Given a triangle $A B C$. On its sides $A B$ and $B C$, points $C_{1}$ and $A_{1}$ are fixed, respectively. Find a point $P$ on the circumcircle of triangle $A B C$ such that the distance between the centers of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$ is minimized.
Let $K$ be the second intersection point of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$ (see the left figure). We will prove that $K$ lies on the line $A_{1} C_{1}$.

Let point $P$ lie on the arc $A C$. Quadrilaterals $A C_{1} K P$ and $K A_{1} C P$ are cyclic, hence $\angle C_{1} K P = 180^{\circ} - \angle C_{1} A P$ and $\angle A_{1} K P = 180^{\circ} - \angle A_{1} C P$. On the other hand, from the cyclic quadrilateral $A B C P$, we get that $\angle C_{1} K P + \angle A_{1} C P = 180^{\circ}$. Therefore, $\angle C_{1} K P + \angle A_{1} K P = 180^{\circ}$, which means that point $K$ lies on the line $A_{1} C_{1}$. (Note that the proven statement is equivalent to problem $\underline{56623}$ a).

When point $P$ lies on the arc $A B$ or $B C$, the proof is analogous.

Let $M_{A}$ and $M_{C}$ be the midpoints of segments $C_{1} K$ and $A_{1} K$, and $O_{1}$ and $O_{2}$ be the centers of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$, respectively. Then $O_{1} M_{A}$ is the perpendicular bisector of $C_{1} K$, and $O_{2} M_{C}$ is the perpendicular bisector of $A_{1} K$. Therefore, $M_{A} M_{C} = \frac{1}{2} A_{1} C_{1}$, and $O_{1} O_{2} \geq M_{A} M_{C}$, with equality holding if and only if $O_{1} O_{2} \| A_{1} C_{1}$.

The line of centers $O_{1} O_{2}$ is perpendicular to the common chord $K P$, so we need to find the position of point $P$ such that lines $K P$ and $A_{1} C_{1}$ are perpendicular. In this case, $\angle C_{1} K P = 90^{\circ}$. Quadrilateral $A C_{1} K P$ is cyclic, hence $\angle B A P = 90^{\circ}$. Therefore, the distance between the centers of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$ is minimal when point $P$ is diametrically opposite to point $B$.
![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-46.jpg?height=980&width=1942&top_left_y=1128&top_left_x=54)

## Answer

$P$ is diametrically opposite to point $B$.