" Đề thi Olympic sinh viên thế giới năm 2000 " . 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. | Solutions for the first day problems at the IMC 2000 Problem 1. Is it true that if f 0 1 0 1 is a monotone increasing b monotone decreasing then there exists an x 2 0 1 for which f x x Solution. a Yes. Proof Let A x 2 0 1 f x xg. If f 0 0 we are done if not then A is non-empty 0 is in A bounded so it has supremum say a. Let b f a . I. case a b. Then using that f is monotone and a was the sup we get b f a f a b 2 a b 2 which contradicts a b. II. case a b. Then we get b f a f a b 2 a b 2 contradiction. Therefore we must have a b. b No. Let for example f x 1 x 2 if x 1 2 and f x 1 2 - x 2 if x 1 2 This is clearly a good counter-example. Problem 2. Let p x x5 x and q x x5 x2. Find all pairs w z of complex numbers with w z for which p w p z and q w q z . Short solution. Let P x y p y x4 x3 y x2 y2 xy3 y4 1 x - y and Q x y q x4 x3 y x2 y2 xy3 y4 x y. x - y We need those pairs w z which satisfy P w z Q w z 0. From P Q 0 we have w z 1. Let c wz. After a short calculation we obtain c2 3c 2 0 which has the solutions c 1 and c 2. From the system w z 1 wz c we obtain the following pairs 1 V3z 1 TV3A 1 P7z 1 TP7i ------- -------- and -------------- -------- . y 2 2 y y 2 2 y 1 Solutions for the second day problems at the IMC 2000 Problem 1. a Show that the unit square can be partitioned into n smaller squares if n is large enough. b Let d 2. Show that there is a constant N d such that whenever n N d a d-dimensional unit cube can be partitioned into n smaller cubes. Solution. We start with the following lemma If a and b be coprime positive integers then every sufficiently large positive integer m can be expressed in the form ax by with x y non-negative integers. Proof of the lemma. The numbers 0 a 2a . b 1 a give a complete residue system modulo b. Consequently for any m there exists a 0 x b 1 so that ax m mod b . If m b 1 a then y m ax b for which x by m is a non-negative integer too. Now observe that any dissection of a cube into n smaller cubes may be rehned to give a .
