Problem. Let $P$ be a point inside triangle $ABC$ such that $\mathrm{\angle }CPA={90}^{\circ }$ and $\mathrm{\angle }CBP=\mathrm{\angle }CAP$. Let $X$ and $Y$ be the midpoints of $AB$ and $AC$ respectively. Prove that $\mathrm{\angle }PXY={90}^{\circ }$.

Origin: New Zealand, Camp Selection 2018, problem 4.

Solution. Let's consider a homothety with center $A$ and scale factor $2$. This homothety maps $X\to B$, $Y\to C$, $P\to P^{\prime }$, where $P^{\prime }$ is the point on the ray $AP$ such that $AP=P{P}^{\prime }$.

Notice that since $\mathrm{\angle }CPA={90}^{\circ }$, $P$ lies on the circle with diameter $AC$ and center $Y$. Thus, $AY=YP$. These two segments, $AY$ and $YP$, map to the segments $AC$ and $C{P}^{\prime }$ respectively under the homothety considered above. Therefore, $AC=C{P}^{\prime }$. This, in turn, implies that $\mathrm{\angle }{P}^{\prime }AC=\mathrm{\angle }A{P}^{\prime }C$. When we combine this with the condition $\mathrm{\angle }CBP=\mathrm{\angle }CAP$, we obtain $\mathrm{\angle }P{P}^{\prime }C=\mathrm{\angle }CBP$. This implies that the the quadrilateral $CPB{P}^{\prime }$ is inscribed (cyclic). Therefore, $\mathrm{\angle }CB{P}^{\prime }=\mathrm{\angle }CP{P}^{\prime }={90}^{\circ }$. Finally, notice that under the homothety considered above the angle $\mathrm{\angle }PXY$ maps to the angle $\mathrm{\angle }{P}^{\prime }BC$, thus $\mathrm{\angle }PXY=\mathrm{\angle }{P}^{\prime }BC={90}^{\circ }$, and we are done.