Question

Let $O$ be an interior point of $△ABC$ such that $\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=0$. Then the ratio of the area of $△ABC$ to the area of $△AOC$ is

• A. 2
• B. 3 / 2
• C. 3
• D. 5 / 3

Answer 1

Answer: (C)

In the diagram, let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$, respectively. Then we have
$\overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OD}$ ...(1)
and $2(\overrightarrow{OB}+\overrightarrow{OC})=4\overrightarrow{OE}$ ...(2)
From equations $(1)$ and $(2)$, we get
$\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=2(\overrightarrow{OD}+2\overrightarrow{OE})=0$
It follows that $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and $|\overrightarrow{OD}|=2|\overrightarrow{OE}|$.

Consequently, $\frac{S_{△AEC}}{S_{△AOC}}=\frac{3}{2}$
and $\frac{S_{△AEC}}{S_{△AOC}}=\frac{3\times 2}{2}=3$