Product of Diagonals from Point of Regular Polygon

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A_0, A_1, \ldots, A_{n - 1}$ be the vertices of a regular $n$-gon $P = A_0 A_1 \cdots A_{n - 1}$ which is circumscribed by a unit circle.


Then:

$\ds \prod_{k \mathop = 2}^{n - 2} A_0 A_k = \frac {n \csc^2 \frac \pi n} 4$

where $A_0 A_k$ is the length of the line joining $A_0$ to $A_k$.


Proof

First it is worth examining the degenerate case $n = 3$, where there are no such lines.

In this case:

\(\ds \frac {n \csc^2 \frac \pi n} 4\) \(=\) \(\ds \frac 4 3 \csc^2 \frac \pi 3\)
\(\ds \) \(=\) \(\ds \frac 3 4 \paren {\frac {2 \sqrt 3} 3}^2\) Cosecant of $60 \degrees$
\(\ds \) \(=\) \(\ds 1\)

The continued product $\ds \prod_{k \mathop = 2}^{n - 2} A_0 A_k$ is vacuous here.

By convention, such a vacuous product is defined as being equal to $1$.

So the result holds for $n = 3$.


RegularPolygonWithDiagonals.png


Let $n > 3$.

The illustration above is a diagram of the case where $n = 7$.

From Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle, the vertices of $P$ can be identified with the complex $n$th roots of unity.

Thus:

\(\ds \prod_{k \mathop = 2}^{n - 2} A_1 A_k\) \(=\) \(\ds \prod_{k \mathop = 2}^{n - 2} \paren {1 - \alpha^k}\) where $\alpha = e^{2 i \pi / n}$ is the first complex $n$th root of unity
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {1 - \alpha} \paren {1 - \alpha^{n - 1} } } \prod_{k \mathop = 1}^{n - 1} \paren {1 - \alpha^k}\)
\(\ds \) \(=\) \(\ds \dfrac n {\paren {1 - \alpha} \paren {1 - \alpha^{n - 1} } }\) Product of Differences between 1 and Complex Roots of Unity
\(\ds \) \(=\) \(\ds \dfrac n {\paren {1 - \paren {\cos \frac {2 \pi} n + i \sin \frac {2 \pi} n} } \paren {1 - \paren {\cos \frac {2 \paren {n - 1} \pi} n + i \sin \frac {2 \paren {n - 1} \pi} n} } }\) Definition of $\alpha$
\(\ds \) \(=\) \(\ds \dfrac n {\paren {1 - \paren {\cos \frac {2 \pi} n + i \sin \frac {2 \pi} n} } \paren {1 - \paren {\cos \frac {2 \pi} n - i \sin \frac {2 \pi} n} } }\)
\(\ds \) \(=\) \(\ds \dfrac n {\paren {1 - \cos \frac {2 \pi} n}^2 + \paren {\sin \frac {2 \pi} n}^2}\)
\(\ds \) \(=\) \(\ds \dfrac n {1 - 2 \cos \frac {2 \pi} n + \paren {\cos \frac {2 \pi} n}^2 + \paren {\sin \frac {2 \pi} n}^2}\)
\(\ds \) \(=\) \(\ds \dfrac n {2 - 2 \cos \frac {2 \pi} n}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \dfrac n {2 - 2 \paren {1 - 2 \sin^2 \frac \pi n} }\) Double Angle Formula for Cosine: Corollary $2$
\(\ds \) \(=\) \(\ds \dfrac n {2 - 2 + 4 \sin^2 \frac \pi n}\)
\(\ds \) \(=\) \(\ds \frac {n \csc^2 \frac \pi n} 4\) Definition of Cosecant

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Product_of_Diagonals_from_Point_of_Regular_Polygon&oldid=662650"