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Iranian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Iran

Given triangle $ABC$ , variable points $X$ and $Y$ are chosen on segments $AB$ and $AC$ , respectively. Let $Z$ be a point on the line $BC$ such that $ZX = ZY$ . The circumcircle of $XYZ$ intersects the line $BC$ at $T$ , for the second time. Point $P$ is chosen on line $XY$ such that $\angle PTZ = 90^\circ$ . Let $Q$ be a point on the same side of line $XY$ as $A$ , satisfying $\angle QXY = \angle ACP$ and $\angle QYX = \angle ABP$ . Prove that the circumcircle of triangle $QXY$ passes through a fixed point (as $X$ and $Y$ vary).

Solution

Let $F'$ be the intersection of $CX$ with the circumcircle of $ABC$ and $G$ be the intersection of $XY, BC$ . Note that $TZ$ is the external angle bisector of $\angle XTY$ and $PT$ is perpendicular to $TZ$ , so $PT$ is the angle bisector of $\angle XTZ$ . Thus $(GP, XY) = -1$ and $BY, CX, AP$ are concurrent. Let $R$ be the intersection of these lines, and $D$ be the intersection of $AP, BC$ . By looking through point $C$ , we have $(DP, RA) = (GP, XY) = -1$ . By projecting $X$ to the circumcircle of $ABC$ and also projecting this circle onto $AP$ through $C$ we have
$(BC, SA) = (AF', FB) = (AR, PD) = -1.$
This shows that $S$ is the intersection of the circumcircle of $ABC$ with symmedian and the proof is complete as the circumcircle of triangle $QXY$ passes through this fixed point $S$ .

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