Let P Q R be a triangle. The points A, B and C are on the sides Q R, R P and P Q respectively such that (Q A)/(A R)=(R B)/(B P)=(P C)/(C Q)=1/2. Then…

Let $\displaystyle P Q R$ be a triangle. The points $\displaystyle A, B$ and $\displaystyle C$ are on the sides $\displaystyle Q R, R P$ and $\displaystyle P Q$ respectively such that $\displaystyle \frac{Q A}{A R}=\frac{R B}{B P}=\frac{P C}{C Q}=\frac{1}{2}$. Then $\displaystyle \frac{\operatorname{Area}(\triangle P Q R)}{\operatorname{Area}(\triangle A B C)}$ is equal to