See all problems here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=528313.

Since chord $AB$ of $\Omega$ is tangent to $\omega$ at $C$, $PC$ bisects $\angle{APB}$; that is, $Q$ is the midpoint of $\overarc{AB}$ without $P$. Let $\Gamma$ be the circle centered at $Q$ with radius $QA = QB$ - clearly $A, B \in \Gamma$. Additionally, $I \in \Gamma$, since \begin{align*} \angle{AIB} &= 90^{\circ} + \angle{APB}/2 \\ &= 90^{\circ} + (90^{\circ} - \angle{AQB}/2) \\ &= 180^{\circ} - \angle{AQB}/2 \end{align*} Analogously, $X, Y \in \Gamma$ also (since $\angle{ARB} = \angle{APB} = \angle{ASB}$). Next, remark that $RQ$ is tangent to $\omega$ at $X$ (and similarly $SQ$ is tangent to $\omega$ at $Y$). Since $Q$ is the midpoint of $\overarc{AB}$, $RQ$ bisects $\angle{ARB}$ so $X \in RQ$. Lemma 6 here tells us $X, C$, and the point of tangency of $RQ$ and $\omega$ are colinear, so this tangency point lies simultaneously on lines $CX$ and $RX$, so it must be $X$. Then \begin{align*} \angle{PXI} &= 180^{\circ} - \angle{QPX} - \angle{QIX} - \angle{PQX} \\ &= 180^{\circ} - \angle{QPR}/2 - (180^{\circ} - \angle{PQX})/2 - \angle{PQX} \\ &= 90^{\circ} - \angle{QPR}/2 - \angle{PQX}/2 \end{align*} using the fact that $RQ$ is tangent to $\omega$ at $X$ and $\triangle{IQX}$ is $Q$-isosceles. Similarly, $\angle{PYI} \&= 90^{\circ} - \angle{QPS}/2 - \angle{PQY}/2$ so \begin{align*} \angle{PXI} + \angle{PYI} &= 180^{\circ} - \angle{QPR}/2 - \angle{PQX}/2 - \angle{QPS}/2 - \angle{PQY}/2 \\ &= 180^{\circ} - (\angle{QPR} + \angle{QPS} + \angle{XQY})/2 \\ &= 180^{\circ} - 180^{\circ}/2 = 90^{\circ} \end{align*} as desired.

X, Y are not on gamma, but R and S are. I think you have to check the définitions of the points. [ edit : sorry, i was wrong ]

Lemma: Let $D$ be the midpoint of minor arc $BC$ of the circumcicle of $\triangle ABC$. Let $I$ be a point on segment $AD$. Then $I$ is the incentre of $\triangle ABC$ iff $DB=DC=DI$. Proof: Trivial angle chasing. Let $O_1,O_2$ be the centres of $\omega,\Omega$ respectively. Now the homothety centered at $P$ sending $\omega$ to $\Omega$ sends $O_1$ to $O_2$ and $C$ to $Q$. Hence $O_1C \parallel O_2Q$. But $O_1C \perp AB$, so $O_2Q \perp AB \Rightarrow Q$ is the midpoint of arc $AB$ in $\Omega$. Let $QR \cap \omega=X'$. $\angle QAC=\angle QBA=\angle APC \Rightarrow QA$ is tangent to $\odot APC \Rightarrow QA^2=QC.QP=QX'^2 \Rightarrow QA=QX'$. Since $X'$ lies on $RQ$, the internal bisector of $\angle ARB$ and $QA=QX'$, we conclude from the lemma $X'$ is the incentre of $\triangle ARB \Rightarrow X=X'$. Similarly we can prove $QS \cap \omega=Y$. Now $\angle PXI+\angle PYI=\angle XIY-\angle XPY=\angle XIQ+\angle YIQ-\angle XPY$ $=90^{\circ}-\frac 12\angle XQI+90^{\circ}-\frac 12\angle YQI-\frac 12\angle XO_1Y$ $=180^{\circ}-\frac 12(\angle XQY+\angle XO_1Y)=180^{\circ}-90^{\circ}=90^{\circ}$, done!

