Track / Stage 4 / 9 of 340 #269 of 2000

Problem 269

AMC 12 late, AIME early

Geometry Difficulty 4.3 Prove it Berkeley Math Circle · United States

Problem:

Let $ABC$ be an acute triangle with orthocenter $H$ , circumcenter $O$ , and incenter $I$ . Prove that ray $AI$ bisects $\angle HAO$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:

Without loss of generality, $AB < AC$ . It follows that $\angle BAH = 90^{\circ} - \angle B$ , since the extension of $AH$ is perpendicular to $BC$ . Moreover, we also have $\angle AOC = 2\angle B$ ; but since $OA = OC$ , this implies $\angle OAC = \frac{1}{2}\left(180^{\circ} - \angle AOC\right) = 90^{\circ} - \angle B$ . So we conclude that $\angle BAH = \angle CAO$ . Since $\angle BAI = \angle CAI$ as well, it follows that $\angle HAI = \angle OAI$ , which is what we wanted to prove.

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