First selection test

First selection test Problem 1. Let ABC be a triangle right at C and consider points D, E on the sides BC, CA, respectively such that BD = AE = k. Lines BE and AC CD AD intersect at point O. Show that BOD = 60 if and only if k = 3. Marcel Chirită Problem 2. In a plane 5 points are given such that all triangles having vertices at these points are of area not greater than 1. Show that there exists a trapezoid which contains all point in the interior (or on the sides) and having the area not exceeding 3. Marcel Chirită Problem 3. For any positive integer n let s(n) be the sum of its digits in decimal representation. Find all numbers n for which s(n) is the largest proper divisor of n. Laurenţiu Panaitopol Second selection test Problem 4. Prove that a3 numbers a, b, and c. + b3 + c3 bc ca ba a + b + c, for all positive real Problem 5. Consider a circle C of center O and let A, B be points on the circle with AOB = 90. Circles C 1 (O 1 ) and C 2 (O 2 ) are internally tangent to C at points A, B, respectively, and -moreover- are tangent to themselves. Circle C 3 (O 3 ), located inside the angle AOB, is externally tangent to C 1, C 2 and internally tangent to C. Prove that points O, O 1, O 2, O 3 are vertices of a rectangle. Problem 6. An 7 7 array is divided in 49 unit squares. Find all integers n N for which n checkers can be placed on the unit squares so that each row and each line have an even number of checkers. (0 is an even number, so there may exist empty rows or columns. A square may be occupied by at most 1 checker). Dinu Şerbănescu 1

Third selection test Problem 7. Suppose ABCD is a cyclic quadrilateral of area 8. Prove that if there exists a point O in the plane of the triangle such that OA + OB + OC + OD = 8, then ABCD is an isosceles trapezoid (or a square). Flavian Georgescu Problem 8. Prove that ( a b + b c + c ) 2 3 ( a + b a 2 c for all positive real numbers a, b, and c. + b + c a + c + a ), b Problem 9. Find all real numbers a and b satisfying 2(a 2 + 1)(b 2 + 1) = (a + 1)(b + 1)(ab + 1). Cezar Lupu Valentin Vornicu Problem 10. Show that the set of real numbers can be partitioned into subsets having two elements. Fourth selection test Problem 11. Let A = {1, 2,..., 2006}. Find the maximal number of subsets of A that can be chosen such that the intersection of any 2 such distinct subsets has 2004 elements. Problem 12. Let ABC be a triangle and let A 1, B 1, C 1 be the midpoints of the sides BC, CA, AB, respectively. Show that if M is a point in the plane of the triangle such that MA = MB = MC = 2, MA 1 MB 1 MC 1 then M is the centroid of the triangle. Dinu Şerbănescu 2

Problem 13. Suppose a, b, c are positive real numbers with the sum equal to 1. Prove that: a 2 b + b2 c + c2 a 3(a2 + b 2 + c 2 ). Mircea Lascu Problem 14. The set of positive integers is partitioned in subsets with infinite elements each. The question (in each of the following cases) is if there exists a subset in the partition such that any positive integer has a multiple in this subset. a) Prove that if the number of subsets in the partition is finite the answer is yes. b) Prove that if the number of subsets in the partition is infinite, then the answer can be no (for a certain partition). Fifth selection test Problem 15. Let ABC be a triangle and D a point inside the triangle, located on the median from A. Show that if BDC = 180 BAC, then AB CD = AC BD. Eduard Băzăvan Problem 16. Consider the integers a 1, a 2, a 3, a 4, b 1, b 2, b 3, b 4 with a k b k for all k = 1, 2, 3, 4. If {a 1, b 1 } + {a 2, b 2 } = {a 3, b 3 } + {a 4, b 4 }, show that the number (a 1 b 1 )(a 2 b 2 )(a 3 b 3 )(a 4 b 4 ) is a square. Note. For any sets A and B, we denote A + B = {x + y x A, y B}. Adrian Zahariuc Problem 17. Let x, y, z be positive real numbers such that 1 1 + x + 1 1 + y + 1 1 + z = 2. Prove that 8xyz 1. Mircea Lascu 3

