Mathematical contests 1995 -1996
Mathematical Contests 1995-1996 Olympiad Problems and Solutions from around the World Published by the American Mathematics Competitions, 1997 Preface This book is a supplement to Mathematical Olympiads 1995: Olympiad Problems from Around the World, published by the American Mathematics Competitions. It contains solutions to the problems from 28 national and regional contests featured in the earlier pamphlet, together with selected problems (without solutions) from national and regional contests given during the academic year 1995-1996. This collection is intended as practice for the serious student who wishes to improve his or her performance on the USAMO. Some of the problems are comparable to the USAMO in that they came from national contests. Others are harder, as some country's first have a national olympiad, and later one or more exams to select a team for the IMO. Some problems come from regional international contests ("mini-IMO'S"). Different nations have different mathematical cultures, so you will find some of these problems extremely hard and some rather easy. We have tried to present a wide variety of problems, especially from those countries that have often done well at the IMO. Each contest has its own time limit. We have not furnished this information, because we have not always included complete contests. As a rule of thumb, most contests allow a time limit ranging between one-half to one full hour per problem. This book was prepared by Titu Andreescu (Illinois Mathematics and Science Academy), Kiran Kedlaya (Princeton University) and Paul Zeitz (University of San Francisco). Thanks also to Walter Mientka unending support, and to the students at the 1996 USA Mathematical Olympiad Summer Program for their help in preparing some of the solutions. The problems in this publication are copyrighted by the International Mathematical Olympiad (IMO). Requests for reproduction permissions should be directed to: Dr. Walter Mientka Secretary, IMO Advisory Broad University of Nebraska @ Lincoln 1740 Vine Street Lincoln, NE 68588-0658, USA. Contents 1 1995 Contests: Problems and Solutions 3 1.1 Austria . . . . . . 3 1.2Bulgaria. . . 6 1.3 China . . . . . . . . 13 1.4 Czech and Slovak Republics . . . 19 1.5 France . . . . . . . . . . 23 1.6 Germany . . . . 27 1.7 Greece . . . . . . . 31 1.8 Hungary. . . . . . . . . . . . . 34 1.9India. 42 1.10 Iran . . . . . . . . . . . . 48 1.11 Ireland . . . . 54 1.12 Israel . . . . . . . . 60 1.13 Japan . . . . . . 65 1.14 Korea . . . . . 70 1.15 Poland . . . . 79 1.16 Romania . . . . 83 1.17 Russia . . . . 96 1.18 Singapore . . . . . . . . .117 1.19 Taiwan . . . . . . . . . . .120 1.20 Turkey . . . . . . . . 124 1.21 United Kingdom . . . . . . . . 128 1.22 United States of America . . . .132 1.23 Vietnam . . . . . . . . .138 1.24 Austrian-Polish Mathematics Competition. .142 1.25 Balkan Mathematical Olympiad. . .149 1.26 Czech-Slovak Match . . . . . . . . . 152 1.27 Iberoamerican Olympiad . . . . . . . . . 156 1.28 UNESCO . . . . . . . . . . . . . . . . . .160 21996 National Contests: Problems 163 2.1Bulgaria . . . 163 2.2 Canada . . . 165 2.3 China . 166 2.4 Czech and Slovak Republics . . . . . 167 2.5 France . . . . . . . . 168 2.6 Germany . . . . 169 1 2.7 Greece . . . . . . . . . . . . . 170 2.8Iran . . . 171 2.9 Ireland . . . . . . 173 2.10 Italy. . . . .175 2.11 Japan . . . . . . . . . . . . . .176 2.12 Poland . . . . .177 2.13 Romania . . .178 2.14 Russia . . . . .181 2.15 Spain . . . 184 2.16 Turkey . . . . . . . . . . . 185 2.17 United Kingdom . . 