Tài liệu Đề thi Olympic sinh viên thế giới năm 2004 pdf

11 th International Mathematical Competition for University Students Skopje, 25–26 July 2004 Solutions for problems on Day 2 1. Let A be a real 4 × 2 matrix and B be a real 2 × 4 matrix such that AB =     1 0 −1 0 0 1 0 −1 −1 0 1 0 0 −1 0 1     . Find BA. [20 points] Solution. Let A =  A 1 A 2  and B =  B 1 B 2  where A 1 , A 2 , B 1 , B 2 are 2 × 2 matrices. Then     1 0 −1 0 0 1 0 −1 −1 0 1 0 0 −1 0 1     =  A 1 A 2   B 1 B 2  =  A 1 B 1 A 1 B 2 A 2 B 1 A 2 B 2  therefore, A 1 B 1 = A 2 B 2 = I 2 and A 1 B 2 = A 2 B 1 = −I 2 . Then B 1 = A −1 1 , B 2 = −A −1 1 and A 2 = B −1 2 = −A 1 . Finally, BA =  B 1 B 2   A 1 A 2  = B 1 A 1 + B 2 A 2 = 2I 2 =  2 0 0 2  2. Let f, g : [a, b] → [0, ∞) be continuous and non-decreasing functions such that for each x ∈ [a, b] we have  x a  f(t) dt ≤  x a  g(t) dt and  b a  f(t) dt =  b a  g(t) dt. Prove that  b a  1 + f(t) dt ≥  b a  1 + g(t) dt. [20 points] Solution. Let F (x) =  x a  f(t) dt and G(x) =  x a  g(t) dt. The functions F, G are convex, F (a) = 0 = G(a) and F (b) = G(b) by the hypothesis. We are supposed to show that  b a  1 +  F  (t)  2 dt ≥  b a  1 +  G  (t)  2 dt i.e. The length ot the graph of F is ≥ the length of the graph of G. This is clear since both functions are convex, their graphs have common ends and the graph of F is below the graph of G — the length of the graph of F is the least upper bound of the lengths of the graphs of piecewise linear functions whose values at the points of non-differentiability coincide with the values of F , if a convex polygon P 1 is contained in a polygon P 2 then the perimeter of P 1 is ≤ the perimeter of P 2 . 3. Let D be the closed unit disk in the plane, and let p 1 , p 2 , . . . , p n be fixed points in D. Show that there exists a point p in D such that the sum of the distances of p to each of p 1 , p 2 , . . . , p n is greater than or equal to 1. [20 points] Solution. considering as vectors, thoose p to be the unit vector which points into the opposite direction as n  i=1 p i . Then, by the triangle inequality, n  i=1 |p − p i | ≥      np − n  i=1 p i      = n +      n  i=1 p i      ≥ n 4. For n ≥ 1 let M be an n × n complex matrix with distinct eigenvalues λ 1 , λ 2 , . . . , λ k , with multiplicities m 1 , m 2 , . . . , m k , respectively. Consider the linear operator L M defined by L M (X) = MX + XM T , for any complex n × n matrix X. Find its eigenvalues and their multiplicities. (M T denotes the transpose of M; that is, if M = (m k,l ), then M T = (m l,k ).) [20 points] Solution. We first solve the problem for the special case when the eigenvalues of M are distinct and all sums λ r + λ s are different. Let λ r and λ s be two eigenvalues of M and v r , v s eigenvectors associated to them, i.e. Mv j = λv j for j = r, s. We have Mv r (v s ) T +v r (v s ) T M T = (Mv r )(v s ) T +v r  Mv s  T = λ r v r (v s ) T +λ s v r (v s ) T , so v r (v s ) is an eigenmatrix of L M with the eigenvalue λ r + λ s . Notice that if λ r = λ s then vectors u, w are linearly independent and matrices u(w) T and w(u) T are linearly independent, too. This implies that the eigenvalue λ r + λ s is double if r = s. The map L M maps n 2 –dimensional linear space into itself, so it has at most n 2 eigenvalues. We already found n 2 eigenvalues, so there exists no more and the problem is solved for the special case. In the general case, matrix M is a limit of matrices M 1 , M 2 , . . . such that each of them belongs to the special case above. By the continuity of the eigenvalues we obtain that the eigenvalues of L M are • 2λ r with multiplicity m 2 r (r = 1, . . . , k); • λ r + λ s with multiplicity 2m r m s (1 ≤ r < s ≤ k). (It