Problem 18. For a positive integer n denote r(n) the number having the digits of n in reverse order- for example, r(2006) = 6002. Prove that for any positive integers a and b the numbers 4a 2 + r(b) and 4b 2 + r(a) can not be simultaneously squares. Marius Ghergu SOLUTIONS Problem 1. Consider the rectangle ACDP. The hypothesis rewrites as BD = AE = k, so AP E = BP D and AP D = EP B. Moreover, DP AP AP = P D, hence P AD P EB. P E P B It follows that DAP = P EB, so AP OE is cyclic and hence BOD = AOE = AP E. The claim is proved by the following chain of equivalences: BOD = 60 tan BOD = 3 tan AP E = 3 AE = 3 k = 3. AP Problem 2. Denote A, B, C, D, E the given points and suppose ABC is the triangle having the maximal area. The distance from D to BC is not greater than the distance from A to BC, hence D - and similarly E - are located between the parallel line from A to BC and its mirror image across BC. Applying the same reasoning for AB and AC, one obtains a triangular (bounded) region A 1 B 1 C 1 -with ABC as median triangle- in which all points must lie. Points D and E are located in at most 2 of the triangles A 1 BC, AB 1 C, ABC 1, hence one of the trapezoids ABA 1 B 1, BCB 1 C 1 or ABA 1 B 1 contains all points. Since the area of the trapezoids is 3 times the area of ABC, hence not greater than 3, the conclusion is proved. Problem 3. The numbers are 18 and 27. Let k be the number of digits of n in decimal representation. Notice that: (1) n = p s(n), where p is prime so that any prime divisor of s(n) is greater than of equal to p; (2) s(n) 2 n, so 10 k 1 n s(n) 2 (9k) 2, hence k 4. We study the following cases: a) If k = 4, then n = abcd, n s(n) 2 leq36 2 = 1296, so a = 1. Then s(n) 28, thus n 28 2 < 1000, false. b) If k 3, then abc, so 9(11 a + b) = (p 1)(a + b + c). 4

If 9 divides p 1, since p < a+b+c = 27 we get p = 19. Next 9a = b+2c, hence a 3. As a + b + c 23 - see (1)- we have no solution. If 9 do not divide p 1, from 3 a + b + c and (1) we get p = 2 or p = 3. For p = 3 we have n = 3(a + b + c), so a = 0 and 10 b + c = 3(b + c). Consequently, 7b = 2c and n = 27. For p = 2 we get n = 2(a + b + c), so a = 0 and 8b = c, hence n = 18. Problem 4. The inequality rewrites as a 4 + b 4 + c 4 abc(a + b + c). Using successively the inequality x 2 + y 2 + z 2 xy + yz + zx we get a 4 + b 4 + c 4 a 2 b 2 + b 2 c 2 + c 2 a 2 = (ab) 2 + (bc) 2 + (ca) 2 abc(a + b + c), as desired. Problem 5. Let R, r 1, r 2 be the radii of the circles C, C 1, C 2 and let r = R r 1 r 2. Consider the point P so that OO 1 P O 2 is a rectangle. From the tangency conditions we get OO 1 = R r 1, OO 2 = R r 2 and O 1 O 2 = r 1 + r 2 = R r. It is sufficient to prove that the circle centered at P with radius r is C 3. To prove this, notice that O 1 P = OO 2 = R r 2 = r + r 1, O 2 P = OO 1 = R r 1 = r + r 2, and OP = O 1 O 2 = R r, so the 3 tangency conditions are fulfilled. Problem 6. One can place 4, 6,..., 40, 42 checkers in the given conditions. We start by noticing that n is the sum of 7 even numbers, hence n is also even. On a row one can place at most 6 checkers, hence n 6 7 = 42.. The key step is to use 2k 2k squares filled completely with checkers and 2k + 1 2k + 1 squares having checkers on each unit square except for one diagonal. Notice that these types of squares satisfies the conditions, and moreover, we can glue together several such squares within the problem conditions. Below we describe the disposal of n checkers for any even n between 4 and 42. For 4,8,12,16,20,24,28,32 or 36 checkers 1,2,3,4,5,6,7,8 or 9 2 2 squares; notice that all fit inside the 7 7 array!. For 6 checkers consider a 3 3 square -except for one diagonal-; then adding 2 2 squares we get the disposal of 10,14,18,22,..., 38 checkers. For 40 checkers we use a 5 5 and five 2 2 squares. 5

