In a circular segment with a central angle of $2 \alpha$ and radius $r$, inscribe a rectangle of maximum area.
We distinguish three cases:
a)
$0^{\circ}<\alpha \leq 90^{\circ}$
b)
$90^{\circ}<\alpha<135^{\circ}$,
c)
$135^{\circ} \leq \alpha \leq 180^{\circ}$.

a) The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_0eb9596e6306618cd18dg-1.jpg?height=423&width=599&top_left_y=463&top_left_x=752)

Figure 1

The area of the rectangle inscribed in the circular segment is

$$
T=2 x(y-r \cos \alpha)=2 \sqrt{r^{2}-y^{2}}(y-r \cos \alpha)
$$

Consider the function $t=\frac{T^{2}}{4}$ instead of $T$, which clearly has its maximum at the same point:

$$
t=(r-y)(r+y)(y-r \cos \alpha)(y-r \cos \alpha)
$$

Multiplying the first factor by $m$ and the second by $n$, the sum of the factors is:

$$
\begin{gathered}
m(r-y)+n(r+y)+(y-r \cos \alpha)+(y-r \cos \alpha)= \\
=(m+n-2 \cos \alpha) r-(m-n-2) y
\end{gathered}
$$

which is independent of $y$ if

$$
m-n-2=0
$$

By the geometric mean theorem, the product of the four factors is maximized when the factors are equal, i.e.,

$$
m(r-y)=y-r \cos \alpha \quad \text { and } \quad n(r+y)=y-r \cos \alpha
$$

Solving the equations under (2) for the constants $m, n$, and substituting the obtained values into (1), we get

$$
\frac{y-r \cos \alpha}{r-y}-\frac{y-r \cos \alpha}{r+y}-2=0
$$

or

$$
2 y^{2}-y r \cos \alpha-r^{2}=0
$$

From this, taking only the positive root,

$$
y=\frac{r \cos \alpha+r \sqrt{\cos ^{2} \alpha+8}}{4}=\frac{r}{4}\left(\cos \alpha+\sqrt{8+\cos ^{2} \alpha}\right)
$$

The corresponding value of $x$ is

$$
x=\sqrt{r^{2}-y^{2}}
$$

From (3),

$$
y^{2}=\frac{r^{2}}{16}\left(8+2 \cos ^{2} \alpha+2 \cos \alpha \sqrt{8+\cos ^{2} \alpha}\right)
$$

and thus

$$
r^{2}-y^{2}=\frac{r^{2}}{16}\left(8-2 \cos ^{2} \alpha-2 \cos \alpha \sqrt{8+\cos ^{2} \alpha}\right)
$$

so

$$
x=\sqrt{r^{2}-y^{2}}=\frac{r}{4} \sqrt{8-2 \cos ^{2} \alpha-2 \cos \alpha \sqrt{8+\cos ^{2} \alpha}}
$$

The area of the rectangle with the maximum area is

$$
T_{\max }=2 x(y-r \cos \alpha)
$$

From (3),

$$
y-r \cos \alpha=\frac{-3 r \cos \alpha+r \sqrt{8+\cos ^{2} \alpha}}{4}=\frac{r}{4}\left(-3 \cos \alpha+\sqrt{8+\cos ^{2} \alpha}\right)
$$

therefore

$$
T_{\max }=\frac{r^{2}}{8}\left(-3 \cos \alpha+\sqrt{8+\cos ^{2} \alpha}\right) \sqrt{8-2 \cos ^{2} \alpha-2 \cos \alpha \sqrt{8+\cos ^{2} \alpha}}
$$

b) If $90^{\circ}<\alpha<135^{\circ}$, there are rectangles inscribed in the circular segment whose all four vertices lie on the arc of the segment (e.g., the dashed rectangle in Figure 2). The area of such a rectangle is

![](https://cdn.mathpix.com/cropped/2024_05_02_0eb9596e6306618cd18dg-2.jpg?height=564&width=655&top_left_y=691&top_left_x=716)

Figure 2

$$
T^{\prime}=2 r \sin \beta \cdot 2 r \cos \beta=2 r^{2} \sin 2 \beta
$$

where $0<\beta \leq \alpha-90^{\circ}<45^{\circ}$.

$T^{\prime}$ is maximized when $\beta=\alpha-90^{\circ}<45^{\circ}$, meaning we need to find the rectangle with the maximum area among those whose one side lies on the chord. In this case, the procedure used in (a) can be followed. We will arrive at the same results, but $\cos \alpha$ will naturally be negative everywhere.

c) If $135^{\circ} \leq \alpha \leq 180^{\circ}$, the circular segment contains the square inscribed in the circle, which, as is known and can be read from (4), is the rectangle with the maximum area inscribed in the circle. In this case, generally, an infinite number of squares can be drawn in the circular segment, except for the boundary case $\alpha=135^{\circ}$, when the square drawn on the chord as a side is the only solution.