AUT ABooklet

Geometry Difficulty 5.6 AIME, harder Prove it Austria

Let $ABC$ be a triangle. Let $P$ be the point on the extension of $BC$ beyond $B$ such that $BP = BA$ . Let $Q$ be the point on the extension of $BC$ beyond $C$ such that $CQ = CA$ . Prove that the circumcenter $O$ of the triangle $APQ$ lies on the angle bisector of the angle $\angle BAC$ .

Figure 3: Problem 10

Solution

Since $ACQ$ is an isosceles triangle, the perpendicular bisector of $AQ$ is the angle bisector of $\angle QCA$ . But the perpendicular bisector of $AQ$ also passes through the circumcenter $O$ of the triangle $APQ$ .

Therefore, $O$ lies on the angle bisector of $\angle QCA$ which is the exterior angle bisector of $\angle ACB$ by definition of $Q$ .

Analogously, the point $O$ lies also on the exterior angle bisector of $\angle CBA$ . Therefore, the point $O$ is the intersection of the two exterior angle bisectors which makes it the excenter of the excircle of $ABC$ tangent to $BC$ . This excenter lies on the angle bisector of $\angle BAC$ as desired.

(Theresia Eisenkölbl) $\square$

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