Tham khảo tài liệu 'đề thi toán apmo (châu á thái bình dương)_đề 14', khoa học tự nhiên, toán học phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | XXI Asian Pacific Mathematics Olympiad March 2009 Problem 1. Consider the following operation on positive real numbers written on a blackboard Choose a number r written on the blackboard erase that number and then write a pair of positive real numbers a and b satisfying the condition 2r2 ab on the board. Assume that you start out with just one positive real number r on the blackboard and apply this operation k2 1 times to end up with k2 positive real numbers not necessarily distinct. Show that there exists a number on the board which does not exceed kr. Solution Using AM-GM inequality we obtain 1 2 2ab a2 b2 1 r2 ab a2b2 a2b2 a2 b2 Consequently if we let Si be the sum of the squares of the reciprocals of the numbers written on the board after I operations then Si increases as I increases that is So S1 Sk2-1. Therefore if we let s be the smallest real number written on the board after k2 1 operations then -2 t2 for any number t among k2 numbers on the board and hence k x s2 Sk2-1 So r2 which implies that s kr as desired. Remark. The nature of the problem does not change at all if the numbers on the board are restricted to be positive integers. But that may mislead some contestants to think the problem is a number theoretic problem rather than a combinatorial problem. 1 Problem 2. Let a1 a2 a3 a4 a5 be real numbers satisfying the following equations ai I a2 I a3 I a4 I a5 1 - 12. 7 2 I o 7 2 I TO I . TO I r T2 for k 1 2 3 4 5. k 2 1 k 2 2 k2 3 k 2 4 k 2 5 k2 a1 a2 a3 a4 a5 Find the value of 37 38 39 40 41 Express the value in a single fraction. Solution1 Let R x J - 2 X2 3 x . R 1 1 R 2 4 R 3 9 R 4 16 R 5 2-5 and R 6 is the value to be found. Let s put P x x2 1 x2 2 x2 3 x2 4 x2 5 and Q x R x P x . Then for k 1 2 3 4 5 we get Q k R k P k that is P k k2Q k 0. Since k2 P x x2Q x is a polynomial of degree 10 with roots 1 2 3 4 5 we get P x x2Q x A x2 1 x2 4 x2 9 x2 16 x2 25 . Putting x 0 we get of by P x yields A .P 0 . . 1 4 9 16 25 12o . Finally dividing both sides 1
