Iranian Mathematical Olympiad

i) Notice that $$\angle QEB=\angle QPB=\angle QDA⟹\angle QEX=\angle QDX$$ So quadrilateral $XDEQ$ is cyclic and similarly quadrilateral $YCFQ$ is cyclic. Let $S$ be the second intersection point of these two circles. It's clear that $\angle QSX=\angle QEB=\angle QCY$ so $X,S,Y$ are collinear. Let $R^{\prime }$ be the second intersection point of the circumcircle of triangle $DSC$ with line $XY$. We have $$\angle R^{\prime }DC=\angle R^{\prime }SC=\angle YSC=\angle YFC=\angle FAP=\angle FDP=\angle CDP$$ Similarly it's obtained that $\angle R^{\prime }CD=\angle DCP$. So $R^{\prime }$ is the reflection of $P$ onto $CD$, and so $R\equiv R^{\prime }$. □

ii) It's easy to see that $FR=FP=AD$ and $RD=PD=AF$. Thus $AFRD$ is a parallelogram and therefore $RD\parallel AF$ and $RF\parallel AD$. Hence, $$\frac{RY}{RX}=\frac{AD}{DX}=\frac{PF}{DX}$$ So the problem becomes equivalent to showing that $\frac{RY}{RX}=\frac{PY}{PX}$. Thus it suffices to show that $\frac{PY}{PX}=\frac{PF}{DX}$. Note that $\angle ADP=\angle AFP$, so $\angle PDX=\angle PFY$ is enough to show that $$\frac{PD}{DX}=\frac{FY}{FP}⟺\frac{AF}{FY}=\frac{DX}{AD}$$ From $RD\parallel AF$ and $RF\parallel AD$ both sides of the last equation are found to be equal to $\frac{RX}{RY}$ and hence the claim. $\square$