If the radius of a circle is $r$, and the sides of an arbitrary right-angled quadrilateral are $x$ and $y$, then the perimeter of this quadrilateral is:

$$
k=2(x+y)
$$

but

$$
x^{2}+y^{2}=4 r^{2}
$$

so

$$
\frac{k}{2}=x+\sqrt{4 r^{2}-x^{2}}
$$

or

$$
\begin{aligned}
& \frac{k^{2}}{4}-k x+x^{2}=4 r^{2}-x^{2} \\
& 2 x^{2}-k x+\frac{k^{2}}{4}-4 r^{2}=0
\end{aligned}
$$

from which

$$
x=\frac{k \pm \sqrt{32 r^{2}-k^{2}}}{4}
$$

$x$ can only be real if

$$
32 r^{2} \geqq k^{2}
$$

so the maximum value of $k^{2}$ is $32 r^{2}$, and thus the maximum value of $k$ is: $4 r \sqrt{2}$; in this case

$$
x=\frac{k}{4}=r \sqrt{2}
$$

and

$$
y=\frac{k}{2}-\frac{k}{4}=\frac{k}{4}=x
$$

that is, among the right-angled quadrilaterals inscribed in a circle, the square has the largest perimeter.

(Szabó István.)

Number of solutions: 39.
Among all the rectangles inscribed in a circle, which one has the largest perimeter?