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 January. 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 circle). 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 1 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 www.mathsolympiad.org.nz 2