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 $\sqrt{2+\sqrt{2}}$ cm. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle if the angle between the tangents is $45^{\circ}$, and it is known that $\sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-02.jpg?height=446&width=574&top_left_y=1613&top_left_x=815)

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 $A B C D A C=\sqrt{4+2 \sqrt{2}}$. If radii of the circle are drawn to the points of tangency, a square with side $R$ is formed. Then $O A=R \sqrt{2}, \quad O C=\frac{O K}{\sin 22.5^{\circ}}=\frac{2 R}{\sqrt{2-\sqrt{2}}}$. As a result, we get the equation $\frac{2 R}{\sqrt{2-\sqrt{2}}}=R \sqrt{2}+\sqrt{4+2 \sqrt{2}}$, from which

$$
R=\frac{\sqrt{4+2 \sqrt{2}} \cdot \sqrt{2-\sqrt{2}}}{2-\sqrt{2} \sqrt{2-\sqrt{2}}}=\frac{\sqrt{2+\sqrt{2}} \cdot \sqrt{2-\sqrt{2}}}{\sqrt{2}-\sqrt{2-\sqrt{2}}}=\sqrt{2}+\sqrt{2-\sqrt{2}}
$$

Answer. $\sqrt{2}+\sqrt{2-\sqrt{2}}$.