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Geometry Difficulty 4.7 AIME Prove it South Africa

Let $A$ , $B$ , $C$ be three distinct points in the plane for which $AB = AC$ . Describe the locus of the points $P$ for which $\angle APB = \angle APC$ .

Solution

Clearly every point on the perpendicular bisector of $BC$ satisfies the requirements of the problem. We now show that no other point $P$ does.
Let $D$ be the midpoint of $BC$ , and suppose that there exists a point $P$ not on the line $AD$ such that $\angle APB = \angle APC$ . Let $Q$ be the reflection of $P$ about the line $AD$ ; then $Q$ also satisfies $\angle AQB = \angle AQC$ . Hence $A$ , $P$ , $Q$ and $C$ are concyclic, and so are $A$ , $P$ , $Q$ and $B$ . However, this means that both $B$ and $C$ lie on the circumcircle of triangle $APQ$ , which is clearly impossible.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.