62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

First note, that as $\angle ABC>90^{\circ }$, point $K$ lies on $AC$ (fig. 1). Also, it's clear that $PA=PB$. We will show that $K$ lies on the circle $\omega$ with a center $P$ and radius $PA$. It's enough to prove, that $\angle APB=2(180^{\circ }-\angle AKB)$. Indeed, take on the larger arc of circle $\omega$ any point $X$. Then $\angle AXB=\frac{1}{2}\angle APB$. If this condition holds, then $\angle AKB+\angle AXB=180^{\circ }$, so points $A,K,B$ and $X$ lie on a circle $\omega$. As $P$ is the center of this circle, $PK=PA$.

So, let's prove that $\angle APB=2(180^{\circ }-\angle AKB)$. As $PA=PB$, and also using the theorem about the angle between the chord and a tangent, we get the following: $\angle APB=180^{\circ }-2\angle ABP=180^{\circ }-2\angle ACB=180^{\circ }-2(90^{\circ }-\angle CKB)=2\angle CKB=2(180^{\circ }-\angle AKB)$.