JEE Advanced 2023 — Paper 2 — Question 11

Represent vertices as complex numbers on circle of radius 2: $A_k = 2e^{i\frac{2k\pi}{8}} = 2e^{i\frac{k\pi}{4}}$, $k=1,2,\dots,8$. Let point P be $P = 2e^{i\theta}$ on the same circle. Distance $PA_k = |2e^{i\theta} - 2e^{i\frac{k\pi}{4}}| = 2|e^{i\theta} - e^{i\frac{k\pi}{4}}|$. Using $|e^{i\alpha} - e^{i\beta}| = 2\left|\sin\frac{\alpha-\beta}{2}\right|$, we get $PA_k = 4\left|\sin\left(\frac{\theta}{2} - \frac{k\pi}{8}\right)\right|$. Thus the product $P = \prod_{k=1}^{8} PA_k = 4^8 \prod_{k=1}^{8} \left|\sin\left(\frac{\theta}{2} - \frac{k\pi}{8}\right)\right| = 2^{16} \prod_{k=1}^{8} \left|\sin\left(\frac{\theta}{2} - \frac{k\pi}{8}\right)\right|$. Use the identity $\prod_{k=1}^{8} \sin\left(x - \frac{k\pi}{8}\right) = \frac{\sin(8x)}{2^7}$. For $x = \frac{\theta}{2}$, $\prod_{k=1}^{8} \sin\left(\frac{\theta}{2} - \frac{k\pi}{8}\right) = \frac{\sin(4\theta)}{2^7}$, and absolute value gives $\frac{|\sin(4\theta)|}{2^7}$. Substitute: $P = 2^{16} \cdot \frac{|\sin(4\theta)|}{2^7} = 2^9 |\sin(4\theta)|$. Since $|\sin(4\theta)| \le 1$, the maximum value of $P$ is $2^9 = 512$.