Mathematical Reflections 2012, Issue 6 - Problem S248

Problem:
Let $\mathcal{C}(O,R)$ be a circle and let $P$ be a point in its plane. Consider a pair of diametrically
opposite points $A$ and $B$ lying on $\mathcal{C}$. Prove that while points $A$ and $B$ vary on the
circumference of $\mathcal{C}$, the circumcircles of triangles $ABP$ pass through another fixed point.

Proposed by Ivan Borsenco.

Solution:
Let us consider another pair of diametrically opposite points $A'$ and $B'$ on $\mathcal{C}$. Then, $O$ is the midpoint of $AB$ and $A'B'$, so in the triangles $ABP$ and $A'B'P$ the points $O$ and $P$ are fixed, and this implies that the line $PO$ is fixed. Therefore, the line $PO$ intersects the circumcircle of the triangle $A'B'P$ in another point $Q$, which is fixed. By the arbitrarity of the pair of diametrically opposite points $A',B'$, we obtain that all the circumcircles of triangles $ABP$ pass through the point $Q$.