Problem
Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1. Then find the inradius of triangle ABC.
Solution
Since we need to find the inradius of the triangle, we’ll start with using one of the standard formulas.
Here’s one: r = Δ/s. Why this one? As we’ve found the value of Δ already in the previous problem, and all we need is the value of s (or the semi-perimeter).
We have s = (a + b + c)/2. Since the sides are in AP, we have 2b = a + c.
So, we get s = 3b/2. And, using the sine rule, this relation changes to s = 3sinB.
Finally, since B = π/2 – 2A (derived previously), we get s = 3cos2A. Coming back to the inradius, we have r = Δ/(3cos2A).
Substituting the values of Δ and cos2A from the previous problem, we get r = (3√7/16)/(3√7/4) or 1/4.
Comments
This one’s easier than the previous one, since the heavylifting was done already. Once again, you’ll need the formulas handy – then this problem is just a direct application, plus some simple manipulations.