Karolína drew a square with a side of $6 \mathrm{~cm}$. On each side of the square, she marked two points in blue, dividing the corresponding side into three equal parts. Then she constructed a quadrilateral with all vertices blue and no two vertices lying on the same side of the square.

What areas of quadrilaterals could Karolína have obtained? List all possibilities.

(L. Hozová)
Quadrilaterals with different areas can be distinguished as follows:

a) One diagonal of the quadrilateral is parallel to a side of the square. Let's denote the vertices of the square and the quadrilateral such that the diagonal $L N$ is parallel to the sides $A B$ and $C D$:

![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-2.jpg?height=448&width=460&top_left_y=781&top_left_x=798)

The area of triangle $L N K$ is half the area of rectangle $L N A B$, and the area of triangle $L N M$ is half the area of rectangle $L N D C$. Triangles $L N K$ and $L N M$ together form the entire quadrilateral, and rectangles $L N A B$ and $L N D C$ together form the given square. The area of the square is $6 \cdot 6 = 36 \, (\mathrm{cm}^2)$, and the area of quadrilateral $K L M N$ is half of that:

$$
36 : 2 = 18 \, (\mathrm{cm}^2)
$$

It is not significant whether the parallel segments are considered horizontally or vertically. The placement of points $K$ and $M$ on the sides of the square also does not play a significant role in the previous reasoning.

b) No diagonal of the quadrilateral is parallel to a side of the square. In this case, we distinguish two possibilities:
![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-2.jpg?height=368&width=778&top_left_y=1868&top_left_x=640)

Indeed, swapping any vertex in any of these quadrilaterals results in a quadrilateral whose diagonal is parallel to a side of the square (i.e., the case discussed above). The given quadrilaterals have different areas, which we express as the difference between the area of the entire square and the areas of the four triangular corners. The area of the square is $6 \cdot 6 = 36 \, (\mathrm{cm}^2)$. The corners are of three types, and their areas are

$$
\frac{1}{2} \cdot 2 \cdot 2 = 2 \, (\mathrm{cm}^2), \quad \frac{1}{2} \cdot 2 \cdot 4 = 4 \, (\mathrm{cm}^2), \quad \frac{1}{2} \cdot 4 \cdot 4 = 8 \, (\mathrm{cm}^2)
$$

The areas of the first and second quadrilaterals are

$$
36 - 4 \cdot 4 = 20 \, (\mathrm{cm}^2), \quad 36 - 2 \cdot 2 - 2 \cdot 8 = 16 \, (\mathrm{cm}^2)
$$

Karolína could have obtained quadrilaterals with areas of 16, 18, and $20 \, (\mathrm{cm}^2)$.

Another solution. On each side of the square, Karolína chose one of two blue points as a vertex of the quadrilateral. This way, she could construct a total of $2 \cdot 2 \cdot 2 \cdot 2 = 16$ quadrilaterals. However, many of them are congruent, meaning they have the same areas. The non-congruent quadrilaterals that Karolína could construct are four:
![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-3.jpg?height=372&width=1606&top_left_y=1000&top_left_x=224)

Indeed, swapping any vertex in any quadrilateral results in a quadrilateral that is congruent to one of these four quadrilaterals.

The area of each quadrilateral can be expressed as the difference between the area of the entire square and the areas of the four triangular corners. For the third and fourth quadrilaterals, we subtract the same corners (just in a different way), so their areas are the same. Therefore, it is sufficient to check the first three quadrilaterals. Their areas are calculated in the previous solution.

Karolína could have obtained quadrilaterals with areas of 16, 18, and $20 \, (\mathrm{cm}^2)$.

Notes. The congruences in the previous reasoning belong to the symmetries of the square, i.e., transformations that preserve the given square. These include axial symmetries, central symmetry, and rotations by $90^{\circ}$. The square has a total of eight symmetries.