" Đề thi Olympic sinh viên thế giới năm 1994 " . 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. Từ đó đến nay, các kỳ thi Olympic sinh viênthế giới đã liên tục được mở rộng quy mô rất lớn. Kỳ thi này là một sự kiện quan trọng đối với phong trào học toán của sinh viên thế giới trong trường đại. | International Competition in Mathematics for Universtiy Students in Plovdiv Bulgaria 1994 1 PROBLEMS AND SOLUTIONS First day July 29 1994 Problem 1. 13 points a Let A be a n x n n 2 symmetric invertible matrix with real positive elements. Show that zn n2 2n where zn is the number of zero elements in A-1. b How many zero elements are there in the inverse of the n x n matrix 1 1 1 1 1 . .1 C 1 2 2 2 . .2 A 1 2 1 1 . .1 1 2 1 2 . .2 Solution. Denote by ay and by the elements of A and A 1 respectively. Then for k m we have PP akibim 0 and from the positivity of ay we i 0 conclude that at least one of bim i 1 2 . ng is positive and at least one is negative. Hence we have at least two non-zero elements in every column of A-1. This proves part a . For part b all by are zero except bi i 2 bnn 1 n bi i i bi i i - 1 i for i 1 2 . n - 1. Problem 2. 13 points Let f 2 C1 a b lim f x 1 lim f x 1 and x a x b f0 x f2 x 1 for x 2 a b . Prove that b a and give an example where b a . Solution. From the inequality we get arcfo- f x x f x 1 0 dx arctg f x x 1 f2 x 1 0 for x 2 a b . Thus arctg f x x is non-decreasing in the interval and using the limits we get a b. Hence b a . One has equality for f x cotg x a 0 b . Problem 3. 13 points 2 Given a set S of 2n 1 n 2 N different irrational numbers. Prove that there are n different elements xi x2 . xn 2 S such that for all nonnegative rational numbers ai a2 . an with ai a2 --------- an 0 we have that ai xi a2x2 anxn is an irrational number. Solution. Let I be the set of irrational numbers Q - the set of rational numbers Q Q 0 1 . We work by induction. For n 1 the statement is trivial. Let it be true for n 1. We start to prove it for n. From the induction argument there are n 1 different elements x i x2 . xn_i 2 S such that 1 aixi a2x2 an ixri i 2 I for all ai a2 . an 2 Q with ai a2 an_i 0. Denote the other elements of S by xn xn i . x2n_i. Assume the statement is not true for n. Then for k 0 1 . n 1 there are rk 2 Q such that n i n i 2 T .
