<s>[INST] In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$ 
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is the triangle $KLM$ also an equilateral side?  In what relation are the area of ​​triangle $KLM$ the area of ​​triangle$ ABC$?



My attempt: 
I can see that $KLM$ is an equilateral triangle.  But, why the fraction $\frac{1}{7}$ is the area ratio of ​​the triangle $KLM$ to the ​​triangle $ABC$?
 [/INST] According to Routh’s theorem
the ratio of the areas in general case is
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}
.
\end{align}
The question is a special case, for which
$x=y=z=\tfrac12$, so
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(\tfrac18-1)^2}{(\tfrac14+\tfrac12+1)^3}
=\tfrac17
.
\end{align}
</s>