Question

In a $△ABC$ a point $P$ is taken such that $\frac{AP}{BP}=\frac{1}{3}$ and a point $Q$ is taken on $BC$ such that $\frac{CQ}{BQ}=\frac{3}{1}.$ If $R$ is the point of intersection of the lines $AQ$ and $CP$. The area of $△BRC$ is $1$ unit and area of $△ABC$ is $\frac{a}{b},a\text{and}b$ is coprime, then $a+b$ is

Answer 1

Let $A$ be the origin and the position vectors of
$B:\overrightarrow{b},C:\overrightarrow{c}$


Now, the position vectors of
$Q:\frac{3\overrightarrow{b}+\overrightarrow{c}}{4}P:\frac{\overrightarrow{b}}{4}$
Let $\frac{AR}{QR}=\frac{\lambda }{1}$, then position vector of
$R:\lambda \left(\frac{3\overrightarrow{b}+\overrightarrow{c}}{4(\lambda +1)}\right)\cdots \left(1\right)$
Let $\frac{CR}{RP}=\frac{\mu }{1}$, then position vector of
$R:\left(\frac{\mu \frac{\overrightarrow{b}}{4}+\overrightarrow{c}}{\mu +1}\right)\cdots \left(2\right)$
From equation $1$ and $2$,
$\frac{3\lambda }{4(\lambda +1)}=\frac{\mu }{4(\mu +1)}\frac{\lambda }{4(\lambda +1)}=\frac{1}{\mu +1}\Rightarrow 3\times \frac{1}{\mu +1}=\frac{\mu }{4(\mu +1)}\Rightarrow \mu =12$
So,
$R:\left(\frac{3\overrightarrow{b}+\overrightarrow{c}}{13}\right)$
$△ABC$ and $△BRC$ have same base.
Therefore areas will be proportional to length of $AQ$ and $RQ$,
$\overrightarrow{AQ}=\frac{3\overrightarrow{b}+\overrightarrow{c}}{4}\overrightarrow{RQ}=\overrightarrow{AQ}-\overrightarrow{AR}=\frac{3\overrightarrow{b}+\overrightarrow{c}}{4}-\left(\frac{3\overrightarrow{b}+\overrightarrow{c}}{13}\right)=\frac{9}{13}\left(\frac{3\overrightarrow{b}+\overrightarrow{c}}{4}\right)$

$\frac{Area\left(△ABC\right)}{Area\left(△BRC\right)}=\frac{\left|\overrightarrow{AQ}\right|}{\left|\overrightarrow{RQ}\right|}\Rightarrow \frac{Area\left(△ABC\right)}{Area\left(△BRC\right)}=\frac{13}{9}=\frac{a}{b}\Rightarrow a+b=22$

Alternate solution:
$△BQR\text{and}△CQR$ has same base and one common vertex, therefore ratio of area of two triangle will be ratio of their bases.
$\frac{Area\left(△BQR\right)}{Area\left(△CQR\right)}=\frac{1}{3}$
Let the
$Area\left(△BQR\right)=x\Rightarrow Area\left(△CQR\right)=3x$
Similarly
$Area\left(△ARP\right)=y\Rightarrow Area\left(△BRP\right)=3y$
Assuming the $Area\left(△ARC\right)=a$
Now,
$\frac{Area\left(△ABQ\right)}{Area\left(△AQC\right)}=\frac{1}{3}\Rightarrow \frac{4y+x}{3x+a}=\frac{1}{3}\Rightarrow a=12y$
Now,
$\frac{Area\left(△APC\right)}{Area\left(△BPC\right)}=\frac{1}{3}\Rightarrow \frac{12y+y}{3y+4x}=\frac{1}{3}\Rightarrow x=9y$
Given $Area\left(△BRC\right)=1$
$4x=1\Rightarrow x=\frac{1}{4}$
Therefore,
$Area\left(△ABC\right)=y+12y+4x+3y=16y+4x=\frac{16}{9}x+4x=\frac{4}{9}+1=\frac{13}{9}$
Hence,
$a+b=13+9=22$