For 42 checkers we complete the 7 7 array but a diagonal. Finally, notice that 2 checkers cannot be placed within these rules. Problem 7. Let α be the measure of the angle determined by the diagonals. Since 8 = OA + OB + OC + OD AC + BD 2 AC BD 2 AC BD sin α = 2 2S = 8, we get AC = BD = 1 and α = 90. The claim follows from a simple arch subtraction. Problem 8. Let a = x, b = y and c = z. The inequality rewrites b c a ( ) successively: x 2 + y 2 + z 2 + 2xy + 2yz + 2zx 3 x + y + z + 1 + 1 + 1 2 x y z ( ) ( ) x 2 + y 2 + z 2 1 + 2 + 1 + 1 3 x + y + z + 1 + 1 + 1 2(x 2 + y 2 + x y z 2 x y z z 2 ) + 1 + 1 + 1 3(x + y + z). x y z From AM-GM inequality we get 2x 2 + 1 x = x2 + x 2 + 1 x 3 x 2 x 2 1 x = 3x. Summing with the analogously relations we get the conclusion. Problem 9. First solution. Consider the given equation as quadratic in a: a 2 (b 2 b + 2) a(b + 1) 2 + 2b 2 b + 1 = 0. The discriminant is = (b 1) 2 (7b 2 2b + 7), hence we have solutions only for b = 1. It follows that a = 1. Second solution. Using Cauchy-Schwarz inequality, we get 2(a 2 + 1) (a+) 2, 2(b 2 + 1) (b + 1) 2 and (a 2 + 1)(b 2 + 1) (ab + 1) 2. Multiplying, we obtain 2(a 2 +1)(b 2 +1) (a+1)(b+1)(ab+1), hence the equality case occur in all the inequalities, so a = b = 1. Problem 10. For example, consider R\Z partitionated in { x, x} and Z in {2n, 2n+1}. Another example: split R into disjoint intervals [2n, 2n + 1), with n Z. Then take the pairs (x, x + 1) from each interval [2n, 2n + 1), with x [2n, 2n + 1). 6

Problem 11. The required number is 2006, the number of the subsets having 2005 elements. For start, notice that each subset must have at least 2004 elements. If there exist a set with exactly 2004 elements, then this is unique and moreover, only 2 other subsets may be chosen. If no set has 2004 elements, then we can choose among the 2006 subsets with 2005 elements and the set A with 2006 elements. But if A is among the chosen subsets, then any intersection will have more than 2004 elements, false. The claim is now justified. Problem 12. Let A 2, B 2, C 2 be the mirror images of the point M with respect to points A 1, B 1, C 1. The given condition shows that MA = MA 2, MB = MB 2, MC = MC 2. From the parallelograms AMBC 2, BMCA 2, AMCB 2 we derive that MA = MA 2 = BC 2 = B 2 C, MB = MB 2 = AC 2 = CA 1 and MC = MC 2 = BA 2 = AB 2. It follows that MA 2 BC 2, MA 2 CB 2 and MB 2 AC 2 are also parallelograms, therefore A, M and A 2 are collinear. The conclusion is now obvious. Problem 13. First solution. Using Cauchy-Schwarz inequality we get: a 2 b + b2 c + c2 a = a4 ba + b4 2 cb + c4 2 ac (a2 + b 2 + c 2 ) 2 2 a 2 b + b 2 c + c 2 a. The inequality is reduced to a 2 + b 2 + c 2 3(a 2 b + b 2 c + c 2 a) or (a + b + c)(a 2 + b 2 + c 2 ) 3(a 2 b + b 2 c + c 2 a). The last one rewrites as a(a b) 2 0, which is obvious. Second solution. Since a + b + c = (a + b + c) 2 = 1, the inequality gives successively: or a 2 b + b2 c + c2 a (a + b + c) 3(a2 + b 2 + c 2 ) (a + b + c) 2, ( a 2 b 2a + b) (a b) 2, 7

