## Task A-2.2. (4 points)

Let $A B C D$ be a square with side length 1. Circle $k$ has a radius of 1 and its center is at point $C$. Determine the radius of circle $k_{1}$ which is tangent to circle $k$ and the segments $\overline{A B}$ and $\overline{A D}$.
## Solution.

Let $C_{1}$ be the center of the circle $k_{1}$, and let $A_{1}, B_{1}, D_{1}$ be the points of tangency of $k_{1}$ with the circle $k$, the line segment $\overline{A B}$, and the line segment $\overline{A D}$, respectively. Let $r$ be the desired radius of the circle $k_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-09.jpg?height=694&width=717&top_left_y=1472&top_left_x=675)

Then,

$$
\left|A_{1} C_{1}\right|=\left|B_{1} C_{1}\right|=\left|D_{1} C_{1}\right|=r
$$

and $A B_{1} C_{1} D_{1}$ is a square with side length $r$.

Since $|A C|=\left|A C_{1}\right|+\left|C_{1} A_{1}\right|+\left|A_{1} C\right|$

we get $\sqrt{2}=r \sqrt{2}+r+1$

and finally

$$
r=\frac{\sqrt{2}-1}{\sqrt{2}+1}=3-2 \sqrt{2}
$$