A circle of radius $1+\sqrt{2}$ is circumscribed around an isosceles right triangle. Find the radius of the circle that touches the legs of this triangle and internally touches the circle circumscribed around it.
Let $M$ and $N$ be the points of tangency of the desired circle with the legs $A C$ and $B C$ of triangle $A B C$, and $Q$ be the center of this circle. Then the quadrilateral $Q M C N$ is a square.

## Solution

Let $r$ be the radius of the desired circle, $Q$ its center, $M, N$ the points of tangency with the legs $A C$ and $B C$ of triangle $A B C$, $O$ the center of the circumscribed circle, and $P$ the point of tangency of the circles. Then the points $O$ and $Q$ lie on the diameter $C P$, and the quadrilateral $Q M C N$ is a square with side $r$. Therefore,

$$
C Q = r \sqrt{2}, \quad C Q + Q P = C P, \quad r \sqrt{2} + r = 2(1 + \sqrt{2})
$$

From this, we find that

$$
r = \frac{2(1 + \sqrt{2})}{\sqrt{2} + 1} = 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-01.jpg?height=538&width=580&top_left_y=1409&top_left_x=745)

## Answer

2.

## Problem