Solutions - New Zealand Maths Olympiad Committee online

Document technical information

Format pdf
Size 130.1 kB
First found May 22, 2018

Document content analysis

Category Also themed
not defined
no text concepts found





New Zealand Mathematical Olympiad Committee
2011 April Problems — Solutions
These problems are intended to help students prepare for the 2011 camp selection problems,
used to choose students to attend our week-long residential training camp in Christchurch in
The solutions will be posted in about one month’s time, but can be obtained before then by
email if you write to one of us with evidence that you’ve tried the problems seriously.
Good luck!
Chris Tuffley, 2011 NZ IMO team leader
[email protected]
Ilya Chevyrev, 2011 NZ IMO team deputy leader
[email protected]
1. Points A,B,C lie on a circle. Line P B is tangent to the circle at B. Let P A1 and P C1
be the perpendiculars from P to lines AB and BC respectively (A1 and C1 lie on lines
AB and BC respectively). Prove that A1 C1 is perpendicular to AC.
Solution: Since ∠P C1 B = ∠P A1 B = 90◦ , the points P ,A1 ,C1 ,B lie on a circle. There
are two cases, when they are in order P A1 C1 B or P A1 BC1 . We will consider the order
P A1 C1 B and the other case is very similar (the main observation is that they all lie on a
We have that
∠CC1 A1 = 180◦ − ∠A1 C1 B = ∠A1 P B = 90◦ − ∠A1 BP.
On the other hand ∠A1 BP = ∠ACB. Hence ∠CC1 A1 = ∠A1 P B = 90◦ − ∠ACC1 , which
shows A1 C1 and AC are perpendicular. The other case is done similarly.
2. Does there exist a positive integer k such that all the positive integers from 1 to k can
be partitioned into two sets, and all the numbers in each set can be written down one
after another (in some order without spaces) to form two new numbers, so that these two
numbers are equal.
Solution: There does not exist such an integer. Assume such a k exists. It is clear
that k ≥ 10, since the numbers from 1 to 9 never have digits repeating. Consider 10n ,
the largest power of 10 before k, i.e. 10n ≤ k < 10n+1 . Then 10n must be in one of the
two sets and hence the constructed number from this set must have n consecutive zeros
preceded by a 1.
Since natural numbers are not written with zeros at the front, this string of n zeros
preceded by a 1 could not have formed in the other number by joining two or more other
natural numbers, and so it must have come from one natural number. But the next
smallest natural number containing such a string is 10n+1 , so if the two obtained numbers
are equal, we must have used a number greater than or equal to 10n+1 in the other set.
But this implies 10n+1 ≤ k, and so contradicts the maximality of n.
3. Nine skiers participated in a race. They started one by one, and each skier completed
the race with a constant speed (which could be different for different skiers). Determine
if it could happen that each skier participated in an overtaking exactly four times. (In
each overtaking, exactly two skiers participated: the one who overtakes and the one who
is overtaken.)
Solution: This could not happen. Since the speed of each skier is constant, the skier
who started first could not overtake anyone. Hence he must have been overtaken four
times. Hence he arrived in the fifth position. On the other hand, the skier who started
last could not be overtaken by anyone, hence he overtook four other skiers and again
arrived in the fifth position, which gives a contradiction.
4. Let 100 pairwise distinct real numbers be arranged in a circle. Prove that there exist
four consecutive numbers along the circle such that the sum of the middle two numbers is
strictly less than the sum of the other two numbers.
Solution: Solution 1. Assume the contrary. Let the numbers be a1 , a2 , . . . , a100 and
use the notation a100+n = an (i.e. we work with indices modulo 100). Then for n =
1, 2, . . . , 100 we have the inequality an + an+3 ≤ an+1 + an+2 , which is equivalent to
an+3 − an+2 ≤ an+1 − an . This implies a100 − a99 ≤ a98 − a97 ≤ . . . ≤ a2 − a1 ≤
a100 − a99 . So all of these inequalities are equalities. Hence for k = 1, 2, . . . , 50 we have
a2k − a2k−1 = b for a fixed b, and similarly for k = 1, 2, . . . , 50 we have a2k+1 − a2k = c
for a fixed c. Summing these equalities gives 50b + 50c = 0, so b = −c. But then
a2 − a1 = b = −c = −(a3 − a2 ) = a2 − a3 =⇒ a1 = a3 , which contradicts that all numbers
are distinct.
Solution 2. Let a be the smallest number in the circle, b and c its neighbours (b < c) and
d the other neighbour of b (different from a). Then these numbers go in the order d, b, a, c
(or reversed) and a < d, b < c, so we have a + b < d + c, which is what we wanted.
March 28, 2011

Report this document