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Problem 893

AMC 10/12, early questions

Geometry Difficulty 4.0 Multiple choice China Mathematical Competition (Hainan) · China

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

Pick one

Official solution

In the diagram, let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$ , respectively. Then we have
$\overrightarrow{OA} + \overrightarrow{OC} = 2 \overrightarrow{OD}, \qquad (1)$
and
$2(\overrightarrow{OB} + \overrightarrow{OC}) = 4 \overrightarrow{OE}. \qquad (2)$
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, $\frac{S_{\triangle AEC}}{S_{\triangle AOC}} = \frac{3}{2}$
and $\frac{S_{\triangle ABC}}{S_{\triangle AOC}} = \frac{3 \times 2}{2} = 3$ . Answer: C.

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