TAILIEUCHUNG - Đề thi Olympic Toán Sinh viên Quốc tế 2012 - IMC2011 Day 1

Đề thi Olympic toán sinh viên quốc tế năm 2011 . Đây là một sân chơi lớn để sinh viên thế giới có dịp gặp gỡ, trao đổi, giao lưu và thể hiện khả năng học toán, làm toán của mình. | IMC2011 Blagoevgrad Bulgaria Day 1 July 30 2011 Problem 1. Let f R R be a continuous function. A point x is called a shadow point if there exists a point y G R with y x such that f y f x . Let a b be real numbers and suppose that all the points of the open interval I a b are shadow points a and b are not shadow points. Prove that a f x f b for all a x b b f a f b . Jose Luis Diaz-Barrero Barcelona Solution. a We prove by contradiction. Suppose that exists a point c G a b such that f c f b . By Weierstrass theorem f has a maximal value m on c b this value is attained at some point d G c b . Since f d max f f c f b we have d b so d G c b c a b . The point d lying in a b is a shadow point therefore c. f y f d for some y d. From combining our inequalities we get f y f d f b . Case 1 y b. Then f y f b contradicts the assumption that b is not a shadow point. Case 2 y b. Then y G d b c c b therefore f y f d m maxf f y contradiction again. c b Since a b and a is not a shadow point we have f a f b . By part a we already have f x f b for all x G a b . By the continuity at a we have f a Ji o f x .Ji o f b f b x a U x a U Hence we have both f a f b and f a f b so f a f b . Problem 2. Does there exist a real 3 X 3 matrix A such that tr A 0 and A2 A I tr A denotes the trace of A A is the transpose of A and I is the identity matrix. Moubinool Omarjee Paris Solution. The answer is NO. Suppose that tr A 0 and A2 A I. Taking the transpose we have A I - A2 I - A 2 I - I - A2 2 2A2 - A4 A4 - 2A2 A 0. The roots of the polynomial x4 - 2x2 x x x - 1 x2 x - 1 are 0 1 1 p so these numbers can be the eigenvalues of A the eigenvalues of A2 can be 0 1 1 2 5. By tr A 0 the sum of the eigenvalues is 0 and by tr A2 tr I - A4 3 the sum of squares of the eigenvalues is 3. It is easy to check that this two conditions cannot be satisfied simultaneously. Problem 3. Let p be a prime number. Call a positive integer n interesting if xn - 1 xp - x 1 f x pg x for some polynomials f and g with integer .

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