A regular six-pointed star $A B C D E F G H I J K L$ with center $S$, as shown in the figure, was formed by the union of two equilateral triangles, each with an area of $72 \mathrm{~cm}^{2}$. Calculate the area of the quadrilateral $A B C S$.

( $S$. Bednářová)

![](https://cdn.mathpix.com/cropped/2024_04_17_55b3e082e70f80bb120bg-3.jpg?height=840&width=757&top_left_y=231&top_left_x=684)
The star is symmetrical with respect to six axes of symmetry, and the hexagon $B D F H J L$ must also be symmetrical with respect to the same axes. This implies that it has all sides of equal length, all internal angles of equal size, and is therefore regular. We will add segments $L S, B S, D S, F S, H S$, and $J S$ to the diagram, which divide this regular hexagon into six congruent equilateral triangles.

![](https://cdn.mathpix.com/cropped/2024_04_17_55b3e082e70f80bb120bg-3.jpg?height=840&width=757&top_left_y=1442&top_left_x=684)

The triangles $L A B, B C D, D E F, F G H, H I J$, and $J K L$ are equilateral because all their internal angles are clearly $60^{\circ}$. The triangles mentioned above always share one side with $S$. Now we see that the star has been divided into a total of twelve congruent triangles.

We will calculate the area of one of these small triangles. We know that each of the original equilateral triangles (i.e., $A E I$ and $C G K$) had an area of $72 \mathrm{~cm}^{2}$. Furthermore, we know that each is composed of nine small triangles. Therefore, one small triangle has an area of $72: 9 = 8 \left(\mathrm{~cm}^{2}\right)$. The area of quadrilateral $A B C S$ corresponds to the area of two such triangles, which is $2 \cdot 8 = 16 \left(\mathrm{~cm}^{2}\right)$.

Another solution. Similarly to the previous method, we divide the star into twelve congruent equilateral triangles, each with an area of $8 \mathrm{~cm}^{2}$. Therefore, the entire star has an area of $12 \cdot 8 = 96 \left(\mathrm{~cm}^{2}\right)$. The segments $A S, C S, E S, G S, I S$, and $K S$ divide it into six quadrilaterals that are congruent to each other: they have corresponding sides of equal length and corresponding angles of equal size (which also follows from the symmetries of the star). One of these quadrilaterals is $A B C S$. Its area is therefore one-sixth of the area of the star, i.e., $96: 6 = 16 \left(\mathrm{~cm}^{2}\right)$.

Evaluation. 2 points for a justified division of the star into 12 congruent equilateral triangles or an analogous insight; 2 points for the area of $8 \mathrm{~cm}^{2}$ of one small triangle; 2 points for deriving the area of quadrilateral $A B C S$.