17th Chinese 2002 Problem A1 ABC is a triangle. What conditions must angles A, B, C satisfy if angle B < angle C and we can find X on AB and Y on AC such that BX = CY and angle BZX = angle CZY, where AZ is the angle bisector of angle A? If this condition is satisfied, express BX in terms of the sides of the triangle. Solution
: A < 2B; BX = BC 2 /(AB + AC). Solution by Aleksander Ilic Put x = ∠BZX = ∠CZY, and y = ∠A/2. The sine rule gives CY/sin x = CZ/sin(C + x), BX/sin x = BZ/sin(B + x). Hence CZ/BZ = sin(C + x)/sin(B + x). But CZ/BZ = AC/AB (since AZ is the angle bisector) = sin B/sin C. Hence sin B sin(B + x) = sin C sin(C + x). We have cos(B + (B + x)) = cos B cos(B + x) - sin B sin(B + x), cos( (B + x) - B) = cos(B + x) cos B + sin B sin(B + x). So cos(2B + x) = cos(2C + x). Now x <= angle AZC < 90 o and B < 90 o , so 2B + x < 270 o . Also x + C < 180 o , so 2C + x < 360 o . Hence (2B + x) + (2C + x) = 360 o . But A + B + C = 180 o , so x = A. Y must lie on the side AC, so x ≤ ∠AZC = 180 o - y - C = A/2 + B, so A/2 <= B is a necessary condition. Now suppose A/2 ≤ B. Then we can find X on AB and Y on AC such that ∠BZX = ∠CZY = A. Hence triangles BZX and YZC are similar to BAC. Hence BZ/BX = AB/BC and CY/CZ = BC/CA. Hence CY/BX = (BC/CA)(AB/BC)(CZ/BZ) = 1. So A/2 ≤ B is also a sufficient condition. We have BZ/CZ = AB/AC, so BZ/BC = AB/(AB + AC). But BX/sin A = BZ/sin C, so BX = (BC/AB) BZ = BC 2 /(AB + AC). 17th Chinese 2002 © John Scholes jscholes@kalva.demon.co.uk 20 Dec 2002