For an equilateral triangle $A B C$, a regular hexagon $K L M N O P$ is inscribed such that points $K, L$ lie on side $A B$, points $M, N$ lie on side $B C$, and points $O, P$ lie on side $A C$. 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_8a25ab3d9d3d7a609f3bg-4.jpg?height=722&width=757&top_left_y=810&top_left_x=678)

The angle $A K P$ is the adjacent angle to the angle $P K L$. The angle $P K L$ is an internal angle of a regular hexagon, i.e., it measures $120^{\circ}$. Therefore, the measure of angle $A K P$ is $180^{\circ}-120^{\circ}=60^{\circ}$. Then, triangle $A K P$ is equilateral because its internal angles $P A K$ and $A K P$ (and thus also $K P A$) measure $60^{\circ}$. The same reasoning can be used to verify the equilateral nature of triangles $L B M$ and $O N C$.

If we divide the hexagon $K L M N O P$ into six congruent equilateral triangles, we find that they are congruent to the equilateral triangles $A K P, L B M$, and $O N C$ (due to the shared sides $P K, L M, N O$). Therefore, they all have the same area. The hexagon consists of six such triangles, while triangle $A B C$ consists of nine, so the ratio of their areas is $6: 9=2: 3$. Thus, the area of the hexagon $K L M N O P$ is $\frac{2}{3} \cdot 60=40\left(\mathrm{~cm}^{2}\right)$.

Evaluation. 2 points for explaining that triangles $A K P, L B M, O N C$ are equilateral; 2 points for explaining that these triangles and the six triangles forming the hexagon are congruent; 1 point for comparing the areas of both figures; 1 point for the result.

Another solution. First, as in the previous solution, we prove that triangles $A K P$, $L B M$, and $O N C$ are equilateral. Since one side of these triangles is always a side of the regular hexagon $K L M N O P$, triangles $A K P, L B M, O N C$ are congruent and it holds that $|A K|=|K L|=|L B|=\frac{1}{3}|A B|$, i.e., points $K, L$ divide the segment $A B$ into thirds. Similarly, points $M, N$ (respectively $O, P$) divide the segment $B C$ (respectively $A C$) into thirds.

Let the side length of the hexagon $K L M N O P$ be denoted as $a$. Then the area $S_{1}$ of this hexagon can be calculated using the formula

$$
S_{1}=6 \cdot \frac{\sqrt{3}}{4} a^{2}=\frac{3 \sqrt{3}}{2} a^{2} .
$$

The area of the equilateral triangle $A B C$ is then

$$
S_{2}=\frac{\sqrt{3}}{4}(3 a)^{2}=\frac{9 \sqrt{3}}{4} a^{2}
$$

From this, it follows that $S_{1}=\frac{2}{3} S_{2}$, and by substitution, $S_{1}=\frac{2}{3} \cdot 60=40\left(\mathrm{~cm}^{2}\right)$.

Evaluation. 2 points for explaining that triangles $A K P, L B M, O N C$ are equilateral; 1 point for justifying that $|K L|=\frac{1}{3}|A B| ; 1$ point for expressing the areas of both figures in terms of one unknown; 1 point for the ratio of the areas of both figures or an analogous insight; 1 point for the result.

Yet another solution. Another possibility essentially copies the previous procedure, but it uses the given area of the triangle to approximately calculate all necessary data. All calculations are performed using tables and without a calculator.

First, we calculate the side length of the triangle from the formula for the area of an equilateral triangle:

$$
60=\frac{\sqrt{3}}{4} b^{2}
$$

from which $b \doteq 11.8(\mathrm{~cm})$. Analogously to the previous solutions, we prove that triangles $A K P, L B M, O N C$ are equilateral and that points $K, L, M, N, O, P$ divide the respective sides of the triangle into thirds. Then the side length of the hexagon $K L M N O P$ measures $a=b: 3 \doteq 3.93(\mathrm{~cm})$. From this, using the formula for the area of a regular hexagon, we calculate the area of $K L M N O P$:

$$
S=6 \cdot \frac{\sqrt{3}}{4} a^{2} \doteq 40.1\left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 1 point for calculating the side length of triangle $A B C ; 2$ points for explaining that triangles $A K P, L B M, O N C$ are equilateral; 2 points for calculating the side length of the hexagon and justifying the reasoning; 1 point for calculating the area of the hexagon.