Singapore International Mathematical Olympiad 2008 Senior Team Training. Take Home Test Solutions. 15x 2 7y 2 = 9 y 2 0 (mod 3) x 0 (mod 3).

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1 Singapore International Mathematical Olympiad 2008 Senior Team Training Take Home Test Solutions. Show that the equation 5x 2 7y 2 = 9 has no solution in integers. If the equation has a solution in integer, then 5x 2 7y 2 = 9 y 2 0 (mod 3) y 0 (mod 3). Hence y = 3y for some integer y. This implies that 5x 2 7(3y ) 2 = 9 5x 2 2y 2 = 3 2x 2 0 (mod 3) x 0 (mod 3). Hence x = 3x for some integer x. This implies that 5(3x ) 2 7(3y ) 2 = 9 5x 2 7y 2 = y 2 (mod 3) y 2 2 (mod 3). The last congruence is impossible. Hence the given equation has no solution in integers. 2. Let n and k be positive integers. Prove that is divisible by n 5 +. We prove by induction on k: When k =, we have (n 4 )(n 3 n 2 + n ) k + (n + )n 4k (n 4 )(n 3 n 2 + n ) + (n + )n 3 = n 7 n 6 + n 5 + n 2 n + = n 2 (n 5 + ) n(n 5 + ) + n 5 + = (n 5 + )(n 2 n + ).

2 2 So the statement is true for k =. Now assume that the statement is true for k and consider the case k +, we have (n 4 )(n 3 n 2 + n ) k+ + (n + )n 4(k+) = (n 4 )(n 3 n 2 + n ) k (n 3 n 2 + n ) + (n + )n 4k n 4 = [(n 4 )(n 3 n 2 + n ) k (n 3 n 2 + n ) + (n + )n 4k (n 3 n 2 + n )] + [(n + )n 4k n 4 (n + )n 4k (n 3 n 2 + n )] = (n 3 n 2 + n )[(n 4 )(n 3 n 2 + n ) k + (n + )n 4k ] + n 4k (n + )(n 4 n 3 + n 2 n + ) = (n 3 n 2 + n )[(n 4 )(n 3 n 2 + n ) k + (n + )n 4k ] + n 4k (n 5 + ). This is divisible by n 5 + by induction hypothesis. Hence the statement is true for k +. This completes the induction. 3. Let f(x) = (x + ) p (x 3) q where p and q are positive integers. = x n + a x n + a 2 x n a n x + a n, () Given that a = a 2, prove that 3n is a perfect square. (2) (b) Prove that there exist infinitely many pairs (p, q) of positive integers p and q such that the equality a = a 2 is valid for the polynomial p(x). (Belarussian 2003) (a) Since (x + ) p = x p + px p p(p ) + x p (x 3) q = x q 3qx q 9q(q ) + x q 2 +, 2 we have n = p + q, a = p 3q, a 2 = 9q2 9q+p 2 p 6pq 2. Therefore a = a 2 Since n = p + q, we are done. 2p 6q = 9q 2 9q + p 2 p 6pq (3q p) 2 = 3(p + q). (b) This is equivalent to showing that the equation (3q p) 2 = 3(p + q) has an infinite family of solutions in positive integers. Treating this as a

3 quadratic equation in p, 9p 2 (6q + 3)p + (9q 2 3q) = 0 we have 48q + 9 p = 6q + 3 ±. 2 Thus 48q + 9 is an odd square. Hence 48q + 9 = (2k + ) 2, or q = k2 + k 2 2 Let k = 2t +, then q = 2t 2 + 3t and p = 36t 2 3t, where t N is the required infinite family. 4. Show that if m < n, then 2 2m + divides 2 2n. Hence deduce that 2 2m + and 2 2n + are relatively prime. Conclude that there are infinitely many primes. If m < n then n = m + k for some integer k, so we have 2 2n = 2 2m 2 k = (2 2m + )(2 2m (2 k ) 2 2m (2 k 2) m ). Hence 2 2m + divides 2 2n. Let d = gcd(2 2m +, 2 2n + ). By above, we have 3 2 2n + = (2 2n ) + 2 = l(2 2m + ) + 2 for some integer l d 2 (since d (2 2m + ) and d (2 2n + )) d = or d = 2 d = (since 2 2m + is odd) Thus gcd(2 2m +, 2 2n + ) =, i.e. 2 2m + and 2 2n + are relatively prime. For any positive integer n, let p n be a prime divisor of 2 2n +. For any m, n with n m, we have p n p m since 2 2n + and 2 2m + are relatively prime. Hence {p, p 2, p 3...} is an infinite set of primes. 5. Let x, y, z be positive numbers so that xyz =. Prove that x + y + z 3 z x + 3 x y + 3 y z. By AM-GM inequality, y + z + z 3 3 yz 2 = 3 3 z x, x + x + z 3 3 x 2 z = 3 3 x y,

