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Iranian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Iran

Given triangle $ABC$ and line $\ell$ passing through point $A$ , point $X$ on $\ell$ is considered to be variable. Circles $\omega_b$ and $\omega_c$ pass through both of the points $A$ and $X$ and are tangent to sides $AB$ and $AC$ , respectively. Tangents $BY$ and $CZ$ are drawn from vertices $B$ and $C$ to circles $\omega_b$ and $\omega_c$ respectively. Prove that as $X$ varies, the circumcircle of triangle $ZXY$ would be passing through two fixed points.

Solution

Let $D$ be intersection of $\ell$ and $BC$ . Let $P$ and $Q$ be the reflections of $A$ with respect to $BC$ and $D$ , respectively. We claim that the circumcircle of $XYZ$ passes through $P$ and $Q$ .
Since $PQ \parallel BC$ , we have $\angle XQP = \angle XDB$ , in line with $BX = BY = BP$ , we have:
$\begin{aligned} \angle PYA &= \frac{1}{2} \angle PBA = \angle ABC \\ \implies \angle XYP &= \angle AYP - \angle AYX = \angle ABC - \angle DAB = \angle ADB = \angle XQP \end{aligned}$
So $Y$ lies on the circumcircle of $XPQ$ . Similarly, $Z$ also lies on this circle, and so we are done.

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