186 2.18 United States of America . . . . .188 2.19 Vietnam . . . 189 31996 Regional Contests: Problems 190 3.1 Asian Pacific Mathematics Olympiad . . . . . . .190 3.2 Austrian-Polish Mathematics Competition 191 3.3 Balkan Mathematical Olympiad . . . . . . .193 3.4 Czech-Slovak Match . . . . . . 194 3.5 Iberoamerican Olympiad . . . . . . . . 195 3.6 St. Petersburg City Mathematical Olympiad . .197 2 1 1995 Contests: Problems and Solutions 1.1 Austria 1. For how many a) even and b) odd numbers n, does n divide 312 - 1, yet n does not divide 3k - 1 for k = 1,2, ,11. Solution: We note 312 - 1 = (36 -1)(3 6 + 1) = (32 - 1)(34 + 32 + 1)(32 + 1)(34 -3 2 + 1) = (23)(7. 13)(2 -5)(73). Recall that the number of divisors of pi' pk' is (e1 + 1) (ek + 1). Therefore 312 -1 has 2 - 2 - 2 - 2 = 16 odd divisors and 4 . - .16 = 64 even divisors. If 312 - 1 (mod m) for some integer m, then the smallest integer d such that 3' - 1 (mod m) divides 12. (Otherwise we could write 12 = pq + r with 0 < r < d and find 3' =- 1 (mod m).) Hence to ensure n %3k -1 for k = 1, ,11, we need only check k = 1,2,3,4,6. But 2.13 24 .5 23.7.13. The odd divisors we throw out are 1, 5, 7, 13, 91, while the even divi- sorsare2' for 1 <i < 4, 2'-5for 1<i<4,andeachof2'.7,2j-13, and 23 - 7 - 13 for 1 < i < 3. As we are discarding 17 even divisors and 5 odd ones, we remain with 47 even divisors and 11 odd ones. 2. Consider a triangle ABC. For each circle K which passes through A and B, define PK, QK to be the intersections of K with BC and AC, respectively. (a) Show that for any circles K, K' passing through A and B, the lines PKQK and PK'QK, are parallel. 3 (b) Determine the locus of the circumcircle of CPKQK Solution: (a) We show that for all K, the line PKQK is parallel to the reflection of AB across the internal angle bisector at C. If PK and QK lie on the segments BC and CA, respectively, we have LQKPKC = 7r - LBPKQK = LQKAB = LCAB, which implies the claim. The proof in case PK or LK lies outside its corresponding segment is similar (or can be subsumed by viewing the angles above as directed angles modulo 7r). (b) From (a) we deduce that the circles CPKQK are all homothetic; hence they form a family of circles having a common tangent at C. 3. The positive real numbers x1,. , xn (n > 3) satisfy x1 = 19, x2 = 95 and x3 < xk_lxk+1 for 2 < k < n - 1. Determine the largest of the numbers x1, , xn. Solution: Note that if xk_1 < xk, then Xk+1 > xk(xk/xk-1)2 > xk. Hence by induction, x1 < x2 < .< xn and xn is the largest. 4. A cube is inscribed in a regular tetrahedron of unit length in such a way that each of the 8 vertices of the cube lie on faces of the tetrahedron. Determine the possible side lengths for such a cube. Solution: Let A, B, C, D be the vertices of the tetrahedron and s the length of the side of the cube. We consider two cases, each leading to a unique side length s. In the first case, some three vertices lie on a single face of the tetrahedron, say ABC. This means an entire face of the cube must lie on ABC. The cross-section along the plane of the opposite face of the cube is an equilateral triangle, whose side length we call t, with a square inside having each vertex on some side of the triangle. Let PQR be the vertices of the triangle. One of the sides, say PQ, of the triangle must contain two vertices of 4 the square. Then a homothety through P with ratio t/s carries the square to a square erected externally on PQ. The ratio t/s must also equal the ratio of the distances from P to PQ and the side of the external square opposite PQ, which is (t + tf /2)/t. We