[ Tangent Circles ]

In a right-angled triangle, the hypotenuse is equal to c. The centers of three circles with radius $\frac{c}{5}$ are located at its vertices. Find the radius of the fourth circle that is tangent to the three given circles and does not contain them inside itself.
The center of the desired circle is equidistant from the vertices of the triangle.

## Solution

Let \(ABC\) be the given triangle, \(\angle C = 90^\circ\), \(O\) be the center of the desired circle, and \(R\) be its radius. Then,

\[
O A = O B = O C = R + \frac{c}{5}
\]

Therefore, \(O\) is the midpoint of the hypotenuse \(AB\). Consequently,

\[
R + \frac{c}{5} = \frac{c}{2}, \quad R = \frac{3c}{10}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_76207e7feb35677daaaeg-48.jpg?height=363&width=544&top_left_y=2551&top_left_x=757)

## Problem