hence (a b) 2 b Since a, b, c 1, we are done. (a b) 2. Problem 14. a) Let A k be the partition classes, with k = 1, 2,... r. Assuming that the answer is no, there exist n k, k = 1, 2,... r, such that no multiples of n k is in A k. But n 1 n 2... n r lies in one of the sets A k and is multiple of any n k, false. b) We exhibit a partition for which the answer is no. Let A k be the set of all numbers written only with the first k primes at any positive power; moreover, put 1 A 1. Fixing k, the number p 1 p 2... p k+1 have no multiples in A k. Problem 15. Let E be the mirror image of D across the midpoint of the side BC. We notice that DBEC is a parallelogram and ABEC is cyclic. The equality of the areas of triangles ABE and ACE implies AB BE = AC CE. We are left only to notice that CE = BD and BE = CD. Problem 16. Without loss of generality, assume a k > b k, k = 1, 4. Then a 1 + a 2 = a 3 + a 4 and b 1 + b 2 = b 3 + b 4. We analyze 2 cases: i) a 1 + b 2 = a 3 + b 4 and a 2 + b 1 = a 4 + b 3. Subtracting we get a 2 b 2 = a 4 b 4, a 1 b 1 = a 3 b 3 and the claim is obvious. ii) a 1 + b 2 = a 4 + b 3 and a 2 + b 1 = a 3 + b 4. By subtraction we obtain a 2 b 2 = a 3 b 3, a 1 b 1 = a 4 b 4, as needed. Problem 17. Canceling the denominators we get 1 = xy + yz + zx + 2xyz. From AM-GM inequality we get 1 4 4 2x 3 y 3 z 3, so 1 (8xyz) 3. The conclusion follows. Problem 18. Assume by contradiction that the claim holds and let b a. The number r(b) has at most as many digits as b, so r(b) < 10b 10a. It follows that (2a) 2 < 4a 2 + 10a < (2a + 3) 2, hence 4a 2 + r(b) = (2a + 1) 2 or (2a + 2) 2, thus r(b) = 4a + 1 or 8a + 4. Notice that r(b) > a b, implying that a and b have the same number of digits. Then, as above, we get r(a) {4b + 1, 8b + 4}. 8

We analyze 3 cases: 1. r(a) = 4b+1 and r(b) = 4a+1. Subtracting we get (r(a) a)+(r(b) b) = 3(b a) + 2, which is false since 9 divides r(n) n for any positive integer n. 2. r(a) = 8b+4 and r(b) = 4a+1; (the same reasoning is to be applied for r(b) = 8a+4 and r(a) = 4b+1.) Subtracting we obtain (r(a) a)+(r(b) b) = 7b + 3a + 3, so 3 divides b. Then 3 divides also r(b) = 4a + 1, so a and r(a) have the remainder 2 when divided at 3. This leads to a contradiction with r(a) = (8b + 3) + 1. 3. r(a) = 8b + 4 and r(b) = 8a + 4. Both r(a) and r(b) have the last digit even, so at least 2. Then a and b have the first digit greater than of equal to 2, so 8a + 4 and 8b + 4 have more digits than a and b. It follows that r(a) < 8b + 4 and r(b) < 8a + 4, a contradiction. 9