Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024)

Circles with centers $O_1$ and $O_2$ have a common chord $PQ$. The line $O_1{O}_2$ therefore intersects the segment $PQ$ in its midpoint since it is the perpendicular bisector of $PQ$. Thus, the line $O_1{O}_2$ cannot pass through the vertex $A$, since it lies on the line $PQ$, but not inside the segment $PQ$. Let us note that since the triangle $ABC$ is acute, both centers $O_1,O_2$ lie inside the half-plane $BCP$.

To show the implication from the problem statement, assume that the line $O_1{O}_2$ passes through vertex $C$. In triangle $PQC$ the vertex $C$ lies on the perpendicular bisector of $PQ$, thus this triangle is isosceles. Therefore $$\angle BQA=\angle CQP=\angle CPQ=\angle CPA=\angle CBA,$$ where in the last step we used the equality of angles over the arc $AC$ of the circle $k$. The triangle $BQA$ is isosceles with apex $A$. Both points $A$ and $O_1$ lie on the perpendicular bisector of $BQ$, which yields $$\angle BA{O}_1=90^{\circ }-\angle ABQ=90^{\circ }-\angle CQP=\angle BC{O}_1.$$ The segment $B{O}_1$ can be seen from points $A$ and $C$ under the same angle, and the points $B,O_1,C$ and $A$ are conicyclic. Thus we showed that the point $O_1$ lies on the circle $k$. In the second case, where the line $O_1{O}_2$ passes through vertex $B$, we analogously get, that the point $O_2$ lies on $k$. This concludes the proof of the implication from the problem statement.