Determine Inradius of Obtuse Triangle with Vertices on Unit Circle

To find the inradius of the obtuse-angled triangle $ABC$, we proceed with the following steps:

Step 1: Set up the angle relations

Let the angles of $\triangle ABC$ be $A, B, C$ such that $A > B > C$. Given that $ABC$ is an obtuse-angled triangle and the difference between the largest and smallest angle is $\frac{\pi}{2}$, we have: $A - C = \frac{\pi}{2} \implies A = C + \frac{\pi}{2}$

Since the sum of the angles in a triangle is $\pi$: $A + B + C = \pi \implies \left(C + \frac{\pi}{2}\right) + B + C = \pi \implies B = \frac{\pi}{2} - 2C$

Step 2: Apply the Sine Rule and AP condition

The sides $a, b, c$ opposite to angles $A, B, C$ are in an arithmetic progression (A.P.). Since $A > B > C$, we have $a > b > c$, so: $2b = a + c$

Using the Sine Rule with circumradius $R = 1$: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = 2$

Thus, the side lengths are: $a = 2\sin A, \quad b = 2\sin B, \quad c = 2\sin C$

Substitute these into the arithmetic progression condition $2b = a + c$: $2(2\sin B) = 2\sin A + 2\sin C \implies 2\sin B = \sin A + \sin C$

Using the expressions for $A$ and $B$ in terms of $C$: $\sin A = \sin\left(C + \frac{\pi}{2}\right) = \cos C$ $\sin B = \sin\left(\frac{\pi}{2} - 2C\right) = \cos 2C$

Substituting these back yields: $2\cos 2C = \cos C + \sin C$

Step 3: Solve for the trigonometric values

We know that $\cos 2C = \cos^2 C - \sin^2 C = (\cos C - \sin C)(\cos C + \sin C)$. Therefore: $2(\cos C - \sin C)(\cos C + \sin C) = \cos C + \sin C$

Since $C \in \left(0, \frac{\pi}{4}\right)$, we have $\cos C + \sin C \neq 0$. Dividing both sides by $(\cos C + \sin C)$: $\cos C - \sin C = \frac{1}{2}$

Squaring both sides: $(\cos C - \sin C)^2 = \frac{1}{4} \implies 1 - 2\sin C \cos C = \frac{1}{4} \implies \sin 2C = \frac{3}{4}$

Then, $\cos 2C = \sqrt{1 - \sin^2 2C} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$.

Using the identity $(\cos C + \sin C)^2 = 1 + \sin 2C = 1 + \frac{3}{4} = \frac{7}{4}$, we obtain: $\cos C + \sin C = \frac{\sqrt{7}}{2}$

Now we can solve for $\cos C$ and $\sin C$: $\cos C = \frac{\sqrt{7} + 1}{4}, \quad \sin C = \frac{\sqrt{7} - 1}{4}$

Step 4: Calculate the semi-perimeter and area of $\triangle ABC$

The side lengths of the triangle are: $a = 2\cos C = \frac{\sqrt{7} + 1}{2}$ $b = 2\cos 2C = \frac{\sqrt{7}}{2}$ $c = 2\sin C = \frac{\sqrt{7} - 1}{2}$

The semi-perimeter $s$ is: $s = \frac{a + b + c}{2} = \frac{\frac{\sqrt{7} + 1}{2} + \frac{\sqrt{7}}{2} + \frac{\sqrt{7} - 1}{2}}{2} = \frac{3\sqrt{7}}{4}$

The area $\Delta$ of $\triangle ABC$ using $R = 1$ is: $\Delta = \frac{abc}{4R} = \frac{\left(\frac{\sqrt{7}+1}{2}\right) \left(\frac{\sqrt{7}-1}{2}\right) \left(\frac{\sqrt{7}}{2}\right)}{4} = \frac{\left(\frac{6}{4}\right) \left(\frac{\sqrt{7}}{2}\right)}{4} = \frac{3\sqrt{7}}{16}$

Step 5: Compute the inradius $r$

The inradius $r$ is given by the formula: $r = \frac{\Delta}{s} = \frac{\frac{3\sqrt{7}}{16}}{\frac{3\sqrt{7}}{4}} = \frac{4}{16} = \frac{1}{4} = 0.25$

Thus, the inradius of the triangle $ABC$ is $0.25$ (or $\frac{1}{4}$).