The vertices of the n-gon are the roots of x^n-1, so the vertices other than 1 are the roots of (x^n-1)/(x-1) = 1 + x + x^2 + ... + x^(n-1). This is the product of x-root for the n-1 other roots, so plug in x=1 and you get the product of the complex-number difference from 1 to each other root; the norm of the result is the product of the distances that you asked about. But evaluating the polynomial at x=1 you get n (and so no taking-the-norm step is needed). --Michael On Sun, Aug 19, 2018 at 3:33 PM Adam P. Goucher <apgoucher@gmx.com> wrote:

The first product is, by definition, equal to the magnitude of the Vandermonde determinant (where omega is an Nth root of unity):

Prod[0 <= i < j < N] (omega^i - omega^j)

which is equal to the determinant of the Discrete Fourier Transform matrix M_ij := (omega^(ij)). Now, if we prepend the scaling factor of 1/sqrt(N), the DFT is unitary. Without this scaling factor, it therefore has a determinant of 1/sqrt(N)^N = N^(N/2).

Is there a more elementary way to see this (not relying on the Vandermonde determinant)?

Best wishes,

Adam P. Goucher

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Sent: Sunday, August 19, 2018 at 10:42 PM From: "Mike Speciner" <ms@alum.mit.edu> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Chord length product

If you inscribe a regular N-gon in a unit circle, then the product of all the vertex-to-vertex chord lengths is N**(N/2), and the product of all the fixed-vertex-to-other-vertex chord lengths is N. Is there some relatively intuitive reason that either of these should be true?

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-- Forewarned is worth an octopus in the bush.

Thank you. These are all lovely elegant proofs (e.g., DFT Vandermonde determinant for all chords and polynomial ((x+1)**N-1)/x for chords from fixed vertex, perhaps also noting that the product of the roots of a polynomial is given by the constant term) of the facts I asked about, and I was actually aware of similar proofs, but I don't really find any intuition in these proofs. To me, they don't answer the question of "why should this be?" I was hoping for something perhaps geometric, although a physics interpretation would also be interesting. I admit that interpreting the product naturally as a hypervolume seems unlikely to provide much help. --ma On 20-Aug-18 07:39, Veit Elser wrote:

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On Aug 20, 2018, at 1:08 AM, Michael Kleber <michael.kleber@gmail.com> wrote:

The vertices of the n-gon are the roots of x^n-1, so the vertices other than 1 are the roots of (x^n-1)/(x-1) = 1 + x + x^2 + ... + x^(n-1). This is the product of x-root for the n-1 other roots, so plug in x=1 and you get the product of the complex-number difference from 1 to each other root; the norm of the result is the product of the distances that you asked about. But evaluating the polynomial at x=1 you get n (and so no taking-the-norm step is needed).

--Michael I lost you somewhere in the middle. I would have said the chords you want are the roots of ((x+1)^n-1)/x = 1 + (x+1) + … +(x+1)^(n-1), and plugging in x=0 gives their product (which is the product of the distances up to a phase).

But is it intuitive? Was Mike looking for a physics interpretation?

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