Solutions of Benelux Mathematical Olympiad 2010

Transcription

1 Solutions of Benelux Mathematical Olymiad 200 Problem. A finite set of integers is called bad if its elements add u to 200. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer n such that the set {502, 503, 504,..., 2009} can be artitioned into n Benelux-sets. (A artition of a set S into n subsets is a collection of n airwise disjoint subsets of S, the union of which equals S.) Solution. As = 200, the set S = {502, 503,..., 2009} is not a Benelux-set, so n = does not work. We will rove that n = 2 does work, i.e. that S can be artitioned into 2 Benelux-sets. Define the following subsets of S: A = {502, 503,..., 670}, B = {67, 672,..., 005}, C = {006, 007,..., 339}, D = {340, 34,..., 508}, E = {509, 50,..., 2009}. We will show that A C E and B D are both Benelux-sets. Note that there does not exist a bad subset of S of one element, since that element would have to be 200. Also, there does not exist a bad subset of S of more than three elements, since the sum of four or more elements would be at least = 204 > 200. So any ossible bad subset of S contains two or three elements. Consider a bad subset of two elements a and b. As a, b 502 and a + b = 200, we have a, b = 508. Furthermore, exactly one of a and b is smaller than 005 and one is larger than 005. So one of them, say a, is an element of A B, and the other is an element of C D. Suose a A, then b = 340, so b D. On the other hand, suose a B, then b = 339, so b C. Hence {a, b} cannot be a subset of A C E, nor of B D. Now consider a bad subset of three elements a, b and c. As a, b, c 502, a + b + c = 200, and the three elements are airwise distinct, we have a, b, c = 005. So a, b, c A B. At least one of the elements, say a, is smaller than = 670, and at least one of the elements, say b, is larger than 670. So a A and b B. We conclude that {a, b, c} cannot be a subset of A C E, nor of B D. This roves that A C E and B D are Benelux-sets, and therefore the smallest n for which S can be artitioned into n Benelux-sets is n = 2. Remark. Observe that A C E and B D E 2 are also Benelux-sets, where {E, E 2 } is any artition of E.

2 Problem 2. Find all olynomials (x) with real coefficients such that (a + b 2c) + (b + c 2a) + (c + a 2b) = 3(a b) + 3(b c) + 3(c a) for all a, b, c R. Solution. For a = b = c, we have 3(0) = 9(0), hence (0) = 0. Now set b = c = 0, then we have (a) + ( 2a) + (a) = 3(a) + 3( a) for all a R. So we find a olynomial equation ( 2x) = (x) + 3( x). () Note that the zero olynomial is a solution to this equation. Now suose that is not the zero olynomial, and let n 0 be the degree of. Let a n 0 be the coefficient of x n in (x). At the left-hand side of (), the coefficient of x n is ( 2) n a n, while at the right-hand side the coefficient of x n is a n + 3 ( ) n a n. Hence ( 2) n = + 3 ( ) n. For n even, we find 2 n = 4, so n = 2, and for n odd, we find 2 n = 2, so n =. As we already know that (0) = 0, we must have (x) = a 2 x 2 + a x, where a and a 2 are real numbers (ossibly zero). The olynomial (x) = x is a solution to our roblem, as (a + b 2c) + (b + c 2a) + (c + a 2b) = 0 = 3(a b) + 3(b c) + 3(c a) for all a, b, c R. Also, (x) = x 2 is a solution, since (a + b 2c) 2 + (b + c 2a) 2 + (c + a 2b) 2 = 6(a 2 + b 2 + c 2 ) 6(ab + bc + ca) = 3(a b) 2 + 3(b c) 2 + 3(c a) 2 for all a, b, c R. Now note that if (x) is a solution to our roblem, then so is λ(x) for all λ R. Also, if (x) and q(x) are both solutions, then so is (x) + q(x). We conclude that for all real numbers a 2 and a the olynomial a 2 x 2 + a x is a solution. Since we have already shown that there can be no other solutions, these are the only solutions. Solution 2. For a = b = c, we have 3(0) = 9(0), hence (0) = 0. Now set b = c = 0, then we have (a) + ( 2a) + (a) = 3(a) + 3( a) for all a R. So we find a olynomial equation ( 2x) = (x) + 3( x). (2)

