In a given circle, let $AB$ be a side of an inscribed square. Extend $AB$ to $C$ so that $BC = AB$. If we connect $C$ with the center $O$ of the circle and $OC$ intersects the circle at $D$, then $CD$ is equal to twice the side of a regular decagon inscribed in the circle.
First solution. If the extension of $CO$ intersects the circle at $E$ and $CD = x$, then according to a known theorem:

$$
CA \cdot CB = CE \cdot CD
$$

or

$$
2r\sqrt{2} \cdot r\sqrt{2} = (x + 2r) \cdot x
$$

which gives

$$
4r^2 = x^2 + 2rx
$$

and

$$
x = -r \pm r\sqrt{5}
$$

If we take the positive value of the square root, we get

$$
x = r(\sqrt{5} - 1)
$$

(Szobotha Dezsó, Esztergom.)

Second solution. Drop a perpendicular from the center $O$ of the circle to the chord $AB$ and denote the foot of this perpendicular by $F$. Then

$$
\overline{OC}^2 = \overline{OF}^2 + \overline{FC}^2
$$

or

$$
(r + x)^2 = \left(\frac{r\sqrt{2}}{2}\right)^2 + \left(\frac{3r\sqrt{2}}{2}\right)^2
$$

which again gives

$$
x = r(\sqrt{5} - 1)
$$

(Szóllös Hermann, Esztergom.)

Third solution.

$$
\begin{gathered}
\overline{OC}^2 = \overline{OB}^2 + \overline{BC}^2 + 2OB \cdot BC \cdot \cos 135^\circ \\
\overline{OC}^2 = r^2 + 2r^2 + 2r \cdot r\sqrt{2} \cdot \frac{1}{2}\sqrt{2} \\
\overline{OC}^2 = r^2 + 2r^2 + 2r^2 = 5r^2
\end{gathered}
$$

Thus

$$
OC = r\sqrt{5}
$$

and therefore

$$
CD = OC - r = r(\sqrt{5} - 1)
$$

(Pálos Tibor, Budapest.)