IMC 2015, Blagoevgrad, Bulgaria
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1 IMC 05, Blagoevgrad, Bulgaria Day, July 9, 05 Problem. For any integer n and two n n matrices with real entries, B that satisfy the equation + B ( + B prove that det( det(b. Does the same conclusion follow for matrices with complex entries? (Proposed by Zbigniew Skoczylas, Wrocªaw University of Technology Solution. Multiplying the equation by ( + B we get I ( + B( + B ( + B( + B + B + B + BB I + B + B + I B + B + I 0. Let X B ; then XB and B X, so we have X +X +I 0; multiplying by (X IX, Hence, 0 (X IX (X + X + I (X I (X + X + I X 3 I. X 3 I (det X 3 det(x 3 det I det X det det(xb det X det B det B. In case of complex matrices the statement is false. Let ω ( + i 3. Obviously ω / R and ω 3, so 0 + ω + ω + ω + ω. Let I and let B be a diagonal matrix with all entries along the diagonal equal to either ω or ω ω such a way that det(b (if n is not divisible by 3 then one may set B ωi. Then I, B B. Obviously I + B + B 0 and ( + B ( B B I + B + B. By the choice of and B, det det B.
2 Problem. For a positive integer n, let f(n be the number obtained by writing n in binary and replacing every 0 with and vice versa. For example, n 3 is 0 in binary, so f(n is 000 in binary, therefore f(3 8. Prove that n k f(k n 4. When does equality hold? (Proposed by Stephan Wagner, Stellenbosch University Solution. If r and k are positive integers with r k < r then k has r binary digits, so k + f(k }{{... } ( r. r ssume that s n s. Then and therefore s r n(n + + n f(k k r k< r (k + f(k + n (k + f(k k s k n (k + f(k s r ( r + (n s + ( s r s s r r + (n s + ( s r r 3 (4s ( s + ( s n s + 3 s n 4 n k ( s n 3 4s + s 3 ( f(k n 4 ( s n 3 4s + s n(n n ( s 3n + 3 4s s ( ( n s+ n s Notice that the dierence of the last two factors is less than, and one of them must be an integer: s+ is integer if s is even, and s+ 4 is integer if s is odd. Therefore, 3 3 either one of them is 0, resulting a zero product, or both factors have the same sign, so the product is strictly positive. This solves the problem and shows that equality occurs if n s+ 3 (s is even or n s+ 4 3 (s is odd.
3 Problem 3. Let F (0 0, F ( 3, and F (n 5 F (n F (n for n. Determine whether or not is a rational number. F ( n (Proposed by Gerhard Woeginger, Eindhoven University of Technology Solution. The characteristic equation of our linear recurrence is x 5 x + 0, with roots x and x. So F (n a n + b ( n with some constants a, b. By F (0 0 and F ( 3, these constants satisfy a + b 0 and a + b 3. So a and b, and therefore F (n n n. so Observe that F ( n n ( n n ( n n n+, F ( n ( n n+ 0. Hence the sum takes the value, which is rational. Solution. s in the rst solution we nd that F (n n n. Then F ( n n n ( n ( n+ ( (( n n+ k ( n k0 k0 ( n (k+ ( m. m k0 ( k n (Here we used the fact that every positive integer m has a unique representation m n (k + with non-negative integers n and k. This shows that the series converges to. Problem 4. Determine whether or not there exist 5 integers m,..., m 5 such that 5 k m k arctan(k arctan(6. ( (Proposed by Gerhard Woeginger, Eindhoven University of Technology Solution. We show that such integers m,..., m 5 do not exist. Suppose that ( is satised by some integers m,..., m 5. Then the argument of the complex number z + 6i coincides with the argument of the complex number z ( + i m ( + i m ( + 3i m3 ( + 5i m 5. Therefore the ratio R z /z is real (and not zero. s Re z and Re z is an integer, R is a nonzero integer.
