Dolmatov $C$.

Point $M$ lies on the side $A C$ of an acute-angled triangle $A B C$. Circles are circumscribed around triangles $A B M$ and $C B M$. For what position of point $M$ will the area of the common part of the circles bounded by them be the smallest?
The angles under which the segment $BM$ is seen from the centers of the circles have a constant magnitude.

## Solution

Let $O$ and $O_{1}$ be the centers of the circumcircles of triangles $ABM$ and $CBM$. The common part of the specified circles is the union of two segments with a common chord $BM$.

Since

$$
\angle BOM = 2 \angle BAC, \angle BO_{1}M = 2 \angle BCA,
$$

the magnitudes of the angles $BOM$ and $BO_{1}M$ are constant. Therefore, the area of each segment is smaller the shorter the chord $BM$. Consequently, the area of the common part of the circles is minimal when $BM$ is the altitude of triangle $ABC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-25.jpg?height=383&width=920&top_left_y=31&top_left_x=572)

## Answer

The point $M$ is the foot of the altitude of triangle $ABC$.