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Bulgarian National Olympiad - Final Round

Geometry Difficulty 5.2 AIME, harder Prove it Bulgaria

Given is a triangle $ABC$ and the points $M$ , $P$ lie on the segments $AB$ , $BC$ , respectively, such that $AM = BC$ and $CP = BM$ . If $AP$ and $CM$ meet at $O$ and $2\angle AOM = \angle ABC$ , find the measure of $\angle ABC$ .

Solution

Let $D$ be the reflection of $P$ across $C$ , so $AB = BD$ and $O'$ be the circumcenter of $\triangle ABD$ . As $\triangle AO'B \cong \triangle BO'D$ and $MB = CD$ , we have $\angle O'CB = \angle O'MA$ , so $MBCO'$ is cyclic. Now $\angle O'CM = \angle O'BM = \angle AOM$ , hence $CO' \parallel AP$ . Therefore $O'$ must be the midpoint of $AD$ , implying that $\angle ABC = 90^\circ$ .

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