Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National Olympiad Prove it Canada

Problem:
Let $ABC$ be a triangle with circumcircle $\Gamma$ and $AB \neq AC$ . Let $D$ and $E$ lie on the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle BAE = \angle DAC$ . Let the incenters of $BAE$ and $CAD$ be $X$ and $Y$ respectively, and let the external tangents of the incircles of $BAE$ and $CAD$ intersect at $Z$ . Prove that $Z$ lies on the common chord of $\Gamma$ and the circumcircle of $AXY$ .

Solution

Solution:
Let $AX$ and $AY$ intersect $(ABC)$ again at $P$ and $Q$ , let the inradii of $ABE$ and $ACD$ be $r_B$ and $r_C$ , and let $(AXY)$ intersect $(ABC)$ again at $N$ .

First note that $\angle BAP = \frac{1}{2}\angle BAE = \frac{1}{2}\angle CAD = \angle QAC$ , so $XP = BP = CQ = CY$ . This thus implies that, since $NXP$ and $NYQ$ are spirally similar, $NX = NY$ and $NP = NQ$ (in particular, $N$ is the midpoint of arc $XAY$ on $(AXY)$ ).

Now, note that
$\frac{ZX}{ZY} = \frac{r_b}{r_c} = \frac{AX\sin(\angle PAE)}{AY\sin(\angle QAD)} = \frac{AX}{AY}.$
Since $Z$ lies on the ray $YX$ , we have that $Z$ lies on the external angle bisector of $\angle XAY$ . But so does $N$ , hence $Z$ , $A$ , $N$ are collinear, as desired.

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