Calculate Area Expression for Obtuse Triangle with Sides in AP

To find the value of $(64a)^2$, we analyze the properties of triangle $ABC$:

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

Using the angle sum property of a triangle $A + B + C = \pi$: $\left(C + \frac{\pi}{2}\right) + B + C = \pi \implies B = \frac{\pi}{2} - 2C$

1. Sides in Arithmetic Progression (AP): Since $A > B > C$, the sides opposite to these angles satisfy $a > b > c$. Given that the sides $a, b, c$ are in AP, we have: $2b = a + c$

By the Sine Rule, the sides of the triangle inscribed in a circle of radius $R = 1$ are given by $a = 2R\sin A = 2\sin A$, $b = 2\sin B$, and $c = 2\sin C$. Thus: $2\sin B = \sin A + \sin C$

Substitute $A = C + \frac{\pi}{2}$ and $B = \frac{\pi}{2} - 2C$ into the equation: $2\sin\left(\frac{\pi}{2} - 2C\right) = \sin\left(C + \frac{\pi}{2}\right) + \sin C$ $2\cos 2C = \cos C + \sin C$

1. Solving for Trigonometric Ratios of $C$: Using the identity $\cos 2C = \cos^2 C - \sin^2 C = (\cos C - \sin C)(\cos C + \sin C)$: $2(\cos C - \sin C)(\cos C + \sin C) = \cos C + \sin C$

Since $0 < C < \frac{\pi}{4}$, $\cos C + \sin C \neq 0$. Dividing both sides by $(\cos C + \sin C)$: $2(\cos C - \sin C) = 1 \implies \cos C - \sin C = \frac{1}{2}$

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

Since $2C \in \left(0, \frac{\pi}{2}\right)$, $\cos 2C > 0$: $\cos 2C = \sqrt{1 - \sin^2 2C} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \frac{\sqrt{7}}{4}$

1. Calculating Area $a$: The area $a$ of a triangle inscribed in a circumcircle of radius $R = 1$ is given by: $a = \frac{abc}{4R} = \frac{(2\sin A)(2\sin B)(2\sin C)}{4(1)} = 2\sin A \sin B \sin C$

Substituting $A = C + \frac{\pi}{2}$ and $B = \frac{\pi}{2} - 2C$: $a = 2 \cos C \cdot \cos 2C \cdot \sin C = (2\sin C \cos C) \cos 2C = \sin 2C \cos 2C$

Substitute the values of $\sin 2C$ and $\cos 2C$: $a = \frac{3}{4} \times \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{16}$

1. Calculating $(64a)^2$: $64a = 64 \times \frac{3\sqrt{7}}{16} = 12\sqrt{7}$

$(64a)^2 = (12\sqrt{7})^2 = 144 \times 7 = 1008$