10.407 The sides of triangle $A B C$ are divided by points $M, N$ and $P$ such that $A M: M B=B N: N C=C P: P A=1: 4$. Find the ratio of the area of the triangle bounded by the lines $A N, B P$ and $C M$ to the area of triangle $\boldsymbol{A} \boldsymbol{B} C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-246.jpg?height=302&width=362&top_left_y=1431&top_left_x=101)

Let $x = S_{A M K}$, then $S_{B M K} = 4x$ (Theorem 10).

\[
\begin{aligned}
& S_{A B K} = 5x \\
& S_{A N C} = 4 S_{A N B}, S_{K N C} = 4 S_{K N B} (\text{Theorem } 10) \Rightarrow \\
& \Rightarrow S_{A N C} - S_{K N C} = S_{A K C} = 4 (S_{A N B} - S_{K N B}) \Rightarrow \\
& \Rightarrow S_{A K C} = 4 S_{A B K} = 20x \Rightarrow \\
& \Rightarrow S_{A M C} = 21x, S_{B C M} = 4 S_{A M C} = 84x \Rightarrow
\end{aligned}
\]

\[
\Rightarrow S_{A B C} = 105x, \text{ i.e., } \frac{S_{A M K}}{S_{A B C}} = \frac{1}{105}. \text{ Similarly, } S_{B N Q} = S_{C P L} = \frac{1}{105} S_{A B C} = x.
\]

From this, $S_{A K L P} = S_{B Q \text{ KM}} = S_{C L Q N} = 20x - x = 19x \Rightarrow$

\[
S_{K Q L} = S_{A B C} - 3(S_{A K L P} + S_{A M K}) = 105x - 3 \cdot 20x = 45x, \frac{S_{K Q L}}{S_{A B C}} = \frac{3}{7}.
\]

Answer: $\frac{3}{7}$.