JEE Advanced 2020 Paper 2 · Mathematics · Trigonometry and Inverse Trigonometric Functions · hard

In a triangle ABC, AB=5, AC=8 and ∠ A=60^°. Let O and I be the circumcentre and the incentre of the triangle. Find the distance OI.

Numerical answer

In a triangle

ABC

AB=5

AC=8

and

\angle A=60^{\circ}

. Let

O

and

I

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

OI

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 1.53

By the cosine rule,

BC^{2}=25+64-2\cdot5\cdot8\cdot\tfrac12=49

, so

BC=7

. The semi-perimeter is

s=\frac{5+8+7}{2}=10

and the area is

\Delta=\tfrac12\cdot5\cdot8\sin60^{\circ}=10\sqrt3

. Hence the inradius is

r=\Delta/s=\sqrt3

and the circumradius is

R=\frac{BC}{2\sin A}=\frac{7}{\sqrt3}

. Euler's relation gives

OI^{2}=R^{2}-2Rr=\frac{49}{3}-2\cdot\frac{7}{\sqrt3}\cdot\sqrt3=\frac{49}{3}-14=\frac{7}{3}

. Therefore

OI=\sqrt{7/3}\approx1.5275

, i.e.

1.53

. A direct coordinate computation of both centres gives the same value.

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