Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to $B'$ so that $BB'=2\cdot AB$. Similarly extend $\overline{BC}$ beyond $C$ to $C'$ ($CC'=2\cdot BC$), and $\overline{CA}$ beyond $A$ to $A'$ ($AA'=2\cdot CA$). What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$?

1. Let $[ABC]=1$ and suppose each side is extended beyond its endpoint by $m$ times its length. As in the problem, $[BB'C]=m$ and $[B'CC']=m\cdot[B'BC]=m^2$, so $[BB'C']=m+m^2$, and the same holds at the other two corners. Hence $[A'B'C']:[ABC]=1+3(m+m^2)=1+3m+3m^2$.
2. With $m=2$: ratio $=1+6+12=\boxed{19:1}$.