Let $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots, \mathrm{A}_{8}$ be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let $\mathrm{PA}_{\mathrm{i}}$ denote the distance between the points P and $\mathrm{A}_{\mathrm{i}}$ for $\mathrm{i}=1,2, \ldots, 8$. If P varies over the circle, then the maximum value of the product $\mathrm{PA}_{1} \cdot \mathrm{PA}_{2} \cdot \cdots \mathrm{PA}_{8}$, is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Complex Numbers

Let

A_{1}, A_{2}, A_{3}, \dots, A_{8}

be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let

PA_{i}

denote the distance between the points P and

A_{i}

for

i = 1, 2, \dots, 8

. If P varies over the circle, then the maximum value of the product

PA_{1} \cdot PA_{2} \cdot \dots PA_{8}

, is

Answer: 512

Place the octagon vertices at

A_{k} = 2 e^{iπ k /4}

for

k = 0, 1, \dots, 7

(using complex numbers).

These are the 8th roots of

2^{8} = 256

z^{8} = 256

.

For a point

P = 2 e^{i θ}

on the circle, the distances are:

P A_{k} = ∣2 e^{i θ} - 2 e^{iπ k /4} ∣ = 2∣ e^{i θ} - e^{iπ k /4} ∣

The product:

k = 0 \prod 7 P A_{k} = 2^{8} k = 0 \prod 7 ∣ e^{i θ} - e^{iπ k /4} ∣

Using the identity for roots: if

α

is an 8th root of unity, then

k = 0 \prod 7 (z - α^{k}) = z^{8} - 1

Setting

z = e^{i θ}

k = 0 \prod 7 ∣ e^{i θ} - e^{iπ k /4} ∣ = ∣ e^{i 8 θ} - 1∣ = 2∣ sin 4 θ ∣

Maximum when

∣ sin 4 θ ∣ = 1

Maximum = 2^{8} \cdot 2 = 512

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer