Advanced Vectors - Part 3

Geometry Level 4

$O$ is a point in $\triangle ABC$ such that $2\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\mathbf 0$ , the area of $\triangle ABC$ and $\triangle OBC$ are $S_{\triangle ABC}$ , $S_{\triangle OBC}$ respectively. Find $\dfrac{S_{\triangle OBC}}{S_{\triangle ABC}}$ .

The answer can be expressed as $\dfrac{p}{q}$ , where $p,q$ are positive coprime integers. Submit $p+q$ .

The answer is 3.

1 solution

David Vreken
Aug 5, 2019

Place $\triangle ABC$ on a coordinate system so that $A(t, u)$ , $B(0, 0)$ , $C(v, 0)$ , and $O(r, s)$ .

Then $\overrightarrow{OA} = (t - r, u - s)$ , $\overrightarrow{OB} = (-r, -s)$ , and $\overrightarrow{OC} = (v - r, -s)$ .

Since $2\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = 0$ , by the $y$ -components $2(u - s) + -s + -s = 0$ or $u = 2s$ .

The ratio of the areas is $\frac{S_{\triangle OBC}}{S_{\triangle ABC}} = \frac{\frac{1}{2}vs}{\frac{1}{2}vu} = \frac{s}{u}$ , but since $u = 2s$ , the ratio of the areas is $\frac{s}{2s} = \frac{1}{2}$ .

Therefore, $p = 1$ , $q = 2$ , and $p + q = \boxed{3}$ .

Advanced Vectors - Part 3

The answer is 3.

1 solution

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