For a square with side $a$, draw quarter-circle arcs with radius $a$ around two adjacent vertices. What is the area that is not covered by the two quarter-circles within the square?
I. Solution: Subtract the area of the two quarter-circles from the area of the square with side length $a$. In this case, we subtract the area of the curved region bounded by the two $60^{\circ}$ arcs $A P B$ twice. To get the desired area $t$, we need to add the area of the $A P B$ region to the previous area difference (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_98fafdcd2c2ddb0cd74ag-1.jpg?height=507&width=510&top_left_y=305&top_left_x=775)

The area of the curved region $A P B$ is obtained by subtracting the area of the equilateral triangle $A B P$, $\frac{a \cdot a \cdot \sin 60^{\circ}}{2}=\frac{a^{2} \sqrt{3}}{4}$, from the area of the two sixth-circles, $2 \cdot \frac{a^{2} \pi}{6}$.

Based on this, the desired area is

$$
\begin{gathered}
t=a^{2}-2 \cdot \frac{a^{2} \pi}{4}+\left(2 \cdot \frac{a^{2} \pi}{6}-\frac{a^{2} \sqrt{3}}{4}\right)=a^{2}\left(1-\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)= \\
=a^{2} \cdot \frac{12-2 \pi-3 \sqrt{3}}{12} \sim 0.043 a^{2}
\end{gathered}
$$

From this, it can be seen that the desired area is approximately 4.3% of the area of the entire square.

András Pinkert (Bp., VIII., Széchenyi g. I. o. t.)

II. Solution: The area $t$ can be obtained more simply by subtracting the area of the equilateral triangle $A P B$ and the areas of the twelfth-circles $A P D$ and $B P C$ from the area of the square:

$$
t=a^{2}-a^{2} \frac{\sqrt{3}}{4}-2 a^{2} \frac{\pi}{12}=a^{2} \frac{12-3 \sqrt{3}-2 \pi}{12}
$$

Erzsébet Hild (Békéscsaba, Lorántffy Zs. lg. II. o. t.)