[Pythagorean Theorem (direct and inverse).] Circle, sector, segment, etc. $\quad]$

The radius of the circle with center $O$ is $2 \sqrt{5}$. In the sector $A O B$ with an angle of $45^{\circ}$, a rectangle $K L M N$ is inscribed. Side $K L$ is located on segment $O A$, vertex $M$ is on the arc $A B$, and vertex $N$ is on segment $O B$. Find the sides of the rectangle if one of them is twice the other. radius.
Let $K L=2 K N$. Denote $K N=L M=x$. From the right triangles $O K N$ and $O L M$, we find that

$$
O K=K N=x, O M 2=O L 2+L M 2=(O K+K L) 2+L M 2=(x+2 x) 2+x 2=9 \times 2+x 2=10 \times 2,
$$

and since $O M 2=(2 \sqrt{5}) 2=40$, then $10 x 2=40$. From this, we find that $x=2$. Therefore, $K N=L M=2$ and $K L=M N=4$.

If, however, $K N=2 K L$, then similarly we get that $K N=L M=4 \sqrt{\frac{5}{13}}$ and $K L=M N=2 \sqrt{\frac{5}{13}}$.

## Answer

$2 ; 4$ or $4 \sqrt{\frac{5}{13}} ; 2 \sqrt{\frac{5}{13}}$.