersect at points A and B. Lines p and q passing through
point A intersect circle S1 at points P1 and Q1 and circle S2 at points P2 and Q2 . Prove
that the angle between lines P1 Q1 and P2 Q2 is equal to the angle between circles S1 and S2 .
19.27. Circles S1 and S2 intersect at points A and B. Under the rotational homothety
P with center A that sends S1 into S2 point M1 from circle S1 turns into M2 . Prove that
line M1 M2 passes through B.
19.28. Circles S1 , . . . , Sn pass through point O. A grasshopper hops from point Xi
on circle Si to point Xi+1 on circle Si+1 so that line Xi Xi+1 passes through the intersection
point of circles Si and Si+1 distinct from O. Prove that after n hops (from S1 to S2 from S2
to S3 , . . . , from Sn to S1 ) the grasshopper returns to the initial position.
19.29. Two circles intersect at points A and B and chords AM and AN are tangent to
these circles. Let us complete triangle M AN to parallelogram M AN C and divide segments
BN and M C by points P and Q in equal proportions. Prove then that ∠AP Q = ∠AN C.
19.30. Consider two nonconcentric circles S1 and S2 . Prove that there exist precisely
two rotational homotheties with the angle of rotation of 90◦ that send S1 into S2 .
362
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
***
19.31. Consider square ABCD and points P and Q on sides AB and BC, respectively,
so that BP = BQ. Let H be the base of the perpendicular dropped from B on P C. Prove
that ∠DHQ = 90◦ .
19.32. On the sides of triangle ABC similar triangles are constructed outwards: △A1 BC ∼
△B1 CA ∼ △C1 AB. Prove that the intersection points of medians of triangles ABC and
A1 B1 C1 coincide.
19.33. The midpoints of sides BC and B1 C1 of equilateral triangles ABC and A1 B1 C1
coincide (the vertices of both triangles are listed clockwise). Find the value of the angle
between lines AA1 and BB1 and also the ratio of the lengths of segments AA1 and BB1 .
19.34. Triangle ABC turns under a rotational homothety into triangle A1 B1 C1 ; let O
be an arbitrary point. Let A2 be the vertex of parallelogram OAA1 A2 ; let points B2 and C2
be similarly defined. Prove that △A2 B2 C2 ∼ △ABC.
19.35. On top of a rectangular map lies a map of the same locality but of lesser scale.
Prove that it is possible to pierce by a needle both maps so that the points where both maps
are pierced depict the same point of the locality.
19.36. Rotational homotheties P1 and P2 with centers A1 and A2 have the same angle
of rotation and the product of their coefficients is equal to 1. Prove that the composition
P2 ◦ P1 is a rotation and its center coincides with the center of another rotation that sends
−−−→ −−→
A1 into A2 and whose angle of rotation is equal to 2∠(M A1 , M N ), where M is an arbitrary
point and N = P1 (M ).
19.37. Triangles M AB and M CD are similar but have opposite orientations. Let O1 be
−→ −−→
the center of rotation through an angle of 2∠(AB, BM ) that sends A to C and O2 the center
−→ −−→
of rotation through an angle of 2∠(AB, AM ) that sends B to D. Prove that O1 = O2 .
***
19.38. Consider a half circle with diameter AB. For every point X on this half circle a
point Y is placed on ray XA so that XY = kXB. Find the locus of points Y .
19.39. Consider point P on side AB of (unknown?) triangle ABC and triangle LM N .
Inscribe triangle P XY similar to LM N into triangle ABC.
19.40. Construct quadrilateral ABCD given ∠B+∠D and the lengths a = AB, b = BC,
c = CD and d = DA.
See also Problem 5.122.
§6. The center of a rotational homothety
19.41. a) Let P be the intersection point of lines AB and A1 B1 . Prove that if no points
among A, B, A1 , B1 and P coincide, then the common point of circles circumscribed about
triangles P AA1 and P BB1 is the center of a rotational homothety that sends A to A1 and
B to B1 and that such a rotational homothety is unique.
b) Prove that the center of a rotational homothety that sends segment AB to segment
BC is the intersection point of circles passing through point A and tangent to line BC at
point B and the circle passing through C and tangent to line AB at point B.
19.42. Points A and B move along two intersecting lines with constant but distinct
speeds. Prove that there exists a point, P , such that at any moment AP : BP = k, where
k is the ratio of the speeds.
19.43. Construct the center O of a rotational homothety with a given coefficient k 6= 1
that sends line l1 into line l2 and point A1 that belongs to l1 into point A2 . (?)
§7. THE SIMILARITY CIRCLE OF THREE FIGURES
363
19.44. Prove that the center of a rotational homothety that sends segment AB into
segment A1 B1 coincides with the center of a rotational homothety that sends segment AA1
into segment BB1 .
19.45. Four intersecting lines form four triangles. Prove that the four circles circumscribed about these triangles have one common point.
19.46. Parallelogram ABCD is not a rhombus. Lines symmetric to lines AB and CD
through diagonals AC and DB, respectively, intersect at point Q. Prove that Q is the center
of a rotational homothety that sends segment AO into segment OD, where O is the center
of the parallelogram.
19.47. Consider wo regular pentagons with a common vertex. The vertices of each
pentagon are numbered 1 to 5 clockwise so that the common vertex has number 1. Vertices
with equal numbers are connected by straight lines. Prove that the four lines thus obtained
intersect at one point.
19.48. On sides BC, CA and AB of triangle ABC points A1 , B1 and C1 are taken so
that △ABC ∼ △A1 B1 C1 . Pairs of segments BB1 and CC1 , CC1 and AA1 , AA1 and BB1
intersect at points A2 , B2 and C2 , respectively. Prove that the circles circumscribed about
triangles ABC2 , BCA2 , CAB2 , A1 B1 C2 , B1 C1 A2 and C1 A1 B2 intersect at one point.
§7. The similarity circle of three figures
Let F1 , F2 and F3 be three similar figures, O1 the center of a rotational homothety that
sends F2 to F3 . Let points O2 and O3 be similarly defined. If O1 , O2 and O3 do not belong
to one line, then triangle O1 O2 O3 is called the similarity triangle of figures F1 , F2 and F3 and
its circumscribed circle is called the similarity circle of these figures. In case points O1 , O2
and O3 coincide the similarity circle degenerates into the center of similarity and in case
when not all these points coincide but belong to one line the similarity circle degenerates
into the axis of similarity.
In the problems of this section we assume that the similarity circle of the figures considered is not degenerate.
19.49. Lines A2 B2 and A3 B3 , A3 B3 and A1 B1 , A1 B1 and A2 B2 intersect at points P1 ,
P2 , P3 , respectively.
a) Prove that the circumscribed circles of triangles A1 A2 P3 , A1 A3 P2 and A2 A3 P1 intersect
at one point that belongs to the similarity circle of segments A1 B1 , A2 B2 and A3 B3 .
b) Let O1 be the center of rotational homothety that sends segment A2 B2 into segment
A3 B3 ; points O2 and O3 be similarly defined. Prove that lines P1 O1 , P2 O2 and P3 O3 intersect
at one point that belongs to the similarity circle of segments A1 B1 , A2 B2 and A3 B3 .
Points A1 and A2 are called correspondent points of similar figures F1 and F2 if the
rotational symmetry that sends F1 to F2 transforms A1 into A2 . Correspondent lines and
correspondent segments are analogously defined.
19.50. Let A1 B1 , A2 B2 and A3 B3 and also A1 C1 , A2 C2 and A3 C3 be correspondent
segments of similar figures F1 , F2 and F3 . Prove that the triangle formed by lines A1 B1 ,
A2 B2 and A3 B3 is similar to the triangle formed by lines A1 C1 , A2 C2 and A3 C3 and the
center of the rotational homothety that sends one of these triangles into another one belongs
to the similarity circle of figures F1 , F2 and F3 .
19.51. Let l1 , l2 and l3 be the correspondent lines of similar figures F1 , F2 and F3 and
let the lines intersect at point W .
a) Prove that W belongs to the similarity circle of F1 , F2 and F3 .
364
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
b) Let J1 , J2 and J3 be distinct from W intersection points of lines l1 , l2 and l3 with the
similarity circle. Prove that these points only depend on figures F1 , F2 and F3 and do not
depend on the choice of lines l1 , l2 and l3 .
Points J1 , J2 and J3 are called constant points of similar figures F1 , F2 and F3 and
triangle J1 J2 J3 is called the constant triangle of similar figures.
19.52. Prove that the constant triangle of three similar figures is similar to the triangle
formed by their correspondent lines and these triangles have opposite orientations.
19.53. Prove that constant points of three similar figures are their correspondent points.
The similarity circle of triangle ABC is the similarity circle of segments AB, BC and
CA (or of any three similar triangles constructed from these segments). Constant points of
a triangle are the constant points of the three figures considered.
19.54. Prove that the similarity circle of triangle ABC is the circle with diameter KO,
where K is Lemoin’s point and O is the center of the circumscribed circle.
19.55. Let O be the center of the circumscribed circle of triangle ABC, K Lemoin’s
point, P and Q Brokar’s points, ϕ Brokar’s angle (see Problems 5.115 and 5.117). Prove
that points P and Q belong to the circle of diameter KO and OP = OQ and ∠P OQ = 2ϕ.
Problems for independent study
19.56. Given triangles ABC and KLM . Inscribe triangle A1 B1 C1 into triangle ABC
so that the sides of A1 B1 C1 wre parallel to the respective sides of triangle KLM .
19.57. On the plane, there are given points A and E. Construct a rhombus ABCD
with a given height for which E is the midpoint of BC.
19.58. Consider a quadrilateral. Inscribe a rombus in it so that the sides of the rombus
are parallel to the diagonals of the quandrangle.
19.59. Consider acute angle ∠AOB and point C inside it. Find point M on leg OB
equidistant from leg OA and from point C.
19.60. Consider acute triangle ABC. Let O be the intersection point of its heights; ω the
circle with center O situated inside the triangle. Construct triangle A1 B1 C1 circumscribed
about ω and inscribed in triangle ABC.
19.61. Consider three lines a, b, c and three points A, B, C each on the respective line.
Construct points X, Y , Z on lines a, b, c, respectively, so that BY : AX = 2, CZ : AX = 3
and so that X, Y , Z are all on one line.
Solutions
19.1. A homothety with the center at the intersection point of the diagonals of the
quadrilateral and with coefficient 3/2 sends the intersection points of the medians of the
triangles in question into the midpoints of the sides of the quadrilateral. It remains to make
use of the result of Problem 1.2.
19.2. The homothety with center K that sends △KBC into △KAD sends point M into
N and, therefore, K belongs to line M N . The homothety with center L that sends △LBC
into △LDA sends M into N . Therefore, L belongs to line M N .
19.3. Suppose the continuations of the lateral sides AB and CD intersect at point K
and the diagonals of the trapezoid intersect at point L. By the preceding problem line KL
passes through the midpoint of segment AD and by the hypothesis this line divides angle
∠AKD in halves. Therefore, triangle AKD is an isosceles one (see Problem 16.1); hence, so
is trapezoid ABCD.
SOLUTIONS
365
19.4. The homothety with center M and coefficient −2 sends lines P A1 , P B1 and P C1
into lines la , lb and lc , respectively, and, therefore, the point Q to be found is the image of
P under this homothety.
