Baltic Way 2019

Denote by $P^{\prime }$ and $P^{\prime \prime }$ reflections of $P$ with respect to $AB$ and $AC$. Easy to see that $P$ is an anti-Steiner point of the line $P^{\prime }{P}^{\prime \prime }$, so $P\equiv B$ or $P\equiv C$— therefore $\angle BAC=90^{\circ }$.

---

Alternative solution.

The implication $⟺$ is very easy. We shall prove $⟹$. Suppose that triangle $ABC$ is acute. Let $P^{\prime }$ and $P^{\prime \prime }$ be the reflections of $P$ with respect to $AB$ and $AC$, respectively. Then $\angle PA{P}^{\prime \prime }=2\angle BAC$. Let $H^{\prime }$, $H^{\prime \prime }$ be the reflections of $H$ with respect to $AB$ and $AC$, respectively. Since $H^{\prime }$ lies on the circumcircle of $ABC$ and $P$ lies inside this circle, we have $\angle AP{H}^{\prime }>\angle AC{H}^{\prime }$. Hence $\angle A{P}^{\prime }H=\angle AP{H}^{\prime }>\angle AC{H}^{\prime }=\frac{\pi }{2}-\angle BAC$. Similarly, $\angle A{P}^{\prime \prime }H>\frac{\pi }{2}-\angle BAC$. Since $H$ lies on $P^{\prime }{P}^{\prime \prime }$, we get $$2\angle BAC=\angle P^{\prime }A{P}^{\prime \prime }=\pi -\angle AP{H}^{\prime }-\angle A{P}^{\prime \prime }H<\pi -(\frac{\pi }{2}-\angle BAC)-(\frac{\pi }{2}-\angle BAC)=2\angle BAC$$ a contradiction. We obtain similar contradiction when triangle $ABC$ is obtuse.