conclude that t = (1 + f12)s. On the other hand, the ratio of the height of the cube to the height of the tetrahedron is 1 - t. The latter height is 213 (an easy computation), so 1 - t =312s. Adding these expressions for t yields s = 1/(1 + f/2 + 3/2). In the second case, no more than two vertices lie on each face. Hence each face must contain an edge of the tetrahedron. Of these four edges, two must be parallel, and so must be parallel to the inter- section of the two corresponding faces of the tetrahedron, which is simply an edge of the tetrahedron. The only direction perpendicular to this edge which lies in either of the other two faces is along the opposite edge of the tetrahedron, so the edges of the cube lying on those faces must be parallel to that opposite edge. By an easy calculation (or by inscribing the tetrahedron in a cube), the segment joining the midpoints of opposite edges has length f/2, and by symmetry, two faces of the cube must be perpendicular to this segment. Then the distance from one of these faces to the closer edge parallel to it is f/4 - s/2. However, the ratio of this dis- tance to ,F2/2 must equal the ratio of the edges of the cube and the tetrahedron (by a homothety), which is s. We solve for s and find s = (f - 1)/2 in this case. 5 1.2Bulgaria 1. Let p and q be positive numbers such that the parabola y = x2 - 2px + q has no common point with the x-axis. Let 0 denote the origin (0, 0). Prove that there exist points A and B on the parabola such that the segment AB is parallel to the x-axis and LAOB is a right angle if and only if p2 < q < 1/4. Find the values of p and q for which the points A and B are defined uniquely. Solution: Since the parabola has no common point with the x-axis, the roots of the equation x2 - 2px + q = 0 are not real and hence q < p2. Let the points A(xi,yo) and B(x2i yo) have the required properties. Then xl and x2 are the roots of the equation x2 - 2px + q - yo = 0 and yo > q - p2, because the vertex of the parabola has coordinates (p, q - p2). On the other hand OA2 = xi + yo, OB2 = x2 + yo, AB2 = (x1 - x2)2 and it follows from the Pythagorean theorem that yo+x1x2 = 0. But x1x2 = q-yo and thus yo -yo+q = 0. Consequently the existence of the points A and B is equivalent to the assertion that the equation f (y) = y2 - y + q = 0 has a solution yo > q - p2. (A and B are defined in a unique way if this is the only solution.) A necessary condition is that the discriminant of the equation is not negative, i.e. q < 4. The last condition is sufficient because f(q - p2) = q - p2 + p2 > 0 and12 > 4 > q - p2. The corresponding solution yo is unique if q = 4. 2. Let A1A2A3A4A5A6A7,B1B2B3B4B5B6B7,C1C2C3C4C5C6C7 be regular heptagons with areas SA, SB and Sc, respectively. Suppose that A1A2 = B1B3 = C1C4. Prove that 1 SB + Sc 2 < SA <2-f. Solution: Let A1A2 = a,A1A3 = b,A1A4 = c. By Ptolemys theo- rem for the quadrangle A1A3A4A5 it follows that ab + ac = bc, i.e. n + a = 1. Since AA1A2A3 = AB1B2B3i then BBB = b and hence B1 B2 = a2 . Analogously C1 C2 = 1 2. Therefore SB+s° a2 + a2 b c SA 6 c 2 Then b + a > 2 (6 + n )2 = 2 (equality is not possible because 2 6 b # a ). On the other hand a2 a2 a a 2 2a2 2a2 T2 + c2 b +c) be 1 be . (1) 2 sin2 1 By the sine theorem we get abe = sin2 = 4cos (1+cos 7 7 Since cos 7 < cos a = 2 , then6c > 2 1 2 = f - 1. From 42(1+2) here and from (1) we get the right hand side inequality of the prob- lem. 3. Let n be an integer greater than 1. Find the number of permutations (al, a2, , an) of the numbers 1, 2, , n such that there exists only one index i E { 1, 2, , n - 11 with ai > ai+i. Solution: Denote by Pn the number of the permutations we are looking for. Obviously pl = 0 and p2 = 1. Let n > 2. The number of permutations with an = n is equal to pn-1. Consider all the permutations (al, a2i , an) with ai = n, where 1 < i < n - 1 is fixed. There are (i-i) of these. Consequently n-1 (:) pn-1 + 2 1- 1 pnPn-1+ 1 = i 1 From here - pn=(2ri-1-1)+(2n'-2-1)+ +(2-1)=2n-n-1 . 