In the diagram, a square window design is shown. Each of the four congruent circles passes through one corner of the small square, and two adjacent circles may touch each other. The four circular areas and the small square area are dark, while the rest of the large square is transparent. What should be the side length of the small square so that the total area of the dark parts is a) minimal b) maximal?

![](https://cdn.mathpix.com/cropped/2024_05_02_c68fea59e07dbf36a620g-1.jpg?height=328&width=331&top_left_y=318&top_left_x=886)
Let the side of the larger square be a unit, the side of the smaller one be $x$, and the radius of the circles be $r$. Based on the diagram, it is clear that $\sqrt{2}=2(r \sqrt{2}+r)+x \sqrt{2}$, from which

$$
r=\frac{\sqrt{2}}{2} \cdot \frac{1-x}{\sqrt{2}+1}=\frac{\sqrt{2}}{2} \cdot(\sqrt{2}-1)(1-x)
$$

The total area of the dark regions is:

$$
T(x)=x^{2}+4 \cdot r^{2} \cdot \pi=x^{2}+2 \cdot \pi \cdot(\sqrt{2}-1)^{2} \cdot(1-x)^{2}
$$

and thus

$$
T(x)=(6 \cdot \pi-4 \cdot \pi \cdot \sqrt{2}+1) \cdot x^{2}+(8 \cdot \pi \cdot \sqrt{2}-12 \cdot \pi) \cdot x+(6 \cdot \pi-4 \cdot \pi \cdot \sqrt{2})
$$

According to the conditions of the problem:

$$
\begin{gathered}
0<x<1 \quad \text { and } \\
4 r \leq 1
\end{gathered}
$$

It is easy to see that the coefficient of $x^{2}$ in (2) is positive, so a local minimum is possible in the interval specified by (3). Using the identity

$$
a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a}
$$

the minimum of $T(x)$ is at

$$
x_{0}=-\frac{8 \pi \cdot \sqrt{2}-12 \pi}{2 \cdot(6 \pi-4 \pi \cdot \sqrt{2}+1)} \approx 0.5187
$$

Considering that with this value of $x$ - calculating $r$ from (1) - (3) and (4) are satisfied, there is indeed a minimum at $x_{0}$.

Since the interval in (3) is open, there will be no maximum.

Based on the work of Varga Áron (Budapest, Trefort Ágoston Kéttannyelvű Műsz. Szki. és Gimn., IV. o.t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_3ab6f0525ec894ab411fg-1.jpg?height=345&width=455&top_left_y=1535&top_left_x=824)