8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 11, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=5$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$.
Solution:

Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{2}=S_{A C_{1} D}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_dfc970d9130b2433ec72g-16.jpg?height=410&width=475&top_left_y=2262&top_left_x=1410)
$S_{0}=S_{\text {DEF }} \cdot$ Then

$A A_{1}=\sqrt{A B^{2}+B A_{1}^{2}-2 A B \cdot B C_{1} \cos 60^{\circ}}=\sqrt{91}$,

$$
\frac{S_{2}}{S_{1}}=\left(\frac{A C_{1}}{A A_{1}}\right)^{2}=\frac{25}{91}, \frac{S_{1}}{S}=\frac{B A_{1}}{B C}=\frac{5}{11}
$$

$S_{0}=S-3 S_{1}+3 S_{2}=S-3 \cdot \frac{5}{11} S+3 \cdot \frac{25}{91} \cdot \frac{5}{11} S=\frac{S}{91}, \frac{S}{S_{0}}=91$.

Answer: 91.