Problem 3
Two circles in a plane intersect. Let be one of their points of intersection. Starting simultaneously from , two points move with constant speeds, each travelling along its own circle in the same sense. The two points return to simultaneously after one revolution. Prove that there exists a fixed point in the plane such that, at any time, the distances from to the moving points are equal.
Step 3 of 5: Match the fixed side lengths
Detailed analysis
The reflection exchanges O and O′, so OP=O′A and O′P=OA. Since X and X′ lie on their respective circles, OX=OA and O′X′=O′A.