Problem 3

Two circles in a plane intersect. Let

A

be one of their points of intersection. Starting simultaneously from

A

, two points move with constant speeds, each travelling along its own circle in the same sense. The two points return to

A

simultaneously after one revolution. Prove that there exists a fixed point

P

in the plane such that, at any time, the distances from

P

to the moving points are equal.

Step 4 of 5: Match the swept angles

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\angle AOX=\angle AO\prime X\prime,\quad \angle AOP=\angle AO\prime P

Detailed analysis

Equal revolution times and equal senses give equal angular displacements, so angle AOX equals angle AO′X′. The reflection also gives angle AOP equal to angle AO′P. Consequently the included angles POX and PO′X′ are equal.