We connect the vertices of a square with side length $a$ to the midpoints of the opposite two sides. What is the area of the octagon determined by these eight lines?
I. solution: Given the construction of the defined octagon, it is axially symmetric to the diagonals and midlines of the square, and centrally symmetric to the center of the square. The four axes of symmetry divide the octagon of area $T$ into 8 congruent - each with an area of $t$ - general triangles. (Among these, 4-4 have the same direction of traversal. Thus, our octagon is equilateral, with opposite parallel sides, but not regular.)

![](https://cdn.mathpix.com/cropped/2024_05_02_d51b31f508a988f601f1g-1.jpg?height=510&width=507&top_left_y=314&top_left_x=798)

The labeling is shown in the figure. (The center of the square is point $O$.) From the construction, it follows that $O E = E B_{1} = \frac{a}{4}$, and thus, on the one hand, the $\triangle E B_{1} F = \triangle O E F = t$, or

$$
\triangle O B_{1} G = 3 t
$$

on the other hand, $\triangle O B_{1} G$ is a quarter of the square $O B_{1} C C_{1}$, or

$$
\triangle O B_{1} C = \frac{1}{4} \cdot \frac{a^{2}}{4} = \frac{a^{2}}{16}
$$

Thus,

$$
3 t = \frac{a^{2}}{16}
$$

from which

$$
t = \frac{a^{2}}{48}
$$

and thus

$$
T = 8 t = \frac{a^{2}}{6}
$$

András Orosz (Bp. XVI., Corvin Mátyás g. II. o. t.)

II. solution: Since point $F$ is the centroid of $\triangle B C D$, $O C = 3 \cdot O F$ and thus, if we enlarge the octagon from point $O$ as the center of similarity by a ratio of $3:1$, we get the octagon $B E^{\prime} C G^{\prime} D \ldots$, whose area (since $B_{1} E^{\prime} = \frac{a}{4}$) is clearly $a^{2} + 4 \cdot \frac{1}{2} a \cdot \frac{a}{4} = a^{2} + \frac{a^{2}}{2} = \frac{3 a^{2}}{2}$.

Thus,

$$
T = \frac{1}{9} \cdot \frac{3 a^{2}}{2} = \frac{a^{2}}{6}
$$

Remark: Similarly, it can be shown that the area of the octagon enlarged by a ratio of $2:1$ is

$$
a^{2} - 4 \cdot \frac{a^{2}}{12} = \frac{2 a^{2}}{3}
$$

and therefore

$$
T = \frac{1}{4} \cdot \frac{2 a^{2}}{3} = \frac{a^{2}}{6}
$$

Pál Bártfai (Bp. I., Petófi g. II. o. t.)

III. solution: Since the diagonals of the square are axes of symmetry, the area of the kite $O F G H$ is $\frac{T}{4}$. The area of this kite, based on the above, is

$$
\frac{1}{2} \cdot O G \cdot H F = \frac{1}{2} \cdot \frac{a}{4} \cdot \frac{a}{3} = \frac{a^{2}}{24}
$$

and thus

$$
T = 4 \cdot \frac{a^{2}}{24} = \frac{a^{2}}{6}
$$

Similarly, since the midlines are also axes of symmetry, the area of the kite $O E F G$ is also $\frac{T}{4}$. The area of this kite is

$$
\begin{aligned}
& \frac{1}{2} \cdot \frac{a \sqrt{2}}{6} \cdot \frac{a \sqrt{2}}{4} = \frac{a^{2}}{24} \\
& \text{ Györgyi Lackner (Bp. V., 1. sz. textilip, techn. II. o. t.) }
\end{aligned}
$$