Question

In a triangle $PQR,P$ is the largest angle and $\cos P=\frac{1}{3}.$ Further the incircle of the triangle touches the sides $PQ,QR$ and $RP$ at $N,L$ and $M$ respectively, such that the length of $PN,QL$ and $RM$ are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)

• A. $16$
• B. $18$
• C. $24$
• D. $22$

Answer 1

Answer: (B), (D)

$22$

The largest side is $QR$.
$PM=PN=2kRM=RL=2k+4QL=QN=2k+2$

The sides of the triangle is given as,
$QR=4k+6RP=4k+4QP=4k+2\cos P=\frac{(PQ{)}^2+(PR{)}^2-(QR{)}^2}{2\left(PQ\right)\left(PR\right)}$

$\Rightarrow \frac{1}{3}=\frac{(4k+2{)}^2+(4k+4{)}^2-(4k+6{)}^2}{2(4k+2)(4k+4)}\Rightarrow \frac{1}{3}=\frac{(2k+1{)}^2+4(k+1{)}^2-(2k+3{)}^2}{4(2k+1)(k+1)}\Rightarrow \frac{1}{3}=\frac{4(k+1{)}^2-(4k+4\left)\right(2)}{4(2k+1)(k+1)}$

Further simplify for the value of $k.$

$\frac{1}{3}=\frac{(k+1{)}^2-2(k+1)}{(2k+1)(k+1)}\Rightarrow (k+1)(2k+1)=3(k-1)(k+1)\Rightarrow k=-1\&4$

Consider the positive value of $k$

$\therefore QR=4\times 4+6=22\therefore RP=4\times 4+4=20\therefore QR=4\times 4+2=18$