JEE Advanced 2023 Paper 2, Mathematics Section 4 Q1: Solution of Triangles

Let the angles be $A<C<B$ with $B-A=\frac{\pi }{2}$, and write $A=\alpha$, so $B=\frac{\pi }{2}+\alpha$ and $C=\frac{\pi }{2}-2\alpha$ (using $A+B+C=\pi$).

The sides opposite are in AP, say $n-d,n,n+d$ (opposite to $A,C,B$). With circumradius $R=1$, the law of sines gives:

$n-d=2\sin \alpha ,n+d=2\sin \left(\frac{\pi }{2}+\alpha \right)=2\cos \alpha ,n=2\sin \left(\frac{\pi }{2}-2\alpha \right)=2\cos 2\alpha .$

Adding the first two: $2n=2(\sin \alpha +\cos \alpha )$, so $\cos 2\alpha =\frac{\sin \alpha +\cos \alpha }{2}$. Using $\cos 2\alpha =(\cos \alpha -\sin \alpha )(\cos \alpha +\sin \alpha )$,

$2(\cos \alpha -\sin \alpha )=1\Rightarrow 1-\sin 2\alpha =\frac{1}{4}\Rightarrow \sin 2\alpha =\frac{3}{4}.$

Then $\cos 2\alpha =\frac{\sqrt{7}}{4}$ (positive since $\alpha$ is small). The area is

$a=\frac{1}{2}n(n+d)\sin \alpha =\frac{1}{2}(2\cos 2\alpha )(2\cos \alpha )\sin \alpha =\sin 2\alpha \cdot \cos 2\alpha =\frac{3}{4}\cdot \frac{\sqrt{7}}{4}=\frac{3\sqrt{7}}{16}.$

Therefore $(64a{)}^2={\left(64\cdot \frac{3\sqrt{7}}{16}\right)}^2=(12\sqrt{7}{)}^2=144\cdot 7=1008.$