4 4 and x + y + y 3 3 xy 2 = 3 3 y z. Summing the three inequalities gives the desired result. 6. Let p, p 2,..., p n (n 2) be any rearrangement of, 2,... n. Show that p + p 2 + p 2 + p > n p n + p n n + 2. Since AM HM, Thus n [(p + p 2 ) + (p 2 + p 3 ) + + (p n + p n )] { [ ]}. + + n p + p 2 p n + p n + + p + p 2 p n + p n (n ) 2 2(p + + p n ) p p n (n )2 n(n + ) 3 = (n ) 2 (n )(n + 2) (n ) 2 > (n )(n + 2) = n n + 2.

5 7. From a point P outside a circle, tangent lines P A and P B are drawn with A and B on the circle. A third line P CD meets the circle at C and D, with C lying in between P and D. A point Q is chosen on the chord CD so that DAQ = P BC. Show that DBQ = P AC. 5 Since ABC = ADQ BAC = P BC = DAQ, ADQ ABC. Thus BC AD = AB DQ. Also, P CA P AD. Hence P C P A = AC AD. Similarly, P C P B = BC BD. But P A = P B. So we have AC AD = BC BD, and thus AC BD = BC AD = AB DQ. By Ptolemy s Theorem, AC BD + BC AD = AB CD. Therefore, AB CD = 2AB DQ, or DQ = 2CD. So Q is the midpoint of CD. Now AD AB = DQ BC = CQ BC and BCQ = BAD. It follows that CBQ ABD. Hence CBQ = ABD. Thus DBQ = ABC = P AC.

6 6 8. In triangle ABC, A = 60 and AB > AC. The altitudes BE and CF intersect at H. Points M and N are chosen on the segments BH and HF so that BM = CN. If O is the circumcircle of ABC, find the ratio MH + NH. OH Note that BOC = 2 A = 20 and BHC = 80 A = 20 (since A, E, H, F are concyclic). Hence B, O, H, C are concyclic. Thus OBH = OCH. As BO = CO and BM = CN as well, OBM is congruent to OCN. Hence OM = ON and BM O = CN O. It follows that O, M, H, N are concyclic. Therefore, NOM = NHE = 20. Also, ONM = OHM = OHB = OCB = 30. Thus MN sin 20 = OM sin 30 = 3. Finally, from Ptolemy s Theorem, we have Therefore, MH ON + NH OM = OH MN. MH + NH OH = MN OM = 3.

7 7 9. On the plane, there are 3 mutually and externally disjoint circles Γ, Γ 2 and Γ 3 centred at X, X 2 and X 3 respectively. The two internal common tangents of Γ 2 and Γ 3, (Γ 3 and Γ, Γ and Γ 2 ) meet at P, (Q, R respectively). Prove that X P, X 2 Q and X 3 Z are concurrent. Let the radii of Γ, Γ 2 and Γ 3 be r, r 2 and r 3 respectively. Then X R : RX 2 := r : r 2, X 2 P : P X 3 := r 2 : r 3 and X 3 Q : QX := r 3 : r. Thus X R X2P X3Q =. RX 2 P X 3 QX By the converse of Ceva s Theorem, X P, X 2 Q and X 3 Z are concurrent.

8 8 0. The excircle centred at I a with respect to A of ABC touches the sides AB, BC and AC or their extensions at E, D and F respectively. Let H be the foot of the perpendicular from B onto I a C. Prove that E, H, F are collinear. Join EH, F H, DH, EI a, BI a. Then E, I a, H, B are concylic. Also D, B, E, I a are concylic. Thus B, E, I a, H, D all lie on a circle with diameter BI a. As BD = BE, we have BHD = BHE. On the other hand, DHC is congruent to CHF so that DHC = CHF. As BHC = 90, we have DHE + DHF = 2 BHC = 80. Therefore, E, H, F are collinear.