26th CanMO 1994 Problem 4 AB is the diameter of a circle. C is a point not on the line AB. The line AC cuts the circle again at X and the line BC cuts the circle again at Y. Find cos ACB in terms of CX/CA and CY/CB. Solution
: cos 2 ACB = (CX/CA).(CY/CB). It is the positive root for C outside the circle and the negative root for C inside the circle. For C outside the circle, we have cos ACB = CX/CB = CY/CA (one should check that this is true in all configurations). Hence result. For C on the circle it is still true because cos ACB = 0 and CX = CY = 0. For C inside the circle, cos ACB = - cos ACY = - CY/CA and also = - cos BCX = - CX/CB. 26th CanMO 1994 © John Scholes jscholes@kalva.demon.co.uk 18 Aug 2002