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Romanian Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Romania

The points $M$ , $N$ , and $P$ are chosen on the sides $BC$ , $CA$ and $AB$ of the triangle $ABC$ such that $BM = BP$ and $CM = CN$ . The perpendicular dropped from $B$ onto $MP$ and the perpendicular dropped from $C$ onto $MN$ intersect at $I$ . Prove that the angles $\widehat{IPA}$ and $\widehat{INC}$ are congruent.

Gabriel Popa

Solution

Since $CM = CN$ and $CI \perp MN$ , the line $CI$ is the perpendicular bisector of the line segment $MN$ , hence $IM = IN$ . Similarly, we have $IM = IP$ . Triangles $IMC$ and $INC$ are equal, so $\widehat{IMC} \equiv \widehat{INC}$ , and, in a similar way, we deduce that $\widehat{IMB} \equiv \widehat{IPB}$ . It follows that $\widehat{IPA} = \widehat{IMC}$ , and, finally, that $\widehat{IPA} \equiv \widehat{INC}$ .

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