On the sides of the square $ABCD$ - outside the square - we construct the equilateral triangles $ABE, BCF, CDG, DAH$, and then over the diagonal $AC$ the equilateral triangles $ACM$ and $ACP$, and over the diagonal $BD$ the equilateral triangles $BDN$ and $BDQ$. Calculate the areas of the figures $EFGH$, $MNPQ$, and the area of their common part!
$1^{0}$. The $E F G H$ is a square. Indeed, its diagonals, $E G$ and $F H$, are equal and bisect each other perpendicularly at the center $O$ of the square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d3b068c183d6d343966cg-1.jpg?height=637&width=637&top_left_y=256&top_left_x=733)

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$$
O E=\frac{a}{2}+\frac{a}{2} \sqrt{3}=\frac{a}{2}(\sqrt{3}+1)
$$

The area of the square $EFGH$ is the sum of the areas of four congruent right-angled isosceles triangles, that is,

$$
\begin{gathered}
t_{1}=4 \cdot \frac{\overline{O E}^{2}}{2}=2 \overline{O E}^{2}=2 \cdot \frac{a^{2}}{4}(\sqrt{3}+1)^{2}= \\
=a^{2}(2+\sqrt{3}) \sim 3.732 a^{2}
\end{gathered}
$$

$2^{0}$. The $M N P Q$ is also a square, whose diagonals $M P$ and $N Q$ are equal and bisect each other perpendicularly at the point $O$.

Half of the diagonal:

$$
O M=A C \frac{\sqrt{3}}{2}=A B \sqrt{2} \frac{\sqrt{3}}{2}=\frac{a}{2} \sqrt{6}
$$

The area of $M N P Q$ is

$$
t_{2}=4 \cdot \frac{\overline{O M}^{2}}{2}=2 \overline{O M}^{2}=2\left(\frac{a}{2} \sqrt{6}\right)^{2}=2 \frac{a^{2}}{4} \cdot 6=3 a^{2}
$$

$3^{0}$. The area of the common part of the two squares can be obtained by subtracting the sum of the areas of the four congruent right-angled isosceles triangles that lie outside the other square from the area of the square $M N P Q$. If the height of one such triangle, for example, $M R T$, is $h$, then its area is $2 \frac{h^{2}}{2}=h^{2}$.

The distance of the side $E F$ from $O$ $=\frac{E F}{2}=\frac{O E \sqrt{2}}{2}=\frac{a(\sqrt{3}+1)}{2} \cdot \frac{\sqrt{2}}{2}$.

Thus $\quad h=\frac{a \sqrt{6}}{2}-\frac{a(\sqrt{3}+1)}{2 \sqrt{2}}=\frac{a}{2 \sqrt{2}}(2 \sqrt{3}-\sqrt{3}-1)=\frac{a}{2 \sqrt{2}}(\sqrt{3}-1)$.

The sum of the areas of the four triangles:

$$
4 h^{2}=\frac{a^{2}}{2}(\sqrt{3}-1)^{2}=\frac{a^{2}}{2}(3-2 \sqrt{3}+1)=a^{2}(2-\sqrt{3})
$$

The area of the common part of the two squares:

$$
t_{3}=t_{2}-4 h^{2}=3 a^{2}-a^{2}(2-\sqrt{3})=a^{2}(\sqrt{3}+1) \sim 2.732 a^{2}
$$

Lajos Répás (Baross Gábor g. VI. o. Szeged.)