Golden A.

Each side of the triangle is divided into three equal parts. The division points serve as vertices of two triangles, the intersection of which is a hexagon. Find the area of this hexagon if the area of the given triangle is $S$.

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The vertices of the given hexagon divide the sides of each of the resulting triangles into three equal parts.

## Solution

Let $A B C$ be the given triangle. Denote the points of division as shown in the figure. Then $S_{A_{1} B_{1} C_{1}}=1 / 3 S$.

Let $F$ be the point of intersection of the lines $C_{2} A_{2}$ and $A C, M N K L P Q$ be the hexagon mentioned in the problem.

From the equality of triangles $F A_{2} C$ and $C_{2} A_{2} A_{1}$, it follows that $C F=C_{2} A_{1}=1 / 3 A C$.

From the similarity of triangles $A_{1} N C_{2}$ and $B_{1} N F$, we find that $A_{1} N: N B_{1}=C_{2} A_{1}: B_{1} F=1: 2$. Similarly, $A_{1} M: M C_{1}=1: 2$. Therefore, $S_{A_{1} M N}=1 / 9 S_{A_{1} B_{1} C_{1}}=S / 27$.

Similarly, $S_{B_{1} K L}=S_{C_{1} P Q}=S / 27$. Therefore, $S_{M N K L P Q}=S / 3-3 \cdot S / 27=2 S / 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-28.jpg?height=452&width=966&top_left_y=1832&top_left_x=546)

## Answer

$2 S / 9$.