Iranian Mathematical Olympiad

Let $F^{\prime }$ 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 $$(BC,SA)=(A{F}^{\prime },FB)=(AR,PD)=-1.$$ 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 $S$.