Let $ABC$ be a triangle. On its sides $AB$ and $BC$ are fixed points $C_1$ and $A_1$, respectively. Find a point $ P$ on the circumscribed circle of triangle $ABC$ such that the distance between the centers of the circumscribed circles of the triangles $APC_1$ and $CPA_1$ is minimal.
Let $K$ be the second intersection point of circumcircles of triangles $BPC_1$ and $PA_1C$. Also, let $O_1$ and $O_2$ are circumcenters of triangles $C_1BP$ and $A_1CP$
First we prove that the points $A_1$, $K$ and $C_1$ are collinear. $$\angle C_1KA_1=\angle C_1KP+\angle A_1KP=180-\angle C_1BP+180-\angle PCA_1=180$$. Now, let $M$ and $N$ be the midpoints of $C_1K$ and $A_1K$ respectively. Obviously, $O_1MNO_2$ is a right angled trapezoid. So $O_1O_2 \geq MN= \frac{A_1C_1}{2}$. We have found the minimal distance between $O_1$ and $O_2$ which is $\frac{A_1C_1}{2}$ and the equality occurs if and only if $O_1O_2$ is parallel to $MN$.
Now let's determined where point $P$ should be. We know that $O_1O_2\perp PK$ therefore we get $\angle MKP=90$. But quadrilateral $C_1KBP$ hence we have $\angle ABP=90$.Now it's obvious that P is the intersection of $BO$ with circumcircle of $ABC$ where $O$ is the circumcircle of $ABC$