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Problem 841

AMC 12 late, AIME early

Geometry Difficulty 4.5 Prove it NMO Selection Tests For The Junior Balkan Mathematical Olympiad · Romania

Let $ABC$ be an acute triangle such that $AB \neq AC$ . Let $M$ be the midpoint of $[BC]$ , $H$ be the orthocenter of $ABC$ , $O_1$ be the midpoint of $[AH]$ and $O_2$ be the circumcenter of $BCH$ . Prove that $O_1AMO_2$ is a parallelogram.

JBMO ShortList 2014

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We use the following well-known facts:
(1) The reflection of $H$ in line $BC$ lies on the circumcircle of triangle $ABC$ .
(2) $AH = 2MO$ , where $O$ is the circumcenter of $ABC$ .
From (1) it follows that the reflection of the circumcenter of $ABC$ is the circumcenter of $HBC$ , hence $O_2$ is the reflection of $O$ in $BC$ . We have
$MO_2 = MO \stackrel{(2)}{=} \frac{AH}{2} = AO_1$
and, as $AO_1 \parallel MO_2$ , $O_1AMO_2$ is a parallelogram.

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