, 2024

Circles with centers $O_1$ and $O_2$ have a common chord $PQ$. The line $O_1O_2$ therefore intersects the segment $PQ$ in its midpoint since it is the perpendicular bisector of $PQ$. Thus, the line $O_1O_2$ 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 $O_1O_2$ 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
$$\angle BAO_1 = 90^\circ - \angle ABQ = 90^\circ - \angle CQP = \angle BCO_1.$$
The segment $BO_1$ 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 $O_1O_2$ 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.