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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:

$AB = BC$ and $M$ is the midpoint of $AC$ . $H$ is chosen on $BC$ so that $MH$ is perpendicular to $BC$ . $P$ is the midpoint of $MH$ . Prove that $AH$ is perpendicular to $BP$ .

Solution

Solution:

Take $X$ on $AH$ so that $BX$ is perpendicular to $AH$ . Extend to meet $HM$ at $P'$ . Let $N$ be the midpoint of $AB$ . $A$ , $B$ , $M$ and $X$ are on the circle center $N$ radius $NA$ (because angles $AMB$ and $AXB$ are $90^{\circ}$ ). Also $MN$ is parallel to $BC$ (because $AMN$ , $ACB$ are similar), so $NM$ is perpendicular to $MH$ , in other words $HM$ is a tangent to the circle. Hence $P'M = P'X$ . $P'B$ . Triangles $P'XH$ and $P'HB$ are similar (angles at $P'$ same and both have a right angle), so $P'H / P'X = P'B / P'H$ , so $P'H \cdot P'H = P'X \cdot P'B$ . Hence $P'H = P'M$ and $P'$ coincides with $P$ .

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