Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Find the answer

4. As shown in Figure 1, let point $O$ be inside $\triangle A B C$ , and $O A + 2 O B + 3 O C = 0$ . Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is:

Pick one

Solution

4.C.

Let $D$ and $E$ be the midpoints of sides $AC$ and $BC$ , respectively, then
$\begin{array}{l} O A+O C=2 O D, \\ 2(O B+O C)=4 O E . \end{array}$

From (1) and (2), we get
$O A+2 O B+3 O C=2(O D+2 O E)=0,$

which means $O D$ and $O E$ are collinear, and $|O D|=2|O E|$ .
Therefore, $\frac{S_{\triangle A B C}}{S_{\triangle A D C}}=\frac{3}{2}$ . Thus, $\frac{S_{\triangle A B C}}{S_{\triangle A D C}}=\frac{3 \times 2}{2}=3$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.