4. Let n > 2 and 0 < xi < l for all i = 1, 2, , n. Show that (x1+x2+ +xn)-(x1x2+x2x3+ +xn_lxn+xnxl)< n 2 and determine when there is equality. Solution: Denote by S(x1i x2, , xn) the left hand side of the in- equality. This function is linear with respect to each of the variables xi. In particular, S(x1,x2, ,xn) < max(S(0,x2, ,xn),S(1,x2, ,xn)) . 7 From here it follows by induction that it is enough to prove the inequality when the xi's are equal to 0 or 1. On the other hand, for arbitrary xi we have 2S(x1i x2, ,xn)= n-(1-X1)(1-X2) -(1-x2)(1-x3) (1-xn)(1-x1) - x1x2 - x2x3 - - xnxl, i.e. S(x1, x2, , xn) < 2 , when xi E [0,1].In the case when the xi's are equal to 0 or 1, the left hand side of the last inequality is an integer. Consequently S(xl, x2, , xn) < [111. 2 It follows that when n is even, the equality is satisfied if (x1i x2, , xn) = (0, 1, 0,1, , 0,1) or (x17 x2, , xn) = (1, 0, 1, 0, ,1, 0) where x E [0,1] is arbitrary. 5. On sides BC, CA, AB of triangle ABC choose points At, B1, C1 in such a way that the lines AA1i BB1, CCl have a common point M. Prove that if M is the centroid of triangle A1B1C1i then M is the centroid of triangle ABC. Solution: Let M be the medcenter of AA1B1C1. Take a point A2 on MA- in such a way that B1 Al C1 A2 is a parallelogram. Take points B2 and C2 analogously. Since A1C1IIA1B1jIC1B2i then the points A2i C1, B2 are colinear and C1 is the midpoint of A2B2. The same is true for the points A2, B1, C2 and C2, A1, B2. We shall prove that A2 = A, B2 = B and C2 = C, which will solve the problem. Assume that A2 # A and let A be between A2 and M. Then C2 is between C and M, B is between B2 and M, and consequently A2 is between A and M, which is a contradiction. x2 + 2 6. Find all pairs of positive integers (x, y) for which y is an in- x - y teger which divides 1995. Solution: It is enough to find all pairs (x, y) for which x > y and x2 + y2 = k(x - y), where k divides 1995 = 3.5.7. 19. We shall use the following well-known fact: if p is prime of the kind 4q + 3 and if it divides x2 + y2 then p divides x and y. (For p = 3, 7,19 the last statement can be proved directly.) If k is divisible by 3, then x and y 8 [...]... + 6r) Square both sides again and simplify: 11r2 + 60r - 36 = 0 The unique positive root of this equation is r = 6/11 5 Suppose that al, a2i , alo are arbitrary distinct natural numbers whose sum is 1995 What is the minimal value of ala2 + a2a3 + a9alo + alo + al? Solution: The minimum is 4069, achieved by (a1, ,alo) = (1950,1,9,2,8,3, 7,4,6,5) We prove this by starting from an arbitrary configuration, . Mathematical Contests 199 5- 1996 Olympiad Problems and Solutions from around the World Published by the American Mathematics Competitions, 1997 Preface This book is a supplement to Mathematical. Mathematical Olympiads 1995: Olympiad Problems from Around the World, published by the American Mathematics Competitions. It contains solutions to the problems from 28 national and regional contests featured. which passes through A and B, define PK, QK to be the intersections of K with BC and AC, respectively. (a) Show that for any circles K, K& apos; passing through A and B, the lines PKQK and PK'QK,