In a triangle PQR,P is the largest angle and cosP=31. 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)16
(B)18✓ Correct
(C)24
(D)22✓ Correct
Answer: (B), (D)
Solution
Let the tangent lengths be PN=2x, QL=2x+2, and RM=2x+4 (consecutive even integers).
Since tangents from an external point to a circle are equal:
PN=PM=2x
QL=QN=2x+2
RM=RL=2x+4
The sides are:
PQ=PN+NQ=2x+(2x+2)=4x+2
QR=QL+LR=(2x+2)+(2x+4)=4x+6
RP=RM+MP=(2x+4)+2x=4x+4
Using the cosine rule with cosP=1/3:
cosP=2(4x+2)(4x+4)(4x+2)2+(4x+4)2−(4x+6)2=31
Simplifying:
2(16x2+24x+8)16x2−16=31
8x2+12x+44x2−4=31
12x2−12=8x2+12x+4
4x2−12x−16=0
x2−3x−4=0
(x−4)(x+1)=0
So x=4 (taking positive value).
The sides are: 4(4)+2=18, 4(4)+4=20, 4(4)+6=22.
Therefore, possible lengths are 18 and 22.
JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options