Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Notice that a homothety centred at $P$ mapping $\omega$ to $\Omega$ maps $C$ to $Q$, and maps the line $AB$ to the tangent to $\Omega$ at $Q$. Thus this tangent is parallel to $AB$, and hence $Q$ is the midpoint of arc $AB$ (not containing $P$). So the points $I, X$, and $Y$ lie on the segments $PQ, RQ$, and $SQ$, respectively.

Recall that for any triangle $KLM$ with the circumcircle $\Gamma$ and incentre $J$, the points $K, L$, and $J$ are equidistant from the midpoint of arc $KL$ of $\Gamma$ not containing $M$. Applying this to triangles $APB, ARB$, and $ASB$ we obtain that
$$QA = QB = QX = QY = QI.$$
Since $Q$ is the midpoint of arc $AB$, we get that $\angle QPA = \angle QPB = \angle QAB$. Thus the triangles $QAC$ and $QPA$ are similar, and $QC \cdot QP = QA^2 = QX^2$. Since $QX$ is tangent to $\omega$, it follows that $X$ is their point of tangency; analogously, $Y$ is the point of tangency of $QS$ with $\omega$.

Finally, from isosceles triangles $QXI$ and $QYI$ we get $\angle QXI = \angle QIX = 90^{\circ} - \angle IQX/2$ and $\angle QYI = \angle QIY = 90^{\circ} - \angle IQY/2$. Denoting by $O$ the centre of $\omega$, we obtain $\angle QIX + \angle QIY = 180^{\circ} - \angle XQY/2 = 180^{\circ} - (180^{\circ} - \angle XOY)/2 = 90^{\circ} + \angle XPY$. Thus,
$$\angle PXI + \angle PYI = \angle XIY - \angle XPY = (90^{\circ} + \angle XPY) - \angle XPY = 90^{\circ},$$
as required.