3 Define q(x) = (x) + ( x), then we find that q(2x) = (2x) + ( 2x) = (( x) + 3(x)) + ((x) + 3( x)) = 4q(x). (3) Note that the zero olynomial is a solution to this equation. Now suose that q is not the zero olynomial, and let m 0 be the degree of q. Let b m 0 be the coefficient of x m in q(x). At the left-hand side of (3), the coefficient of x m is 2 m b m, while at the right-hand side the coefficient of x m is 4b m. Hence m = 2. As q(x) = (x) + ( x), the olynomial q(x) does not contain any nonzero terms with odd exonent of x. Since also q(0) = 2(0) = 0, we conclude that q(x) = b 2 x 2, where b 2 is a real number (ossibly zero). From (2) we now deduce that (2x) = ( x) + 3(x) = 2(x) + q(x), so (2x) 2(x) = b 2 x 2. (4) Suose that that degree n of is greater than 2. Let a n 0 be the coefficient of x n in (x). At the left-hand side of (4), the coefficient of x n is (2 n 2) a n 0. But the coefficient of x n at the right-hand side vanishes, yielding a contradiction. So the degree of is at most 2. As we already know that (0) = 0, we must have (x) = a 2 x 2 + a x, where a and a 2 are real numbers (ossibly zero). We finally check that every olynomial of this form is indeed a solution (see solution ).

4 Problem 3. On a line l there are three different oints A, B and P in that order. Let a be the line through A erendicular to l, and let b be the line through B erendicular to l. A line through P, not coinciding with l, intersects a in Q and b in R. The line through A erendicular to BQ intersects BQ in L and BR in T. The line through B erendicular to AR intersects AR in K and AQ in S. (a) Prove that P, T, S are collinear. (b) Prove that P, K, L are collinear. Solution. (a) Since P, R and Q are collinear, we have P AQ P BR, hence AQ BR = AP BP. Conversely, P, T and S are collinear if it holds that AS BT = AP BP. So it suffices to rove BT BR = AS AQ. Since ABT = 90 = ALB and T AB = BAL, we have ABT ALB. And since ALB = 90 = QAB and LBA = ABQ, we have ALB QAB. Hence ABT QAB, so BT BA = AB AQ. Similarly, we have ABR AKB SAB, so Combining both results, we get BT BR which had to be roved. BR BA = AB AS. = BT / BA BR / BA = AB / AQ AB / AS = AS AQ,

5 (b) Let the line P K intersect BR in B and AQ in A and let the line P L intersect BR in B 2 and AQ in A 2. Consider the oints A, A and S on the line AQ, and the oints B, B and T on the line BR. As AQ BR and the three lines A B, AB and ST are concurrent (in P ), we have A A : AS = B B : BT, where all lengths are directed. Similarly, as A B, AR and SB are concurrent (in K), we have A A : AS = B R : RB. This gives so BB BT = RB RB = RB + BB RB BB = = + BB RB = BB BR, BT + BR Similary, using the lines A 2 B 2, AB and QR (concurrent in P ) and the lines A 2 B 2, AT and QB (concurrent in L), we find This gives so B 2 B : BR = A 2 A : AQ = B 2 T : T B. BB 2 BR = T B 2 T B = T B + BB 2 T B BB 2 =. = + BB 2 T B = BB 2 BT, + BR BT We conclude that B = B 2, which imlies that P, K and L are collinear.. Solution 2. (a) Define X as the intersection of AT and BS, and Y as the intersection of AR and BQ. To rove that P, S and T are collinear, we will use Menelaos theorem in ABX, so we have to rove AP BS XT P B SX T A =. Note that B is between P and A, X is between S and B, and X is between T and A, so it suffices to rove that AP BS XT P B SX T A =.

6 Because AQ and BR are arallel, we have AQP BRP, hence AP BP = QA RB. (5) Also, since ASB = KBR and BAS = 90 = BKR, we have ASB KBR, hence BS RB = AS AS, so BS = RB. (6) KB KB Similarly, we have AT B QAL, hence T A AQ T B T B =, so T A = AQ. (7) AL AL As ASX = ASB = 90 ABS = 90 ABK = KAB = Y AB, and SAX = 90 XAB = 90 LAB = ABL = ABY, we have SXA AY B, hence SX AY = AS AS, so SX = AY. (8) BA BA Similarly, we have BXT AY B, hence By combining (5) - (9), we find XT Y B = BT BT, so XT = Y B. (9) AB AB AP BS XT P B SX T A = QA RB AS KB RB BA AS AY BT AL Y B AB T B AQ = AL Y B KB AY. (0) Since Y LA = 90 = Y KB and AY L = BY K, we have AY L BY K, hence AL BK = AY BY, so AL BY =. () BK AY By combining (0) and (), we find as we wanted to rove. AP BS XT P B SX T A =, (b) Again, we will use Menelaos theorem in ABX, so we have to rove AP BK XL P B KX LA =.

