Animated Solution for Mathematics - Trigonometry: Comprehension Passage

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a geometric puzzle that feels like a symphony of trigonometry.

We are given an obtuse-angled triangle $ABC$ inscribed in a circle of radius $R=1$. The sides are in arithmetic progression, and the angles have a specific, elegant relationship.

Imagine the triangle sitting inside the unit circle. We are told the difference between the largest angle $B$ and the smallest angle $A$ is $\frac{\pi }{2}$.

We write $B-A=\frac{\pi }{2}$, or $B=A+\frac{\pi }{2}$. Now, consider the sum of angles in any triangle: $A+B+C=\pi$.

By substituting our expression for $B$, we find $A+(A+\frac{\pi }{2})+C=\pi$, which simplifies beautifully to:

Just like that, we have reduced the entire angular complexity of the triangle to a single variable, $A$. This is the power of reduction—taking a complex system and finding its heartbeat.

Next, we tackle the arithmetic progression of the sides. If the sides $a,c,b$ are in AP, then $a+b=2c$.

But how do we connect sides to angles? The Sine Rule is our bridge:

Since $R=1$, we have $a=2\sin A$, $b=2\sin B$, and $c=2\sin C$. Substituting these into our AP condition, we get $2\sin A+2\sin B=2(2\sin C)$, or simply:

This is where the magic happens. We substitute our expressions for $B$ and $C$:

Now, let us simplify. Note that $\sin (A+\frac{\pi }{2})=\cos A$ and $\sin (\frac{\pi }{2}-2A)=\cos 2A$. Our equation becomes:

Recall the double angle identity: $\cos 2A={\cos }^{2}A-{\sin }^{2}A=(\cos A-\sin A)(\cos A+\sin A)$. Substituting this, we have:

Since $\sin A+\cos A$ is non-zero, we divide both sides by it to get $1=2(\cos A-\sin A)$. To solve for $A$, we square both sides:

Since ${\cos }^{2}A+{\sin }^{2}A=1$ and $2\sin A\cos A=\sin 2A$, we get $1=4(1-\sin 2A)$. Solving this yields:

The area $\Delta$ of a triangle is given by $2R^2\sin A\sin B\sin C$. With $R=1$, this becomes $2\sin A\cos A\cos 2A$, which simplifies to $\sin 2A\cos 2A$.

We know $\sin 2A=\frac{3}{4}$, so $\cos 2A=\sqrt{1-(\frac{3}{4}{)}^2}=\frac{\sqrt{7}}{4}$. Thus:

The question asks for $(64\Delta {)}^2$. Calculating $64\times \frac{3\sqrt{7}}{16}$ gives $12\sqrt{7}$. Squaring this, we get $144\times 7=1008$.

Finally, for the inradius $r=\frac{\Delta }{s}$, where $s=\frac{3c}{2}$, we find $r=\frac{2\Delta }{3c}$. Using $c=2\sin C=2\cos 2A$, the terms cancel out to leave:

We have conquered the problem, not by brute force, but by following the elegant path of geometry. The final result is 1008.