Let the angles of the triangle be $A, B, C$ and the sides opposite to these angles be $a, b, c$. The triangle is inscribed in a circle of radius $R=1$.

Step 1: Formulate equations from the given conditions

1. Angles: The sides are in arithmetic progression (AP). Let's assume the side lengths are in increasing order, $a < b < c$. This implies the angles opposite to them are also in increasing order, $A < B < C$. The triangle is obtuse, so the largest angle $C > \frac{\pi}{2}$. The difference between the largest and smallest angle is $\frac{\pi}{2}$, so we have: $C - A = \frac{\pi}{2} \implies C = A + \frac{\pi}{2}$
2. Angle Sum Property: The sum of angles in a triangle is $\pi$: $A + B + C = \pi$ Substituting $C = A + \frac{\pi}{2}$, we get: $A + B + (A + \frac{\pi}{2}) = \pi \implies 2A + B = \frac{\pi}{2} \implies B = \frac{\pi}{2} - 2A$
3. Sides in AP: The sides $a, b, c$ are in AP, which means $a+c = 2b$.
4. Sine Rule: According to the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Since the circumradius $R=1$, we have $a = 2\sin A$, $b = 2\sin B$, and $c = 2\sin C$.

Step 2: Solve for the angles

Substitute the expressions for sides from the Sine Rule into the AP condition: $2\sin A + 2\sin C = 2(2\sin B) \implies \sin A + \sin C = 2\sin B$ Now, substitute the relations for angles $B$ and $C$ in terms of $A$: $\sin A + \sin(A + \frac{\pi}{2}) = 2\sin(\frac{\pi}{2} - 2A)$ Using the trigonometric identities $\sin(x+\frac{\pi}{2})=\cos x$ and $\sin(\frac{\pi}{2}-x)=\cos x$, the equation becomes: $\sin A + \cos A = 2\cos(2A)$ Using the double angle identity $\cos(2A) = \cos^2 A - \sin^2 A = (\cos A - \sin A)(\cos A + \sin A)$: $\sin A + \cos A = 2(\cos A - \sin A)(\cos A + \sin A)$ Since $A$ is an angle of a triangle, $\sin A + \cos A = \sqrt{2}\sin(A+\pi/4) eq 0$. We can divide both sides by $(\sin A + \cos A)$: $1 = 2(\cos A - \sin A) \implies \cos A - \sin A = \frac{1}{2}$ To find the values of trigonometric functions, we can use the relation $\sin A + \cos A = 2\cos(2A)$. We need to find $\cos(2A)$ first. Squaring $\cos A - \sin A = \frac{1}{2}$: $(\cos A - \sin A)^2 = (\frac{1}{2})^2$ $\cos^2 A + \sin^2 A - 2\sin A \cos A = \frac{1}{4}$ $1 - \sin(2A) = \frac{1}{4} \implies \sin(2A) = \frac{3}{4}$ Since $B = \frac{\pi}{2} - 2A$ and $B>0$, we have $2A < \frac{\pi}{2}$, which means $\cos(2A)$ must be positive. $\cos(2A) = \sqrt{1 - \sin^2(2A)} = \sqrt{1 - (\frac{3}{4})^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$ Now we can find the sines of all the angles:

• $\sin B = \sin(\frac{\pi}{2} - 2A) = \cos(2A) = \frac{\sqrt{7}}{4}$.
• From $\sin A + \cos A = 2\cos(2A)$, we have $\sin A + \cos A = 2(\frac{\sqrt{7}}{4}) = \frac{\sqrt{7}}{2}$. We now have a system of equations:
  1. $\cos A - \sin A = \frac{1}{2}$
  2. $\cos A + \sin A = \frac{\sqrt{7}}{2}$ Adding the two equations gives $2\cos A = \frac{1+\sqrt{7}}{2} \implies \cos A = \frac{1+\sqrt{7}}{4}$. Subtracting the first from the second gives $2\sin A = \frac{\sqrt{7}-1}{2} \implies \sin A = \frac{\sqrt{7}-1}{4}$.
• $\sin C = \sin(A + \frac{\pi}{2}) = \cos A = \frac{1+\sqrt{7}}{4}$.

We verify that the triangle is obtuse. The largest angle is $C$. $\cos C = \cos(A+\pi/2) = -\sin A = -\frac{\sqrt{7}-1}{4} < 0$, so $C > \frac{\pi}{2}$. The condition is satisfied.

Step 3: Calculate the area of the triangle

The area of a triangle, let's call it $a_{area}$ to avoid confusion with side $a$, inscribed in a circle of radius $R$ is given by the formula: $a_{area} = 2R^2 \sin A \sin B \sin C$ The problem denotes the area as $a$. With $R=1$: $a = 2(1)^2 \left(\frac{\sqrt{7}-1}{4}\right) \left(\frac{\sqrt{7}}{4}\right) \left(\frac{\sqrt{7}+1}{4}\right)$ $a = 2 \frac{(\sqrt{7}-1)(\sqrt{7}+1)\sqrt{7}}{4 \times 4 \times 4} = 2 \frac{(7-1)\sqrt{7}}{64} = 2 \frac{6\sqrt{7}}{64} = \frac{12\sqrt{7}}{64} = \frac{3\sqrt{7}}{16}$

Step 4: Calculate the final value

We need to find the value of $(64a)^2$. First, calculate $64a$: $64a = 64 \times \frac{3\sqrt{7}}{16} = 4 \times 3\sqrt{7} = 12\sqrt{7}$ Now, square this result: $(64a)^2 = (12\sqrt{7})^2 = 12^2 \times (\sqrt{7})^2 = 144 \times 7$ $144 \times 7 = 1008$ Thus, the value of $(64a)^2$ is 1008.