Maximum Product of Distances from Point on Circle to Regular Octagon Vertices

To find the maximum value of the product of distances $PA_1 \cdot PA_2 \dots PA_8$, we can use the complex plane representation.

Let the center of the circle of radius $R = 2$ be placed at the origin of the complex plane.

The vertices $A_1, A_2, \dots, A_8$ of the regular octagon lie on the circle $|z| = 2$. Without loss of generality, we can orient the octagon such that its vertices are represented by the complex numbers: $z_k = 2 e^{i \frac{2k\pi}{8}} = 2 \omega^k, \quad \text{for } k = 1, 2, \dots, 8$ where $\omega = e^{i \frac{2\pi}{8}} = e^{i \frac{\pi}{4}}$ is a primitive $8^{\text{th}}$ root of unity.

The complex numbers $z_1, z_2, \dots, z_8$ are the eight roots of the equation: $z^8 - 2^8 = 0 \implies z^8 - 256 = 0$

Therefore, the polynomial factors as: $\prod_{k=1}^{8} (z - z_k) = z^8 - 256$

Since the point $P$ lies on the circle of radius $2$, its position can be represented by the complex number $z = 2 e^{i\theta}$ for some $\theta \in \mathbb{R}$.

The distance between $P$ and $A_k$ is given by $PA_k = |z - z_k|$. The product of these distances is: $\prod_{k=1}^{8} PA_k = \prod_{k=1}^{8} |z - z_k| = \left| \prod_{k=1}^{8} (z - z_k) \right| = |z^8 - 256|$

Substituting $z = 2 e^{i\theta}$ into the expression: $z^8 = (2 e^{i\theta})^8 = 256 e^{i 8\theta}$

So, the product of the distances becomes: $\prod_{k=1}^{8} PA_k = |256 e^{i 8\theta} - 256| = 256 \left| e^{i 8\theta} - 1 \right|$

Using the identity $|e^{i\phi} - 1| = \sqrt{(\cos\phi - 1)^2 + \sin^2\phi} = \sqrt{2 - 2\cos\phi} = 2\left|\sin\left(\frac{\phi}{2}\right)\right|$: $\prod_{k=1}^{8} PA_k = 256 \times 2 |\sin(4\theta)| = 512 |\sin(4\theta)|$

The maximum value of $|\sin(4\theta)|$ is $1$, which occurs when $e^{i 8\theta} = -1$ (i.e., when $P$ lies at the midpoint of the arc between two adjacent vertices).

Thus, the maximum value of the product $PA_1 \cdot PA_2 \dots PA_8$ is: $512 \times 1 = 512$