can happen that the sums λ r + λ s are not pairwise different; for those multiple values the multiplicities should be summed up.) 5. Prove that  1 0  1 0 dx dy x −1 + | ln y| − 1 ≤ 1. [20 points] Solution 1. First we use the inequality x −1 − 1 ≥ | ln x|, x ∈ (0, 1], which follows from (x −1 − 1)   x=1 = | ln x|| x=1 = 0, (x −1 − 1)  = − 1 x 2 ≤ − 1 x = | ln x|  , x ∈ (0, 1]. Therefore  1 0  1 0 dx dy x −1 + | ln y| − 1 ≤  1 0  1 0 dx dy | ln x| + | ln y| =  1 0  1 0 dx dy | ln(x · y)| . Substituting y = u/x, we obtain  1 0  1 0 dx dy | ln(x · y)| =  1 0   1 u dx x  du | ln u| =  1 0 | ln u| · du | ln u| = 1. Solution 2. Substituting s = x −1 − 1 and u = s − ln y,  1 0  1 0 dx dy x −1 + | ln y| − 1 =  ∞ 0  ∞ s e s−u (s + 1) 2 u duds =  ∞ 0   u 0 e s (s + 1) 2 ds  e −u u dsdu. Since the function e s (s+1) 2 is convex,  u 0 e s (s + 1) 2 ds ≤ u 2  e u (u + 1) 2 + 1  so  1 0  1 0 dx dy x −1 + | ln y| − 1 ≤  ∞ 0 u 2  e u (u + 1) 2 + 1  e −u u du = 1 2   ∞ 0 du (u + 1) 2 +  ∞ 0 e −u du  = 1. 6. For n ≥ 0 define matrices A n and B n as follows: A 0 = B 0 = (1) and for every n > 0 A n =  A n−1 A n−1 A n−1 B n−1  and B n =  A n−1 A n−1 A n−1 0  . Denote the sum of all elements of a matrix M by S(M). Prove that S(A k−1 n ) = S(A n−1 k ) for every n, k ≥ 1. [20 points] Solution. The quantity S(A k−1 n ) has a special combinatorical meaning. Consider an n × k table filled with 0’s and 1’s such that no 2 × 2 c ontains only 1’s. Denote the number of such fillings by F nk . The filling of each row of the table corresponds to some integer ranging from 0 to 2 n − 1 written in base 2. F nk equals to the number of k-tuples of integers such that every two consecutive integers correspond to the filling of n × 2 table without 2 × 2 squares filled with 1’s. Consider binary expansions of integers i and j i n i n−1 . . . i 1 and j n j n−1 . . . j 1 . There are two cases: 1. If i n j n = 0 then i and j can be consecutive iff i n−1 . . . i 1 and j n−1 . . . j 1 can be consequtive. 2. If i n = j n = 1 then i and j can be consecutive iff i n−1 j n−1 = 0 and i n−2 . . . i 1 and j n−2 . . . j 1 can be consecutive. Hence i and j can be consecutive iff (i + 1, j + 1)-th entry of A n is 1. Denoting this entry by a i,j , the sum S(A k−1 n ) =  2 n −1 i 1 =0 · · ·  2 n −1 i k =0 a i 1 i 2 a i 2 i 3 · · · a i k−1 i k counts the possible fillings. Therefore F nk = S(A k−1 n ). The the obvious statement F nk = F kn completes the proof. 11 th International Mathematical Competition for University Students Skopje, 25–26 July 2004 Solutions for problems on Day 1 Problem 1. Let S be an infinite set of real numbers such that |s 1 + s 2 + ··· + s k | < 1 for every finite subset {s 1 , s 2 , . . . , s k } ⊂ S. Show that S is countable. [20 points] Solution. Let S n = S ∩( 1 n , ∞) for any integer n > 0. It follows from the inequality that |S n | < n. Similarly, if we define S −n = S ∩ (−∞, − 1 n ), then |S −n | < n. Any nonzero x ∈ S is an element of some S n or S −n , because there exists an n such that x > 1 n , or x < − 1 n . Then S ⊂ {0}∪  n∈N (S n ∪S −n ), S is a countable union of finite sets, and hence countable. Problem 2. Let P (x) = x 2 − 1. How many distinct real solutions does the following equation have: P (P (. . . (P    2004 (x)) . . . )) = 0 ? [20 points] Solution. Put P n (x) = P (P ( (P    n (x)) )). As P 1 (x) ≥ −1, for each x ∈ R, it must be that P n+1 (x) = P 1 (P n (x)) ≥ −1, for each n ∈ N and each x ∈ R . Therefore the equation P n (x) = a, where a < −1 has no real solutions. Let us prove that the equation P n (x) = a, where a > 0, has exactly two distinct real solutions. To this end we use mathematical induction by n. If n = 1 the assertion follows directly. Assuming that