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: $\angle APB = 180^\circ - 2\angle ABP = 180^\circ - 2\angle ACB = 180^\circ - 2(90^\circ - \angle CKB) = 2\angle CKB = 2(180^\circ - \angle AKB)$.