Iranian Mathematical Olympiad

i) Notice that
$$\angle QEB = \angle QPB = \angle QDA \implies \angle QEX = \angle QDX$$
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 $R'$ be the second intersection point of the circumcircle of triangle $DSC$ with line $XY$. We have
$$\angle R'DC = \angle R'SC = \angle YSC = \angle YFC = \angle FAP = \angle FDP = \angle CDP$$
Similarly it's obtained that $\angle R'CD = \angle DCP$. So $R'$ is the reflection of $P$ onto $CD$, and so $R \equiv R'$. □

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
$$\frac{PD}{DX} = \frac{FY}{FP} \iff \frac{AF}{FY} = \frac{DX}{AD}$$
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. $\square$