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.
Best Answer
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.