Sharygin I.F.

Given an acute-angled triangle $A B C$. A line parallel to $B C$ intersects sides $A B$ and $A C$ at points $M$ and $P$ respectively. For what position of points $M$ and $P$ will the radius of the circumcircle of triangle $B M P$ be the smallest?
For any position of the line $M P \angle B M P=180^{\circ}-\angle B$.

Using the corollary of the Law of Sines, we find the radius of the circumcircle of $\triangle B M P: R=\frac{B P}{2 \sin \angle B}$. It takes the smallest value simultaneously with the length of the segment $B P$, that is, if $B P \perp A C$ (see figure).

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