Divide the sides of the unit square $ABCD$ into 5 equal parts. The 2nd division point from $A$ on the side $AB$ is denoted by $D'$, and similarly, the 2nd division point from $B$ on the side $BC$, from $C$ on the side $CD$, and from $D$ on the side $DA$ are denoted by $A'$, $B'$, and $C'$, respectively. The lines $AA'$, $BB'$, $CC'$, and $DD'$ form a quadrilateral.

What is the area of the quadrilateral?
We denote the vertices of the resulting quadrilateral as $E, F, G, H$, and the center of the original square as $O$. A $90^{\circ}$ rotation around $O$ maps the points $A, B, C, D, A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ to $B, C, D, A, B^{\prime}, C^{\prime}, D^{\prime}, A^{\prime}$, respectively. For example, the intersection point $F$ of segments $A A^{\prime}$ and $B B^{\prime}$ is mapped to the intersection point $G$ of segments $B B^{\prime}$ and $C C^{\prime}$. Similarly, it can be shown that under this rotation, the points $G, H, E$ are mapped to $H, E$, and $F$, respectively, meaning that the quadrilateral $E F G H$ is mapped onto itself. This implies that $E F G H$ is a square, and its center is $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_95ab49d400f6c2b435b9g-1.jpg?height=485&width=499&top_left_y=392&top_left_x=802)

To calculate the area of the square, it is sufficient to determine the length of one of its sides. In the right triangle $A B A^{\prime}$, $B F$ is the altitude to the hypotenuse $A A^{\prime}$ (since $\angle E F G = 90^{\circ}$), so the triangles $A B A^{\prime}$, $A F B$, and $B F A^{\prime}$ are similar. Therefore, the ratios of corresponding sides of these triangles are equal:

$$
\frac{F A}{A B} = \frac{A B}{A A^{\prime}}
$$

that is, using the fact that by the Pythagorean theorem $A A^{\prime} = \sqrt{1^2 + \left(\frac{2}{5}\right)^2}$,

$$
F A = \frac{1 \cdot 1}{\sqrt{\frac{29}{25}}} = \frac{5 \sqrt{29}}{29}
$$

On the other hand:

$$
\frac{F B}{B A^{\prime}} = \frac{A B}{A A^{\prime}}
$$

so

$$
F B = \frac{\frac{2}{5} \cdot 1}{\sqrt{\frac{29}{25}}} = \frac{2 \sqrt{29}}{29}
$$

But $F B = A E$, so the side length of the small square is:

$$
E F = F A - F B = \frac{3 \sqrt{29}}{29}
$$

Thus, the area of the small square is $\frac{9}{29}$.

Remark. The more general problem, where each side of the original square is divided into $n$ equal parts and the $k$-th division point is chosen on each side, can be solved in a similar manner. In this case, the area of the small square is:

$$
T = \frac{(n-k)^2}{n^2 + k^2}
$$