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Romanian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Romania

A triangle $ABC$ with $AB = AC$ is obtuse at $A$ . Let $M$ be the mirror image of $A$ across $C$ . The perpendicular bisector of the line segment $AM$ meet the line $AB$ at point $P$ . Given that lines $PM$ and $BC$ are perpendicular, prove that $APM$ is an equilateral triangle.

Marcel Neferu

Solution

Lines $BC$ and $PM$ meet at $D$ . Denote $x$ the measure of $\angle ABC$ . Then $\angle MCD = \angle ACB = \angle ABC = x$ and $\angle PMC = 90^\circ - x$ . The triangle $PAM$ is isosceles, since $PC$ is median as well as perpendicular bisector of the line segment $AM$ , hence $\angle PMC = \angle PAC$ . On the other hand, $\angle PAC = \angle ABC + \angle ACB = 2x$ , implying $x = 30^\circ$ . Then $\angle PMC = \angle PAC = 60^\circ$ , whence the claim.

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