19.5. Consider homothety HBk with center B that sends segment AC into segment A′ C ′
tangent to the circumscribed circle of triangle ABC. Denote the midpoints of segments P K
and A′ C ′ by O1 and D, respectively, and the center of S by O.
Circle S is the inscribed circle of triangle A′ BC ′ and, therefore, it suffices to show that
homothety HBk sends O1 to O. To this end it suffices to verify that BO1 : BO = BA : BA′ .
This equality follows from the fact that P O1 and DA are heights of similar right triangles
BP O and BDA′ .
19.6. Let k be the similarity coefficient of polygons and k < 1. Shifting the sides of
the initial polygon inside consecutively by k, k 2 , k 3 , . . . units of length we get a contracting
system of embedded convex polygons similar to the initial one with coefficients k, k 2 , k 3 ,
. . . . The only common point of these polygons is the center of the inscribed circle of the
initial polygon.
19.7. Let A1 , B1 and C1 be the midpoints of sides BC, AC and AB, respectively. The
homothety with center at the intersection point of the medians of triangle ABC and with
coefficient − 12 sends the circumscribed circle S of triangle ABC into the circumscribed circle
S1 of triangle A1 B1 C1 . Since S1 passes through all the vertices of triangle ABC, we can
construct triangle A′ B ′ C ′ whose sides are parallel to the respective sides of triangle ABC
and for which S1 is the inscribed circle, see Fig. 26.
Figure 68 (Sol. 19.7)
Let r and r′ be the radii of the inscribed circles of triangles ABC and A′ B ′ C ′ ; let R
and R1 be the radii of S and S1 , respectively. Clearly, r ≤ r′ = R1 = R/2. The equality is
attained if triangles A′ B ′ C ′ and ABC coincide, i.e., if S1 is the inscribed circle of triangle
ABC. In this case AB1 = AC1 and, therefore, AB = AC. Similarly, AB = BC.
19.8. Since
−−−→
−−−→ −−→
−−→
−−−→ M A1 + · · · + M An − M Ai
M Ai
M Mi =
=−
,
n−1
n−1
1
it follows that the homothety with center M and coefficient − n−1
sends Ai into Mi .
19.9. Let A and B be a pair of most distant from each other points of polygon Φ. Then
1/2
1/2
Φ1 = HA (Φ) and Φ2 = HB (Φ) are the required figures.
Indeed, Φ1 and Φ2 do not intersect because they lie on different sides of the midperpendicular to segment AB. Moreover, Φ1 and Φ2 are contained in Φ because Φ is a convex
polygon.
19.10. Let M be the intersection point of the medians of triangle ABC, O the midpoints
−−→
−→
of segment AB. Clearly, 3OM = OC and, therefore, points M fill in the circle obtained
from the initial circle under the homothety with coefficient 31 and center O.
366
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
19.11. a) The homothety with center B that sends the inscribed circle into the escribed
circle tangent to side AC sends point M into point M ′ . Point M ′ is the endpoint of the
diameter perpendicular to AC and, therefore, M ′ is the tangent point of the inscribed circle
with AC, hence, it is the intersection point of BM with AC. Therefore, K = M ′ and K
is the tangent point of the escribed circle with side AC. Now it is easy to compute that
AK = 12 (a + b − c) = CD, where a, b and c are the lengths of the sides of triangle ABC.
b) Consider a homothety with center M that sends line EH into a line tangent to the
given circle. This homothety sends points E, F , K and H into points E ′ , F ′ , K ′ and H ′ ,
respectively. By heading a) E ′ F ′ = K ′ H ′ ; hence, EF = KH.
19.12. Let us make use of the solution and notations of Problem 19.11 a). Since
AK = DC, then B1 K = B1 D and, therefore, B1 O is the midline of triangle M KD.
19.13. Let Oα , Oβ , Oγ and Oδ be the centers of circles α, β, γ and δ, respectively, O1
and O2 the centers of the inscribed and circumscribed circles, respectively, of triangle ABC.
A homothety with center O1 sends triangle Oα Oβ Oγ into triangle ABC. This homothety
sends point O2 into the center of the circumscribed circle of triangle Oα Oβ Oγ ; this latter
center coincides with Oδ . Therefore, points O1 , O2 and Oδ belong to one line.
19.14. Let A1 , B1 and C1 be the centers of the given circles tangent to the sides of
the triangle, O the center of the circle tangent to these circles, O1 and O2 the centers of
the inscribed and circumscribed circles of triangle ABC. Lines AA1 , BB1 and CC1 are the
bisectors of triangle ABC and, therefore, they intersect at point O1 . It follows that triangle
A1 B1 C1 turns into triangle ABC under a homothety with center O1 and the coefficient of
the homothety is equal to the ratio of distances from O1 to the sides of triangles ABC and
.
A1 B1 C1 , i.e., is equal to r−ρ
r
Under this homothety the circumscribed circle of triangle ABC turns into the circumscribed circle of triangle A1 B1 C1 . Since OA1 = OB1 = OC1 = 2ρ, the radius of the
rR
circumscribed circle of triangle A1 B1 C1 is equal to 2ρ. Hence, R r−ρ
= 2ρ, i.e., ρ = 2r+R
.
r
19.15. On the bisector of angle ∠ABC take an arbitrary point O and construct a circle
S with center O tangent to the legs of the angle. Line BM intersects circle S at points M1
and M2 . The problem has two solutions: circle S turns into the circles passing through M
and tangent to the legs of the angle under the homothety with center B that sends M1 into
M and under the homothety with center B that sends M2 into M .
19.16. Clearly, both circles are tangent to one of the triangle’s sides. Let us show how to
construct circles tangent to side AB. Let us take line c′ parallel to line AB. Let us construct
circles S1′ and S2′ of the same radius tangent to each other and to line c′ . Let us construct
tangents a′ and b′ to these circles parallel to lines BC and AC, respectively. The sides of
triangle A′ B ′ C ′ formed by lines a′ , b′ and c′ are parallel to respective sides of triangle ABC.
Therefore, there exists a homothety sending triangle A′ B ′ C ′ into triangle ABC. The desired
circles are the images of circles S1′ and S2′ with respect to this homothety.
19.17. a) On sides AB and BC of triangle ABC fix segments AX1 and CY1 of equal
length a.Through point Y1 draw a line l parallel to side AC. Let Y2 be the intersection point
of l and the circle of radius a with center X1 situated(who?) inside the triangle. Then point
Y to be found is the intersection point of line AY2 with side BC and X is a point on ray
AB such that AX = CY .
b) On side AB, take an arbitrary point X1 distinct from B. The circle of radius BX1
with center X1 intersects ray BC at points B and Y1 . Construct point C1 on line BC such
that Y1 C1 = BX1 and such that Y1 lies between B and C1 . The homothety with center B
that sends point C1 into C sends X1 and Y1 into points X and Y to be found.
SOLUTIONS
367
19.18. Take segment AD and draw circles S1 and S2 with center A and radii AB and
AC, respectively. Vertex B is the intersection point of S1 with the image of S2 under the
homothety with center D and coefficient − DB
= − AB
.
DC
AC
19.19. On the great circle S2 take an arbitrary point X. Let S2′ be the image of S2
under the homothety with center X and coefficient 31 , let Y be the intersection point of S2′
and S1 . Then XY is the line to be found.
19.20. From points B and C draw perpendiculars to lines AB and AC and let P be
their intersection point. Then the intersection point of lines AP and BC is the desired one.
19.21. Let us draw common exterior tangents l1 and l2 to circles S1 and S2 , respectively.
Lines l1 and l2 intersect at a point K which is the center of a homothety H that sends
S1 to S2 . Let A1 = H(A). Points A and K lie on a line that connects the centers of the
circles and, therefore, AA1 is a diameter of S2 , i.e., ∠ACA1 = 90◦ and A1 C k AB. It
follows that segment AB goes into A1 C under H. Therefore, line BC passes through K and
∠ADK = 90◦ . Point D belongs to circle S with diameter AK. It is also clear that point D
lies inside the angle formed by lines l1 and l2 . Therefore, the locus of points D is the arc of
S cut off by l1 and l2 .
19.22. The hypothesis of the problem implies that the map f is one-to-one.
a) Suppose f sends point A to point A′ and B to B ′ . Then
→
−−→′ −→ −−→′ −−
−→ −−→ −→ −−→
BB = BA + AA + A′ B ′ = −AB + AA′ + AB = AA′ ,
i.e., f is a parallel translation.
b) Consider three points A, B and C not on one line. Let A′ , B ′ and C ′ be their images
under f . Lines AB, BC and CA cannot coincide with lines A′ B ′ , B ′ C ′ and C ′ A′ , respectively,
since in this case A = A′ , B = B ′ and C = C ′ . Let AB 6= A′ B ′ . Lines AA′ and BB ′ are
not parallel because otherwise quandrilateral ABB ′ A′ would have been a parallelogram and
→
−→ −−
AB = A′ B ′ . Let O be the intersection point of AA′ and BB ′ . Triangles AOB and A′ OB ′
−−→
−→
are similar with similarity coefficient k and, therefore, OA′ = k OA, i.e., O is a fixed point
of the transformation f . Therefore,
−−−−→ −−−−−−−→
−−→
Of (X) = f (O)f (X) = k OX
for any X which means that f is a homothety with coefficient k and center O.
19.23. Let H = H2 ◦ H1 , where H1 and H2 are homotheties with centers O1 and O2 and
coefficients k1 and k2 , respectively. Denote:
A′ = H1 (A), B ′ = H1 (B), A′′ = H2 (A′ ), B ′′ = H2 (B ′ ).
−−→
−−−→
−−→
−−−→
−→
−→
Then A′ B ′ = k1 AB and A′′ B ′′ = k2 A′ B ′ , i.e., A′′ B ′′ = k1 k2 AB. With the help of the
preceding problem this implies that for k1 k2 6= 1 the transformation H is a homothety with
coefficient k1 k2 and if k1 k2 = 1, then H is a parallel translation.
It remains to verify that the fixed point of H belongs to the line that connects the centers
−−−→
−−−→
−−−→
−−→
of homotheties H1 and H2 . Since O1 A′ = k1 O1 A and O2 A′′ = k2 O2 A′ , it follows that
−−−→′′
−−−→ −−−→
−−−→
−−→
O2 A = k2 (O2 O1 + O1 A′ ) = k2 (O2 O1 + k1 O1 A) =
−−−→
−−−→
−−→
k2 O2 O1 + k1 k2 O1 O2 + k1 k2 O2 A.
For a fixed point X we get the equation
−−→
−−−→
−−→
O2 X = (k1 k2 − k2 )O1 O2 + k1 k2 O2 X
368
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
−−→
−−−→
1 k2 −k2
and, therefore, O2 X = λO1 O2 , where λ = k1−k
.
1 k2
19.24. Point A is the center of homothety that sends S1 to S2 and B is the center of
homothety that sends S2 to S3 . The composition of these homotheties sends S1 to S3 and
its center belongs to line AB. On the other hand, the center of homothety that sends S1
to S3 is point C. Indeed, to the intersection point of the outer tangents there corresponds
a homothety with any positive coefficient and a composition of homotheties with positive
coefficients is a homothety with a positive coefficient.
