Task 2. (10 points) A circle touches the extensions of two sides $A B$ and $A D$ of square $A B C D$ with side $2-\sqrt{5-\sqrt{5}}$ cm. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle if the angle between the tangents is $72^{\circ}$, and it is known that $\sin 36^{\circ}=\frac{\sqrt{5-\sqrt{5}}}{2 \sqrt{2}}$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-20.jpg?height=451&width=576&top_left_y=1579&top_left_x=817)

Fig. 1

The segment cut off from vertex $A$ by the point of tangency of the circle is equal to the radius of this circle. The diagonal of the square $A B C D \quad A C=\sqrt{2}(2-\sqrt{5-\sqrt{5}})$. If we draw radii of the circle to the points of tangency, a square with side $R$ is formed. Then $\quad O A=R \sqrt{2}, \quad O C=\frac{O K}{\sin 36^{\circ}}=\frac{2 \sqrt{2} R}{\sqrt{5-\sqrt{5}}}$. As a result, we get the equation $\frac{2 \sqrt{2} R}{\sqrt{5-\sqrt{5}}}=R \sqrt{2}+\sqrt{2}(2-\sqrt{5-\sqrt{5}})$, from which

$$
R=\frac{2-\sqrt{5-\sqrt{5}}}{\frac{2}{\sqrt{5-\sqrt{5}}}-1}=\frac{\sqrt{5-\sqrt{5}}(2-\sqrt{5-\sqrt{5}})}{2-\sqrt{5-\sqrt{5}}}=\sqrt{5-\sqrt{5}}
$$

Answer. $\sqrt{5-\sqrt{5}}$.