Diagonals of an n-gon

What happens if we take the product of the lengths of all of the diagonals which stem from a single vertex of a regular $n$ -gon? For example, for this regular pentagon (below), we’d want to calculate the product of all of the green segments.

Now, we can set the length from the center of the $n$ -gon to be anything, but we’ll normalize it to 1 (and it will make sense why later). We can try to calculate this by hand for a few values of $n$ to try to find the pattern that I’m getting at…

Let’s start with a regular pentagon, since we have the picture so conveniently at hand. For the longer diagonal, we can draw segments connecting the vertices to the center of the pentagon. Since we know the angle between these blue segments and that the lengths of the blue segments are both $1$ , we can use the law of cosines to calculate the length of the longer diagonal: $\sqrt{1^2+1^2-2\cdot 1\cdot 1\cdot \cos(144^{\circ})}=\sqrt{\frac12(5+\sqrt{5})}.$

We perform the same operation with the smaller diagonal. Using law of cosines again, we find the length of the smaller diagonal to be $\sqrt{1^2+1^2-2\cdot 1\cdot 1\cdot \cos(72^{\circ})}=\sqrt{\frac12(5-\sqrt{5})}.$

Immediately, we notice a symmetry between these two answers. Moreover, when we take the product of all of the green segments, the square-roots will disappear, since every diagonal’s length is repeated. So, the product of these diagonals is $\left(\sqrt{\frac12(5+\sqrt{5})}\right)^2\left(\sqrt{\frac12(5-\sqrt{5})}\right)^2=\frac12(5+\sqrt{5})\cdot \frac12(5-\sqrt{5})=5.$ Such a nice answer!

A regular hexagon yields much simpler values when we perform the same process: We get that the shortest diagonal is has length $1,$ the medium diagonal has length $\sqrt3,$ and the longest diagonal has length $2.$ Taking the product of all diagonals gives us $1^2\cdot (\sqrt3)^2\cdot 2=6.$

So perhaps you have noticed the pattern by now. The product of all the lengths of a diagonals from one vertex of a regular $n$ -gon seems to be $n$ !

Another curious thing one might notice is that if we take the sum of the squares of each length, we also get a nice number. (I accidentally stumbled upon this by typing in the wrong expression in WolframAlpha.) For the pentagon, we get $2\cdot \left(\sqrt{\frac12(5+\sqrt{5})}\right)^2+2\cdot \left(\sqrt{\frac12(5-\sqrt{5})}\right)^2=(5+\sqrt5)+(5-\sqrt5)=10.$ For the hexagon, we get $2\cdot 1^2+1\cdot (\sqrt3)^2 + 2^2=12.$ Yes, as expected, the sum of the squares of these lengths is indeed $2n,$ for a regular $n$ -gon.

But why are both of these patterns true? To answer this, we look to the complex plane.

We situate our $n$ -gon such that the center lies at the origin and the vertex from which our diagonals originate lies at $(1,0).$

If $\omega$ is the $n$ th root of unity, then the product of the diagonals is the product of the complex magnitudes: $|\omega-1|\cdot|\omega^2-1|\cdots|\omega^{n-1}-1|.$ Since all of the $n$ roots of unity are roots of the polynomial $x^n-1,$ we have

$\begin{aligned} x^n-1&=(x-1)(x-\omega)\cdots (x-\omega^{n-1}) \\ \frac{x^n-1}{x-1}&=(x-\omega)\cdots (x-\omega^{n-1}) \\ x^n+\cdots x+1&=(x-\omega)\cdots (x-\omega^{n-1}).\end{aligned}$

Plugging in $1$ to the polynomial, we get $n=(1-\omega)\cdots (1-\omega^{n-1})=|(\omega-1)\cdots(\omega^{n-1}-1)|=|\omega-1|\cdots|\omega^{n-1}-1|$ (complex magnitudes multiply), exactly as desired.

For the other part of the puzzle, that is, that the sum of the squares of the diagonals equals $2n,$ we must first recall a fact about complex numbers: $z\bar{z}=|z|^2,$ where $z\in\mathbb{C}.$ For our purposes, this means that $(\omega^k-1)(\omega^{n-k}-1)=|\omega^k-1|^2.$ As a result, we can write the sum of the products of the diagonals $|\omega-1|^2+|\omega^2-1|^2+\cdots +|\omega^{n-1}-1|^2$ as

$\begin{aligned} \sum_{k=1}^{n-1}|\omega^k-1|^2&=\sum_{k=1}^{n-1}(\omega^k-1)(\omega^{n-k}-1)\\ &=\sum_{k=1}^{n-1}(\omega^k\omega^{n-k}-\omega^k-\omega^{n-k}+1) \\ &=\sum_{k=1}^{n-1}(2-\omega^k-\omega^{n-k}) \\ &=(2n-2)-\sum_{k=1}^{n-1}(\omega^k+\omega^{n-k}) \\ &=2n-\sum_{k=0}^{n-1}(\omega^k+\omega^{n-k}) \\ &=2n-2\sum_{k=0}^{n-1}\omega^k \\ &=2n.\end{aligned}$

(The last equivalence we get by observing that $\omega$ is a root of $\frac{x^n-1}{x-1},$ a polynomial which can be rewritten as $x^{n-1}+\cdots+x+1.$ )

Now we have proved both of these patterns! I think this is a super neat problem, and kudos to Dr.V/Eric to introducing the first pattern to me.

If you want to do some experimentation of your own (and maybe find some other nifty patterns), here’s a Geogebra widget to play around with!

Source: Appeared in Dr.V’s “Cracking the AIME” minicourse + Eric’s Complex Analysis A class.

Diagonals of an n-gon

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