7 Note that AP P B < 0, and BK KX < 0 if and only if XL LA AP BK XL P B KX LA =. < 0, so it suffices to rove that As BXL = AXK and BLX = 90 = AKX, we have BLX AKX, hence XL XK = BL AK. (2) Since ALB = 90 = QAB, we have ALB QAB, hence LA AQ = LB LB, so LA = AQ. (3) AB AB Similarly, we have AKB ABR, hence BK RB = AK AK, so BK = RB. (4) AB AB By combining (5) and (2) - (4), we find AP BK XL P B KX LA = QA RB BL AK AB LB AQ AK RB =, AB which is what we wanted to rove. Solution 3. As AKB = ALB = 90, the oints K and L belong to the circle with diameter AB. Since QAB = ABR = 90, the lines AQ and BR are tangents to this circle. Aly Pascal s theorem to the oints A, A, K, L, B and B, all on the same circle. This yields that the intersection Q of the tangent in A and the line BL, the intersection R of the tangent in B and the line AK, and the intersection of KL and AB are collinear. So KL asses through the intersection of AB and QR, which is oint P. Hence P, K and L are collinear. This roves art b. Now aly Pascal s theorem to the oints A, A, L, K, B and B. This yields that the intersection S of the tangent in A and the line BK, the intersection T of the tangent in B and the line AL, and the intersection P of KL and AB are collinear. This roves art a. Solution 4.

8 (a) W.l.o.g. we may assume that A = (0, 0) and B = (, 0) and the line through P is in the uer half lane, so l is the x-axis, a is the y-axis and b is the line x =. Take P = (, 0) ( > ) and Q = (0, q) (q > 0). Since P Q is given by x + y =, we find q R = (, q( ) ). Now AR is given by y = q( ) through B = (, 0), is given by y = x, hence BS, the line erendicular to AR and assing (x ). We find S = (0, q( ) q( ) ). Moreover BQ is given by y = q(x ), hence AT, the line erendicular to BQ and assing through A = (0, 0), is given by y = x. We find T = (, ). Since q q BT = /q = BP q( ) = AS, we conclude that P, T and S are collinear. AP (b) Point K is the intersection of AR and BS. Solving for x yields so q( ) x = (x ) q( ) ( ) q( ) + x = q( ) q( ) K = ( q( ) q( ) + q( ) x =, q( ) q( ) + q( ) q( ) + q( ) Point L is the oint of intersection of AT and BQ. Solving for x yields so x = q(x ) q ( ) q + q x = q x = q + q q ( ) q L = + q, + q. q q Let K 0 and L 0 be the rojections of K and L on the x-axis. We have to show that the following fractions are equal: ). K 0 K K 0 P = q( ) + q( ) q( ) q( ) + q( ) and L 0 L L 0 P = q +q q q +q

9 Working out cross roducts twice, this comes down to ( ) ( q? q( ) + + q = + q q( ) q( ) q q ( ) ( ) q + q q (? q( ) + q = + q q( ) + q( ) q( ) ) ) ( q + q q =? 2 q( ) + q( ) q( ) q + q q =? ( ) q( ) + q( ), q( ) q( ) + q( ) ) which is clearly true.

