The 26 th Annual Vojtěch Jarník International Mathematical Competition Ostrava, 8 th April 2016 Category I. f(0) e f (ξ) = = ln f(1) ln f(0). - PDF

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Ostrava, 8 th April 26 Problem Let f : R (, ) be a continuously differentiable function Prove that there exists ξ (, ) such that e f (ξ) f() f(ξ) f() f(ξ) Solution The equality is equivalent to and e f

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Ostrava, 8 th April 26 Problem Let f : R (, ) be a continuously differentiable function Prove that there exists ξ (, ) such that e f (ξ) f() f(ξ) f() f(ξ) Solution The equality is equivalent to and e f (ξ) f (ξ) f(ξ) ( ) f(ξ) f() f() ln f() ln f() [ points] The existence of ξ (, ) satisfying the last equality follows from Lagrange s mean value theorem with the function φ(x) ln f(x), x [, ] Ostrava, 8 th April 26 Problem 2 Find all positive integers n such that ϕ(n) divides n (ϕ(n) denotes Euler s totient function, ie the number of positive integers k n coprime to n) [ points] Solution Answer: n, 2, 3, 5, 9, 2, If n is prime, then ϕ(n) n divides n (n )(n + ) + 4, thus n divides 4 and we get answers n 2, 3, 5 Let now n be composite and n k i pαi i has k distinct prime factors Then ϕ(n) k i pαi i (p i ) is divisible by 2 k, thus n is divisible by 2 k It implies that n is odd and k 2, since 8 never divides n Next, if α i 2, then p i divides ϕ(n), thus p i divides n and p i 3 In this case 9 does not divide n 2 + 3, hence 9 does not divide ϕ(n), ie α i 2 So, we have three cases: n 9 (this is another answer), n 9p for prime p 3 and n pq for different odd primes p, q ) n 9p Then ϕ(n) 6(p ), n (p )(p + ) + 84 So, 3(p ) divides 84, thus p divides 28, but for possible p 5, 29 we again get that 8 divides ϕ(n) 2) n pq If q 3 we see that ϕ(n) 2(p ) divides n (p )(p + ) + 2, so 2(p ) divides 2, for p 7 we get the answer n 2 Now assume that p 3, q 3 Then 3 does not divide n 2 + 3, thus 3 does not divide ϕ(n) (p )(q ) It implies that both p and q are congruent to 2 modulo 3, and we may write p 2a +, q 2b +, where a, b are congruent to 2 modulo 3 We get that ϕ(n) 4ab divides n (4ab + 2a + 2b + ) ab(4ab + ) + 4(a 2 + b 2 + a + b + ), ie ab divides a 2 + b 2 + a + b + Let us prove that it is impossible when a, b are congruent to 2 modulo 3 Assume the contrary and choose a pair of such a, b with minimal value of a + b Obviously a, b are coprime, so we may suppose that a b (here we use that a b is forbidden modulo 3) The number b 2 + b + is divisible by a, denote b 2 + b + ma The number m is congruent to 2 modulo 3 and m a since ma b 2 + b + (b + ) 2 a 2 Note that m divides b 2 + b + and b divides a 2 (m 2 + m + ) (b 2 + b + ) 2 + a(b 2 + b + ) + a 2 (a 2 + a + ) + bx for some integer x Since a and b are relatively prime, we get that b divides m 2 + m + Thus both m and b divide Q : b 2 + m 2 + b + m +, therefore bm divides Q and (b, m) is smaller pair then (a, b) This contradiction finishes the proof Ostrava, 8 th April 26 Problem 3 Let d 3 and let A A d+ be a simplex in R d (A simplex is the convex hull of d + points not lying in a common hyperplane) For every i,, d + let O i be the circumcentre of the face A A i A i+ A d+, ie O i lies in the hyperplane A A i A i+ A d+ and it has the same distance from all points A,, A i, A i+,, A d+ For each i draw a line through A i perpendicular to the hyperplane O O i O i+ O d+ Prove that either these lines are parallel or they have a common point [ points] Solution If O,, O d+ lie in the hyperplane, our lines are parallel If not, they form a simplex which has a circumcentre Q Let O be circumcentre of A A d+, let P be symmetric to O in a point Q Let s prove that all our lines pass through P That is, P A i must be perpendicular to O j O k if i, j, k are different Denote by M i a midpoint OA i Then A i P is parallel to a middle line M i Q of triangle OA i P i We have M i O j M i O k OA i /2, thus both points M i, Q lie in a perpendicular bisector to the segment O j O k, hence M i Q O j O k as desired Ostrava, 8 th April 26 Problem 4 Find the value of the sum A n, where n A n k k k 2 n k 2 + k2 2 k k2 n Solution We will show more general fact: [ points] Theorem Let ( ) be an increasing sequence of real numbers greater or equal than, such that then series converges to S Then n n n k k k! Sk for every k, 2, Proof [I] by induction on k For k the equality n and denote We have S is obvious Assume now that the equality holds for k, 2,, m n + n k S k k + + m The last term sums to S m and the first equals to ( + ) + + m m + + m m ( ) m m Once again the last term sums to S m Repeating the above transformations leads to So adding the above equality gives hence + + m m + + m m + S m S m S (m )S m, S m m S m S m (m )! Sm S m! Sm + + m + + m Proof [II] in the case n Let F (x) x an As n, the function F is well-defined and continuous for x (, ) We have F (x) dx x n n S Moreover F (x x 2 x k ) F (x 2 x k ) F (x k ) n n k x an x an +an 2 2 x an + +an k k, hence n n k Let now S(x) k x F (t) dt t d We have of course dx S(αx)n n F (αx) n S(αx) So x F (x x 2 x k ) F (x 2 x k ) F (x k ) dx dx 2 dx k x x 2 x k ( ( ( ( k! Sk F (x x 2 x k ) dx ) ) dxk F (xk ) x x k S(x x 2 x k ) x x F (x x 2 x k ) F (x 2 x k ) F (x k ) dx dx 2 dx k x x 2 x k F (x 2 x k ) dx 2 ) ) dxk F (xk ) x 2 x k Corollary In the notation of the theorem we have A n e S, where n A k n n k k Solution: equal to e π2 /6 It is known that n n π2 2 6 From the Corollary we get directly that the sum in question is
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