Let PQR be a triangle.

Question

Let $PQR$ be a triangle. The points $A, B$ and $C$ are on the sides $QR, RP$ and $PQ$ respectively such that $\dfrac{QA}{AR}=\dfrac{RB}{BP}=\dfrac{PC}{CQ}=\dfrac12$. Then $\dfrac{	ext{Area}(	riangle PQR)}{	ext{Area}(	riangle ABC)}$ is equal to:

Accepted Answer

Let $P=\vec0,\ Q=\vec q,\ R=\vec r$. Since $QA:AR=1:2$, $A = \dfrac{2\vec q+\vec r}3$. Since $RB:BP=1:2$, $B=\dfrac{2\vec r}3$. Since $PC:CQ=1:2$, $C=\dfrac{\vec q}3$.

Area of $\triangle PQR = \dfrac12|\vec q\times\vec r|$.

$\vec{AB} = B-A = \dfrac{2\vec r}3-\dfrac{2\vec q+\vec r}3 = \dfrac{\vec r-2\vec q}3$, $\vec{AC}=C-A = \dfrac{\vec q}3-\dfrac{2\vec q+\vec r}3=\dfrac{-\vec q-\vec r}3$.

Area of $\triangle ABC = \dfrac12|\vec{AB}\times\vec{AC}| = \dfrac12\cdot\dfrac19|(\vec r-2\vec q)\times(-\vec q-\vec r)| = \dfrac1{18}|3(\vec q\times\vec r)| = \dfrac16|\vec q\times\vec r|$

$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{\frac12}{\frac16}=3$$

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