19.25. a) Let K, L, M be the intersection points of lines AB and CD, AP and DQ,
BP and CQ, respectively. These points are the centers of homotheties HK , HL and HM
with positive coefficients that consequtively send segments BC to AD, AD to P Q and BC
to P Q. Clearly, HL ◦ HK = HM . Therefore, points K, L and M belong to one line.
b) Let K, L, M be the intersection points of lines AB and CD, AQ and DP , BQ
and CP , respectively. These points are the centers of homotheties, HK , HL and HM that
consequtively send segments BC to AD, AD to QP , BC to QP ; the coefficient of the
first homothety is a positive one those of two other homotheties are negative ones. Clearly,
HL ◦ HK = HM . Therefore, points K, L and M belong to one line.
19.26. Since ∠(P1 A, AB) = ∠(P2 A, AB), the oriented angle values of arcs ⌣ BP1 and
⌣ BP2 are equal. Therefore, the rotational homothety with center B that sends S1 to S2
sends point P1 to P2 and line P1 Q1 into line P2 Q2 .
19.27. Oriented angle values of arcs ⌣ AM1 and ⌣ AM2 are equal, consequently,
∠(M1 B, BA) = ∠(M2 B, BA) and, therefore, points M1 , M2 and B belong to one line.
19.28. Let Pi be a rotational homothety with center O that sends circle Si to Si+1 .
Then Xi+1 = Pi (Xi ) (see Problem 19.27). It remains to observe that the composition
Pn ◦ · · · ◦ P2 ◦ P1 is a rotational homothety with center O that sends S1 to S1 , i.e., is an
identity transformation.
19.29. Since ∠AM B = ∠N AB and ∠BAM = ∠BN A, we have △AM B ∼ △N AB
and, therefore, AN : AB = M A : M B = CN : M B. Moreover, ∠ABM = 180◦ − ∠M AN =
∠AN C. It follows that △AM B ∼ △ACN , i.e., the rotational homothety with center A
sending M to B sends C to N and, therefore, it maps Q to P .
19.30. Let O1 and O2 be the centers of given circles, r1 and r2 be their radii. The
coefficient k of the rotational homothety which maps S1 to S2 is equal to r1 /r2 and its
center O belongs to the circle with diameter O1 O2 . Moreover, OO1 : OO2 = k = r1 /r2 . It
remains to verify that the circle with diameter O1 O2 and the locus of points O such that
OO1 : OO2 = k have precisely two common points. For k = 1 it is obvious and for k 6= 1 the
locus in question is described in the solution of Problem 7.14: it is the(A?) circle and one of
its intersection points with line O1 O2 is an inner point of segment O1 O2 whereas the other
intersection point lies outside the segment.
19.31. Consider a transformation which sends triangle BHC to triangle P HB, i.e., the
composition of the rotation through an angle of 90◦ about point H and the homothety with
coefficient BP : CB and center H. Since this transformation maps the vertices of any square
into vertices of a square, it maps points C and B to points B and P , respectively. Then it
maps point D to Q, i.e., ∠DHQ = 90◦ .
−−→
−−→
19.32. Let P be a rotational homothety that sends CB to CA1 . Then
−−→ −−→ −−→
AA1 + BB1 + CC1 =
−→
−−→
−−→
−→
−→
−→
−
→
AC + P (CB) + CB + P (BA) + BA + P (AC) = 0 .
SOLUTIONS
369
Hence, if M is the center of mass of triangle ABC, then
−−−→ −−−→ −−−→
M A 1 + M B1 + M C 1 =
−−→ −−→ −−→
−−→ −−→ −−→
−
→
(M A + M B + M C) + (AA1 + BB1 + CC1 ) = 0 .
−−→
−−−→
19.33. Let M be the common midpoint of sides BC and B1 C1 , x = M B and y = M B1 .
Further, let P be the rotational homothety with center M , the angle of rotation 90◦ and
√
−−→
−−→
coefficient 3 that sends B to A and B1 to A1 . Then BB1 = y − x and AA1 = P (y) −
−−→
−−→
−−→
P (x) = P (BB√1 ). Therefore, the angle between vectors AA1 and BB1 is equal to 90◦ and
AA1 : BB1 = 3.
19.34. Let P be the rotational homothety that sends triangle ABC to triangle A1 B1 C1 .
Then
−−−→ −−→ −−→ −−→ −−→
A2 B2 = A2 O + OB2 = A1 A + BB1 =
−→ −−−→
BA + A1 B1 =
−→
−→
− AB + P (AB).
Similarly, the transformation f (a) = −a + P (a) sends the other vectors of the sides of
triangle ABC to the vectors of the sides of triangle A2 B2 C2 .
19.35. Let the initial map be rectangle K0 on the plane, the smaller map rectangle
K1 contained in K0 . Let us consider a rotational homothety f that maps K0 to K1 . Let
Ki+1 = f (Ki ) for i > 1. Since the sequence Ki for i = 1, 2, . . . is a contracting sequence of
embedded polygons, there exists (by Helly’s theorem) a unique fixed point X that belongs
to all the rectangles Ki .
Let us prove that X is the required point, i.e., f (X) = X. Indeed, since X belongs to
Ki , point f (X) belongs to Ki+1 , i.e., point f (X) belongs also to all rectangles Ki . Since
there is just one point that belongs to all rectangles, we deduce that f (X) = X.
19.36. Since the product of coefficients of rotational homotheties P1 and P2 is equal to
1, their composition is a rotation (cf. Problem 17.36). Let O be the center of rotation P2 ◦ P1
and R = P1 (O). Since P2 ◦ P1 (O) = O, it follows that P2 (R) = O. Therefore, by hypothesis
A1 O : A1 R = A2 O : A2 R and ∠OA1 R = ∠OA2 R, i.e., △OA1 R ∼ △OA2 R. Moreover, OR is
a common side of these similar triangles; hence, △OA1 R = △OA2 R. Therefore, OA1 = OA2
and
−−→ −−→
−−→ −→
−−−→ −−→
∠(OA1 , OA2 ) = 2∠(OA1 , OR) = 2∠(M A1 , M N ),
−−−→ −−→
i.e., O is the center of rotation through an angle of 2∠(M A1 , M N ) that maps A1 to A2 .
19.37. Let P1 be the rotational homothety with center B sending A to M and P2 be
rotational homothety with center D sending M to C. Since the product of coefficients of
these rotational homotheties is equal to (BM : BA) · (DC : DM ) = 1, their composition
P2 ◦ P1 is a rotation (sending A to C) through an angle of
−→ −−→
−−→ −−→
−→ −−→
∠(AB, BM ) + ∠(DM , DC) = 2∠(AB, BM ).
On the other hand, the center of the rotation P2 ◦ P1 coincides with the center of the
−→ −−→
rotation through an angle of 2∠(AB, AM ) that sends B to D (cf. Problem
√ 19.36).
19.38. It is easy to verify that tan ∠XBY = k and BY : BX = k 2 +√1, i.e., Y is
obtained from X under the rotational homothety with center B and coefficient k 2 + 1, the
370
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
angle of rotation being of value arctan k. The locus to be found is the image of the given
half circle under this rotational homothety.
19.39. Suppose that triangle P XY is constructed and points X and Y belong to sides
AC and CB, respectively. We know a transformation that maps X to Y , namely, the
rotational homothety with center P , the angle of rotation ϕ = ∠XP Y = ∠M LN and the
homothety coefficient k = P Y : P X = LN · LM . Point Y to be found is the intersection
point of segment BC and the image of segment AC under this transformation.
19.40. Suppose that rectangle ABCD is constructed. Consider the rotational homothety
with center A that sends B to D. Let C ′ be the image of point C under this homothety.
Then ∠CDC ′ = ∠B + ∠D and DC ′ = BC·AD
= bd
.
AB
a
′
We can recover triangle CDC from CD, DC ′ and ∠CDC ′ . Point A is the intersection
point of the circle of radius d with center D and the locus of points X such that C ′ X : CX =
d : a (this locus is a circle, see Problem 7.14). The further construction is obvious.
19.41. a) If O is the center of a rotational homothety that sends segment AB to segment
A1 B1 , then
∠(P A, AO) = ∠(P A1 , A1 O) and ∠(P B, BO) = ∠(P B1 , B1 O)
(1)
and, therefore, point O is the intersection point of the inscribed circles of triangles P AA1
and P BB1 .
The case when these circles have only one common point P is clear: this is when segment
AB turns into segment A1 B1 under a homothety with center P .
If P and O are two intersection points of the circles considered, then equalities (1) imply
that △OAB ∼ △OA1 B1 and, therefore, O is the center of a rotational homothety that maps
segment AB into segment A1 B1 .
b) It suffices to notice that point O is the center of a rotational homothety that maps
segment AB to segment BC if and only if ∠(BA, AO) = ∠(CB, BO) and ∠(AB, BO) =
∠(BC, CO).
19.42. Let A1 and B1 be the positions of the points at one moment, A2 and B2 the
position of the points at another moment. Then for point P we can take the center of a
rotational homothety that maps segment A1 A2 to segment B1 B2 .
19.43. Let P be the intersection point of lines l1 and l2 . By Problem 19.41 point O
belongs to the circumscribed circle S1 of triangle A1 A2 P . On the other hand, OA2 : OA1 = k.
The locus of points X such that XA2 : XA1 = k is circle S2 (by Problem 7.14). Point O is
the intersection point of circles S1 and S2 (there are two such points).
19.44. Let O be the center of a rotational homothety that maps segment AB to segment
A1 B1 . Then △ABO ∼ △A1 B1 O, i.e., ∠AOB = ∠A1 OB1 and AO : BO = A1 O : B1 O.
Therefore, ∠AOA1 = ∠BOB1 and AO : A1 O = BO : B1 O, i.e., △AA1 O ∼ △BB1 O. Hence,
point O is the center of the rotational homothety that maps segment AA1 to segment BB1 .
19.45. Let lines AB and DE intersect at point C and lines BD and AE intersect at
point F . The center of rotational homothety that maps segment AB to segment ED is
the distinct from C intersection point of the circumscribed circles of triangles AEC and
BDC (see Problem 19.41) and the center of rotational homothety sending AE to BD is
the intersection point of circles circumscribed about triangles ABF and EDF . By Problem
19.44 the centers of these rotational homotheties coincide, i.e., all the four circumscribed
circles have a common point.
19.46. The center O of parallelogram ABCD is equidistant from the following pairs of
lines: AQ and AB, AB and CD, CD and DQ and, therefore, QO is the bisector of angle
∠AQD. Let α = ∠BAO, β = ∠CDO and ϕ = ∠AQO = ∠DQO. Then α + β = ∠AOD =
360◦ − α − β − 2ϕ, i.e., α + β + ϕ = 180◦ and, therefore, △QAO ∼ △QOD.
SOLUTIONS
371
19.47. Let us solve a slightly more general problem. Suppose point O is taken on circle
S and H is a rotational homothety with center O. Let us prove that then all lines XX ′ ,
where X is a point from S and X ′ = H(X), intersect at one point.
Let P be the intersection point of lines X1 X1′ and X2 X2′ . By Problem 19.41 points O, P ,
X1 and X2 lie on one circle and points O, P , X1′ and X2′ also belong to one circle. Therefore,
P is an intersection point of circles S and H(S), i.e., all lines XX ′ pass through the distinct
from O intersection point of circles S and H(S).
