Let's write down the area of the largest square that can be inscribed in a circular sector with a unit radius and an opening angle of $\varphi$, as a function of $\varphi$! $\left(0<\varphi \leq 180^{\circ}\right.$; the vertices of the square should lie on the boundary lines of the circular sector.)
Solution: It is more obvious to inscribe a square $KLMN$ such that the midpoint of its side $PQ$ lies on the axis of symmetry of the circular segment $OAB$ (Figure 1), but when $\varphi \leq 90^{\circ}$, it is also possible that one vertex of the square falls on the arc of the circular segment, and two or one vertex falls on the bounding radii (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_4fa764ea28eeb4eeefddg-1.jpg?height=1094&width=638&top_left_y=298&top_left_x=732)

Accordingly, after writing the area $t_{1}(\varphi)$ and $t_{2}(\varphi)$ of the square with side lengths $a_{1}$ and $a_{2}$ as functions of $\varphi$ for these two possibilities, for $0^{\circ} < \varphi \leq 30^{\circ}$, or $90^{\circ} \geq \varphi > 30^{\circ}$, $k_{1}(x) = k(\varphi)$ is positive, so $t_{1}$ is smaller than $t_{2}$, meaning the area of the non-symmetrically inscribed square is larger for a circular segment with such an opening angle. - However, for $x = \operatorname{ctg} \frac{\varphi}{2} > 2 + \sqrt{3}$, or $\varphi / 2 < 90^{\circ}$, only the first method of inscribing can be used, in which case $t_{1}$ gives the desired function. - Therefore, the desired function $t(\varphi)$ can be written as follows:

$$
t(\varphi)=\left\{\begin{array}{l}
\frac{4}{\left(\operatorname{ctg} \frac{\varphi}{2} + 2\right)^{2} + 1}, \text{ if } 0^{\circ} < \varphi \leq 30^{\circ} \text{ and if } 90^{\circ} < \varphi \leq 180^{\circ} \\
\frac{1}{(\operatorname{ctg} \varphi + 1)^{2} + 1}, \text{ if } 30^{\circ} \leq \varphi \leq 90^{\circ}
\end{array}\right.
$$

For $\varphi = 30^{\circ}$, both formulas are valid.

Gábor Várady (Győr, Révai M. g. III. o. t.)

Note. Observe that at $90^{\circ}$, $t\left(90^{\circ}\right) = t_{2}\left(90^{\circ}\right)$ is larger than $t_{1}\left(90^{\circ}\right)$, and for larger angles, $t(\varphi)$ connects to this smaller value. At first glance, it may seem surprising that a larger square can be cut from a $90^{\circ}$ circular segment than from a $100^{\circ}$ one; however, do not forget the requirement that the vertices of the square must lie on the perimeter of the circular segment.

[^0]:    ${ }^{1}$ It can be shown that no other suitable method of inscribing exists.