QJEEAdv13Maa1ab: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Maa1ab Solution: In a triangle , is the largest angle and . Further the incircle …

For the above circle PN and PM are its pair of tangents. So, its length are equal i.e. $|\mathrm{PN}|=|\mathrm{PM}| \text {. }$

Similarly $|\mathrm{QL}|=|\mathrm{QN}|$ and $|\mathrm{RL}|=|\mathrm{RM}|$

Let $|\mathrm{PN}|=|\mathrm{PM}|=x,|\mathrm{QL}|=|\mathrm{QN}|=y$ and $|R L|=|R M|=z$

Given $|\mathrm{PN}|,|\mathrm{QL}|$ and $|\mathrm{RM}|$ are consecutive even integer

$\therefore \quad x=2 n, y=2 n+2 \text { and } z=2 n+4$

Given, $\cos P=\frac{1}{3}$

$\begin{aligned}& \Rightarrow \quad \frac{(4 n+4)^2+(4 n+2)^2-(4 n+6)^2}{2(4 n+4)(4 n+2)}=\frac{1}{3} \\& \left(16 n^2+16+32 n\right)+\left(16 n^2+4+16 n\right) \\& \Rightarrow \frac{-\left(16 n^2+36+48 n\right)}{32 n^2+48 n+16}=\frac{1}{3} \\& \Rightarrow \quad 48 n^2-48=32 n^2+48 n+16 \\& \Rightarrow \quad 16 n^2-48 n-64=0 \\& \Rightarrow \quad n^2-3 n-4=0 \\& \Rightarrow \quad n^2-4 n+n-4=0 \\& \Rightarrow \quad(n-4)(n+1)=0 \\& \Rightarrow \quad n=4 \\\end{aligned}$

$\begin{aligned}& \text { Now }|P Q|=4 n+2=18,|Q R|=4 n+6=22 \\& \text { and }|R P|=4 n+4=20\end{aligned}$

Hints :

(i) Recall cosine rule in a triangle $\triangle \mathrm{ABC}$ ,

$\begin{aligned}& \cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}, \cos \mathrm{B}=\frac{c^2+a^2-b^2}{2 c a} \\\text{and} \quad & \cos \mathrm{C}=\frac{a^2+b^2-c^2}{2 a b}\end{aligned}$

(ii) Three consecutive even integers are $2 n, 2 n +2,2 n+4$ where $n$ is an integer.