19.48. Let O be the center of a rotational homothety sending triangle A1 B1 C1 to triangle
ABC. Let us prove that, for instance, the circumscribed circles of triangles ABC2 and
A1 B1 C2 pass through point O. Under the considered homothety segment AB goes into
segment A1 B1 ; therefore, point O coincides with the center of the rotational homothety that
maps segment AA1 to segment BB1 (see Problem 19.44). By problem 19.41 the center of the
latter homothety is the second intersection point of the circles circumscribed about triangles
ABC2 and A1 B1 C2 (or is their tangent point).
Figure 169 (Sol. 19.48)
19.49. Points A1 , A2 and A3 belong to lines P2 P3 , P3 P1 and P1 P2 (Fig. 27). Therefore,
the circles circumscribed about triangles A1 A2 P3 , A1 A3 P2 and A2 A3 P1 have a common point
V (see Problem 2.80 a)), and points O3 , O2 and O1 lie on these circles (see Problem 19.41).
Similarly, the circles circumscribed about triangles B1 B2 P3 , B1 B3 P2 and B2 B3 P1 have a
common point V ′ . Let U be the intersection point of lines P2 O2 and P3 O3 . Let us prove
that point V belongs to the circle circumscribed about triangle O2 O3 U . Indeed,
∠(O2 V, V O3 ) = ∠(V O2 , O2 P2 ) + ∠(O2 P2 , P3 O3 ) + ∠(P3 O3 , O3 V ) =
∠(V A1 , A1 P2 ) + ∠(O2 U, U O3 ) + ∠(P3 A1 , A1 V ) = ∠(O2 U, U O3 ).
Analogous arguments show that point V ′ belongs to the circle circumscribed about triangle
O2 O3 U . In particular, points O2 , O3 , V and V ′ belong to one circle. Similarly, points O1 ,
O2 , V and V ′ belong to one circle and, therefore, points V and V ′ belong to the circle
circumscribed about triangle O1 O2 O3 ; point U also belongs to this circle.
We can similarly prove that lines P1 O1 and P2 O2 intersect at one point that belongs
to the similarity circle. Line P2 O2 intersects the similarity circle at points U and O2 and,
therefore, line P1 O1 passes through point U .
19.50. Let P1 be the intersection point of lines A2 B2 and A3 B3 , let P1′ be the intersection
point of lines A2 C2 and A3 C3 ; let points P2 , P3 , P2′ and P3′ be similarly defined. The
rotational homothety that sends F1 to F2 sends lines A1 B1 and A1 C1 to lines A2 B2 and
A2 C2 , respectively, and, therefore, ∠(A1 B1 , A2 B2 ) = ∠(A1 C1 , A2 C2 ). Similar arguments
show that △P1 P2 P3 ∼ △P1′ P2′ P3′ .
372
CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY
The center of the rotational homothety that maps segment P2 P3 to P2′ P3′ belongs to the
circle circumscribed about triangle A1 P3 P3′ (see Problem 19.41). Since
∠(P3 A1 , A1 P3′ ) = ∠(A1 B1 , A1 C1 ) = ∠(A2 B2 , A2 C2 ) = ∠(P3 A2 , A2 P3′ ),
the circle circumscribed about triangle A1 P3 P3′ coincides with the circle circumscribed about
triangle A1 A2 P3 . Similar arguments show that the center of the considered rotational homothety is the intersection point of the circles circumscribed about triangles A1 A2 P3 , A1 A3 P2
and A2 A3 P1 ; this point belongs to the similarity circle of figures F1 , F2 and F3 (see Problem
19.49 a)).
19.51. a) Let l1′ , l2′ and l3′ be the corresponding lines of figures F1 , F2 and F3 such
that li′ k li . These lines form triangle P1 P2 P3 . The rotational homothety with center O3
that maps F1 to F2 sends lines l1 and l1′ to lines l2 and l2′ , respectively, and, therefore, the
homothety with center O3 that maps l1 to l1′ sends line l2 to l2′ . Therefore, line P3 O3 passes
through point W .
Similarly, lines P1 O1 and P2 O2 pass through point W ; hence, W belongs to the similarity
circle of figures F1 , F2 and F3 (see Problem 19.49 b)).
Figure 170 (Sol. 19.51 a))
b) The ratio of the distances from point O1 to lines l2′ and l3′ is equal to the coefficient
of the rotational homothety that maps F2 to F3 and the angle ∠P1 of triangle P1 P2 P3 is
equal to the angle of the rotation. Therefore, ∠(O1 P1 , P1 P2 ) only depends on figures F2 and
F3 . Since ∠(O1 W, W J3 ) = ∠(O1 P1 , P1 P2 ), arc ⌣ O1 J3 is fixed (see Fig. 28) and, therefore,
point J3 is fixed. We similarly prove that points J1 and J2 are fixed.
19.52. Let us make use of notations from Problem 19.51. Clearly,
∠(J1 J2 , J2 J3 ) = ∠(J1 W, W J3 ) = ∠(P3 P2 , P2 P1 ).
For the other angles of the triangle the proof is similar.
19.53. Let us prove, for instance, that under the rotational homothety with center O1
that maps F2 to F3 point J2 goes to J3 . Indeed, ∠(J2 O1 , O1 J3 ) = ∠(J2 W, W J3 ). Moreover,
lines J2 W and J3 W are the corresponding lines of figures F2 and F3 and, therefore, the
distance from lines J2 W and J3 W to point O1 is equal to the similarity coefficient k1 ; hence,
O1 J 2
= k1 .
O1 J 3
19.54. Let Oa be the intersection point of the circle passing through point B and tangent
to line AC at point A and the circle passing through point C and tangent to line AB at
point A.
By Problem 19.41 b) point Oa is the center of rotational homothety that sends segment
BA to segment AC. Having similarly defined points Ob and Oc and making use of the result
of Problem 19.49 b) we see that lines AOa , BOb and COc intersect at a point that belongs
SOLUTIONS
373
to the similarity circle S. On the other hand, these lines intersect at Lemoin’s point K (see
Problem 5.128).
The midperpendiculars to the sides of the triangle are the corresponding lines of the
considered similar figures. The midperpendiculars intersect at point O; hence, O belongs to
the similarity circle S (see Problem 19.51 a)). Moreover, the midperpendiculars intersect S
at fixed points A1 , B1 and C1 of triangle ABC (see Problem 19.51 b)). On the other hand,
the lines passing through point K parallel to BC, CA and AB are also corresponding lines
of the considered figures (see solution to Problem 5.132), therefore, they also intersect circle
S at points A1 , B1 and C1 . Hence, OA1 ⊥ A1 K, i.e., OK is a diameter of S.
19.55. If P is the first of Brokar’s points of triangle ABC, then CP , AP and BP are the
corresponding lines for similar figures constructed on segments BC, CA and AB. Therefore,
point P belongs to the similarity circle S (see Problem 19.51 a)). Similarly, point Q belongs
to S. Moreover, lines CP , AP and BP intersect S at fixed points A1 , B1 and C1 of triangle
ABC (cf. Problem 19.51 b)). Since KA1 k BC (see the solution of Problem 19.54), it
follows that ∠(P A1 , A1 K) = ∠(P C, CB) = ϕ, i.e., ⌣ P K = 2ϕ. Similarly, ⌣ KQ = 2ϕ.
Therefore, P Q ⊥ KO; hence, OP = OQ and ∠P OQ = 21 ⌣ P Q = 2ϕ.
Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
Background
1. Solving various problems it is often convenient to consider a certain extremal or
“boundary” element, i.e., an element at which a certain function takes its maximal or minimal
value. For instance, the longest or the shortest side a triangle, the greatest or the smallest
angle, etc. This method for solving problems is sometimes called the principle (or the rule)
of an extremal element; this term, however, is not conventional.
Figure 171
2. Let O be the intersection point of the diagonals of a convex quadrilateral. Its vertices
can be denoted so that CO ≤ AO and BO ≤ DO (see Fig. *). Then under symmetries
with respect to point O triangle BOC is mapped inside triangle AOD, i.e., in a certain sense
triangle BOC is the smallest and triangle AOD is the greatest (see §4).
3. The vertices of the convex hull and the basic lines are also extremal elements; to an
extent these notions are used in §5 where they are defined and where their main properties
are listed.
§1. The least and the greatest angles
20.1. Prove that if√the lengths of all the sides of a triangle are smaller than 1, then its
area is smaller than 14 3.
20.2. Prove that the disks constructed on the sides of a convex quadrilateral as on
diameters completely cover this quadrilateral.
20.3. In a country, there are 100 airports such that all the pairwise distances between
them are distinct. From each airport a plane lifts up and flies to the nearest airport. Prove
that there is no airport to which more than five planes can arrive.
20.4. Inside a disk of radius 1, eight points are placed. Prove that the distance between
some two of them is smaller than 1.
20.5. Six disks are placed on the plane so that point O is inside each of them. Prove
that one of these disks contains the center of some other disk.
20.6. Inside an acute triangle point P is taken. Prove that the greatest distance from
P to the vertices of this triangle is smaller than twice the shortest of the distances from P
to the sides of the triangle.
375
376
CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
20.7. The lengths of a triangle’s bisectors do not exceed 1. Prove that the area of the
triangle does not exceed √13 .
§2. The least and the greatest distances
20.8. Given n ≥ 3 points on the plane not all of them on one line. Prove that there is a
circle passing through three of the given points such that none of the remaining points lies
inside the circle.
20.9. Several points are placed on the plane so that all the pairwise distances between
them are distinct. Each of these points is connected with the nearest one by a line segment.
Do some of these segments constitute a closed broken line?
20.10. Prove that at least one of the bases of perpendiculars dropped from an interior
point of a convex polygon to its sides is on the side itself and not on its extension.
20.11. Prove that in any convex pentagon there are three diagonals from which one can
construct a triangle.
20.12. Prove that it is impossible to cover a polygon with two polygons which are
homothetic to the given one with coefficient k for 0 < k < 1.
20.13. Given finitely many points on the plane such that any line passing through two
of the given points contains one more of the given points. Prove that all the given points
belong to one line.
20.14. In plane, there are given finitely many pairwise non-parallel lines such that
through the intersection point of any two of them one more of the given lines passes. Prove
that all these lines pass through one point.
20.15. In plane, there are given n points. The midpoints of all the segments with both
endpoints in these points are marked, the given points are also marked. Prove that there
are not less than 2n − 3 marked points.
See also Problems 9.17, 9.19.
§3. The least and the greatest areas
20.16. In plane, there are n points. The area of any triangle with vertices in these points
does not exceed 1. Prove that all these points can be placed in a triangle whose area is equal
to 4.
20.17. Polygon M ′ is homothetic to a polygon M with homothety coefficient equal to
1
− 2 . Prove that there exists a parallel translation that sends M ′ inside M .
§4. The greatest triangle
20.18. Let O be the intersection point of diagonals of convex quadrilateral ABCD. Prove
that if the perimeters of triangles ABO, BCO, CDO and DAO are equal, then ABCD is a
rhombus.
20.19. Prove that if the center of the inscribed circle of a quadrilateral coincides with
the intersection point of the diagonals, then this quadrilateral is a rhombus.
20.20. Let O be the intersection point of the diagonals of convex quadrilateral ABCD.
Prove that if the radii of inscribed circles of triangles ABO, BCO, CDO and DAO are
equal, then ABCD is a rhombus.
§5. The convex hull and the base lines
While solving problems of this section we will consider convex hulls of systems of points
and base lines of convex polygons.
