AoPS - Problem

Source: 2020 China North Mathematical Olympiad Basic Level P4

Tags: geometry, incenter, orthocenter, geometry solved, Circumcenter, cyclic quadrilateral, Spiral Similarity

Miku3D

04.08.2020 10:36

In $\triangle ABC$, $\angle BAC = 60^{\circ}$, point $D$ lies on side $BC$, $O_1$ and $O_2$ are the centers of the circumcircles of $\triangle ABD$ and $\triangle ACD$, respectively. Lines $BO_1$ and $CO_2$ intersect at point $P$. If $I$ is the incenter of $\triangle ABC$ and $H$ is the orthocenter of $\triangle PBC$, then prove that the four points $B,C,I,H$ are on the same circle.

WLOG, suppose $ABC$ is acute, and let $H_1$ be the orthocenter of $ABC$. It's well-known that $AO_1O_2 \overset{+}{\sim} ABC$, so the Spiral Center Lemma yields $P \in (ABC)$. Now, the Orthocenter Reflection Lemma implies $BH_1HC$ is cyclic. In addition, we have $$\angle BIC = 90^{\circ} + \frac{\angle A}{2} = 120^{\circ} = 180^{\circ} - \angle A = \angle BH_1C$$so $BH_1IC$ is cyclic, i.e. $BH_1IHC$ is a cyclic pentagon. $\blacksquare$ Remark: If $ABC$ was not acute, then $H_1$ and $I$ would be on opposite sides of $BC$. Quadrilateral $BICH_1$ would still be cyclic, however, as this configuration yields $$\angle BH_1C = \angle BAC = 60^{\circ}.$$