Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 5.4 AIME, harder Prove it Turkey

Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$ , respectively, of an acute triangle $ABC$ with orthocenter $H$ and circumcenter $O$ . The rays $HA_1, HB_1, HC_1$ cut the circumcircle at the points $A_0, B_0, C_0$ , respectively. Show that $O, H,$ and $H_0$ are collinear where $H_0$ is the orthocenter of $A_0B_0C_0$ .

Solution

Since $H$ is the orthocenter, $\angle BHC = 180^\circ - \angle BAC = \angle BA_0C$ . As $A_1$ is the midpoint of $BC$ , this is possible only if $BHCA_0$ is a parallelogram. Then $\angle ACA_0 = 90^\circ$ , and $AA_0$ is a diameter of the circumcircle of $ABC$ . Therefore, the reflection across $O$ takes $ABC$ to $A_0B_0C_0$ , and takes $H$ to $H_0$ . In particular, $O$ is the midpoint of $HH_0$ .

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