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.
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 + 3 3 4 n ( s 3n + 3 4s s + 3 3 ( ( n s+ n s+ 4. 4 3 3 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.
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.
By considering the squares of the absolute values of z and z, we get ( + 6 R 5 k ( + k m k. Notice that p + 6 57 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: http://math.stackexchange.com/questions/476640/n -simplex-in-an-intersection-of-n-balls/789390
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.
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 3 75. (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
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
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 0.3679 is not sharp. It is known that the best constant is between 0.43 and 0.43. (See I. E. Pritsker, The GelfondSchnirelman method in prime number theory, Canad. J. Math. 57 (005, 0800.