## Task 2 - 191242

Let $M$ be the set of all square areas $Q$ that lie in a given plane $\epsilon$, have a given point $Z$ of the plane $\epsilon$ as their center, and have a given segment length $a$ as their side length.

For arbitrary square areas $Q, Q^{\prime}$ from this set $M$, let $u\left(Q \cup Q^{\prime}\right)$ denote the perimeter of the polygonal area that results from the intersection of the square areas $Q$ and $Q^{\prime}$.

Investigate whether there are square areas $Q, Q^{\prime}$ in $M$ with the smallest possible $u\left(Q \cap Q^{\prime}\right)$.

If this is the case, determine (in dependence on $a$) this smallest possible value of $u\left(Q \cap Q^{\prime}\right)$.
For arbitrary $Q, Q^{\prime} \in M$, the following holds: If $Q=Q^{\prime}$, then $Q \cap Q^{\prime}$ is the square area $Q$ with perimeter $4a$. If $Q \neq Q^{\prime}$, then for any side $AB$ of the square area $Q$, the vertices of $Q$ and $Q^{\prime}$ can be labeled as $A, B, C, D$ and $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$, respectively, such that the points $A, A^{\prime}, B, D^{\prime}, C, C^{\prime}, D, B^{\prime}$ are arranged in this order on the common circumcircle of $Q$ and $Q^{\prime}$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2082.jpg?height=645&width=654&top_left_y=871&top_left_x=701)

According to this, the angle $\angle A Z A^{\prime}$ has a size $\alpha$ with $0^{\circ}<\alpha<90^{\circ}$. By reflecting at the angle bisector $w$ of this angle, $A$ is mapped to $A^{\prime}$, and thus $Q^{\prime}$ is mapped to $Q$. Consequently, due to the direction of traversal, $B$ is mapped to $B^{\prime}$, $C$ to $C^{\prime}$, $D$ to $D^{\prime}$, and the intersection point $P_{1}$ of $AB$ and $A^{\prime}B^{\prime}$, which exists because $0^{\circ}<\alpha<90^{\circ}$, is mapped to itself. Therefore, $P_{1}$ lies on $w$.

The segments $AB$ and $A^{\prime}B^{\prime}$ intersect at the point $P_{1}$ on $AB$ for which $\angle A Z P_{1}=\frac{\alpha}{2}$. Similarly, it follows that:

$AB$ and $D^{\prime}A^{\prime}$ intersect at the point $P_{2}$ on $AB$ for which $\angle B Z P_{2}=\frac{90^{\circ}-\alpha}{2}=45^{\circ}-\frac{\alpha}{2}$.

Since $\angle A Z P_{2}=\frac{\alpha}{2}+45^{\circ}$, and because $0^{\circ}<\frac{\alpha}{2}<\frac{\alpha}{2}+45^{\circ}<90^{\circ}$, the segment $P_{1}P_{2}$ is thus a side of the polygonal area $Q \cap Q^{\prime}$.

Similarly, we find the further intersection points $P_{3}, P_{4}$ of $BC$ with $D^{\prime}A^{\prime}$ and $C^{\prime}D^{\prime}$, respectively, $P_{5}, P_{6}$ of $CD$ with $C^{\prime}D^{\prime}$ and $B^{\prime}C^{\prime}$, and $P_{7}, P_{8}$ of $DA$ with $B^{\prime}C^{\prime}$ and $A^{\prime}B^{\prime}$, and thus $Q \cap Q^{\prime}$ as the octagonal area $P_{1}P_{2} \ldots P_{8}$.

By reflecting at $w$, the intersection point $P_{2}$ of $AB$ and $D^{\prime}A^{\prime}$ is mapped to the intersection point of $A^{\prime}B^{\prime}$ and $DA$, i.e., to $P_{8}$. Therefore, $P_{1}P_{2}=P_{1}P_{8}$.

Similarly, it follows that any two adjacent sides of $Q \cap Q^{\prime}$ have the same length $s=P_{1}P_{2}$, so $u(Q \cap Q^{\prime})=8s$.

Because $\angle A^{\prime}Z^{\prime}B^{\prime}=90^{\circ}$ and $0^{\circ}<\alpha<90^{\circ}$, the segments $ZA$ and $A^{\prime}B^{\prime}$ intersect at a point $R$, and it holds that $\angle A R P_{8}=\angle A^{\prime} R Z$ (vertical angles) and $\angle R A P_{8}=45^{\circ}=\angle R A^{\prime} Z$. Therefore, $\angle P_{1} P_{8} A=\angle R P_{8} A=\alpha$ (angle sum in the triangle), and because $\angle P_{1} A P_{8}=90^{\circ}$, we have $A P_{1}=s \cdot \sin \alpha$.

Similarly, we obtain $\angle P_{2} P_{3} B=90^{\circ}-\alpha$, so $P_{2} B=s \cdot \cos \alpha$. From

$$
\begin{gathered}
a=A P_{1}+P_{1} P_{2}+P_{2} B=s(\sin \alpha+1+\cos \alpha) \quad \text { it follows that } \\
s=\frac{a}{1+\sin \alpha+\cos \alpha}
\end{gathered}
$$

Now,

$$
\sin \alpha+\cos \alpha=\sqrt{2}\left(\cos \alpha \cdot \cos 45^{\circ}+\sin \alpha \cdot \sin 45^{\circ}\right)=\sqrt{2} \cos \left(\alpha-45^{\circ}\right)
$$

because $-45^{\circ}<\alpha-45^{\circ}<45^{\circ}$, it attains its maximum value for exactly $\alpha-45^{\circ}=0$, and this value is $\sqrt{2}$. Therefore, for all $Q, Q^{\prime} \in M$ with $Q \neq Q^{\prime}$, there exists a smallest possible value of $u(Q \cap Q^{\prime})=8s$, and this value is

$$
\frac{8a}{1+\sqrt{2}}=8a(\sqrt{2}-1)
$$

Since $\sqrt{2}-1<\frac{1}{2}$, and thus $8a(\sqrt{2}-1)<4a$, this is also the smallest possible value of $u(Q \cap Q^{\prime})$ for all $Q, Q^{\prime} \in M$ even when $Q=Q^{\prime}$ is allowed.

## Adapted from $[3]$