The point $O$ placed inside triangle so that $\vec{OA}+2\vec{OB}+3\vec{OC}=0$

Question

From the triangle $\triangle ABC$ we have $AB=3$, $BC=5$, $AC=7$. If the point $O$ placed inside the triangle $\triangle ABC$ so that $\vec{OA}+2\vec{OB}+3\vec{OC}=0$ , then what is the ratio of the area of $\triangle ABC$ to the area of $\triangle AOC$ ?

$1)\frac32\qquad\qquad2)\frac53\qquad\qquad3)2\qquad\qquad4)3\qquad\qquad5)\frac72$

By knowing the length of the sides of $\triangle ABC$ I concluded it is an obtuse triangle (Because $3^2+5^2<7^2 $ ). I'm not sure how to use $\vec{OA}+2\vec{OB}+3\vec{OC}=0$ to solve the problem, but from the forth choices I realized it happens when the point $O$ be the centroid of $\triangle ABC$ so this might be the answer.

Answer 1

In terms of vectors, we can rewrite the condition on $O$ as: $$(A-O) + 2(B-O) + 3(C-O) = A + 2B + 3C - 6O = 0.$$ The solution to this equation is $O = \frac{1}{6} (A + 2B + 3C)$.

Now, choose a coordinate system such that $AC$ lies along the $x$-axis. Then the area of triangle $ABC$ is equal to $\frac{1}{2} (AC) \cdot B_2$ (where $B_2$ denotes the $y$-coordinate of the vector $B$), whereas the area of triangle $AOC$ is equal to $\frac{1}{2} (AC) \cdot O_2$. On the other hand, from the above, since $A_2 = C_2 = 0$, we have $O_2 = \frac{1}{6} (A_2 + 2 B_2 + 3 C_2) = \frac{1}{6} (0 + 2 B_2 + 3 \cdot 0)$. From here, it should be easy to put the pieces together and get the desired ratio.

Answer 2

The area of the parallelogram supported by two vectors $\vec{AB}$ and $\vec{AC}$ is given by their cross product, therefore

$\begin{cases}\mathcal A_{ABC}=\frac 12\lVert \vec {AB}\wedge\vec {AC}\lVert\\\mathcal A_{AOC}=\frac 12\lVert \vec {AO}\wedge\vec {AC}\lVert\end{cases}$

Using Chasles relation (i.e. vector addition) we get

$\begin{align}\vec{OA}+2\vec{OB}+3\vec{OC}&=\vec{OA}+2(\vec{OA}+\vec{AB})+3(\vec{OA}+\vec{AC})\\&=6\vec{OA}+2\vec{AB}+3\vec{AC}\\&=\vec 0\end{align}$

Now just cross product with $\vec{AC}$ to make area appear.

$$6\,\vec{OA}\wedge\vec{AC}+2\,\vec{AB}\wedge\vec{AC}+\vec 0=\vec 0$$

And the ratio of triangle areas is $3$

Answer 3

Suppose the triangle is placed so that $O$ is the origin. We have that $$\frac{1}{6}\overrightarrow{OA}+\frac{2}{6}\overrightarrow{OB} +\frac{3}{6}\overrightarrow{OC} =0 $$ Then $(1/2,2/6,3/6)$ are the barycentric coordinates of $O$. Then the coefficient of $\overrightarrow{OB}$ gives the ratio of the signed areas of the triangles $AOC$ and $ABC$ which is $1:3$ in this case. So the required ratio of areas of triangles $ABC$ and $AOC$ is $3$.