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The wiki for regular polygon diagonals says:

"For a regular n-gon inscribed in a unit-radius circle, the product of the distances from a given vertex to all other vertices (including adjacent vertices and vertices connected by a diagonal) equals n."

Can someone provide a basic proof?

https://en.wikipedia.org/wiki/Regular_polygon#Diagonals

sambajetson
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1 Answers1

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Let $f = x-1$ and $g = (x^n-1)/(x-1) = 1+x+x^2+\ldots+x^{n-1}$. Together, $f$ and $g$ have as zeros all the vertices of the unit $n$-gon in the complex plane.

The number we are looking for is now the absolute value of the product $R$ of all terms $a-b$, where $a$ is a zero of $f$ and $b$ is a zero of $g$. This $R$ is known as the resultant of $f$ and $g$.

This resultant is also the determinant of the Sylvester matrix of $f$ and $g$, which is constructed from the coefficients of $f$ and $g$. In this case, the Sylvester matrix has the form

$$ \begin{pmatrix} -1 & & & 1 \\ 1 & \ddots & & \vdots \\ & \ddots & -1 & \vdots \\ & & 1 & 1 \end{pmatrix} $$

and thus the determinant $\pm n$. Hence $|R|= n$.

Anon
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