In a given circular sector, construct a rectangle of maximum area. (Two vertices of the rectangle lie on the bounding arc, the other two on the bounding radii.)
I. Solution: Let the central angle of the circular segment $CAB$ of a circle with center $C$ and radius $r$ be $2\alpha$. For the rectangle $DEFG$ inscribed in the segment as required, let the vertices $D$ and $E$ lie on the arc $\widehat{AB}$, and the vertices $F$ and $G$ lie on the radii $CB$ and $CA$, respectively. The perpendicular bisector of segment $DE$ is both one of the symmetry axes of the rectangle and thus also the perpendicular bisector of $FG$, and as the perpendicular bisector of a chord, it passes through $C$. Therefore, $CF = CG$, making triangle $CFG$ isosceles, and $f$ bisects the angle $ACB$. Thus, after choosing vertex $D$, the inscribed rectangle is determined by drawing lines parallel and perpendicular to the angle bisector of $ACB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a3834d642cf77bb08f41g-1.jpg?height=515&width=805&top_left_y=396&top_left_x=649)

We need to determine a constructible characteristic of the desired rectangle. Let's choose the angle $\omega$ that the radius $CD$ makes with $f$ for this purpose.

The area of our rectangle is four times the area of triangle $CDG$, because the height corresponding to the base $DG = b$ of this triangle is equal to $DH = a/2$; it is sufficient to determine $D$ such that the area of this triangle is maximized. In this triangle, the side $CD = r$ and its viewing angle at $G$, $\angle CGD = 180^\circ - \alpha$, are constants. Therefore, for a fixed $CD$, the geometric locus of $G$ is the arc of the circle corresponding to the chord $CD$ and the viewing angle $180^\circ - \alpha$. The area of triangle $CDG$ is maximized when $G$ is at the "highest" point of the arc, i.e., when the triangle is isosceles. Then $\angle GCD = (180^\circ - \angle CGD) / 2$, so $\alpha - \omega = \alpha / 2$, $\omega = \alpha / 2$, meaning the point $D$ that maximizes the area is one quarter of the arc $\widehat{AB}$, and can be cut out by the bisector of the angle $ACH$.

$$
\text{András Sárközy (Gyöngyös, Vak Bottyán g. IV. o. t.)}
$$

II. Solution: As a preparatory step for the construction, we can also determine the inscribed rectangle of the largest area by expressing the area as a function of $\omega$ and finding its maximum. The two sides of the rectangle, from the isosceles triangle $CDE$ and the sine rule in triangle $CDG$, are:

$$
a = 2r \sin \omega, \quad \text{and} \quad b \frac{r \sin (\alpha - \omega)}{\sin (180^\circ - \alpha)} = \frac{r \sin (\alpha - \omega)}{\sin \alpha}
$$

Thus, the area, using the known identity $2 \sin x \sin y = \cos (x - y) - \cos (x + y)$ for the product of the two variable angles, is:

$$
t = ab = \frac{r^2}{\sin \alpha} \cdot 2 \sin \omega \sin (\alpha - \omega) = \frac{r^2}{\sin \alpha} [\cos (2\omega - \alpha) - \cos \alpha]
$$

$t$ is maximized when the single variable part of this expression is maximized: $\cos (2\omega - \alpha) = 1$, $2\omega - \alpha = 0^\circ$, $\omega = \alpha / 2$.

Zsolt Rátkay (Bp. VI., Kölcsey F. g. III. o. t.)

Remark. Our diagram, as is customary, shows a circular segment smaller than a semicircle. However, two radii that do not extend into each other divide the circle into two segments, one of which is concave. (On circular percentage graphs, the socialist sector is often represented in this way.) Our result does not give the rectangle of the largest area that can be cut out from the circular segment for concave segments, but the largest one whose two vertices lie on the arc and the other two on the radii or the rays extending from $C$ (depending on whether $2\alpha \leq 240^\circ$ or $2\alpha > 240^\circ$).