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:

The correct answer is 0.25.

Step-by-step Explanation:

Step 1: Define the angles and sides of the triangle

Let the angles of triangle $ABC$ be $A$, $B$, and $C$ such that $A>B>C$. Since the triangle is obtuse-angled, the largest angle $A$ is obtuse ($A>\frac{\pi }{2}$).

We are given that the difference between the largest angle ($A$) and the smallest angle ($C$) is $\frac{\pi }{2}$:

$A-C=\frac{\pi }{2} \Rightarrow A=\frac{\pi }{2}+C$

Since the sum of angles in a triangle is $\pi$:

$A+B+C=\pi$

Substituting $A=\frac{\pi }{2}+C$:

$\left(\frac{\pi }{2}+C\right)+B+C=\pi \Rightarrow B+2C=\frac{\pi }{2} \Rightarrow B=\frac{\pi }{2}-2C$

Step 2: Apply the Sine Rule

Let the side lengths opposite to angles $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Since $A>B>C$, we have $a>b>c$.

We are given that the sides are in Arithmetic Progression (A.P.). Therefore:

$2b=a+c$

By the Sine Rule, $a=2R\sin A$, $b=2R\sin B$, and $c=2R\sin C$, where $R$ is the circumradius of the triangle. Substituting these into the A.P. relation gives:

$2\sin B=\sin A+\sin C$

Substitute $A=\frac{\pi }{2}+C$ and $B=\frac{\pi }{2}-2C$:

$2\sin \left(\frac{\pi }{2}-2C\right)=\sin \left(\frac{\pi }{2}+C\right)+\sin C$

$2\cos \left(2C\right)=\cos C+\sin C$

Using the identity $\cos \left(2C\right)={\cos }^{2}C-{\sin }^{2}C=\left(\cos C-\sin C\right)\left(\cos C+\sin C\right)$:

$2\left(\cos C-\sin C\right)\left(\cos C+\sin C\right)=\cos C+\sin C$

Since $C$ is an acute angle in a non-degenerate triangle, $\cos C+\sin C\ne 0$. Dividing both sides by $\cos C+\sin C$:

$2\left(\cos C-\sin C\right)=1$

$\cos C-\sin C=\frac{1}{2}$

Squaring both sides:

${\left(\cos C-\sin C\right)}^2=\frac{1}{4}$

${\cos }^{2}C+{\sin }^{2}C-2\sin C\cos C=\frac{1}{4}$

$1-\sin \left(2C\right)=\frac{1}{4} \Rightarrow \sin \left(2C\right)=\frac{3}{4}$

Step 3: Calculate the area and semi-perimeter of triangle ABC

We are given that the vertices lie on a circle of radius $R=1$.

The semi-perimeter $s$ of the triangle is:

$s=\frac{a+b+c}{2}=\frac{3b}{2}=\frac{3}{2}\left(2R\sin B\right)=3\sin B$

Note that $B=\frac{\pi }{2}-2C$, so $\sin B=\sin \left(\frac{\pi }{2}-2C\right)=\cos \left(2C\right)$.

Using $\sin \left(2C\right)=\frac{3}{4}$, we have:

$\cos \left(2C\right)=\sqrt{1-{\sin }^2\left(2C\right)}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

Thus, $\sin B=\frac{\sqrt{7}}{4}$, which gives:

$s=3\sin B=\frac{3\sqrt{7}}{4}$

The area $\Delta$ of triangle $ABC$ is given by:

$\Delta =\frac{1}{2}ac\sin B=\frac{1}{2}\left(2R\sin A\right)\left(2R\sin C\right)\sin B=2\sin A\sin C\sin B$

Since $A=\frac{\pi }{2}+C$, we have $\sin A=\cos C$:

$\Delta =2\cos C\sin C\sin B=\sin \left(2C\right)\sin B$

Substituting $\sin \left(2C\right)=\frac{3}{4}$ and $\sin B=\frac{\sqrt{7}}{4}$:

$\Delta =\left(\frac{3}{4}\right)\left(\frac{\sqrt{7}}{4}\right)=\frac{3\sqrt{7}}{16}$

Step 4: Find the inradius (r)

The inradius $r$ of a triangle is given by the ratio of its area to its semi-perimeter:

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

Substituting the values of $\Delta$ and $s$:

$r=\frac{{\displaystyle \frac{3\sqrt{7}}{16}}}{{\displaystyle \frac{3\sqrt{7}}{4}}}=\frac{4}{16}=0.25$

Thus, the inradius of triangle $ABC$ is 0.25.