4 By considering the squares of the absolute values of z and z, we get ( + 6 R 5 k ( + k m k. Notice that p is a prime (the fourth Fermat prime, which yields an easy contradiction through p-adic valuations: all prime factors in the right hand side are strictly below p (as k < 6 implies + k < p. On the other hand, in the left hand side the prime p occurs with an odd exponent. Problem 5. Let n, let,,..., n+ be n + points in the n-dimensional Euclidean space, not lying on the same hyperplane, and let B be a point strictly inside the convex hull of,,..., n+. Prove that i B j > 90 holds for at least n pairs (i, j with i < j n +. (Proposed by Géza Kós, Eötvös University, Budapest Solution. Let v i B i. The condition i B j > 90 is equivalent with v i v j < 0. Since B is an interior point of the simplex, there are some weights w,..., w n+ > 0 with n+ w i v i 0. i Let us build a graph on the vertices,..., n +. Let the vertices i and j be connected by an edge if v i v j < 0. We show that this graph is connected. Since every connected graph on n + vertices has at least n edges, this will prove the problem statement. Suppose the contrary that the graph is not connected; then the vertices can be split in two disjoint nonempty sets, say V and W such that V W {,,..., n + }. Since there is no edge between the two vertex sets, we have v i v j 0 for all i V and j W. Consider ( ( ( 0 w i v i w i v i + i V W i V i W w i v i + i V w i w j (v i v j. i W Notice that all terms are nonnegative on the right-hand side. Moreover, w i v i 0 and i V w i v i 0, so there are at least two strictly nonzero terms, contradiction. i W Remark. The number n in the statement is sharp; if v n+ (,,..., and v i (0,..., 0,, 0,..., 0 for i,..., n then v }{{}}{{} i v j < 0 holds only when i n + or j n +. i n i Remark. The origin of the problem is here: -simplex-in-an-intersection-of-n-balls/789390
5 IMC 05, Blagoevgrad, Bulgaria Day, July 30, 05 Problem 6. Prove that n n (n + <. (Proposed by Ivan Krijan, University of Zagreb Solution. We prove that <. ( n (n + n n + Multiplying by n(n +, the inequality ( is equivalent with which is true by the M-GM inequality. pplying ( to the terms in the left-hand side, n < (n + n(n + n(n + < n + (n + n (n + < n ( n. n + Problem 7. Compute + x dx. (Proposed by Jan ustek, University of Ostrava Solution. We prove that + For > the integrand is greater than, so x dx > x dx. dx (. In order to nd a tight upper bound, x two real numbers, δ > 0 and K > 0, and split the interval into three parts at the points + δ and K log. Notice that for suciently large (i.e., for > 0 (δ, K with some 0 (δ, K > we have + δ < K log <. For > the integrand is decreasing, so we can estimate it by its value at the starting points of the intervals: x dx ( +δ + K log +δ + < K log (δ + (K log δ +δ + ( K log K log < < (δ + K +δ log + K log δ + K δ +δ log + e K. Hence, for > 0 (δ, K we have < x dx < δ + K δ +δ log + e K.
6 Taking the it we obtain inf Now from δ +0 and K we get so inf inf x dx sup x dx sup x dx sup x dx and therefore + x dx. x dx δ + e K. x dx, Solution. We will employ l'hospital's rule. Let f(, x x, g(, x x, F ( f(, xdx and G( g(, xdx. Since f and x g are continuous, the parametric integrals F ( and G( are dierentiable with respect to, and and Since get so F ( f(, + G ( g(, + f(, xdx + + [ log x g(, xdx + ] x x dx + G(, x x dx log + log., we can see that G G( ( 0. plying l'hospital's rule to G( F ( F ( Now applying l'hospital's rule to + G ( ( + G( we get F ( x dx 0, + 0. F (. we Problem 8. Consider all 6 6 words of length 6 in the Latin alphabet. Dene the weight of a word as /(k +, where k is the number of letters not used in this word. Prove that the sum of the weights of all words is (Proposed by Fedor Petrov, St. Petersburg State University Solution. Let n 6, then 3 75 (n + n. We use the following well-known Lemma. If f(x is a polynomial of degree at most n, then its (n + -st nite dierence vanishes: n+ f(x : n+ i0 ( i( n+ i f(x + i 0. Proof. If is the operator which maps f(x to f(x + f(x, then n+ is indeed (n + -st power of and the claim follows from the observation that decreases the power of a polynomial. In other words, f(x n+ i ( i+( n+ i f(x + i. pplying this for f(x (n x n, substituting x and denoting i j + we get n ( n + n ( n (n + n ( j (n j n (n + ( j j + j j + (n jn. j0 j0
