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Geometry Difficulty 4.6 AIME Prove it United States

Problem:

Let $ABC$ be an equilateral triangle, and let $P$ be a point on minor $\operatorname{arc} BC$ of the circumcircle of $ABC$ . Prove that $PA = PB + PC$ .

Solution

Solution:

Extend line $PC$ through $C$ to point $D$ such that $CD = BP$ . Note that $\angle ACD = \pi - \angle PCA = \angle ABP$ (since quadrilateral $ABPC$ is cyclic), and $AC = AB$ since $\triangle ABC$ is equilateral. Consequently, $\triangle ACD \cong \triangle ABP$ . In particular, we have $\angle PDA = \angle CDA = \angle BPA = \angle BCA$ (again by cyclicity) $= \pi/3$ . But also $\angle APD = \angle APC = \angle ABC$ (cyclicity) $= \pi/3$ . We conclude that triangle $APD$ is equilateral. So, $PA = PD = PC + CD = PC + BP$ .

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