4th Balkan 1987 Problem 4 Two circles have centers a distance 2 apart and radii 1 and √2. X is one of the points on both circles. M lies on the smaller circle, Y lies on the larger circle and M is the midpoint of XY. Find the distance XY. Solution
: √(7/2). Use coordinates with origin at the center of the small circle. So X is (3/4, (√7)/4 ). Take M to be (cos k, sin k). Then Y is (2 cos k - 3/4, 2 sin k - (√7)/4). Y lies on the circle center (2, 0) radius √2, so (2 cos k - 11/4) 2 + (2 sin k - (√7)/4) 2 = 2. Hence 11 cos k + √7 sin k = 10. We have XY 2 = (2 cos k - 3/2) 2 + (2 sin k - (√7)/2 ) 2 = 8 - 6 cos k - 2√7 sin k = 8 - 6 cos k - 2(10 - 11 cos k) = 16 cos k - 12. So we need to find cos k. We have (10 - 11 cos k) 2 = 7 (1 - cos 2 k), so 128 cos 2 k - 220 cos k + 93 = 0. Factorising: (32 cos k - 31)(4 cos k - 3) = 0. The root cos k = 3/4 corresponds to the point X, so we want the other root, cos k = 31/32. Hence XY 2 = 31/2 - 12 = 7/2. 4th Balkan 1987 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002