the assertion holds for a n ∈ N we prove that it must also hold for n + 1. Since P n+1 (x) = a is equivalent to P 1 (P n (x)) = a, we conclude that P n (x) = √ a + 1 or P n (x) = − √ a + 1. The equation P n (x) = √ a + 1, as √ a + 1 > 1, has exactly two distinct real solutions by the inductive hypothesis, while the equation P n (x) = − √ a + 1 has no real solutions (because − √ a + 1 < −1). Hence the equation P n+1 (x) = a, has exactly two distinct real solutions. Let us prove now that the e quation P n (x) = 0 has exactly n + 1 distinct real solutions. Again we use mathematical induction. If n = 1 the solutions are x = ±1, and if n = 2 the solutions are x = 0 and x = ± √ 2, so in both cases the number of solutions is equal to n + 1. Suppose that the assertion holds for some n ∈ N. Note that P n+2 (x) = P 2 (P n (x)) = P 2 n (x)(P 2 n (x) − 2), so the set of all real solutions of the equation P n+2 = 0 is exactly the union of the sets of all real solutions of the equations P n (x) = 0, P n (x) = √ 2 and P n (x) = − √ 2. By the inductive hypothesis the equation P n (x) = 0 has exactly n + 1 distinct real solutions, while the equations P n (x) = √ 2 and P n (x) = − √ 2 have two and no distinct real solutions, respectively. Hence, the sets above being pairwise disjoint, the equation P n+2 (x) = 0 has exactly n + 3 distinct real solutions. Thus we have proved that, for each n ∈ N, the equation P n (x) = 0 has exactly n + 1 distinct real solutions, so the answer to the question posed in this problem is 2005. Problem 3. Let S n be the set of all sums n  k=1 x k , where n ≥ 2, 0 ≤ x 1 , x 2 , . . . , x n ≤ π 2 and n  k=1 sin x k = 1 . a) Show that S n is an interval. [10 points] b) Let l n be the length of S n . Find lim n→∞ l n . [10 points] Solution. (a) Equivalently, we consider the set Y = {y = (y 1 , y 2 , , y n )| 0 ≤ y 1 , y 2 , , y n ≤ 1, y 1 + y 2 + + y n = 1} ⊂ R n and the image f(Y ) of Y under f(y) = arcsin y 1 + arcsin y 2 + + arcsin y n . Note that f(Y ) = S n . Since Y is a connected subspace of R n and f is a continuous function, the image f(Y ) is also connected, and we know that the only connected subspaces of R are intervals. Thus S n is an interval. (b) We prove that n arcsin 1 n ≤ x 1 + x 2 + + x n ≤ π 2 . Since the graph of sin x is concave down for x ∈ [0, π 2 ], the chord joining the points (0, 0) and ( π 2 , 1) lies below the graph. Hence 2x π ≤ sin x for all x ∈ [0, π 2 ] and we can deduce the right-hand side of the claim: 2 π (x 1 + x 2 + + x n ) ≤ sin x 1 + sin x 2 + + sin x n = 1. The value 1 can be reached choosing x 1 = π 2 and x 2 = ··· = x n = 0. The left-hand side follows immediately from Jensen’s inequality, since sin x is concave down for x ∈ [0, π 2 ] and 0 ≤ x 1 +x 2 + +x n n < π 2 1 n = sin x 1 + sin x 2 + + sin x n n ≤ sin x 1 + x 2 + + x n n . Equality holds if x 1 = ··· = x n = arcsin 1 n . Now we have computed the minimum and maximum of interval S n ; we can conclude that S n = [n arcsin 1 n , π 2 ]. Thus l n = π 2 − n arcsin 1 n and lim n→∞ l n = π 2 − lim n→∞ arcsin(1/n) 1/n = π 2 − 1. Problem 4. Suppose n ≥ 4 and let M be a finite set of n points in R 3 , no four of which lie in a plane. Assume that the p oints can be coloured black or white so that any sphere which intersects M in at least four points has the property that exactly half of the points in the intersection of M and the sphere are white. Prove that all of the points in M lie on one sphere. [20 points] Solution. Define f : M → {−1, 1}, f (X) =  −1, if X is white 1, if X is black . The given condition becomes  X∈S f (X) = 0 for any sphere S which passes through at least 4 points of M. For any 3 given points A, B, C in M, denote by S (A, B, C) the set