Question

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

• A. 24
• B. 18
• C. 16
• D. 22

Answer 1

Answer: (B), (D)

22


QR is largest side

$PM=PN=2k$

$RM=RL=2k+4$

$QL=QN=2k+2$

$\therefore QR=4k+6$

$RP=4k+4$

$QP=4k+2$

$\cos P=\frac{(PQ{)}^2+(PR{)}^2-(QR{)}^2}{2\left(PQ\right)\left(PR\right)}$

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

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

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

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

$(k+1)(2k+1)=3(k-1)(k+1)$

$k=-1\text{or}2k+1=3k-3$

$4=k$

So, PQ = 18

QR = 22

RP = 20