Kozhenkov P.A.

Point $M$ lies on side $AB$ of triangle $ABC$, $AM=a$, $BM=b$, $CM=c$, $c<a$, $c<b$. Find the smallest radius of the circumscribed circle of such a triangle.
Let $K$ be the midpoint of $AB$ and $a \geq b$. Since $MK = 1/2(a-b)$, by the triangle inequality, $CK \leq 1/2(a-b) + c < 1/2(a-b) + b = 1/2(a+b) = 1/2 AB$. Therefore, point $C$ lies inside the circle constructed on $AB$ as the diameter; hence, angle $C$ is obtuse.

Let $O$ be the center of the circumcircle $\Omega$ of triangle $ABC$. Since the chord $AB$ is fixed, the radius will be the smallest when angle $AOB$ is the largest. Since $\angle C = 180^\circ - 1/2 \angle AOB$, angle $C$ must be the smallest.

On the other hand, point $C$ lies on the circle with center at point $M$ and radius $c$. We will prove that angle $C$ will be the smallest when this circle is tangent to $\Omega$. Suppose this is not the case and the specified circles intersect at points $C_1$ and $C_2$ (see the right figure). Then, by choosing point $C$ on the smaller arc $C_1C_2$ (outside the larger circle), we get that $\angle C < \angle AC_1B = \angle AC_2B$. Contradiction.
![](https://cdn.mathpix.com/cropped/2024_05_06_a69fc833ac69b1da7555g-37.jpg?height=432&width=1132&top_left_y=1620&top_left_x=470)

Since the circles are tangent, points $O$, $M$, and $C$ lie on the same line (see the left figure). Let the desired radius be $R$. By the theorem on the product of the lengths of the segments of intersecting chords, $c(2R - c) = ab$, from which $R = 1/2(ab/c + c)$.

## Answer

$1/2(ab/c + c)$.