JEE Advanced 2022 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

In a triangle ABC the side lengths a,b,c are in arithmetic progression, the perimeter is 30, and the largest angle is 120^°. Let O and I be the…

Numerical answer

In a triangle

ABC

the side lengths

a,b,c

are in arithmetic progression, the perimeter is

30

, and the largest angle is

120^\circ

. Let

O

and

I

be the circumcentre and the incentre of the triangle. Find the length

OI

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 6.11

Let the sides be

b-d,\,b,\,b+d

with perimeter

3b=30

, so

b=10

. The largest angle is opposite

b+d

(10+d)^2=(10-d)^2+100-2\cdot10(10-d)\cos120^\circ=(10-d)^2+100+10(10-d)

, which gives

d=4

. So the sides are

6,10,14

and

s=15

. Area

=\frac12\cdot6\cdot10\sin120^\circ=15\sqrt3

. Then

R=\frac{abc}{4\Delta}=\frac{840}{60\sqrt3}=\frac{14}{\sqrt3}

and

r=\frac{\Delta}{s}=\sqrt3

. By Euler's formula

OI^2=R(R-2r)=\frac{14}{\sqrt3}\left(\frac{14}{\sqrt3}-2\sqrt3\right)=\frac{196}{3}-28=\frac{112}{3}

. Hence

OI=\sqrt{112/3}\approx6.11

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