Đề thi Olympic sinh viên thế giới năm 2007

... IMC2007, Blagoevgrad, Bulgaria Day 1, August 5, 2007 Problem 1. Let f be a polynomial of degree 2 with integer coefficients. ... first columns of A 1 , . . . , A k are linearly dependent, the first 1 IMC2007, Blagoevgrad, Bulgaria Day 2, August 6, 2007 Problem 1. Let f : R → R be a continuous function. Suppose that for ... using only a translation or a rotation. Does this imply that f(x) = ax...

Ngày tải lên: 20/10/2013, 18:15

... and C n,n are relatively prime, we can choose integers C j,n and B j,n such that this equation is satisfied. Doing this step by step for all j = n − 1, n − 2, . . . , 1, we finally get B and C such ... is not divisible by 3. Now add triangles P 1 P k P k+2 , P k P k+1 P k+2 and P 1 P k+2 P k+3 . This way we introduce two new triangles at vertices P 1 and P k so parity is preserved. The vertices...

Ngày tải lên: 17/10/2013, 23:15

... required properties. For an arbitrary rational q, consider the function g q (x) = f(x+q)−f(x). This is a continuous function which attains only rational values, therefore g q is constant. Set ... we can assume that P (X) = 0. Now we are going to prove that P (X k ) = 0 for all k ≥ 1. Suppose this is true for all k < n. We know that P (X n + e) = 0 for e = −P (X n ). From the induction ......

Ngày tải lên: 20/10/2013, 18:15

... 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 b ij are zero except b 1,1 = 2, b n,n = (−1) n , b i,i+1 = ... −  bk/n b(k−1)/n f(x)dx   b 0 g(x)dx + O(ω(f, b/n)g 1 ) = 1 b  b 0 f(x)dx  b 0 g(x)dx + O(ω(f, b/n)g 1 ). This proves a). For b) we set b = π, f(x) = sin x, g(x) = (1 + 3cos 2 x) −1 . From...

Ngày tải lên: 21/01/2014, 21:20

... − 1 3 . From the hypotheses we have  1 0  1 x f(t)dtdx ≥  1 0 1 − x 2 2 dx or  1 0 tf(t)dt ≥ 1 3 . This completes the proof. Problem 3. (15 points) Let f be twice continuously differentiable on (0, ... x tends to 0+ we obtain −x 0 ≤ lim inf x→0+ f(x) f  (x) ≤ lim sup x→0+ f(x) f  (x) ≤ 0. Since this happens for all x 0 ∈ (0, r) we deduce that lim x→0+ f(x) f  (x) exists and lim x→0+ f(x...

Ngày tải lên: 21/01/2014, 21:20

... consider k ≥ 2. For any m we have (2) cosh θ = cosh ((m + 1)θ − mθ) = = cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ = cosh (m + 1)θ.cosh mθ −  cosh 2 (m + 1)θ − 1. √ cosh 2 mθ − 1 Set cosh kθ = a, ... the first and the third integral tend to − 1 f  (0) as n → ∞, hence so does the second. Also n  1 A (f(x)) n dx ≤ n(f(A)) n −→ n→∞ 0 (f (A) < 1). We get L = − 1 f  (0) in this case....

Ngày tải lên: 21/01/2014, 21:20

... suffices to show that this sequence is strictly decreasing. Now, p k − q k − (p k−1 − q k−1 ) = n k p k−1 − (n k + 1)q k−1 − p k−1 + q k−1 = (n k − 1)p k−1 − n k q k−1 and this is negative because p k−1 q k−1 = ... is the rational number p q . Our aim is to show that for some m, θ m−1 = n m n m − 1 . Suppose this is not the case, so that for every m, (3) θ m−1 < n m n m − 1 . 4 For each k we...

Ngày tải lên: 21/01/2014, 21:20

... then there are axes making k 2π n angle). If A is infinite then we can think that A = Z and f (m) = m + 1 for every m ∈ Z. In this case we define g 1 as a symmetry relative to 1 2 , g 2 as a symmetry ... It is enough to prove the theorem for every such set. Let A = T (x). If A is finite, then we can think that A is the set of all vertices of a regular n polygon and that f is rotation by 2π n .

Ngày tải lên: 21/01/2014, 21:20

... contains only powers of ¯y = yK, i.e., S 4 /K is cyclic. It is easy to see that this factor-group is not comutative (something more this group is not isomorphic to S 3 ). 3) n = 5 a) If x = (12), then for ... vector space. Solution First choose a basis {v 1 , v 2 , v 3 } of U 1 . It is possible to extend this basis with vectors v 4 ,v 5 and v 6 to get a basis of U 2 . In the same way we can e...

Ngày tải lên: 26/01/2014, 16:20

... (1) From this recursion we can compute the probabilities for small values of n and can conjecture that p (r) n = 1 5 + 4 5·6 n if n ≡ r (mod )5 and p (r) n = 1 5 − 1 5·6 n otherwise. From (1), this ... opposite inequality holds ∀m  1. This contradiction shows that g is a constant, i.e. f(x) = Cx, C > 0. Conversely, it is easy to check that the functions of this type verify the conditions...

Ngày tải lên: 26/01/2014, 16:20