10 Problem 4. Find all quadrules (a, b,, n) of ositive integers, such that is a rime and a 3 + b 3 = n. Solution. Let (a, b,, n) be a solution. Note that we can write the given equation as (a + b)(a 2 ab + b 2 ) = n. As a and b are ositive integers, we have a+b 2, so a+b. Furthermore, a 2 ab+b 2 = (a b) 2 + ab, so either a = b = or a 2 ab + b 2 2. Assume that the latter is the case. Then is a divisor of both a+b and a 2 ab+b 2, hence also of (a+b) 2 (a 2 ab+b 2 ) = 3ab. This means that either is equal to 3 or is a divisor of ab. Since is a divisor of a + b, we have a b, hence either = 3, or a and b. If a and b, then we can write a = a, b = b with a and b ositive integers, and we have (a ) 3 + (b ) 3 = n 3, so (a, b,, n 3) then is another solution (note that (a ) 3 + (b ) 3 is a ositive integer greater than, so n 3 is ositive). Now assume that (a 0, b 0, 0, n 0 ) is a solution such that a. From the reasoning above it follows that either a 0 = b 0 =, or 0 = 3. After all, if we do not have a 0 = b 0 = and we have 0 3, then a. Also, given an arbitrary solution (a, b,, n), we can divide everything by reeatedly until there are no factors left in a. Suose a 0 = b 0 =. Then the solution is (,, 2, ). Suose 0 = 3. Assume that 3 2 (a 2 0 a 0 b 0 + b 2 0). As 3 2 (a 0 + b 0 ) 2, we then have 3 2 (a 0 + b 0 ) 2 (a 2 0 a 0 b 0 + b 2 0) = 3a 0 b 0, so 3 a 0 b 0. But 3 a 0 by assumtion, and 3 a 0 + b 0, so 3 b 0, which contradicts 3 a 0 b 0. We conclude that 3 2 (a 2 0 a 0 b 0 + b 2 0). As both a 0 + b 0 and a 2 0 a 0 b 0 + b 2 0 must be owers of 3, we have a 2 0 a 0 b 0 + b 2 0 = 3. Hence (a 0 b 0 ) 2 + a 0 b 0 = 3. We must have (a 0 b 0 ) 2 = 0 or (a 0 b 0 ) 2 =. The former does not give a solution; the latter gives a 0 = 2 and b 0 = or a 0 = and b 0 = 2. So all solutions with a are (,, 2, ), (2,, 3, 2) and (, 2, 3, 2). From the above it follows that all other solutions are of the form ( k 0a 0, k 0b 0, 0, n 0 + 3k), where (a 0, b 0, 0, n 0 ) is one of these three solutions. Hence we find three families of solutions: (2 k, 2 k, 2, 3k + ) with k Z 0, (2 3 k, 3 k, 3, 3k + 2) with k Z 0, (3 k, 2 3 k, 3, 3k + 2) with k Z 0. It is easy to check that all these quadrules are indeed solutions. Solution 2. Let (a, b,, n) be a solution. Note that we can write the given equation as (a + b)(a 2 ab + b 2 ) = n.

11 As a and b are ositive integers, we have a + b 2 and a 2 ab + b 2 = (a b) 2 + ab. So both factors are ositive and therefore must be owers of. Let k be an integer with k n such that a + b = k. Then a 2 ab + b 2 = n k. If we substitute b = k a, we find n k = (a + b) 2 3ab = 2k 3a( k a). We can rewrite this as: 3a 2 3 k a + 2k n k = 0, from which we see that a is a solution of the following quadratic equation in x: The discriminant of (5) is 3x 2 3 k x + 2k n k = 0. (5) D = ( 3 k ) ( 2k n k ) = 3 (4 n k 2k ) = 3 n k (4 3k n ). As n k = (a + b) 2 3ab < (a + b) 2 = 2k, we have n k < 2k, so 3k n > 0. Since a is a solution of (5), the discriminant must be nonnegative. Hence 4 3k n 0. If = 2, this imlies 3k n = or 3k n = 2; if = 3, this imlies 3k n = ; and if > 3, then 5 so 4 3k n can never be true. Suose = 2 and 3k n =. Then D = 3 2 2k (4 2) = 3 2 2k. But this is a not a square, so the solutions of (5) will not be integers, which yields a contradiction. Suose = 2 and 3k n = 2. Then D = 3 2 2k 2 (4 4) = 0, so the only solution of (5) is x = 3 2k = 2 3 2k. Therefore a = 2 k and b = 2 k a = 2 k, and this gives a solution for all k, namely (2 k, 2 k, 2, 3k 2). Suose = 3 and 3k n =. Then D = 3 3 2k (4 3) = 3 2k, so the solutions of (5) are x = 3k+ ±3 k = (3k ± 3 k ). Therefore a = 2 3 k or a = 3 k. For all k we find the solutions (2 3 k, 3 k, 3, 3k ) and (3 k, 2 3 k, 3, 3k ). We conclude that there are three families of solutions: (2 k, 2 k, 2, 3k 2) with k Z, (2 3 k, 3 k, 3, 3k ) with k Z, (3 k, 2 3 k, 3, 3k ) with k Z. It is easy to check that all these quadrules are indeed solutions.