Task 2. (10 points) A circle touches the extensions of two sides $AB$ and $AD$ of square $ABCD$, and the point of tangency cuts off a segment of length $\sqrt{2}+\sqrt{2-\sqrt{2}}$ cm from vertex $A$. Two tangents are drawn from point $C$ to this circle. Find the side of the square if the angle between the tangents is $45^{\circ}$, and it is known that $\sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-31.jpg?height=451&width=576&top_left_y=1459&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. If radii of the circle are drawn to the points of tangency, a square is formed with side $\sqrt{2}+\sqrt{2-\sqrt{2}}$ cm. Then $O A=\sqrt{2}(\sqrt{2}+\sqrt{2-\sqrt{2}})$, $O C=\frac{O K}{\sin 22.5^{\circ}}=\frac{2(\sqrt{2}+\sqrt{2-\sqrt{2}})}{\sqrt{2-\sqrt{2}}}$, then the diagonal of the square $A B C D$

$$
A C=\frac{2(\sqrt{2}+\sqrt{2-\sqrt{2}})}{\sqrt{2-\sqrt{2}}}-\sqrt{2}(\sqrt{2}+\sqrt{2-\sqrt{2}})
$$

and the side of the square $A B=\frac{A C}{\sqrt{2}}=\frac{\sqrt{2}(\sqrt{2}+\sqrt{2-\sqrt{2}})}{\sqrt{2-\sqrt{2}}}-\sqrt{2}-\sqrt{2-\sqrt{2}}=$

$$
=\frac{(\sqrt{2}+\sqrt{2-\sqrt{2}})(\sqrt{2}-\sqrt{2-\sqrt{2}})}{\sqrt{2-\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{2-\sqrt{2}}}=\sqrt{2+\sqrt{2}}
$$

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