The 16th Japanese Mathematical Olympiad - The Final Round

Question

Five distinct points A, M, B, C and D are on a circle O in this order with MA = MB. Let the lines AC and MD intersect at P, the lines BD and MC at Q. Let the line PQ meet the circle O at X and Y. Prove that MX = MY.

Accepted Answer

Since ACM = BDM (because of MA = MB), PCQ = PDQ, so four points C, D, P and Q are concyclic. So PQD = PCD = ACD = ABD. Therefore AB and PQ, hence AB and XY are parallel. Since M is the middle point of the arc AB, it is also that of the arc XY. Thus MX = MY.