tailieunhanh - Đề thi Olympic sinh viên thế giới năm 2004

" Đề thi Olympic sinh viên thế giới năm 2004 " . 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. | 11th International Mathematical Competition for University Students Skopje 25 26 July 2004 Solutions for problems on Day 2 1. Let A be a real 4 x 2 matrix and B be a real 2 x 4 matrix such that AB 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 Find BA. 20 points Solution. Let A 41 j and B A2 J 1 0 0 1 -1 0 0 -1 therefore A1B1 A2B2 I2 and A1. Finally BA B Bi B2 where A1 A2 B1 B2 are 2 x 2 matrices. Then -1 0 0 1 _ iA1A ZR B _ f A1B1 A1B2 1 0 A B B A2B1 A2B2 01 A ID A ID T Tbn-n D A 1 D A 1 A D 1 A1B2 A2B1 12. Then B1 A1 B2 A1 and A2 B2 I B2 B1A1 B2A2 2I2 fn A2 0 2 2. Let f g a b 0 x be continuous and non-decreasing functions such that for each x G a b we have x rx a TW dt y g i dt J a J a and fa V Üdt fa VWdt. Prove that a 1 f t dt f a 1 g t dt. 20 points Solution. Let F x ff f t dt and G x f g t dt. The functions F G are convex F a 0 G a and F b G b by the hypothesis. We are supposed to show that 1 F t 2 dt 1 G t 2 dt . The length ot the graph of F is the length of the graph of G. This is clear since both functions are convex their graphs have common ends and the graph of F is below the graph of G the length of the graph of F is the least upper bound of the lengths of the graphs of piecewise linear functions whose values at the points of non-differentiability coincide with the values of F if a convex polygon P1 is contained in a polygon P2 then the perimeter of P1 is the perimeter of P2. 3. Let D be the closed unit disk in the plane and let p1 p2 . pn be fixed points in D. Show that there exists a point p in D such that the sum of the distances of p to each of p1 p2 . pn is greater than or equal to 1. 20 points Solution. considering as vectors thoose p to be the unit vector which points into the opposite direction as n V pi. Then by the triangle inequality i 1 n J2ip Pi i 1 n n X np Pi n X pi i 1 i 1 n. 4. For n 1 let M be an n x n complex matrix with distinct eigenvalues A1 A2 . Ak with multiplicities m1 m2 . mk respectively. Consider the linear operator LM defined by LM X MX