of all spheres which pass through A, B, C and at least one other point of M and by |S (A, B, C)| the number of these spheres. Also, denote by  the sum  X∈M f (X). We have 0 =  S∈S(A,B,C)  X∈S f (X) = (|S (A, B, C)| −1) (f (A) + f (B) + f (C)) +  (1) since the values of A, B, C appear |S (A, B, C)| times each and the other values appear only once. If there are 3 points A, B, C such that |S (A, B, C)| = 1, the proof is finished. If |S (A, B, C)| > 1 for any distinct points A, B, C in M, we will prove at first that  = 0. Assume that  > 0. From (1) it follows that f (A) + f (B) + f (C) < 0 and summing by all  n 3  possible choices of (A, B, C) we obtain that  n 3   < 0, which means  < 0 (contradicts the starting assumption). The same reasoning is applied when assuming  < 0. Now, from  = 0 and (1), it follows that f (A) + f (B) + f (C) = 0 for any distinct points A, B, C in M . Taking another point D ∈ M , the following equalities take place f (A) + f (B) + f (C) = 0 f (A) + f (B) + f (D) = 0 f (A) + f (C) + f (D) = 0 f (B) + f (C) + f (D) = 0 which easily leads to f (A) = f (B) = f (C) = f (D) = 0, which contradicts the definition of f. Problem 5. Let X be a set of  2k−4 k−2  + 1 real numbers, k ≥ 2. Prove that there exists a monotone sequence {x i } k i=1 ⊆ X such that |x i+1 − x 1 | ≥ 2|x i − x 1 | for all i = 2, . . . , k − 1. [20 points] Solution. We prove a more general statement: Lemma. Let k, l ≥ 2, let X be a set of  k+l−4 k−2  +1 real numbers. Then either X contains an increasing sequence {x i } k i=1 ⊆ X of length k and |x i+1 − x 1 | ≥ 2|x i − x 1 | ∀i = 2, . . . , k − 1, or X contains a decreasing sequence {x i } l i=1 ⊆ X of length l and |x i+1 − x 1 | ≥ 2|x i − x 1 | ∀i = 2, . . . , l − 1. Proof of the lemma. We use induction on k + l. In case k = 2 or l = 2 the lemma is obviously true. Now let us make the induction step. Let m be the minimal element of X, M be its maximal element. Let X m = {x ∈ X : x ≤ m + M 2 }, X M = {x ∈ X : x > m + M 2 }. Since  k+l−4 k−2  =  k+(l−1)−4 k−2  +  (k−1)+l−4 (k−1)−2  , we can see that either |X m | ≥  (k − 1) + l − 4 (k − 1) −2  + 1, or |X M | ≥  k + (l − 1) − 4 k − 2  + 1. In the first case we apply the inductive assumption to X m and either obtain a decreasing sequence of length l with the required properties (in this case the inductive step is made), or obtain an increasing sequence {x i } k−1 i=1 ⊆ X m of length k −1. Then we note that the sequence {x 1 , x 2 , . . . , x k−1 , M} ⊆ X has length k and all the required properties. In the case |X M | ≥  k+(l−1)−4 k−2  + 1 the inductive step is made in a similar way. Thus the lemma is proved. The reader may check that the number  k+l−4 k−2  + 1 cannot be smaller in the lemma. Problem 6. For every complex numb e r z /∈ {0, 1} define f(z) :=  (log z) −4 , where the sum is over all branches of the complex logarithm. a) Show that there are two polynomials P and Q such that f(z) = P (z)/Q(z) for all z ∈ C \ {0, 1}. [10 points] b) Show that for all z ∈ C \ {0, 1} f(z) = z z 2 + 4z + 1 6(z −1) 4 . [10 points] Solution 1. It is clear that the left hand side is well defined and independent of the order of summation, because we have a sum of the type  n −4 , and the branches of the logarithms do not matter because all branches are taken. It is easy to check that the convergence is locally uniform on C \{0, 1}; therefore, f is a holomorphic function on the complex plane, except possibly for isolated singularities at 0 and 1. (We omit the detailed estimates here.) The function log has its only (simple) zero at z = 1, so f has a quadruple pole at z = 1. Now we investigate the behavior near infinity. We have Re(log(z)) = log |z|, hence (with c := log |z|) |  (log z) −4 | ≤  |log z| −4 =  (log |z| + 2πin) −4 + O(1) =  ∞ −∞ (c + 2πix) −4 dx + O(1) = c −4  ∞ −∞ (1 + 2πix/c) −4 dx + O(1) = c −3  ∞ −∞ (1 + 2πit) −4 dt + O(1) ≤ α(log |z|) −3 for a universal constant α. Therefore, the infinite sum tends to 0 as |z| → ∞ . In particular, the isolated singularity at ∞ is not essential, but rather has (at least a single) zero at ∞. The remaining singularity is at z = 0. It is readily verified that f(1/z) = f(z) (because log(1/z) = −log(z)); this implies that f has a zero at z = 0. We conclude that the infinite sum is holomorphic on C with at most one pole and without an essential singularity at ∞, so it is a rational function, i.e. we can write f(z) = P (z)/Q(z) for some polynomials P and Q which we may as well assume coprime. This solves the first part. Since f has a quadruple pole at z = 1 and no other poles, we have Q(z) = (z − 1) 4 up to a constant factor which we can as well set equal to 1, and this determines P uniquely. Since f(z) → 0 as z → ∞, the degree of P is at most 3, and since P(0) = 0, it follows that P(z) = z(az 2 + bz + c) for yet undetermined complex constants a, b, c. There are a number of ways to compute the coefficients a, b, c, which turn out to be a = c = 1/6, b = 2/3. Since f (z) = f(1/z), it follows easily that a = c. Moreover, the fact lim z→1 (z − 1) 4 f(z) = 1 implies a + b + c = 1 (this fact follows from the observation that at z = 1, all summands cancel pairwise, except the principal branch which contributes a quadruple pole). Finally, we can calculate f(−1) = π −4  nodd n −4 = 2π −4  n≥1odd n −4 = 2π −4    n≥1 n −4 −  n≥1even n −4   = 1 48 . This implies a −b + c = −1/3. These three equations easily yield a, b, c. Moreover, the function f satisfies f(z) + f(−z) = 16f(z 2 ): this follows because the branches of log(z 2 ) = log((−z) 2 ) are the numbers 2 log(z) and 2 log(−z). This observation supplies the two equations b = 4a and a = c, which can be used instead of some of the considerations above. Another way is to compute g(z) =  1 (log z) 2 first. In the same way, g(z) = dz (z−1) 2 . The unknown coefficient d can be computed from lim z→1 (z −1) 2 g(z) = 1; it is d = 1. Then the exponent 2 in the denominator can be increased by taking derivatives (see Solution 2). Similarly, one can start with exponent 3 directly. A more straightforward, though tedious way to find the constants is computing the first four terms of the Laurent series of f around z = 1. For that branch of the logarithm which vanishes at 1, for all |w| < 1 2 we have log(1 + w) = w − w 2 2 + w 3 3 − w 4 4 + O(|w| 5 ); after some computation, one can obtain 1 log(1 + w) 4 = w −4 + 2w −2 + 7 6 w −2 + 1 6 w −1 + O(1). The remaining branches of logarithm give a bounded function. So f(1 + w) = w −4 + 2w −2 + 7 6 w −2 + 1 6 w −1 (the remainder vanishes) and f(z) = 1 + 2(z −1) + 7 6 (z −1) 2 + 1 6 (z −1) 3 (z −1) 4 = z(z 2 + 4z + 1) 6(z −1) 4 . Solution 2. ¿From the well-known series for the cotangent function, lim N→∞ N  k=−N 1 w + 2πi ·k = i 2 cot iw 2 and lim N→∞ N  k=−N 1 log z + 2πi ·k = i 2 cot i log z 2 = i 2 · i e 2i· i log z 2 + 1 e 2i· i log z 2 − 1 = 1 2 + 1 z −1 . Taking derivatives we obtain  1 (log z) 2 = −z ·  1 2 + 1 z −1   = z (z −1) 2 ,  1 (log z) 3 = − z 2 ·  z (z −1) 2   = z(z + 1) 2(z −1) 3 and  1 (log z) 4 = − z 3 ·  z(z + 1) 2(z −1) 3   = z(z 2 + 4z + 1) 2(z −1) 4 . . 2π −4    n≥1 n −4 −  n≥1even n −4   = 1 48 . This implies a −b + c = −1/3. These three equations easily yield a, b, c. Moreover, the function f satisfies f(z) + f(−z) = 16f(z 2 ): this follows. +  G  (t)  2 dt i.e. The length ot the graph of F is ≥ the length of the graph of G. This is clear since both functions are convex, their graphs have common ends