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Mongolian Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Mongolia

Let $ABC$ be an isosceles triangle with $AB = BC$ . Let $M$ and $N$ be midpoints of $AC$ and $BM$ , respectively. $P$ is the foot of the altitude from $A$ to $AN$ of triangle $AMN$ . Prove that triangles $APM$ and $CPB$ are similar.
(Khulan Tumenbayar)

Solution

Let us denote $\angle PNM = \alpha$ . Then $\angle PMC = \angle PNB = 180^\circ - \alpha$ and $\angle PAC = \angle PMB$ . Hence $\triangle APM \sim \triangle MPN$ . As we have $AM = MC$ , $MN = NB$ it yields $\triangle APC \sim \triangle MPB$ . It implies $\angle APC = \angle MPB$ and $\angle MPC = \angle NPB$ . Therefore, $\angle BPC = 90^\circ$ .
Since $\angle PBM = \angle PCM$ , quadrilateral $BCMP$ is cyclic. Hence $\angle MPC = \angle MBC$ . Also, $\angle PMC = \angle BNA = \angle BNC$ , which means $\triangle MPC \sim \triangle NBC$ .

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Hence $\angle BCN = \angle PCA$ is true. It implies that $\angle NCA = \angle PCB$ . Also, we have $\angle APM = \angle BPC = 90^\circ$ . By AAA property, we have $\triangle APM \sim \triangle CPB$ .

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