Product of distances from a point on the circle to equidistant points on the circle

Question

Question (IIT 2023) : Let $A_i\ (1 \le i \le 8)$ be the vertices of the regular octagon that lie on the circle of radius $2$. Let $P$ be a point on the circle and let $|PA_i|$ denote the distance between the point $P$ and $A_i$ for $1\le i\le 8$. If $P$ varies over the circle, then the maximum value of the product is $\prod_{i=1}^{8}|PA_i|$, is:

My Solution :

As all the points are of a octagon they are equally aligned at $\dfrac{\pi}{4}$ with each other and taking point $P$ to be at an angle of $\theta$ with the $x$-axis we can rewrite the product as : $$\prod_{i=1}^{8} 4\sin\left(\frac{i\pi}{8}+\frac{\theta}{2}\right)$$ which can be further simplified into $2^9\sin(4\theta)$ which gives us our answer $2^9$ at $\theta = \dfrac{\pi}{8}$. Now there's another solution floating around online which relies on complex numbers and is quite shorter and it goes as following :

what I am not getting here is how does maximizing the product of the roots of $z^8-2^8=0$ produces the same result as asked in the question, as maximizing the roots of $z^8-2^8$ should give the value of product of the distances if $P$ were to be taken at center of circle. Is there any relation between my solution and the other which verifies this...

Answer 1

Indeed, this explanation isn't sufficient !

They do not show that the maximal value is obtained in particular for :

$$\omega := 2e^{i \pi/8}$$

Proof : $(\omega)^8-2^8=2^8(e^{i \pi/8})^8-2^8=2^8(-1)-2^8=-2^9$

which is (in absolute value) the maximal value (as established in the document) which can be taken by the product of distances.

A global understanding is provided by the graphical representation of the surface with equation

$$z=f(x,y)=2^8|(x+iy)^8 -1|.\tag{1}$$

(see below). Caution : letter $z$ in (1) is for the altitude coordinate ; it's not the complex number $z$).

Graphical representation of function $f$. Imagine it as a membrane constrained to pass through a metal thread (the red curve) ; its maximal value(s) $2^9=512$ are reached at points $\omega_k := 2e^{i(2k+1) \pi/8}$ - in particular for $k=0$ as shown above - alternating with minimal values $0$ at points $2e^{i(2k) \pi/8}$). Please note an internal "plateau" at altitude $\approx 2^8=256$.

Matlab program having generated the figure :

clear all;close all;hold on
I=-2.1:0.02:2.1;
[X,Y]=meshgrid(I,I);
Z=X+i*Y;
F=abs((Z.^8-256).*(abs(Z)<1.99));
% surface plot :
surfl(X,Y,F);shading interp;colormap(gray);view([50,20]);
% curve plot : 
t=0:0.01:2*pi;
plot3(2*cos(t),2*sin(t),abs(256*(exp(8*i*t)-1)),
'r','linewidth',10);

Answer 2

Let $\displaystyle P=2e^{i\alpha},\alpha =\frac{2\pi}{4}$ and

$A_{1}(z_{1})\,,\ A_{2}(z_{2})\, ,\cdots ,A_{8}(z_{8})$ be the vertices of a Regular octagon.

Then $\displaystyle z^{8}-2^{8}=(z-z_1)(z-z_2)\cdots (z-z_8)$

$\displaystyle |(z-z_1)(z-z_2)\cdots (z-z_8)\bigg|=|z^8-2^8|$

$\displaystyle \prod^{8}_{i=1}\bigg|z-z_i\bigg|=\bigg|2^8 e^{8i\alpha}-2^8\bigg|=2^8\bigg|e^{8i\alpha}-1\bigg|=2^8|e^{4i\alpha}\bigg(e^{4i\alpha}-e^{-4i\alpha}\bigg)\bigg|$

$\displaystyle \prod^{8}_{i=1}\bigg|PA_{i}\bigg| =2^{8}\cdot 1\cdot 2\cdot \bigg|\sin(4\alpha)\bigg|\leq 2^9$