§5. THE CONVEX HULL AND THE BASE LINES
377
The convex hull of a finite set of points is the least convex polygon which contains all
these points. The word “least” means that the polygon is not contained in any other such
polygon. Any finite system of points possesses a unique convex hull (Fig. 29).
Figure 171
A base line of a convex polygon is a line passing through its vertex and with the property
that the polygon is situated on one side of it. It is easy to verify that for any convex polygon
there exist precisely two base lines parallel to a given line (Fig. 30).
Figure 172
20.21. Solve Problem 20.8 making use of the notion of the convex hull.
20.22. Given 2n + 3 points on a plane no three of which belong to one line and no four
of which belong to one circle. Prove that one can select three points among these so that n
of the remaining points lie inside the circle drawn through the selected points and n of the
points lie outside the circle.
20.23. Prove that any convex polygon of area 1 can be placed inside a rectangle of area
2.
20.24. Given a finite set of points in plane prove that there always exists a point among
them for which not more than three of the given points are the nearest to it.
20.25. On the table lie n cardboard and n plastic squares so that no two cardboard and
no two plastic squares have common points, the boundary points included. It turned out
that the set of vertices of the cardboard squares coincides with that of the plastic squares.
Is it necessarily true that every cardboard square coincides with a plastic one?
20.26. Given n ≥ 4 points in plane so that no three of them belong to one line. Prove
that if for any 3 of them there exists a fourth (among the given ones) together with which
they form vertices of a parallelogram, then n = 4.
378
CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
§6. Miscellaneous problems
20.27. In plane, there are given a finite set of (not necessarily convex) polygons each
two of which have a common point. Prove that there exists a line having a common point
with all these polygons.
20.28. Is it possible to place 1000 segments on the plane so that the endpoints of every
segment are interior points of certain other of these segments?
20.29. Given four points in plane not on one line. Prove that at least one of the triangles
with vertices in these points is not an acute one.
20.30. Given an infinite set of rectangles in plane. The vertices of each of the rectangles
lie in points with coordinates (0, 0), (0, m), (n, 0), (n, m), where n and m are positive integers
(each rectangle has its own numbers). Prove that among these rectangles one can select such
a pair that one is contained inside the other one.
20.31. Given a convex polygon A1 . . . An , prove that the circumscribed circle of triangle
Ai Ai+1 Ai+2 contains the whole polygon.
Solutions
20.1.√ Let α be the least angle of the triangle. Then α ≤ 60◦ . Therefore, S =
sin 60◦
= 43 .
2
20.2. Let X be an arbitrary point inside a convex quadrilateral. Since
bc sin α
2
≤
∠AXB + ∠BXC + ∠CXD + ∠AXD = 360◦ ,
the maximal of these angles is not less than 90◦ . Let, for definiteness sake, ∠AXB ≥ 90◦ .
Then point X is inside the circle with diameter AB.
20.3. If airplanes from points A and B arrived to point O, then AB is the longest side
of triangle AOB, i.e., ∠AOB > 60◦ . Suppose that airplanes from points A1 , . . . , An arrived
◦
◦
to point O. Then one of the angles ∠Ai OAj does not exceed 360
. Therefore, 360
> 60◦ ,
n
n
i.e., n < 6.
20.4. At least seven points are distinct from the center O of the circle. Therefore, the
◦
< 60◦ . If
least of the angles ∠Ai OAj , where Ai and Aj are given points, does not exceed 360
7
A and B are points corresponding to the least angle, then AB < 1 because AO ≤ 1, BO ≤ 1
and angle ∠AOB cannot be the largest angle of triangle AOB.
20.5. One of the angles between the six segments that connect point O with the centers
◦
of the disks does not exceed 360
= 60◦ . Let ∠O1 OO2 ≤ 60◦ , where O1 and O2 are the
6
centers of the disks of radius r1 and r2 , respectively. Since ∠O1 OO2 ≤ 60◦ , this angle is not
the largest angle in triangle O1 OO2 and, therefore, either O1 O2 ≤ O1 O or O1 O2 ≤ O2 O.
Let, for definiteness, O1 O2 ≤ O1 O. Since point O is inside the circles, O1 O < r1 . Therefore,
O1 O2 ≤ O1 O < r1 , i.e., point O2 is inside the disk of radius r1 with center O1 .
20.6. Let us drop perpendiculars P A1 , P B1 and P C1 from point P to sides BC, CA and
AB, respectively, and select the greatest of the angles formed by these perpendiculars and
rays P A, P B and P C. Let, for definiteness sake, this be angle ∠AP C1 . Then ∠AP C1 ≥ 60◦ ;
hence, P C1 : AP = cos AP C1 ≤ cos 60◦ = 21 , i.e., AP ≥ 2P C1 . Clearly, the inequality still
holds if AP is replaced with the greatest of the numbers AP , BP and CP and P C1 is
replaced with the smallest of the numbers P A1 , P B1 and P C1 .
20.7. Let, for definiteness, α be the smallest angle of triangle ABC; let AD be the
bisector. One of sides AB and AC does not exceed AD/ cos(α/2) since otherwise segment
BC does not pass through point D. Let, for definiteness,
AD
2
AD
√
≤
≤
.
AB ≤
cos(α/2)
cos 30◦
3
SOLUTIONS
379
Then SABC = 21 hc AB ≤ 12 lc AB ≤ √13 .
20.8. Let A and B be those of the given points for which the distance between them is
minimal. Then inside the circle with diameter AB there are no given points. Let C be the
remaining point — the vertex of the greatest angle that subtends segment AB. Then inside
the circle passing through points A, B and C there are no given points.
20.9. Suppose that we have obtained a closed broken line. Then AB is the longest link
of this broken line and AC and BD are the links neighbouring to AB. Then AC < AB,
i.e., B is not the point closest to A and BD < AB, i.e., A is not the point closest to B.
Therefore, points A and B cannot be connected. Contradiction.
20.10. Let O be the given point. Let us draw lines containing the sides of the polygon
and select among them the one which is the least distant from point O. Let this line contain
side AB. Let us prove that the base of the perpendicular dropped from O to AB belongs
to side AB itself. Suppose that the base of the perpendicular dropped from O to line AB is
point P lying outside segment AB. Since O belongs to the interior of the convex polygon,
segment OP intersects side CD at point Q. Clearly, OQ < OP and the distance from O to
line CD is smaller than OQ. Therefore, line CD is less distant from point O than line AB.
This contradicts the choice of line AB.
20.11. Let BE be the longest diagonal of pentagon ABCDE. Let us prove then that
from segments BE, EC and BD one can construct a triangle. To this end, it suffices to
verify that BE < EC + BD. Let O be the intersection point of diagonals BD and EC.
Then
BE < BO + OE < BD + EC.
20.12. Let O1 and O2 be the centers of homotheties, each with coefficient k, sending
polygon M to polygons M1 and M2 , respectively. Then a point from M the most distant
from line O1 O2 is not covered by polygons M1 and M2 .
20.13. Suppose that not all of the given points lie on one line. Through every pair of
given points draw a line (there are finitely many of such lines) and select the least nonzero
distance from the given points to these lines. Let the least distance be the one from point
A to line BC, where points B and C are among given ones.
On line BC, there lies one more of the given points, D. Drop perpendicular AQ from
point A to line BC. Two of the points B, C and D lie to one side of point Q, let these be
C and D. Let, for definiteness, CQ < DQ (Fig. 31).
Figure 173 (Sol. 20.13)
Then the distance from point C to line AD is smaller than that from A to line BC which
contradicts to the choice of point A and line BC.
20.14. Suppose that not all lines pass through one point. Consider the intersection
points of lines and select the least nonzero distance from these points to the given lines. Let
the least distance be the one from point A to line l. Through point A at least three of given
lines pass. Let them intersect line l at points B, C and D. From point A drop perpendicular
AQ to line l.
380
CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
Figure 174 (Sol. 20.14)
Two of the points B, C and D lie on one side of point Q, let them be C and D. Let,
for definiteness, CQ < DQ (Fig. 32). Then the distance from point C to line AD is smaller
than the distance from point A to line l which contradicts the choice of A and l.
20.15. Let A and B be the most distant from each other given points. The midpoints
of the segments that connect point A (resp. B) with the other points are all distinct and lie
inside the circle of radius 12 AB with center A (resp. B). The two disks obtained have only
one common point and, therefore, there are no less than 2(n − 1) − 1 = 2n − 3 distinct fixed
points.
20.16. Among all the triangles with vertices in the given points select a triangle of the
greatest area. Let this be triangle ABC. Let us draw through vertex C line lc so that
lc k AB. If points X and A lie on different sides of line lc , then SABX > SABC . Therefore,
all the given points lie on one side of lc .
Similarly, drawing lines lb and la through points B and A so that lb k AC and la k BC
we see that all given points lie inside (or on the boundary of) the triangle formed by lines la ,
lb and lc . The area of this triangle is exactly four times that of triangle ABC and, therefore,
it does not exceed 4.
20.17. Let ABC be the triangle of the greatest area among these with vertices in the
vertices of polygon M . Then M is contained inside triangle A1 B1 C1 the midpoints of whose
sides are points A, B and C. The homothety with center in the center of mass of triangle
ABC and with coefficient − 21 sends triangle A1 B1 C1 to triangle ABC and, therefore, sends
polygon M inside triangle ABC.
20.18. For definiteness, we may assume that AO ≥ CO and DO ≥ BO. Let points B1
and C1 be symmetric to points B and C through point O (Fig. 33).
Figure 175 (Sol. 20.18)
Since triangle B1 OC1 lies inside triangle AOD, it follows that PAOD ≥ PB1 OC1 = PBOC
and the equality is attained only if B1 = D and C1 = A (see Problem 9.27 b)). Therefore,
ABCD is a parallelogram. Therefore, AB − BC = PABO − PBCO = 0, i.e., ABCD is a
rhombus.
SOLUTIONS
381
20.19. Let O be the intersection point of the diagonals of quadrilateral ABCD. For
definiteness, we may assume that AO ≥ CO and DO ≥ BO. Let points B1 and C1 be
symmetric to points B and C, respectively, through point O. Since O is the center of the
circle inscribed into the quadrilateral, we see that segment B1 C1 is tangent to this circle.
Therefore, segment AD can be tangent to this circle only if B1 = D and C1 = A, i.e.,
if ABCD is a parallelogram. One can inscribe a circle into this parallelogram since this
parallelogram is a rhombus.
20.20. For definiteness, we may assume that AO ≥ CO and DO ≥ BO. Let points B1
and C1 be symmetric to points B and C through point O. Then triangle C1 OB1 is contained
inside triangle AOD and, therefore, the inscribed circle S of triangle C1 OB1 is contained
inside triangle AOD. Suppose that segment AD does not coincide with segment C1 B1 . Then
circle S turns into the inscribed circle of triangle AOD under the homothety with center O
and coefficient greater than 1, i.e., rAOD > rC1 OB1 = rCOB . We have got a contradiction;
hence, A = C1 and D = B1 , i.e., ABCD is a parallelogram.
In parallelogram ABCD, the areas of triangles AOB and BOC are equal and, therefore,
if the inscribed circles have equal radii, then they have equal perimeters since S = pr. It
follows that AB = BC, i.e., ABCD is a rhombus.
20.21. Let AB be the side of the convex hull of the given points, B1 be the nearest to
A of all the given points that lie on AB. Select the one of the remaining points that is the
vertex of the greatest angle that subtends segment AB1 . Let this be point C. Then the
circumscribed circle of triangle AB1 C is the one to be found.