7 The j-th summand ( n j ( j (n j+ jn may be interpreted as follows: choose j letters, consider all (n j n words without those letters and sum up ( j over all those words. Now we change the order of j+ summation, counting at rst by words. For any xed word W with k absent letters we get k ( k ( j0 j j j+ k k+ j0 (k+ ( j j+, since the alternating sum of binomial coecients k k+ j (k+ j+ ( j vanishes. That is, after changing order of summation we get exactly initial sum, and it equals (n + n. Problem 9. n n n complex matrix is called t-normal if t t where t is the transpose of. For each n, determine the maximum dimension of a linear space of complex n n matrices consisting of t-normal matrices. (Proposed by Shachar Carmeli, Weizmann Institute of Science Solution. nswer: The maximum dimension of such a space is n(n+. The number n(n+ can be achieved, for example the symmetric matrices are obviously t-normal and they form a linear space with dimension n(n+. We shall show that this is the maximal possible dimension. Let M n denote the space of n n complex matrices, let S n M n be the subspace of all symmetric matrices and let n M n be the subspace of all anti-symmetric matrices, i.e. matrices for which t. Let V M n be a linear subspace consisting of t-normal matrices. We have to show that dim(v dim(s n. Let π : V S n denote the linear map π( + t. We have dim(v dim(ker (π + dim(im (π so we have to prove that dim(ker (π + dim(im (π dim(s n. Notice that Ker (π n. We claim that for every Ker (π and B V, π(b π(b. In other words, Ker (π and Im (π commute. Indeed, if, B V and t then ( + B( + B t ( + B t ( + B t + B t + B t + BB t t + t B + B t + B t B B t B B + B t (B + B t (B + B t π(b π(b. Our bound on the dimension on V follows from the following lemma: Lemma. Let X S n and Y n be linear subspaces such that every element of X commutes with every element of Y. Then dim(x + dim(y dim(s n Proof. Without loss of generality we may assume X Z Sn (Y : {x S n : xy yx y Y }. Dene the bilinear map B : S n n C by B(x, y tr(d[x, y] where [x, y] xy yx and d diag(,..., n is the matrix with diagonal elements,..., n and zeros o the diagonal. Clearly B(X, Y {0}. Furthermore, if y Y satises that B(x, y 0 for all x S n then tr(d[x, y] tr([d, x], y] 0 for every x S n. We claim that {[d, x] : x S n } n. Let E j i denote the matrix with in the entry (i, j and 0 in all other entries. Then a direct computation shows that [d, E j i ] (j iej i and therefore [d, E j i + Ei j] (j i(e j i Ei j and the collection {(j i(e j i Ei j} i<j n span n for i j. It follows that if B(x, y 0 for all x S n then tr(yz 0 for every z n. But then, taking z ȳ, where ȳ is the entry-wise complex conjugate of y, we get 0 tr(yȳ tr(yȳ t which is the sum of squares of all the entries of y. This means that y 0. It follows that if y,..., y k Y are linearly independent then the equations B(x, y 0,..., B(x, y k 0
8 are linearly independent as linear equations in x, otherwise there are a,..., a k such that B(x, a y a k y k 0 for every x S n, a contradiction to the observation above. Since the solution of k linearly independent linear equations is of codimension k, dim({x S n : [x, y i ] 0, for i,.., k} dim(x S n : B(x, y i 0 for i,..., k dim(s n k. The lemma follows by taking y,..., y k to be a basis of Y. Since Ker (π and Im (π commute, by the lemma we deduce that dim(v dim(ker (π + dim(im (π dim(s n n(n +. Problem 0. Let n be a positive integer, and let p(x be a polynomial of degree n with integer coecients. Prove that max p(x > 0 x e. n Solution. Let For every positive integer k, let (Proposed by Géza Kós, Eötvös University, Budapest M max p(x. J k 0 x 0 ( p(x kdx. Obviously 0 < J k < M k is a rational number. If (p(x k kn i0 a k,i x i then J k kn i0 a k,i. Taking the least i+ common denominator, we can see that J k lcm(,,..., kn +. n equivalent form of the prime number theorem is that log lcm(,,..., N N if N. Therefore, for every ε > 0 and suciently large k we have and therefore M k > J k Taking k and then ε +0 we get Since e is transcendent, equality is impossible. lcm(,,..., kn + < e (+ε(kn+ lcm(,,..., kn + > e, (+ε(kn+ M > e. (+ε(n+ k M e n. Remark. The constant e is not sharp. It is known that the best constant is between 0.43 and (See I. E. Pritsker, The GelfondSchnirelman method in prime number theory, Canad. J. Math. 57 (005, 0800.
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