China Mathematical Competition (Hainan)

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 $$2(\overrightarrow{OB}+\overrightarrow{OC})=4\overrightarrow{OE}.(2)$$ By 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 $\frac{S_{△ABC}}{S_{△AOC}}=\frac{3\times 2}{2}=3$. Answer: C.