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 + b for some real numbers a and b ? Solution. No. The function f(x) = e x also has this property since ce x = e x+log c . Problem...

Đề thi Olympic sinh viên thế giới năm 2006

... 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 ... minimum of f on [x, y]. Then f([x, y]) = [f(b), f (a)]; hence y − x = f (a) − f(b) ≤ |a − b| ≤ y − x This implies {a, b} = {x, y}, and therefore f is a monotone function. Suppose f is increasing. Then...

Đề thi Olympic sinh viên thế giới năm 2008

... 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 ... ) = 0 for k ≥ 1 we immediately get P (f ) = f(0) for all f ∈ V . 1 is a Cauchy sequence in H. (This is the crucial observation.) Indeed, for m > n, the norm y m − y n  may be computed by...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1994 pptx

... at leastone is negative. Hence we have at least two non-zero elements in everycolumn of A−1. This proves part a). For part b) all bijare zero exceptb1,1= 2, bn,n= (−1)n, bi,i+1= ... −bk/nb(k−1)/nf(x)dxb0g(x)dx + O(ω(f, b/n)g1)=1bb0f(x)dxb0g(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 + 3cos2x)−1.From a) andπ0sin ... y) such that f(c) = 0 and f (x) > 0 for x ∈ (c, y]. For x ∈ (c, y] wehave |f(x)| ≤ λf(x). This implies that the function g(x) = ln f (x) − λx isnot increasing in (c, y] because of g(x)...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1995 docx

... −13.From the hypotheses we have101xf(t)dtdx ≥101 − x22dx or10tf(t)dt ≥13. This completes the proof.Problem 3. (15 points)Let f be twice continuously diﬀerentiable on (0, ... x tends to 0+ we obtain−x0≤ lim infx→0+f(x)f(x)≤ lim supx→0+f(x)f(x)≤ 0.Since this happens for all x0∈ (0, r) we deduce that limx→0+f(x)f(x)exists andlimx→0+f(x)f(x)= ... |x|p+ |y|p= 2}.Since Dδis compact it is enough to show that f is continuous on Dδ.For this we show that the denominator of f is diﬀerent from zero. Assumethe contrary. Then |x + y|...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1996 doc

... 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 −cosh2(m + 1)θ − 1.√cosh2mθ − 1Set cosh kθ = a, ... the ﬁrst and the third integral tend to −1f(0)as n → ∞, henceso does the second.Also n1A(f(x))ndx ≤ n(f(A))n−→n→∞0 (f (A) < 1). We get L = −1f(0)in this case.If M ... B.Hence B is a projection. Thus there exists a basis of eigenvectors for B, andthe matrix of B in this basis is of the form diag(1, . . . , 1, 0, . . . , 0).Since A = 2B − E the eigenvalues of...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 1 pptx

... suﬃcesto show that this sequence is strictly decreasing. Now,pk− qk− (pk−1− qk−1) = nkpk−1− (nk+ 1)qk−1− pk−1+ qk−1= (nk− 1)pk−1− nkqk−1and this is negative becausepk−1qk−1= ... is the rational numberpq. Our aim is to showthat for some m,θm−1=nmnm− 1.Suppose this is not the case, so that for every m,(3) θm−1<nmnm− 1.4For each k we writeθk=pkqkas ... number anddet ω(BA − AB) = ωndet(BA − AB) and det(BA − AB) = 0, then ωnis areal number. This is possible only when n is divisible by 3.2Problem 4.Let α be a real number, 1 < α <...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1997 ngày 2 docx

... then thereare axes making k2πnangle). If A is inﬁnite then we can think that A = Zand f (m) = m + 1 for every m ∈ Z. In this case we deﬁne g1as a symmetryrelative to12, g2as 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 regularn polygon and that f is rotation by2πn....

Tài liệu Đề thi Olympic sinh viên thế giới năm 1998 ppt

... containsonly powers of ¯y = yK, i.e., S4/K is cyclic. It is easy to see that this factor-group is not comutative(something more this group is not isomorphic to S3).3) n = 5a) If x = (12), then for ... vector space.Solution First choose a basis {v1, v2, v3} of U1. It is possible to extend this basis with vectors v4,v5andv6to get a basis of U2. In the same way we can extend a ... fi(x)fi(ai) = 0, thus f (x − f1(x)a1− · · · − fk(x)ak) = 0. By the linearity of f this impliesf(x) = f1(x)f(a1) + · · · + fk(x)f(ak), which gives f(x) as a linear combination...

Tài liệu Đề thi Olympic sinh viên thế giới năm 1999 doc

... (1)From this recursion we can compute the probabilities for small values of n and can conjecture that p(r)n=15+45·6nif n ≡ r (mod )5 and p(r)n=15−15·6notherwise. From (1), this ... opposite inequality holds ∀m  1. Thiscontradiction 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 ... solution for equation A3= A + I if and only if λ3= λ + 1, because A3− A − I = (λ3− λ − 1)I. Thisequation, being cubic, has real solution.b) It is easy to check that the polynomial p(x) =...

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