A $2 a$-sided square, from each of its vertices as the center, we draw quarter-circle arcs within the square with radius $a$. In the quadrilateral thus formed by the curved lines, we construct a square whose sides are tangents to the aforementioned arcs. The area of this square depends on the direction of its sides. Investigate this accordingly and also illustrate the variation in the area of the square!

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The translation maintains the original text's formatting and structure, including the mathematical notation and the request for investigation and illustration.
Given a square $OB=2a$ with center $C$, construct a perpendicular from $C$ to the side $NP$ of the inscribed square $MNPQ$, and also to the parallel side $OD$, which forms an angle $DOB=x$ with $OB$ and determines the direction of the sides of the square $MNPQ$. Thus, $OD \perp MN$, so $OD$ passes through the point of tangency $A$ of the side $MN$; $OA=a$. The point $C$ is also the center of the square $MNPQ$; therefore, $CD$ bisects $NP$ at point $E$. Let $y$ be the side length of the inscribed square:

\[
y = NP = 2NE = 2AD = 2(OD - OA)
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_52890ba5e0cad85940dag-1.jpg?height=447&width=423&top_left_y=476&top_left_x=840)

Since $COB = \frac{\pi}{4}$, then $COD = \pm\left(\frac{\pi}{4} - x\right)$, and from the right triangle $OCD$:

\[
\begin{aligned}
& OD = OC \cos\left(\frac{\pi}{4} - x\right) = a\sqrt{2} \cos\left(\frac{\pi}{4} - x\right) \\
& y = 2\left[a\sqrt{2} \cos\left(\frac{\pi}{4} - x\right) - a\right] = \\
& = 2a\left[\sqrt{2} \cos\left(\frac{\pi}{4} - x\right) - 1\right]
\end{aligned}
\]

The area of the square:

\[
y^2 = 4a^2\left[\sqrt{2} \cos\left(\frac{\pi}{4} - x\right) - 1\right]^2
\]

where

\[
0 \leq x \leq \frac{\pi}{2}
\]

For $x=0$, $\cos\left(\frac{\pi}{4} - 0\right) = \frac{\sqrt{2}}{2}$ and $y^2 = 0$.

\[
x = \frac{\pi}{2} \quad \text{also gives} \quad y^2 = 0
\]

As $x$ increases, $\frac{\pi}{4} - x$ decreases, and $\cos\left(\frac{\pi}{4} - x\right)$ increases until $\cos\left(\frac{\pi}{4} - x\right) = 1$, i.e., $\frac{\pi}{4} - x = 0$ or $x = \frac{\pi}{4}$.

If $x$ continues to increase, $\cos\left(\frac{\pi}{4} - x\right)$ decreases and takes the same values between $\frac{\pi}{4}$ and $\frac{\pi}{2}$ as between $0$ and $\frac{\pi}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_52890ba5e0cad85940dag-1.jpg?height=374&width=528&top_left_y=2082&top_left_x=793)

Similarly, the area of the square increases from 0 to a maximum value and this value is:

\[
y_{\max}^2 = 4a^2(\sqrt{2} - 1)^2 = 4a^2(3 - 2\sqrt{2}) \sim 0.686a^2
\]

This value is reached when $x = \frac{\pi}{4}$, i.e., when the sides of the inscribed square are parallel to the diagonals of the square $OB$.

If $x$ continues to increase from $\frac{\pi}{4}$ to $\frac{\pi}{2}$, the area of the inscribed square decreases through the same values back to 0.

Géza Freud (Berzsenyi g. VI. o. Bp. V.).