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 5 of 5: Apply SAS congruence
Detailed analysis
The two triangles have the two corresponding side pairs and included angle equal: OX=O′P, OP=O′X′, and the included angles agree. SAS gives congruence, hence PX=PX′ at every time.