Given a triangle $A B C$. Find a point $M$ on the line $A B$ such that the sum of the radii of the circumcircles of triangles $A C M$ and $B C M$ is minimized.
According to the Law of Sines, the radii of the circumcircles of triangles $A C M$ and $B C M$ are $A C /(2 \sin A M C)$ and $B C /(2 \sin B M C)$, respectively. It is easy to verify that $\sin A M C=\sin B M C$. Therefore, $A C /(2 \sin A M C)+B C /(2 \sin B M C)=$ $(A C+B C) /(2 \sin B M C)$. The last expression will be the smallest if $\sin B M C=1$, i.e., $C M \perp A B$.

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