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...
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Đề thi Olympic sinh viên thế giới năm 2004 11th International Mathematical Competition for University Students Skopje, 25–26 July 2004 Solutions for problems on Day 21. Let A be a real 4 × 2 matrix and B be a real 2 × 4 matrix such that   1 0 −1 0 0 1 0 −1 AB = −1 0 . 1 0 0 −1 0 1Find BA. [20 points] A1Solution. Let A = and B = B1 B2 where A1 , A2 , B1 , B2 are 2 × 2 matrices. Then A2   1 0 −1 0 0  1 0 −1 = A1 B1 B2 = A1 B1 A1 B2 −1 0 1 0 A2 A2 B1 A2 B2 0 −1 0 1therefore, A1 B1 = A2 B2 = I2 and A1 B2 = A2 B1 = −I2 . Then B1 = A−1 , B2 = −A−1 and A2 = B2 = 1 1 −1−A1 . Finally, A1 2 0 BA = B1 B2 = B1 A1 + B2 A2 = 2I2 = A2 0 22. Let f, g : [a, b] → [0, ∞) be continuous and non-decreasing functions such that for each x ∈ [a, b] wehave x x f (t) dt ≤ g(t) dt a a b band a f (t) dt = g(t) dt. a b b Prove that a 1 + f (t) dt ≥ a 1 + g(t) dt. [20 points] x xSolution. Let F (x) = a f (t) dt and G(x) = a 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 b b 2 2 1 + F (t) dt ≥ 1 + G (t) dt a ai.e. The length ot the graph of F is ≥ the length of the graph of G. This is clear since both functions areconvex, their graphs have common ends and the graph of F is below the graph of G — the length of thegraph of F is the least upper bound of the lengths of the graphs of piecewise linear functions whose valuesat the points of non-differentiability coincide with the values of F , if a convex polygon P1 is contained ina 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 thereexists a point p in D such that the sum of the distances of p to each of p1 , p2 , . . . , pn is greater than orequal to 1. [20 points]Solution. considering as vectors, thoose p to be the unit vector which points into the opposite direction asn pi . Then, by the triangle inequality,i=1 n n n |p − pi | ≥ np − pi = n + pi ≥ n.. i=1 i=1 i=14. For n ≥ 1 let M be an n × n complex matrix with distinct eigenvalues λ1 , λ2 , . . . , λk , with multiplicitiesm1 , m2 , . . . , mk , respectively. Consider the linear operator LM defined by LM (X) = M X + XM T , for anycomplex n × n matrix X. Find its eigenvalues and their multiplicities. (M T denotes the transpose of M ;that is, if M = (mk,l ), then M T = (ml,k ).) [20 points]Solution. We first solve the problem for the special case when the eigenvalues of M are distinct and all sumsλr + λs are different. Let λr and λs be two eigenvalues of M and vr , vs eigenvectors associated to them, i.e. TM vj = λvj for j ...