JEE Advanced 2023 — Paper 2 — MATH-IV solution

JEE Advanced2023Paper 2MATH-IV

PARAGRAPH “I”

Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{π}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$ .

Q2. Then the inradius of the triangle $A B C$ is

Your answer

Solution

Continuing from the previous problem, with sides $n - d, n, n + d$ and $n = 2 cos 2 α$ , $n + d = 2 cos α$ , the semi-perimeter is $s = \frac{( n - d ) + n + ( n + d )}{2} = \frac{3 n}{2} .$

The area was $Δ = \frac{1}{2} n (n + d) sin α$ . So the inradius is

$r = \frac{Δ}{s} = \frac{\frac{1}{2} n ( n + d ) sin α}{\frac{3 n}{2}} = \frac{( n + d ) sin α}{3} = \frac{2 cos α sin α}{3} = \frac{sin 2 α}{3} .$

Using $sin 2 α = \frac{3}{4}$ ,

$r = \frac{3/4}{3} = \frac{1}{4} = 0.25.$

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