Baltic Way 2017 · Problem 14

As $\triangle P B D$ is an isosceles right triangle $\angle G D P=\angle B D P=45^{\circ}$. Similarly $\angle P E C=45^{\circ}$, and thus $\angle P E G=45^{\circ}$. Therefore $P G D E$ is cyclic. As $\angle G D F$ and $\angle G E F$ are right $E F G D$ is cyclic. Therefore $D G P F E$ is a cyclic pentagon. Therefore $\angle G F P=\angle G E P=45^{\circ}$. Similarly $\angle F G P=45^{\circ}$. Therefore $\triangle F P G$ is a (right) isosceles triangle.

Remark: It can be shown given two intersecting lines $l$ and $m$, not perpendicular to one another and an point $P$. there exist unique points $F$ and $G$ on $l$ and $m$ respectively such that $\triangle F P G$ is an right isosceles triangle using similar constructions to above.