Question

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

Answer 1

Given that the difference between the largest and smallest angles is $\frac{\pi}{2}$, we have:
A − C =$\frac{\pi}{2}$​
Since the sum of the angles in a triangle is π radians:
A +B +C =$\pi$
Now, let's relate the angles to the sides. The Law of Sines states:
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$
Where a , b and c are the sides opposite to angles A , B and C respectively, and R is the radius of the circumcircle.
Given that the vertices of the triangle lie on a circle of radius 1, the circumradius =R =1.
The inradius r of a triangle is related to its sides and angles by the formula:
$r=\frac{\Delta}{s}$
Where Δ is the area of the triangle and s is the semi-perimeter of the triangle.
Let's denote the sides of the triangle in an arithmetic progression as a − d , a , and a +d (where d is the common difference). Given the sides are in arithmetic progression, the angles will also be in arithmetic progression.
The semi-perimeter s can be calculated as:
$s=\frac{(a-d)+a+(a+d)}{2}=\frac{3a}{2}$​
Now, let's use the formula for the area of the triangle, Δ=rs , to find the inradius r.
$r=\frac{\Delta}{s}=\frac{rs}{s}=\frac{\Delta}{s}=\frac{\Delta}{\frac{3a}{2}}=\frac{2\Delta}{3a}$
To proceed further and find the inradius, we need more information about the triangle, specifically the lengths of its sides or angles. Unfortunately, the given information doesn't provide direct values for the sides or angles.