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Geometry Difficulty 4.5 AIME Prove it Romania

Let ABCABC be an acute triangle such that ABACAB \neq AC. Let MM be the midpoint of [BC][BC], HH be the orthocenter of ABCABC, O1O_1 be the midpoint of [AH][AH] and O2O_2 be the circumcenter of BCHBCH. Prove that O1AMO2O_1AMO_2 is a parallelogram.

JBMO ShortList 2014

Solution

We use the following well-known facts:
(1) The reflection of HH in line BCBC lies on the circumcircle of triangle ABCABC.
(2) AH=2MOAH = 2MO, where OO is the circumcenter of ABCABC.
From (1) it follows that the reflection of the circumcenter of ABCABC is the circumcenter of HBCHBC, hence O2O_2 is the reflection of OO in BCBC. We have
MO2=MO=(2)AH2=AO1 MO_2 = MO \stackrel{(2)}{=} \frac{AH}{2} = AO_1
and, as AO1MO2AO_1 \parallel MO_2, O1AMO2O_1AMO_2 is a parallelogram.

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