20.22. Let AB be one of the sides of the convex hull of the set of given points. Let us
enumerate the remaining points in the order of increase of the angles with vertex in these
points that subtend segment AB, i.e., denote them by C1 , C2 , . . . , C2n+1 so that
∠AC1 B < ∠AC2 B ≤ · · · < ∠AC2n+1 B.
Then points C1 , . . . , Cn lie outside the circle circumscribed about triangle ABCn+1 and
points Cn+2 , . . . , C2n+1 lie inside it, i.e., this is the circle to be constructed.
20.23. Let AB be the greatest diagonal (or side) of the polygon. Through points A
and B draw lines a and b perpendicular to line AB. If X is a vertex of the polygon, then
AX ≤ AB and XB ≤ AB, therefore, the polygon lies inside the band formed by lines a and
b.
Figure 176 (Sol. 20.23)
Draw the base lines of the polygon parallel to AB. Let these lines pass through vertices
C and D and together with a and b form rectangle KLM N (see Fig. 34). Then
SKLM N = 2SABC + 2SABD = 2SACBD .
Since quadrilateral ACBD is contained in the initial polygon whose area is equal to 1,
SKLM N ≤ 2.
382
CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
20.24. Select the least of all the distances between the given points and consider points
which have neighbours at this distance. Clearly, it suffices to prove the required statement
for these points. Let P be the vertex of the convex hull of these points. If Ai and Aj are the
points nearest to P , then Ai Aj ≥ Ai P and Ai Aj ≥ Aj P and, therefore, ∠Ai P Aj ≥ 60◦ . It
follows that P cannot have four nearest neighbours since otherwise one of the angles ∠Ai P Aj
◦
would have been smaller than 180
= 60◦ . Therefore, P is the point to be found.
3
20.25. Suppose that there are cardboard squares that do not coincide with the plastic
ones. Let us discard all the coinciding squares and consider the convex hull of the vertices
of the remaining squares. Let A be a vertex of this convex hull. Then A is a vertex of
two distinct squares, a cardboard one and a plastic one. It is easy to verify that one of the
vertices of the smaller of these squares lies inside the larger one (Fig. 35).
Let, for definiteness, vertex B of the cardboard square lie inside the plastic one. Then
point B lies inside a plastic square and is a vertex of another plastic square, which is impossible. This is a contradiction, hence, every cardboard square coincides with a plastic
one.
20.26. Let us consider the convex hull of the given points. The two cases are possible:
1) The convex hull is a parallelogram, ABCD. If point M lies inside parallelogram
ABCD, then the vertices of all three parallelograms with vertices at A, B, and M lie
outside ABCD (Fig. 36). Hence, in this case there can be no other points except A, B, C
and D.
2) The convex hull is not a parallelogram. Let AB and BC be edges of the convex hull.
Let us draw base lines parallel to AB and BC. Let these base lines pass through vertices
P and Q. Then the vertices of all the three parallelograms with vertices at B, P and Q lie
outside the convex hull (Fig. 37).
They even lie outside the parallelogram formed by the base lines except for the case
when P and Q are vertices of this parallelogram. In this last case the fourth vertex of the
parallelogram does not belong to the convex hull since the convex hull is not a parallelogram.
20.27. In plane, take an arbitrary straight line l and project all the polygons to it. We
will get several segments any two of which have a common point. Let us order line l; consider
left endpoints of the segments-projections and select the right-most left endpoint. The point
belongs to all the segments and, therefore, the perpendicular drawn through it to l intersects
all the given polygons.
20.28. Let 1000 segments lie in plane. Take an arbitrary line l not perpendicular to any
of them and consider the projections of the endpoints of all these segments on l. It is clear
that the endpoint of the segment whose projection is the left-most of the obtained points
cannot belong to the interior of another segment.
20.29. Two variants of disposition of these four points are possible:
(1) The points are vertices of a convex quadrilateral, ABCD. Take the largest of the
angles of its vertices. Let this be angle ∠ABC. Then ∠ABC ≥ 90◦ , i.e., triangle ABC is
not an acute one.
(2) Point D lies inside triangle ABC. Select the greatest of the angles ∠ADB, ∠BDC
and ∠ADC. Let this be angle ∠ADB. Then ∠ADB ≥ 120◦ , i.e., triangle ADB is an obtuse
one.
We can prove in the following way that there are no other positions of the four points.
The lines that pass through three of given points divide the plane into seven parts (Fig. 38).
If the fourth given point belongs to the 2nd, 4th or 6th part, then we are in situation (1); if
it belongs to the 1st, 3rd, 5th or 7th part, then we are in situation (2).
20.30. The rectangle with vertices at points (0, 0), (0, m), (n, 0) and (n, m) the horizontal
side is equal to n and vertical side is equal to m. From the given set select a rectangle with
SOLUTIONS
Figure 177 (Sol. 20.25)
383
384
CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT
Figure 178 (Sol. 20.26)
Figure 179 (Sol. 20.26)
Figure 180 (Sol. 20.29)
the least horizontal side. Let the length of its vertical side be equal to m1 . Consider any
side m1 of the remaining rectangles. The two cases are possible:
1) The vertical sides of two of these m1 -rectangles are equal. Then one of them is
contained in another one.
2) The vertical sides of all these rectangles are distinct. Then the vertical side of one of
them is greater than m1 and, therefore, it contains the rectangle with the least horizontal
side.
20.31. Consider all the circles passing through two neighbouring vertices Ai and Ai+1
and a vertex Aj such that ∠Ai Aj Ai+1 < 90◦ . At least one such circle exists. Indeed, one of
the angles ∠Ai Ai+2 Ai+1 and ∠Ai+1 Ai Ai+2 is smaller than 90◦ ; in the first case set Aj = Ai+2
and in the second case set Aj = Ai . Among all such circles (for all i and j) select a circle S
of the largest radius; let, for definiteness, it pass through points A1 , A2 and Ak .
Suppose that vertex Ap lies outside S. Then points Ap and Ak lie on one side of line
A1 A2 and ∠A1 Ap A2 < ∠A1 Ak A2 ≤ 90◦ . The law of sines implies that the radius of the circumscribed circle of triangle A1 Ap A2 is greater than that of A1 Ak A2 . This is a contradiction
and, therefore, S contains the whole polygon A1 . . . An .
Let, for definiteness sake, ∠A2 A1 Ak ≤ ∠A1 A2 Ak . Let us prove then that A2 and Ak are
neighbouring vertices. If Ak 6= A3 , then
180◦ − ∠A2 A3 Ak ≤ ∠A2 A1 Ak ≤ 90◦
and, therefore, the radius of the circumscribed circle of triangle A2 A3 Ak is greater than the
radius of the circumscribed circle of triangle A1 A2 Ak . Contradiction implies that S passes
through neighbouring vertices A1 , A2 and A3 .
Chapter 21. DIRICHLET’S PRINCIPLE
Background
1. The most popular (Russian) formulation of Dirichlet’s or pigeonhole principle is the
following one: “If m rabbits sit in n hatches and m > n, then at least one hatch contains at
least two rabbits.”
It is even unclear at first glance why this absolutely transparent remark is a quite effective
method for solving problems. The point is that in every concrete problem it is sometimes
difficult to see what should we designate as the rabbits and the hatches and why there are
more rabbits than the hatches. The choice of rabbits and hatches is often obscured; and
from the formulation of the problem it is not often clear how to immediately deduce that
one should apply Dirichlet’s principle. What is very important is that this method gives a
nonconstructive proof (naturally, we cannot say which precisely hatch contains two rabbits
and only know that such a hatch exists) and an attempt to give a constructive proof, i.e.,
the proof by explicitly constructing or indicating the desired object can lead to far greater
difficulties (and more profound results).
2. Certain problems are also solved by methods in a way similar to Dirichlet’s principle.
Let us formulate the corresponding statements (all of them are easily proved by the rule of
contraries).
a) If several segments the sum of whose lengths is greater than 1 lie on a segment of
length 1, then at least two of them have a common point.
b) If several arcs the sum of whose lengths is greater than 2π lie on the circle of radius
1, then at least two of them have a common point.
c) If several figures the sum of whose areas is greater than 1 are inside a figure of area 1,
then at least two of them have a common point.
§1. The case when there are finitely many points, lines, etc.
21.1. The nodes of an infinite graph paper are painted two colours. Prove that there
exist two horizontal and two vertical lines on whose intersection lie points of the same colour.
21.2. Inside an equilateral triangle with side 1 five points are placed. Prove that the
distance between certain two of them is shorter than 0.5.
21.3. In a 3 × 4 rectangle there are placed 6 points.√Prove that among them there are
two points the distance between which does not exceed 5.
21.4. On an 8 × 8 checkboard the centers of all the cells are marked. Is it possible to
divide the board by 13 straight lines so that in each part there are not more than 1 of marked
points?
21.5. Given 25 points in plane so that among any three of them there are two the
distance between which is smaller than 1, prove that there exists a circle of radius 1 that
contains not less than 13 of the given points.
21.6. In a unit square, there are 51 points. Prove that certain three of them can be
covered by a disk of radius 71 .
385
386
CHAPTER 21. DIRICHLET’S PRINCIPLE
21.7. Each of two equal disks is divided into 1985 equal sectors and on each of the disks
some 200 sectors are painted (one colour). One of the disks was placed upon the other one
◦
and they began rotating one of the disks through multiples of 360
. Prove that there exists
1985
at least 80 positions for which not more than 20 of the painted sectors of the disks coincide.
21.8. Each of 9 straight lines divides a square into two quadrilaterals the ratio of whose
areas is 2 : 3. Prove that at least three of those nine straight lines pass through one point.
21.9. In a park, there grow 10, 000 trees planted by a so-called square-cluster method
(100 rows of 100 trees each). What is the largest number of trees one has to cut down in
order to satisfy the following condition: if one stands on any stump, then no other stump is
seen (one may assume the trees to be sufficiently thin).
21.10. What is the least number of points one has to mark inside a convex n-gon in
order for the interior of any triangle with the vertices at vertices of the n-gon to contain at
least one of the marked points?
21.11. Point P is taken inside a convex 2n-gon. Through every vertex of the polygon
and P a line is drawn. Prove that there exists a side of the polygon which has no common
interior points with neither of the drawn straight lines.
21.12. Prove that any convex 2n-gon has a diagonal non-parallel to either of its sides.
21.13. The nodes of an infinite graph paper are painted three colours. Prove that there
exists an isosceles right triangle with vertices of one colour.
§2. Angles and lengths
21.14. Given n pairwise nonparallel lines in plane. Prove that the angle between certain
◦
.
two of them does not exceed 180
n
21.15. In a circle of radius 1 several chords are drawn. Prove that if every diameter
intersects not more than k chords, then the sum of the length of the chords is shorter than
kπ.
21.16. In plane, point O is marked. Is it possible to place in plane a) five disks; b) four
disks that do not cover O and so that any ray with the beginning in O would intersect not
less than two disks? (“Intersect” means has a common point.)
21.17. Given a line l and a circle of radius n. Inside the circle lie 4n segments of length
1. Prove that it is possible to draw a line which is either parallel or perpendicular to the
given line and intersects at least two of the given segments.
