Đề thi Olympic sinh viên thế giới năm 2005

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Đề thi Olympic sinh viên thế giới năm 2005

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" Đề thi Olympic sinh viên thế giới năm 2005 " . 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...

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1. 12th International Mathematics Competition for University Students Blagoevgrad, July 22 - July 28, 2005 First Day Problem 1. Let A be the n × n matrix, whose (i, j)th entry is i + j for all i, j = 1, 2, . . . , n. What is the rank of A? Solution 1. For n = 1 the rank is 1. Now assume n ≥ 2. Since A = (i)n + (j)n , matrix A is the sum i,j=1 i,j=1 of two matrixes of rank 1. Therefore, the rank of A is at most 2. The determinant of the top-left 2 × 2 minor is −1, so the rank is exactly 2. Therefore, the rank of A is 1 for n = 1 and 2 for n ≥ 2. Solution 2. Consider the case n ≥ 2. For i = n, n − 1, . . . , 2, subtract the (i − 1)th row from the nth row. Then subtract the second row from all lower rows.       1 2 ... n 2 3 ... n + 1 2 3 ... n + 1 1 1 . . . 1   3 4 . . . n + 2  1 1 . . . 1    rank  . .  = rank  . .  = rank 0 0 . . . 0  = 2.      . .. .  . .. .   . . . . . . . .. . .. . . . n + 1 n + 2 ... 2n 1 1 ... 1 0 0 ... 0 Problem 2. For an integer n ≥ 3 consider the sets Sn = {(x1 , x2 , . . . , xn ) : ∀i xi ∈ {0, 1, 2}} An = {(x1 , x2 , . . . , xn ) ∈ Sn : ∀i ≤ n − 2 |{xi , xi+1 , xi+2 }| = 1} and Bn = {(x1 , x2 , . . . , xn ) ∈ Sn : ∀i ≤ n − 1 (xi = xi+1 ⇒ xi = 0)} . Prove that |An+1 | = 3 · |Bn |. (|A| denotes the number of elements of the set A.) Solution 1. Extend the deﬁnitions also for n = 1, 2. Consider the following sets An = {(x1 , x2 , . . . , xn ) ∈ An : xn−1 = xn } , An = An \ An , Bn = {(x1 , x2 , . . . , xn ) ∈ Bn : xn = 0} , Bn = Bn \ Bn and denote an = |An |, an = |An |, an = |An |, bn = |Bn |, bn = |Bn |, bn = |Bn | . It is easy to observe the following relations between the a–sequences   an = an + an a = an ,  n+1 an+1 = 2an + 2an which lead to an+1 = 2an + 2an−1 . For the b–sequences we have the same relations   bn = bn + bn b = bn ,  n+1 bn+1 = 2bn + 2bn therefore bn+1 = 2bn + 2bn−1 . By computing the ﬁrst values of (an ) and (bn ) we obtain a1 = 3, a2 = 9, a3 = 24 b1 = 3, b2 = 8
2. 12th International Mathematics Competition for University Students Blagoevgrad, July 22 - July 28, 2005 Second Day Problem 1. Let f (x) = x2 + bx + c, where b and c are real numbers, and let M = {x ∈ R : |f (x)| < 1}. Clearly the set M is either empty or consists of disjoint open intervals. Denote the sum of their lengths by |M |. Prove that √ |M | ≤ 2 2. 2 2 b Solution. Write f (x) = x + 2 + d where d = c − b4 . The absolute minimum of f is d. If d ≥ 1 then f (x) ≥ 1 for all x, M = ∅ and |M | = 0. If −1 < d < 1 then f (x) > −1 for all x, 2 √ b b −1 < x+ +d
3. where a, b ∈ R, a1 , . . . , ak , b1 , . . . bl are positive integers and p1 , . . . , pk , q1 , . . . , ql are irre- ducible polynomials with leading coeﬃcients 1. For p3 = q 2 and the factorisation of p3 = q 2 is unique we get that a3 = b2 , k = l and for some (i1 , . . . , ik ) permutation of (1, . . . , k) we have p1 = qi1 , . . . , pk = qik and 3a1 = 2bi1 , . . . , 3ak = 2bik . Hence b1 , . . . , bl are divisible by b /3 b /3 3 let r = b1/3 · q11 · . . . · ql l be a polynomial. Since r3 = q = f 3 we have f = r. p f3 Solution 2. Let q be the simplest form of the rational function f2 . Then the simplest form 2 p2 p2 f3 of its square is q2 . On the other hand q2 = f2 = f 2 is a polynomial therefore q must f3 p be a constant and so f = f2 = q is a polynomial. Problem 3. In the linear space of all real n × n matrices, ﬁnd the maximum possible dimension of a linear subspace V such that ∀X, Y ∈ V trace(XY ) = 0. (The trace of a matrix is the sum of the diagonal entries.) Solution. If A is a nonzero symmetric matrix, then trace(A2 ) = trace(At A) is the sum of the squared entries of A which is positive. So V cannot contain any symmetric matrix but 0. Denote by S the linear space of all real n × n symmetric matrices; dim V = n(n+1) . 2 Since V ∩ S = {0}, we have dim V + dim S ≤ n2 and thus dim V ≤ n2 − n(n+1) = n(n−1) . 2 2 n(n−1) The space of strictly upper triangular matrices has dimension 2 and satisﬁes the condition of the problem. Therefore the maximum dimension of V is n(n−1) . 2 Problem 4. Prove that if f : R → R is three times diﬀerentiable, then there exists a real number ξ ∈ (−1, 1) such that f (ξ) f (1) − f (−1) = − f (0). 6 2 Solution 1. Let f (−1) 2 f (1) 2 g(x) = − x (x − 1) − f (0)(x2 − 1) + x (x + 1) − f (0)x(x − 1)(x + 1). 2 2 It is easy to check that g(±1) = f (±1), g(0) = f (0) and g (0) = f (0). Apply Rolle’s theorem for the function h(x) = f (x) − g(x) and its derivatives. Since h(−1) = h(0) = h(1) = 0, there exist η ∈ (−1, 0) and ϑ ∈ (0, 1) such that h (η) = h (ϑ) = 0. We also have h (0) = 0, so there exist ∈ (η, 0) and σ ∈ (0, ϑ) such that h ( ) = h (σ) = 0. Finally, there exists a ξ ∈ ( , σ) ⊂ (−1, 1) where h (ξ) = 0. Then f (−1) f (1) f (1) − f (−1) f (ξ) = g (ξ) = − · 6 − f (0) · 0 + · 6 − f (0) · 6 = − f (0). 2 2 2