Let A be one of the two distinct points of intersection of two unequal coplanar circles C1 and C2 with centers O1 and O2 respectively. One of the common tangents to the circles touches C1 at P1 and C2 at P2, while the other touches C1 at Q1 and C2 at Q2. Let M1 be the midpoint of P1Q1 and M2 the midpoint of P2Q2. Prove that ∠O1AO2 = ∠M1AM2.
Solution
Let P1P2 and O1O2 meet at O. Let OA meet C2 again at A2. O is the center of similitude for C1 and C2 so ∠M1AO1 = ∠M2A2O2. Hence if we can show that ∠M2AO2 = ∠M2A2O2, then we are home.
Let X be the other point of intersection of the two circles. The key is to show that A2, M2 and X are collinear, for then ∠M2AO2 = ∠M2XO2 (by reflection) and O2A2X is isosceles.
But since O is the center of similitude, M2A2 is parallel to M1A, and by reflection ∠XM2O = ∠AM2O, so we need to show that triangle AM1M2 is isosceles. Extend XA to meet P1P2 at Y. Then YP12 = YA.YX = YP22, so YX is the perpendicular bisector of M1M2, and hence AM1 = AM2 as required.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998