$$\frac{PQ}{\sin\angle{PXQ}}=\frac{PQ}{\sin\angle{PYQ}}=2R. \implies \angle{PXQ}=\angle{PYQ}$$$$\angle{PIQ}=\angle{PJQ}=90+\frac{\angle{PXQ}}{2}$$$J'$ is Reflection of $J$ to $M$ we have $\angle{PJQ}=\angle{PJ'Q}=\angle{PIQ}. \implies $I=J'$ $ $PIQJ$ is parallelogram $\angle{IPQ}=\angle{JQP}$ $\angle{XPQ}=\angle{YQP}$ Similarly $\angle{XQP}=\angle{YPQ}$ we are done

PJQ=PJ'Q=PIQ does not imply that I=J'.this implies that I, J', P and Q are Concyclic. Not this only I=J'.the case happened only when we are given with PI is parallel to JQ or QI//PJ.then only you can prove that I=J'.so, I think the problem cannot be solvable with the given things in the problem.

UTC12 wrote: PJQ=PJ'Q=PIQ does not imply that I=J'.this implies that I, J', P and Q are Concyclic. Not this only I=J'.the case happened only when we are given with PI is parallel to JQ or QI//PJ.then only you can prove that I=J'.so, I think the problem cannot be solvable with the given things in the problem. I don't think so, Arya is correct. Same solution, just another way Let $O_1, O_1$ be the circumcenters of $\triangle XPQ$ and $\triangle YPQ, M$ is the midpoint of $PQ, A := O_1M \cap \odot (XPQ),$ $ B := O_2M \cap \odot (YPQ)$ it's well-known that $A$ is the circumcenter of $\triangle IPQ$ and $B$ is the circumcenter of $\triangle JPQ \implies PA = QA, PB = QB$ and since the circumradius of $\triangle XPQ$ and the circumradius of $\triangle YPQ$ are the same $\implies PA = QA = PB = QB \implies O_1 \in \odot (JPQ)$ and $O_2 \in \odot (IPQ) \implies J \in IM$ and $J \in \odot (XYO_1) \implies $ (easy angle chasing) $\angle{PJQ}=\angle{PIQ}$. Now, let $I' $ be the reflection of $I$ to $M \implies I' \in IM$ and $\angle{PIQ}=\angle{PI'Q} \implies I' \in \odot (XYO_1)$ and that's exactly the same definition of $J$. Therefore $I' \equiv J \implies PIQJ$ is parallelogram $\implies XPYQ$ is a parallelogram

The Problem isn't Solvable. If we pick $P,X,Q,Y$ to be a kite (which isn't a square) with $\angle XPY = \angle XQY = 90^\circ$. The points will be concyclic, therefore the circumcircle and specifically circumradii of $\triangle PQX,\triangle PQY$ will be equal. Since The kite will be symmetric about $XY$, $X,I,M,J,Y$ are collinear. Both conditions of the question are held and $P,X,Q,Y$ is not a parallelogram.

Veham wrote: The Problem isn't Solvable. If we pick $P,X,Q,Y$ to be a kite (which isn't a square) with $\angle XPY = \angle XQY = 90^\circ$. The points will be concyclic, therefore the circumcircle and specifically circumradii of $\triangle PQX,\triangle PQY$ will be equal. Since The kite will be symmetric about $XY$, $X,I,M,J,Y$ are collinear. Both conditions of the question are held and $P,X,Q,Y$ is not a parallelogram. nvm. I think it should be added that $XPQY$ is not concyclic.

Solved with kingu. There seems to be a lot of hot argument on this problem. I myself was shocked at how the solution turned out. A really bad kind of problem and too easy for the Problem 5 position in my opinion. We first show the following key claim. Claim : Quadrilateral $IPJQ$ is a parallelogram. Proof : Note that since $PQ$ subtends equal angles in circles with equal circumradii, $\measuredangle QXP = \measuredangle PYQ$. Thus, it also follows that $\measuredangle QIP = \measuredangle PJQ$, since we know that the angle subtended at the incenter, is dependent only on the angle subtended on the circumference. But now, if $J'$ is the reflection of $I$ across $M$ we know that $J'$ also lies on the line $\overline{IM}$ and thus, $I$, $J$ and $J'$ are collinear. Further, we also have that \[\measuredangle PJ'Q = \measuredangle QIP = \measuredangle PJQ \]from which it follows that $J'=J$, which proves the claim. Now, simply notice that, \[\measuredangle QPY = 2\measuredangle QPJ = 2\measuredangle PQI = \measuredangle PQX\]from which it follows that $XQ \parallel YP$. Similarly we can also show that $XP \parallel YQ$. Thus, it follows that $XPYQ$ is a parallelogram as desired.

This problem makes a contradiction. That XPQ is congruent to YPQ. The argument is proved only when XPQ is congruent to YPQ. We know that anglePXQ =anglePYQ.let the triangle PXQ is fixed. Then it implies that PYQ is also fixed. To prove our argument. And we know that for proving our argument we need to show that X, M and Y is Collinear. So this implies that PYQ is also fixed. And PY'Q is a triangle like this and P, Y, Y' and Q is Concyclic, which satisfies the all the given condition in the problem. So the triangle PYQ is not need only to be fixed . Then this makes a contradiction. Hence it is impossible to prove the argument Note : Y and Y' be same side of PQ..

Let $O_1,O_2$ be circumcenter of $\triangle XPQ$ and $\triangle YPQ$. Let $R_1 = O_1M \cap (XPQ)$ and $R_2=O_2M \cap (YPQ)$. As side $PQ$ and circumradius are equal we have $R_1I = R_2J$ and $R_1M=R_2M$, by using collinearity we have $IM=MJ$ and so $IBJC$ is parallelogram. Now we have equal angles in both triangle hence $XPYQ$ parallelogram