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Let $$A_1, A_2, A_3, \ldots, A_8$$ 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 $$i = 1, 2, \ldots, 8$$. If P varies over the circle, then the maximum value of the product $$PA_1 \cdot PA_2 \cdots PA_8$$ is
Correct Answer: 512
Let the centre of the circle be $$O$$ and take the complex plane with $$O$$ at the origin.
Because the radius is $$2$$, every vertex of the regular octagon can be written as
$$A_k :\; 2\,\zeta^{\,k}, \qquad k = 0,1,\dots ,7$$
where $$\zeta = e^{\,i\pi/4}$$ is a primitive eighth root of unity.
Let the variable point on the circle be $$P :\; 2\,e^{\,i\theta}, \qquad 0\le \theta \lt 2\pi.$$ The distance between $$P$$ and $$A_k$$ is therefore
$$PA_k \;=\; \bigl|2e^{\,i\theta} - 2\zeta^{\,k}\bigr| = 2\,\bigl|e^{\,i\theta} - \zeta^{\,k}\bigr|.$$ Hence the required product is
$$\begin{aligned} PA_1\cdot PA_2\cdots PA_8 &= \prod_{k=0}^{7} 2\,\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| \\ &= 2^{8}\;\prod_{k=0}^{7} \bigl|e^{\,i\theta}-\zeta^{\,k}\bigr|. \end{aligned}$$
To evaluate the remaining product, recall the factorisation identity
$$\prod_{k=0}^{7}\bigl(z-\zeta^{\,k}\bigr) = z^{8}-1.$$
Taking modulus on the unit circle (i.e. $$|z|=1$$, put $$z=e^{\,i\theta}$$) gives
$$\prod_{k=0}^{7}\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| = \bigl|e^{\,i8\theta}-1\bigr|.$$
Using $$e^{\,ix}-1 = 2i\sin\dfrac{x}{2}\,e^{\,ix/2},$$ we have
$$\bigl|e^{\,i8\theta}-1\bigr| = 2\bigl|\sin 4\theta\bigr|.$$ Therefore
$$PA_1\cdot PA_2\cdots PA_8 = 2^{8}\;\cdot 2\bigl|\sin 4\theta\bigr| = 2^{9}\bigl|\sin 4\theta\bigr|.$$
The maximum value of $$|\sin 4\theta|$$ is $$1$$, so the maximum value of the whole product is
$$2^{9}\times 1 = 512.$$
Hence the required maximum value is 512.
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