Let's inscribe a rectangle with the largest possible area in a circular segment, and do so in such a way that its base is parallel to the chord of the circular segment. Let's examine separately the cases where the angle of the circular segment is $2 \alpha=120^{\circ}$ and $2 \alpha=240^{\circ}$.
Let the base of the sought rectangle be $AB = x$, its height $AC = y$; the chord of the circular segment $MN$, the radius perpendicular to it $OE$; furthermore, $\angle AOE = \omega$ and $\angle MOE = \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f9bef98081e5efe38185g-1.jpg?height=515&width=593&top_left_y=228&top_left_x=755)

Then

$$
x = 2r \sin \omega
$$

From triangle $OAC$ we have

$$
\frac{y}{r} = \frac{\sin (\alpha - \omega)}{\sin \alpha}
$$

and thus

$$
y = \frac{r \sin (\alpha - \omega)}{\sin \alpha}
$$

Therefore, the area of the rectangle is:

$$
T = \frac{2r^2 \sin (\alpha - \omega) \sin \omega}{\sin \alpha}
$$

Since $\frac{2r^2}{\sin \alpha}$ is a constant expression, $T$ is maximized when $p = \sin (\alpha - \omega) \sin \omega$ is maximized.

But

$$
\begin{gathered}
p = \frac{1}{2} \cos [(\alpha - \omega) - \omega] - \frac{1}{2} \cos [(\alpha - \omega) + \omega] \\
= \frac{1}{2} [\cos (\alpha - 2\omega) - \cos \alpha]
\end{gathered}
$$

This expression is maximized when $\cos (\alpha - 2\omega)$ is maximized, i.e., when

$$
\cos (\alpha - 2\omega) = 1
$$

and thus

$$
\omega = \frac{\alpha}{2}
$$

Therefore, the intersection of the line bisecting $\alpha$ with the arc of the circular segment gives one vertex of the sought rectangle.

If

$$
2\alpha = 120^\circ, \text{ then } x = r, y = \frac{r}{\sqrt{3}} \text{ and thus } T = \frac{r^2}{3} \sqrt{3}
$$

If

$$
2\alpha = 240^\circ, \text{ then } x = r \sqrt{3}, y = r \text{ and thus } T = r^2 \sqrt{3}
$$

(Géza Sasvári.)

The problem was also solved by: J. Filkorn, E. Freibauer, A. Herzog, B. Kohn, Ö. Kornis, B. Krausz, Gy. Krisztián, Gy. Lukhaub, J. Neumann, R. Obláth, J. Prokoláb, J. Weisz.