Library / HMMT February Team Round /6 of 9

HMMT February Team Round, 2019

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Scalene triangle $ABC$ satisfies $\angle A = 60^{\circ}$ . Let the circumcenter of $ABC$ be $O$ , the orthocenter be $H$ , and the incenter be $I$ . Let $D, T$ be the points where line $BC$ intersects the internal and external angle bisectors of $\angle A$ , respectively. Choose point $X$ on the circumcircle of $\triangle IHO$ such that $HX \parallel AI$ . Prove that $OD \perp TX$ .

Solution

Solution:

Let $I_A$ denote the $A$ -excenter. Because $\angle A = 60^{\circ}$ , $AI$ is the perpendicular bisector of $OH$ and $B, H, O, C$ all lie on the circle with diameter $II_A$ . We are given that $X$ is on this circle as well, and since $HI = OI$ , $XIOI_A$ is also an isosceles trapezoid. But $II_A$ is a diameter, so this means $X$ must be diametrically opposite $O$ on $(BOC)$ and is actually the intersection of the tangents to $(ABC)$ from $B$ and $C$ .

Now $(T, D; B, C) = -1$ , so $T$ is on the polar of $D$ with respect to $(ABC)$ . $BC$ is the polar of $X$ and $D$ lies on $BC$ , so $X$ must also lie on the polar of $D$ . Therefore $TX$ is the polar of $D$ with respect to $(ABC)$ , and $OD \perp TX$ as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.