HMIC

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $ABC$ be an acute scalene triangle with incenter $I$ . Show that the circumcircle of $BIC$ intersects the Euler line of $ABC$ in two distinct points.

(Recall that the Euler line of a scalene triangle is the line that passes through its circumcenter, centroid, orthocenter, and the nine-point center.)

Solution

Solution:

Let $O$ and $H$ be the circumcenter and orthocenter of $ABC$ . Recall that
$\begin{aligned} \angle BOC & = 2\angle A, \\ \angle BHC & = 180^\circ - \angle A, \\ \angle BIC & = 90^\circ + \frac{1}{2} \angle A. \end{aligned}$
As $ABC$ is acute, $A$ , $I$ , $O$ , $H$ all lie on the same side of $BC$ .

- If $\angle A > 60^\circ$ , then $\angle BOC > \angle BIC$ , so $O$ lies inside $(BIC)$ .
- If $\angle A < 60^\circ$ , then $\angle BHC > \angle BIC$ , so $H$ lies inside $(BIC)$ .
- If $\angle A = 60^\circ$ , then $O$ and $H$ are on $(BIC)$ .

In all cases, two intersections are guaranteed.

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