Computing the maximum area for a N sided polygon whose vertices lie on N concentric circles respectively

Question

As an extension problem which my math teacher gave me, I was asked to solve for the maximum area of triangle, with each of its vertices on a circle of radius 3,4,5 respectively. I got the answer alright (20.495) but the solution is very awkward and the calculus approach I used is definitely not the most convenient one (the solution is sketched in this Desmos file: https://www.desmos.com/calculator/fydjs4wfbm). I am not that interested in this specific problem now and my Current question is how can one solve the problem when it's generalized to an N-sided polygon, with each of its vertex on a different circle, having N concentric circles in total whose radii can be any number as long as they have different values.

I have some code here that uses the method provided by Empy2 (thetaj=arccos(min(1,k/(rjrj+1))) )) (if you have your own idea then ignore it). It looks like a good estimate so I added the code here. But problem still exists when n<6, because k has to be negative for these ns and arccos(x) is undefined when x<-1, so there will be math error when the program does arccos(min(1,k/r1r2)), when k/r1r2 <-1.

import numpy as np
import matplotlib.pyplot as plt

fig, ax = plt.subplots(1, 1, figsize=(8, 8))

ax.axis('equal')

theta = np.linspace(0, 2 * np.pi, 1500)

# number of circles
#works for n>5
n = 3

radius = [i for i in range(1, n+1)]


def asum(k):
    Sum = 0

    for i in range(1, n):
        if i == n - 1:
            a = np.arccos(min(1, k/(radius[i]*(radius[i]+1))))
            if a != 0:
                Sum += a

        else:
            a = np.arccos(min(1, k/(radius[i]*radius[i+1])))

            if a != 0:
                Sum += a
    return Sum

#works for n>5
def findbest(s, e, n):

    real = 2*np.pi
    for i in range(50):
        mid = (s+e)/2
        midV = asum(mid)
        if midV < real:
            e = mid
        if midV > real:
            s = mid
    return (s+e)/2


#k=findbest(startvalue,endvalue,n)  add sensible start and END value                                           -70
s = 0
e = 100000000
k = findbest(s,e, n)

print("aprox k:", k)


angles = [0]
A = []

# update angles

for i in range(1, n):
    if i == n - 1:
        a = np.arccos(min(1, k/(radius[i]*(radius[i]+1))))
        angles.append(a+angles[len(angles)-1])

        if a != 0:
            A.append(a)

    else:
        a = np.arccos(min(1, k/(radius[i]*radius[i+1])))


        angles.append(a+angles[len(angles)-1])
        if a != 0:
            A.append(a)


X = []
Y = []
for i in range(len(angles)):
    X.append(radius[i] * np.cos(angles[i]))
    Y.append(radius[i] * np.sin(angles[i]))


area = 0.5*abs((X[len(X)-1]*Y[0] - Y[len(Y)-1] * X[0]))

for i in range(0, len(X)-1):
    area += 0.5*(X[i] * Y[i+1] - X[i+1] * Y[i])

print("Area:", area)

#####
# plot circle
xs = []
ys = []

for i in range(0, len(radius)):
    xs.append(radius[i] * np.cos(theta))
    ys.append(radius[i] * np.sin(theta))

for i in range(len(xs)):
    ax.plot(xs[i], ys[i], linewidth=0.4, c='black')
######


plt.plot(X, Y, linewidth=1.75, color='b')
plt.plot([radius[0], radius[len(radius)-1]*np.cos(sum(A))],
         [0, radius[len(radius)-1]*np.sin(sum(A))], color='b', linewidth=1.75)

print("Sum of angles:", "R:", sum(A), "D:", sum(A)*180/np.pi)
print(" 2pi - sum of angles:", "R:", sum(A)-2 *
      np.pi, "D:", 180/np.pi * (sum(A)-2*np.pi))

plt.show()

Answer 1

Here is my guess for raddi $(1,2,...,n)$. I think it gives an area $\pi n^2-O(n^{4/3})$.
Let $\phi_j$ be the angle of the $j$th point.
Give them radii in the order $ 1,3,5,...,n-3,n-1,n,n-2,n-4,...,2$. That maximises the sum of the product of adjacent radii.
Let $\phi_j=0$ for $j\lt n/2-n^{1/3}$. Let $\phi_j=2\pi$ for $j\gt n/2+n^{1/3}$.
Let the bulk of the $2\pi$ rotation be within a small set of vertices where the radii are largest.
The straight line between two adjacent vertices comes in a distance $O(n^{1/3})$ from the circumference of a circle of radius $n$ because the angle between verrices is $O(n^{-1/3})$. The radii of the vertices also come in a distance $O(n^{1/3})$ because there are $O(n^{1/3})$ of them. So there is a balance.
I haven't worked out more details.

The following figure is for 100 points, only nine of which are not on the line $\theta = 0$.