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Iranian Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Iran

An arbitrary point $P$ lies on side $BC$ of triangle $ABC$ . Angle bisectors of $\widehat{APB}$ and $\widehat{APC}$ intersect the external angle bisector of $\widehat{A}$ at $X$ and $Y$ , respectively. Circumcircle of triangle $PXY$ meets $BC$ for the second time at $Q$ . Prove that $\widehat{BAP} = \widehat{CAQ}$ .

Solution

Let $Q'$ be a point on side $BC$ such that $\overline{BAP} = \overline{CAQ'} = \alpha$ .

Note that $X, Y$ also lie on the exterior angle bisector of $\overline{PAQ'}$ , that's because
$\overline{PAX} = \overline{Q'AY} = \left(90^\circ - \frac{\hat{A}}{2}\right) + \alpha .$
Also $PX, PY$ are exterior and interior angle bisector of vertex $P$ in triangle $APQ'$ . Therefore $X$ is the $Q'$ -excenter, and $Y$ is the $P$ -excenter of this triangle. So we get
$\overline{XPY} = \overline{XQ'}\overline{Y} = 90^\circ \implies XPQ'Y \text{ is cyclic.}$
Since the intersection of $C_{\overline{XPQ}}$ with $BC$ (other than $P$ ), is a unique point, we get $Q = Q'$ . Therefore $\overline{BAP} = \overline{CAQ'} = \overline{CAQ}$ ■

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