IMO Team Selection Test 2, June 2020

Let $M$ be the circumcentre of $△ABC$ and let $\alpha =\angle BAC$. We first show that $△BYA\sim △BMP$ and then that $△YBM\sim △ABP$.

By the inscribed angle theorem we have $\angle BMC=2\angle BAC=2\alpha$. Quadrilateral $PBMC$ is a kite with axis of symmetry $PM$ (by the equality of radii $|MB|=|MC|$ and equality of tangent segments $|PB|=|PC|$), so $MP$ bisects angle $\angle BMC$. Therefore $\angle BMP=\frac{1}{2}\angle BMC=\alpha$.

Moreover, we have $\angle PBM=90^{\circ }$ (tangent to a circle is perpendicular to its radius), so by the sum of angles of a triangle we have $\angle MPB=90^{\circ }-\alpha$.

On the other hand, we are given that $\angle ABY=90^{\circ }$ and we also have $\angle YAB=\angle YAC-\angle BAC=90^{\circ }-\alpha$. Therefore $\angle ABY=\angle PBM$ and $\angle YAB=\angle MPB$, from which follows that $△BYA\sim △BMP$.

From this similarity it follows that $\frac{|YB|}{|AB|}=\frac{|MB|}{|PB|}$. Combining this with the equality of angles $$\angle YBM=\angle YBA+\angle ABM=90^{\circ }+\angle ABM=\angle ABM+\angle MBP=\angle ABP,$$ we see that $△YBM\sim △ABP$.

Let $T$ now be the intersection of $AP$ and $YM$, then we have $$\angle BYT=\angle BYM=\angle BAP=\angle BAT,$$ from which it follows that $BYAT$ is a cyclic quadrilateral. Therefore $\angle ATY=\angle ABY=90^{\circ }$, so $AT\perp YM$. Analogously, $AP\perp XM$. But from this it now follows that $YM$ and $XM$ coincide and we get that $AP\perp XY$. $\square$