10.385 The area of the triangle is S. Each side of the triangle is divided into three parts in the ratio $m: n: m$. Determine the area of the hexagon whose vertices are the points of division.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-236.jpg?height=303&width=1063&top_left_y=291&top_left_x=101)

Let points $K, M, N, L, Q, P$ divide the sides of the given $\triangle A B C$ in the ratio $m: n: m$.

$$
\begin{gathered}
\text { Then } \frac{B M}{A B}=\frac{m}{2 m+n}=\frac{B N}{B C} \Rightarrow \\
\Rightarrow \triangle M B N \sim \triangle A B C \Rightarrow \frac{S_{M B N}}{S}=\frac{m^{2}}{(2 m+n)^{2}} \text { (Theorem 7). }
\end{gathered}
$$

Similarly, $s_{A K P}=S \cdot \frac{m^{2}}{(2 m+n)^{2}}=s_{Q L C} \Rightarrow$

$$
\Rightarrow S_{M N L Q P K}=S-3 \cdot S_{M B N}=S\left(1-\frac{3 m^{2}}{(2 m+n)^{2}}\right)
$$

Answer: $s\left(1-\frac{3 m^{2}}{(2 m+n)^{2}}\right)$.