For an equilateral triangle $A B C$, a regular hexagon $K L M N O P$ is inscribed such that points $K, M, O$ lie sequentially at the midpoints of sides $A B, B C$, and $A C$. Calculate the area of the hexagon $K L M N O P$ if the area of triangle $A B C$ is $60 \mathrm{~cm}^{2}$. (K. Pazourek)
Let's inscribe a hexagon $K L M N O P$ into the triangle $A B C$ in the prescribed manner.

![](https://cdn.mathpix.com/cropped/2024_04_17_27d9befe2cb3ac9dc509g-3.jpg?height=717&width=817&top_left_y=1315&top_left_x=651)

Given that both figures as a whole are symmetric with respect to three axes of symmetry, the centroid of the hexagon and the centroid of the triangle lie at the same point, which we denote as $T$. The medians of the triangle $A B C$ together with the segments $K T, M T$, and $O T$ divide the hexagon $K L M N O P$ into six congruent isosceles triangles. To justify this claim, it suffices to recognize the congruence of the corresponding sides of these triangles.

Furthermore, the remaining parts of the triangle $A B C$ can also be divided into six triangles congruent to the previous six triangles. As a possible justification for this claim, we can prove the congruence of triangles $P K O$ and $P K A$ using the SAS (Side-Angle-Side) theorem: Both triangles share the side $P K$. The sides $K O$ and $K A$ have the same length because they are the midline of the equilateral triangle $A B C$ and half of its side. The angle $P K O$ is one-fourth of the internal angle $P K L$ of the regular hexagon $K L M N O P$, and thus measures $30^{\circ}$. The angle $P K A$ is the complement of the angle $L K B$ to a right angle, and thus also measures $30^{\circ}$.

Thanks to the division of the triangle $A B C$ into these twelve congruent triangles, we see that the ratio of the areas of the hexagon $K L M N O P$ and the triangle $A B C$ is $6: 12 = 1: 2$, so the area of the hexagon $K L M N O P$ is

$$
S = \frac{1}{2} \cdot 60 = 30 \left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 1 point for dividing the hexagon into the above-mentioned triangles and justifying their congruence; 3 points for any justification that the remaining triangles forming the triangle $A B C$ are congruent to the previous ones; 1 point for comparing the areas of both given figures; 1 point for the result.

Another solution. Let's inscribe the hexagon $K L M N O P$ into the triangle $A B C$ in the prescribed manner.

![](https://cdn.mathpix.com/cropped/2024_04_17_27d9befe2cb3ac9dc509g-4.jpg?height=717&width=811&top_left_y=932&top_left_x=654)

Given that both figures as a whole are symmetric with respect to three axes of symmetry, the centroid of the hexagon and the centroid of the triangle lie at the same point, which we denote as $T$. From the diagram, we see that the regular hexagon $K L M N O P$ consists of six congruent triangles, and the remaining part of the triangle $A B C$ consists of six other congruent triangles. We will prove that these triangles have the same area as the previous ones, for example, by considering triangles $N T O$ and $C N O$:

The segment $K C$ is a median of the triangle $A B C$. From the properties of the centroid and medians, it follows that $|T C| = 2 \cdot |K T|$. Furthermore, in the regular hexagon $K L M N O P$, $|N T| = |K T|$. It is now clear that $|C N| = |N T|$. Triangles $N T O$ and $C N O$ thus have the same lengths of sides $N T$ and $C N$ and also the same corresponding height, so they must have the same area.

Thanks to the division of the triangle $A B C$ into twelve triangles of the same area, we see that the ratio of the areas of the hexagon $K L M N O P$ and the triangle $A B C$ is $6: 12 = 1: 2$, so the area of the hexagon $K L M N O P$ is

$$
S = \frac{1}{2} \cdot 60 = 30 \left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 2 points for explaining that $|C N| = |N T|; 2$ points for explaining that triangles $N T O$ and $C N O$ have the same areas; 1 point for comparing the areas of both given figures; 1 point for the result.

Yet another solution. Given that both figures as a whole are symmetric with respect to three axes of symmetry, the centroid of the hexagon and the centroid of the triangle lie at the same point, which we denote as $T$. The segment $K C$ is a median of the triangle $A B C$ and $K T$ is one-third of it. Let $b$ be the length of the side of the triangle $A B C$. From the Pythagorean theorem applied to the right triangle $K B C$, we get

$$
|K T| = \frac{1}{3} \sqrt{b^{2} - \frac{b^{2}}{4}} = \frac{\sqrt{3}}{6} b
$$

Since the hexagon $K L M N O P$ consists of six congruent equilateral triangles with side length $|K T|$, its area is

$$
S_{1} = 6 \cdot \frac{\sqrt{3}}{4} \left(\frac{\sqrt{3}}{6} b\right)^{2} = \frac{3 \sqrt{3}}{2} \cdot \frac{3}{36} b^{2} = \frac{\sqrt{3}}{8} b^{2}.
$$

The area of the triangle $A B C$ is $S_{2} = \frac{1}{4} \sqrt{3} b^{2}$. By comparing $S_{1}$ and $S_{2}$, we find that the area of the hexagon $K L M N O P$ is half of the area of the triangle $A B C$, so it is equal to $30 \mathrm{~cm}^{2}$.

Evaluation. 1 point for explaining that $K T$ is one-third of $K C$; 1 point for expressing the area of the triangle $A B C$; 2 points for expressing the area of the hexagon using the same unknown; 1 point for the ratio of the areas of both figures or an equivalent insight; 1 point for the result.

Note. Based on the last solution and given the area of the triangle $A B C$, students can sequentially calculate $b \approx 11.8 \mathrm{~cm}, |K T| \approx 3.4 \mathrm{~cm}$, and $S_{2} \approx 30.0 \mathrm{~cm}^{2}$. Such a solution can also be evaluated with the full number of points if the reasoning is correct.