21.18. Inside a unit square there lie several circles the sum of their lengths being equal
to 10. Prove that there exists a straight line intersecting at least four of these circles.
21.19. On a segment of length 1 several segments are marked so that the distance
between any two marked points is not equal to 0.1. Prove that the sum of the lengths of the
marked segments does not exceed 0.5.
21.20. Given two circles the length of each of which is equal to 100 cm. On one of
them 100 points are marked, on the other one there are marked several arcs with the sum
of their lengths less than 1 cm. Prove that these circles can be identified so that no one of
the marked points would be on a marked arc.
21.21. Given are two identical circles; on each of them k arcs are marked, the angle
1
· 180◦ . The circles can be identified(?) so that the
value of each of the arcs is > k2 −k+1
marked arcs of one circle would coincide with the marked arcs of the other one. Prove that
these circles can be identified so that all the marked arcs would lie on unmarked arcs.
SOLUTIONS
387
§3. Area
21.22. In square of side 15 there lie 20 pairwise nonintersecting unit squares. Prove that
it is possible to place in the large square a unit disk so that it would not intersect any of the
small squares.
21.23. Given an infinite graph paper and a figure whose area is smaller than the area of
a small cell prove that it is possible to place this figure on the paper without covering any
of the nodes of the mesh.
21.24. Let us call the figure formed by the diagonals of a unit square (Fig. 39) a cross.
Prove that it is possible to place only a finite number of nonintersecting crosses in a disk of
radius 100.
Figure 181 (21.24)
21.25. Pairwise distances between points A1 , . . . , An is greater than 2. Prove that any
figure whose area is smaller than π can be shifted by a vector not longer than 1 so that it
would not contain points A1 , . . . , An .
21.26. In a circle of radius 16 there are placed 650 points. Prove that there exists a ring
(annulus) of inner radius 2 and outer radius 3 which contains not less than 10 of the given
points.
21.27. There are given n figures in plane. Let Si1 ...ik be the area of the intersection of
figures indexed by i1 , . . . , ik and S be the area of the part of the plane covered by the given
figures; Mk the sum of all the Si1 ...ik . Prove that:
a) S = M1 − M2 + M3 − · · · + (−1)n+1 Mn ;
b) S ≥ M1 − M2 + M3 − · · · + (−1)m+1 Mm for m even and S ≤ M1 − M2 + M3 − · · · +
(−1)m+1 Mm for m odd.
21.28. a) In a square of area 6 there are three polygons of total area 3. Prove that
among them there are two polygons such that the area of their intersection is not less than
1.
b) In a square of area 5 there are nine polygons of total area 1. Prove that among them
there are two polygons the area of whose intersection is not less than 19 .
21.29. On a rug of area 1 there are 5 patches the area of each of them being not less
than 0.5. Prove that there are two patches such that the area of their intersection is not less
than 0.2.
Solutions
21.1. Let us take three vertical lines and nine horizontal lines. Let us consider only
intersection points of these lines. Since there are only 23 = 8 variants to paint three points
two colours, there are two horizontal lines on which lie similarly coloured triples of points.
Among three points painted two colours there are, by Dirichlet’s principle, two similarly
388
CHAPTER 21. DIRICHLET’S PRINCIPLE
coloured points. The vertical lines passing through these points together with the two
horizontal lines selected earlier are the ones to be found.
21.2. The midlines of an equilateral triangle with side 1 separate it into four equilateral
triangles with side 0.5. Therefore, one of the triangles contains at least two of the given points
and these points cannot be vertices of the triangle. The distance between these points is less
than 0.5.
21.3. Let us cut the rectangle into five figures as indicated on Fig. 40. One of the figures
contains at least √
two points and the distance between any two points of each of the figures
does not exceed 5.
Figure 182 (Sol. 21.3)
21.4. 28 fields are adjacent to a side of an 8 × 8 chessboard. Let us draw 28 segments
that connect the centers of neighbouring end(?) fields. Every line can intersect not more
than 2 such segments and, therefore, 13 lines can intersect not more than 26 segments, i.e.,
there are at least 2 segments that do not intersect any of 13 drawn lines. Therefore, it is
impossible to split the chessboard by 13 lines so that in each part there would be not more
than 1 marked point since both endpoints of the segment that does not intersect with the
lines belongs to one of the parts.
21.5. Let A be one of the given points. If all the remaining points lie in disk S1 of radius
1 with center A, then we have nothing more to prove.
Now, let B be a given point that lies outside S1 , i.e., AB > 1. Consider disk S2 of radius
1 with center B. Among points A, B and C, where C is any of the given points, there are
two at a distance less than 1 and these cannot be points A and B. Therefore, disks S1 and
S2 contain all the given points, i.e., one of them contains not less than 13 points.
21.6. Let us divide a given square into 25 similar small squares with side 0.2. By Dirichlet’s principle one of them contains no less than
√ 3 points. The radius of the circumscribed
circle of the square with side 0.2 is equal to 15 2 < 71 and, therefore, it can be covered by a
disk of radius 71 .
21.7. Let us take 1985 disks painted as the second of our disks and place them upon the
first disk so that they would take all possible positions. Then over every painted sector of
the first disk there lie 200 painted sectors, i.e., there are altogether2002 pairs of coinciding
painted sectors. Let there be n positions of the second disk when not less 21 pairs of painted
sectors coincide. Then the number of coincidences of painted sectors is not less than 21n.
Therefore, 21n ≤ 2002 , i.e., n ≤ 1904.8. Since n is an integer, n ≤ 1904. Therefore, at least
for 1985 − 1904 = 81 positions not more than 20 pairs of painted sectors coincide.
21.8. The given lines cannot intersect neighbouring sides of square ABCD since otherwise we would have not two quadrilaterals but a triangle and a pentagon. Let a line intersect
sides BC and AD at points M and N , respectively. Trapezoids ABM N and CDN M have
equal heights, and, therefore, the ratio of their areas is equal to that of their midlines, i.e.,
M N divides the segment that connects the midpoints of sides AB and CD in the ratio of
2 : 3. There are precisely 4 points that divide the midlines of the square in the ratio of 2 : 3.
Since the given nine lines pass through these four points, then through one of the points at
least three lines pass.
SOLUTIONS
389
Figure 183 (Sol. 21.8)
21.9. Let us divide the trees into 2500 quadruples as shown in Fig. 41. In each such
quadruple it is impossible to chop off more than 1 tree. On the other hand, one can chop
off all the trees that grow in the left upper corners of the squares formed by our quadruples.
Therefore, the largest number of trees that can be chopped off is equal to 2500.
21.10. Since any diagonal that goes out of one vertex divides an n-gon into n − 2
triangles, then n − 2 points are necessary.
Figure 184 (Sol. 21.10)
From Fig. 42 one can deduce that n − 2 points are sufficient: it suffices to mark one
points in each shaded triangle. Indeed, inside triangle Ap Aq Ar , where p < q < r, there is
always contained a shaded triangle adjacent to vertex Aq .
21.11. The two cases are possible.
(1) Point P lies on diagonal AB. Then lines P A and P B coincide and do not intersect
the sides. There remain 2n − 2 lines; they intersect not more than 2n − 2 sides.
(2) Point P does not belong to a diagonal of polygon A1 A2 . . . A2n . Let us draw diagonal
A1 An+1 . On both sides of it there lie n sides. Let, for definiteness, point P be inside polygon
A1 . . . An+1 (Fig. 43).
Then lines P An+1 , P An+2 , . . . , P A2n , P A1 (there are n + 1 such lines) cannot intersect
sides An+1 An+2 , An+2 An+3 , . . . , A2n A1 , respectively. Therefore, the remaining straight lines
can intersect not more than n − 1 of these n sides.
21.12. The number of diagonals of a 2n-gon is equal to 2n(2n−3)
= n(2n − 3). It is easy
2
to verify that there are not more than n − 2 diagonals parallel to the given one. Therefore,
there are not more than 2n(n−2) diagonals parallel to the sides. Since 2n(n−2) < n(2n−3),
there exists a diagonal which is not parallel to any side.
390
CHAPTER 21. DIRICHLET’S PRINCIPLE
Figure 185 (Sol. 21.11)
21.13. Suppose that there does not exist an equilateral right triangle whose legs are
parallel to the sides of the cells and with vertices of the same colour. For convenience we
may assume that it is the cells which are painted, not the nodes.
Let us divide the paper into squares of side 4; then on the diagonal of each such square
there are two cells of the same colour. Let n be greater than the number of distinct colorings
of the square of side 4. Consider a square consisting of n2 squares of side 4. On its diagonal
we can find two similarly painted squares of side 4. Finally, take square K on whose diagonal
we can find two similarly painted squares of side 4n.
Figure 186 (Sol. 21.13)
Considering the square with side 4n and in it two similarly painted squares with side 4
we get four cells of the first colour, two cells of the second colour and one cell of the third
colour, see Fig. 44. Similarly, considering square K we get a cell which cannot be of the
first, or second, or third colour.
21.14. In plane, take an arbitrary point and draw through it lines parallel to the given
ones. They divide the plane into 2n angles whose sum is equal to 360◦ . Therefore, one of
◦
these angles does not exceed 180
.
n
21.15. Suppose the sum of the length of the chords is not shorter than πk. Let us prove
that then there exists a diameter which intersects with at least k +1 chords. Since the length
of the arc corresponding to the chord is greater than the length of this chord, the sum of the
lengths of the archs corresponding to given chords is longer than πk. If we add to these arcs
the arcs symmetric to them through the center of the circle, then the sum of the lengths of
all these arcs becomes longer than 2πk. Therefore, there exists a point covered by at least
k + 1 of these arcs. The diameter drawn through this point intersects with at least k + 1
chord.
SOLUTIONS
391
21.16. a) It is possible. Let O be the center of regular pentagon ABCDE. Then the
disks inscribed in angles ∠AOC, ∠BOD, ∠COD, ∠DOA and ∠EOB possess the required
property.
b) It is impossible. For each of the four disks consider the angle formed by the tangents
to the disk drawn through point O. Since each of these four angles is smaller than 180◦ ,
their sum is less than 2 · 360◦ . Therefore, there exists a point on the plane covered by not
more than 1 of these angles. The ray drawn through this point intersects with not more
than one disk.
21.17. Let l1 be an arbitrary line perpendicular to l. Denote the lengths of the projections of the i-th segment to l and l1 by ai and bi , respectively. Since the length of each
segment is equal to 1, we have ai + bi ≥ 1. Therefore,
Let, for definiteness,
(a1 + · · · + a4n ) + (b1 + · · · + b4n ) ≥ 4n.
a1 + · · · + a4n ≥ b1 + · · · + b4n .
Then a1 + · · · + a4n ≥ 2n. The projection of any of the given segment is of length 2n because
all of them lie inside the circle of radius n. If the projections of the given segments to l
would have had no common points, then we would had a1 + · · · + a4n < 2n. Therefore, on
l there exists a point which is the image under the projection of at least two of the given
segments. The perpendicular to l drawn through this point intersects with at least two of
given segments.
21.18. Let us project all the given circles on side AB of square ABCD. The projection
of the circle of length l is a segment of length πl . Therefore, the sum of the lengths of the
projections of all the given circles is equal to 10
. Since 10
> 3 = 3AB, on segment AB there
π
π
is a point which belongs to projections of at least four circles. The perpendicular to AB
drawn through this point intersects at least four circles.