Let $\Gamma$ denote the circle centered at $Q$ with radius $QA=QB$, and note that $X = QR \cap \Gamma$, $Y= QS \cap \Gamma$, $I = PQ \cap \Gamma.$ The inversion with respect to $\Gamma$ fixes $\omega \Longrightarrow \Gamma$ and $\omega$ are orthogonal $\Longrightarrow X, Y \in \omega.$ $P,C$ are inverses with respect to $\Gamma$, so $\angle PXI = \angle CXI$, $\angle PYI = \angle CYI.$ But the sum of these four angles is 180, so $\angle PXI + \angle PYI = 90$, as desired.

proglote wrote: $X = QR \cap \Gamma$, $Y= QS \cap \Gamma$, $I = PQ \cap \Gamma.$ This is wrong,isn't it? Sorry,my fault.

$QA = QB =QX$ and $X$ is in the angle bisector $QR$ of $\angle ARB.$ So $X = QR \cap \Gamma$, and similarly for the others.

This problem was proposed by my friend Regis Prado Barbosa, from Brazil, to Romanian Masters in Mathematics and now appeared in the Romanian TST. Congratulations to my friend, he's very happy with it.

My solution It's obvious that $Q$ is the midpoint of $\overarc{AB}$ We consider the quadrilateral $RAQB$ has the circle $\omega$ tangent to $RQ$, $AB$. By applying Sawayama & Thebault theorem, we have $XC$ passes through the incenter of $\triangle RAB$. But $X$ lies on $RQ$ which is the bisector of $\angle ARB$ so $X$ is the incenter of $\triangle RAB$. Similary with $Y$ $X$ is the incenter of $\triangle RAB$ so $QX = QA = QB$. So $X,Y,I,A,B \in (Q,QI)$ We have : $\angle PXI + \angle PYI = ( \angle QXP - \angle QXI ) + (\angle QYP - \angle QYI)$ $= (180 - \angle PYX - ( 90 -\frac{PQR}{2} ) ) +( 180 - \angle PXY - ( 90 -\frac{PQS}{2} ))$ $= (180 - \angle PYX - \angle PXY ) + \frac{\angle SQR}{2} = \frac{\angle SQR}{2}+\angle XPY = 90$ We are done

Let $T= \omega\cup QR$ and $U= \omega\cup QS$ Given that $Q$ is the midpoint of arc $AB$ not containing $P$ of the circle $\omega$, so, $QA=QX$. By power of point $Q$, we have $QT^2= QC. QP=QA^2$. That gives, $QT=QA=QX$. As the line $RQ$ bisects $\angle ARB$, and point $X$ is the incenter of triangle $ARB$, so $X$ is on the line $RQ$. So, $X=T$ and $Y = U$. As $QA=QI$, so $Q$ is the circumcenter of triangle $YIX$. So, $\angle XIY = 180^{\circ}-\frac{\angle SQR}{2}$. Next, we apply Archimedian lemma, to get the line $PX$ bisects $\angle RPQ$ and $PY$ bisects $\angle SPQ$. That means, $\angle XIY = 90^{\circ}-\frac{\angle RQS}{2}$. This implies, $\angle PXI + \angle PYI= 180^{\circ}-\frac{\angle SQR}{2} - 90^{\circ}+\frac{\angle RQS}{2} = 90^{\circ}$

Let $O$ be the center of $\omega$ and let $X'$ and $Y'$ be the points of tangency of lines $QR$ and $QS$ to $\omega$. Note that from Shooting lemma, we have that $Q$ is the midpoint of arc $AB$. Claim: $X \equiv X'$ and $Y \equiv Y'$. Proof: From shooting lemma and PoP we get that \[ QB^2 = QA^2 = QC \cdot QP \overset{PoP}{=} QX'^2 = QY'^2 \]Hence $A,X',Y',B$ are cyclic with center at $Q$. Since $X'$ and $Y'$ lie on $QR$ and $QS$ respectively, the desired result follows. $\square$ Lastly note that $X,Y,O,Q$ are cyclic. Then we have: \[ 2(\angle PXI + \angle PYI) = 2(\angle XIY - \angle XPY) = 360 - \angle XQY - \angle XOY = 180 \]

using 2above's diagram. Skibidi toilet rizz! Observe by death start lemma that X, Y are tangency points. Also XIY is centered at Q, note too because XCYP is harmonic that XC/XP=YC/YP, now apollonius circle gives XI and YI are angle bisectors. then PXI+PYI=1/2 (PXC+PYC)=90.