USAMO, 2010

Let $T$ be the foot of the perpendicular from $Y$ to line $AB$. We note that $P$, $Q$, $T$ are the feet of the perpendiculars from $Y$ to the sides of triangle $ABX$. Because $Y$ lies on the circumcircle of triangle $ABX$, points $P$, $Q$, and $T$ are collinear by Simson's theorem. Likewise, points $S$, $R$, and $T$ are collinear.

We need to show that $\angle XOZ = 2\angle PTS$. Notice that
$$\frac{\angle XOZ}{2} = \frac{\widehat{XZ}}{2} = \frac{\widehat{XY}}{2} + \frac{\widehat{YZ}}{2} = \angle XAY + \angle ZBY = \angle PAY + \angle SBY$$
and that $\angle PTS = \angle PTY + \angle STY$. Therefore, it suffices to prove that
$$\angle PTY = \angle PAY \text{ and } \angle STY = \angle SBY.$$
For this, it is enough to show that quadrilaterals $APYT$ and $BSYT$ are cyclic. This follows because $\angle APY = \angle ATY = 90^\circ$ and $\angle BTY = \angle BSY = 90^\circ$.

Lines $YQ$ and $YR$ are perpendicular to $BX$ and $AZ$, respectively, so $\angle RYQ$ is equal to the acute angle between lines $BX$ and $AZ$. This angle is $\frac{1}{2}(\widehat{AX} + \widehat{BZ}) = \frac{1}{2}(180^\circ - \widehat{XZ})$ because $X$, $Z$ lie on the circle with diameter $AB$. Also, $\angle AXB = \angle AZB = 90^\circ$ and so $PXQY$ and $SZRY$ are rectangles, whence $\angle PQY = 90^\circ - \angle YXB = 90^\circ - \widehat{YB}/2$ and $\angle YRS = 90^\circ - \angle AZY = 90^\circ - \widehat{AY}/2$. The angle between $PQ$ and $RS$ is therefore
$$\angle PQY + \angle YRS - \angle RYQ = (90^\circ - \frac{\widehat{YB}}{2}) + (90^\circ - \frac{\widehat{AY}}{2}) - (90^\circ - \frac{\widehat{XZ}}{2}) = \frac{\widehat{XZ}}{2} = \frac{\angle XOZ}{2},$$
as desired.