21.19. Let us cut the segment into ten segments of length 0.1, stack them in a pile and
consider their projection to a similar segment as shown on Fig. 45.
Figure 187 (Sol. 21.19)
Since the distance between any two painted points is not equal to 0.1, the painted points
of neighbouring segments cannot be projected into one point. Therefore, neither of the
points can be the image under the projection of painted points of more than 5 segments. It
follows that the sum of the lengths of the projections of the painted segments (equal to the
sum of their lengths) does not exceed 5 · 0.1 = 0.5.
21.20. Let us identify the given circles and let us place a painter in a fixed point of one
of them. Let us rotate this circle and let the painter paint a point of the other circle each
time when it is a marked point that belongs to a marked arc. We have to prove that after a
complete revolution a part of the circle would remain unpainted.
The final result of the painter’s job will be the same as if he were rotated 100 times and
(s)he was asked to paint the other circle on the i-th revolution so that (s)he would have to
392
CHAPTER 21. DIRICHLET’S PRINCIPLE
paint the i-th marked point that belongs to one of the marked arcs. Since in this case at
each revolution less than 1 cm is being painted, it follows that after 100 revolutions there
will be painted less than 100 cm. Therefore, a part of the circle will be unpainted.
21.21. Let us identify(?) our circles and place a painter into a fixed point of one of
them. Let us rotate this circle and let the painter paint the point of the other circle against
which he moves each time when some of the marked arcs intersect. We have to prove that
after a full revolution a part of the circle will be unpainted.
The final result of the painter’s job would be the same as if (s)he were rotated k times
and was asked to paint the circle on the i-th revolution when the i-th marked arc on which
the painter resides would intersect with a marked arc of the other circle.
Let ϕ1 , . . . , ϕn be the angle parameters of the marked arcs. By the hypothesis ϕ1 < α ,
◦
. . . , ϕn < α, where α = k2180
. During the time when the marked arcs with counters i and
−k+1
j intersect the painter paints an arc of length ϕi + ϕj .
Therefore, the sum of the angle values of the arcs painted during the i-th revolution does
not exceed kϕi (ϕ1 + · · · + ϕk ) and the sum of the angle values of the arcs painted during
all k revolutions does not exceed 2k(ϕ1 + · · · + ϕk ). Observe that during all this we have
actually counted the intersection of arcs with similar(?) counters k times.
In particular, point A across which the painter moves at the moment when the marked
arcs coincide has, definitely, k coats of paint. Therefore, it is desirable to disregard the arcs
that the painter paints at the moment when some of the marked arcs with similar counters
intersect. Since all these arcs contain point A, we actually disregard only one arc and the
angle value of this arc does not exceed 2α.
The sum of the angle values of the remaining part of the arcs painted during the i-th
revolution does not exceed (k − 1)ϕ1 + (ϕ1 + · · · + ϕk − ϕi ) and the sum of the angle values
of the remaining part of the arcs painted through all k revolutions does not exceed
(2k − 2) · (ϕ1 + · · · + ϕk ) < (2k 2 − 2k)α.
◦
.
A part of the circle will be unpainted if (2k 2 − 2k)α ≤ 360◦ − 2α, i.e., α ≤ k2180
−k+1
21.22. Let us consider a figure consisting of all the points whose distance from the small
unit square is not greater than 1 (Fig. 46).
Figure 188 (Sol. 21.22)
It is clear that no unit disk whose center is outside this figure intersects the small square.
The area of such a figure is equal to π + 5. The center of the needed disk should also lie
at a distance greater than 1 from the sides of the large square, i.e., inside the square of
side 13. Obviously, 20 figures of total area π + 5 cannot cover a square of side 13 because
20(π + 5) < 132 . The disk with the center in an uncovered point possesses the desired
property.
21.23. Let us paint the figure to(?) the graph paper arbitrarily, cut the paper along
the cells of the mesh and stack them in a pile moving them parallelly with themselves and
SOLUTIONS
393
without turning. Let us consider the projection of this stack on a cell. The projections of
parts of the figure cannot cover the whole cell since their area is smaller. Now, let us recall
how the figure was placed on the graph paper and move the graph paper parallelly with
itself so that its vertices would be in the points whose projection is an uncovered point. As
a result we get the desired position of the figure.
√
21.24. For every cross consider a disk of radius 12 2 with center in the center of the
cross. Let us prove that if two such disks intersect, then the crosses themselves also intersect.
The distance between the centers of equal intersecting disks does not exceed the doubled
radius of any of them and, therefore, the distance between the centers of the corresponding
crosses does not exceed √12 . Let us consider a rectangle given by bars of the first cross and
the center of the second one (Fig. 47).
Figure 189 (Sol. 21.24)
One of the bars of the second cross passes through this rectangle and, therefore, it
intersects the first cross since the length of the bar is equal to √12 and the length of the
diagonal of the rectangle does not exceed √12 . In the disk of a finite radius one can only
√
place finitely many non-intersecting disks of radius 12 2.
21.25. Let Φ be a given figure, S1 , . . . , Sn unit disks with centers at points A1 , . . . , An .
Since disks S1 , . . . , Sn do not intersect, then neither do figures Vi = Φ ∩ Si , consequently,
the sum of their areas does not exceed the area of figure Φ, i.e., it is smaller than π. Let
−−→
O be an arbitrary point and Wi the image of Vi under the translation by vector Ai O. The
figures Wi lie inside the unit disk S centered at O and the sum of their areas is smaller than
the area of this disk. Therefore, point B of disk S does not belong to any of the figures Wi .
−−→
It is clear that the translation by vector BO is the desired one.
21.26. First, notice that point X belongs to the ring with center O if and only if point
O belongs to a similar ring centered at X. Therefore, it suffices to show that if we construct
rings with centers at given points, then not less than 10 rings will cover one of the points of
the considered disk. The considered rings lie inside a disk of radius 16 + 3 = 19 whose area
is equal to 361π. It remains to notice that 9 · 361π = 3249π and the total area of the rings
is equal to 650 · 5π¡ =¢ 3250π.
21.27. a) Let nk be the number of ways to choose k elements from n indistinguishable
ones. One can verify the following Newton binomial formula
n µ ¶
X
n k n−k
n
x y .
(x + y) =
k
k=0
Denote by Wm the area of the part of the plane covered by exactly m figures. This part
consists of pieces each
¡ ¢
¡ ¢ of which is covered by certain m figures. The area of each such piece
has been counted nk times in calculation of Mk because from m figures we can form nk
394
CHAPTER 21. DIRICHLET’S PRINCIPLE
intersections of k figures. Therefore,
¶
µ ¶
µ
µ ¶
n
k+1
n
Wn .
Wk+1 + · · · +
Wk +
Mk =
k
k
k
It follows that
M1 − M2 + M3 − · · · =
µ ¶
µ ¶ µ ¶
µ ¶ µ ¶ µ ¶
1
2
2
n
n
n
W1 + (
−
)W2 + · · · + (
−
+
− . . . )Wn =
1
1
2
1
2
3
W1 + · · · + W n
since
µ ¶
µ ¶ µ ¶ µ ¶
m
m
m
m m
− · · · − (−1)
+
−
=
3
2
1
m
µ ¶ µ ¶
m
m
+ . . . ) + 1) = −(1 − 1)m + 1 = 1.
−
(−1 +
2
1
It remains to observe that S = W1 + · · · + Wn .
b) According to heading a)
S − (M1 − M2 + · · · + (−1)m+1 Mm ) =
(−1)m+2 Mm+1 + (−1)m+3 Mm+2 + . . . · · · + (−1)n+1 Mn =
µ ¶
µ
¶
n
X
i
n+1 i
m+2
)Wi
+ · · · + (−1)
((−1)
n
m
+
1
i=1
¡ ¢
¡ ¢
(it is convenient to assume that nk is defined for k > n so that nk = 0). Therefore, it
suffices to verify that
µ ¶
µ
¶ µ
¶ µ
¶
i
i
i
m+n+1 i
≥ 0 for i ≤ n.
−
+
− · · · + (−1)
n
m+1
m+2
m+3
The identity
(x + y)i = (x + y)i−1 (x + y)
¡ ¢ ¡ i−1 ¢ ¡i−1¢
+ j . Hence,
implies that ji = j−1
¶
¶ µ
µ ¶ µ
¶
¶ µ
µ
i−1
i−1
i
i
m+n+1 i
.
±
=
+ · · · + (−1)
−
n
m
n
m+2
m+1
¡ ¢
= 0 for i ≤ n.
It remains to notice that i−1
n
21.28. a) By Problem 21.27 a) we have
i.e.,
6 = 9 − (S12 + S23 + S13 ) + S123 ,
S12 + S23 + S13 = 3 + S123 ≥ 3.
Hence, one of the numbers S12 , S23 , S13 is not less than 1.
b) By Problem 21.27 b) 5 ≥ 9 − M2 , i.e., M2 ≥ 4. Since from 9 polygons one can form
2
9 · 28 = 36 pairs, the area of the common part of one of such pairs is not less than M
≥ 91 .
36
SOLUTIONS
395
21.29. Let the area of the rug be equal to M , P
the area of the intersection of the patches
indexed by i1 , . . . , ik is equal to Si1 ...ik and Mk = Si1 ...ik . By Problem 21.27 a)
M − M1 + M2 − M3 + M4 − M5 ≥ 0
since M ≥ S. One can write similar inequalities not only for the whole rug but also for every
patch: if we consider the patch S1 as the rug with patches S12 , S13 , S14 , S15 we get
X
X
X
S1 −
S1i +
S1ij −
S1ijk + S12345 ≥ 0.
i
i<j
i<j<k
Adding such inequalities for all five patches we get
M1 − 2M2 + 3M3 − 4M4 + 5M5 ≥ 0
(the summand Si1 ...ik enters the inequality for patches i1 , . . . , ik and, therefore, it enters the
sum of all inequalities with coefficient k). Adding the inequalities
3(M − M1 + M2 − M3 + M4 − M5 ) ≥ 0 and M1 − 2M2 + 3M3 − 4M4 + 5M5 ≥ 0
we get
3M − 2M1 + M2 − M4 + 2M5 ≥ 0.
Adding to this the inequality M4 − 2M5 ≥ 0 (which follows from the fact that S12345 enters
every Si1 i2 i3 i4 , i.e., M4 ≥ 5M5 ≥ 2M5 ) we get 3M − 2M1 + M2 ≥ 0, i.e., M2 ≥ 2M1 − 3M ≥
5 − 3 = 2.
Since from five patches we can form ten pairs, the area of the intersection of patches from
1
one of these pairs is not less than 10
M2 ≥ 0.2.
Chapter 22. CONVEX AND NONCONVEX POLYGONS
Background
1. There are several different (nonequivalent) definitions of a convex polygon. Let us
give the most known and most often encountered definitions. A polygon is called convex if
one of the following conditions is satisfied:
a) the polygon lies on one side of any of its sides (i.e., the intersections of the sides of
the polygon do not intersect its other sides);
b) the polygon is the intersection (i.e., the common part) of several half planes;
c) any segment whose endpoints belong to the polygon wholly belongs to the polygon.
2. A figure is called a convex one if any segment with the endpoints in the points of a
figure belongs to the figure.
3. In solutions of several problems of this chapter we make use of the notion of the convex
hull and the basic line.
§1. Convex polygons
22.1. Given n points in plane such that any four of them are the vertices o```