How to derive a formula for finding the number of diagonals in a n-sided regular polygon.

Question

The formula for finding the number of diagonals in a n-sided convex polygon is:

$$\frac{(n-3)n}{2}$$

But how is this formula derived? How would I want to start deriving this formula?

Answer 1

For each vertex $n $(vertices) you have $(n-3)$ possible vertices to create a diagonal from.

Why $(n-3)$? Because you can't combine with the 2 neighboring vertices and you can't combine with your current vertex. Divide by two because you are counting twice. So you have $\frac{(n-3)n}{2}$ diagonals.

Answer 2

We can solve this by using combination, to make diagonals we need to choose no. of pairs of 2 vertices that can be formed from $n$ vertices i.e $$ \dbinom{n}{2}$$ we will get the answer, but we need to subtract $n$ from it since adjacent vertices cannot make a diagonal. $\therefore$ the final answer is $$\dbinom{n}{2}-n= \dfrac{(n-1)n-2n}{2}=\dfrac{n(n-3)}{2}.$$

Answer 3

Here’s a unique way of deriving the formula. Suppose $p(x)=x^n-1=0$, such that $n \geq 3$. Then $p(x)$ factors as: $x^n-1=\prod_{k=0}^{n-1}\left(x-\zeta_n^k\right)=(x-1)\left(x-\zeta_n\right) \ldots\left(x-\zeta_n^{n-1}\right)=0$. This factorization implies that $x=1, \zeta_n, \ldots, \zeta_n^{n-1}$ where $\zeta_n \neq 1$. Therefore the splitting field of $x^n-1=0$ is $\mathbb{Q}\left(\zeta_n\right)$.

$\zeta_n$ is a primitive $n^{\text {th}}$ root of unity, meaning it generates all other roots of unity, so $\left|Aut\left[\mathbb{Q}\left(\zeta_n\right) / \mathbb{Q}\right]\right|=\phi(n)$, where $\phi$ is Euler’s totient function. Let $\sigma^k$ map $\zeta_n$ to $\zeta_n^{k+1}$. Then $\sigma\left(\zeta_n\right)=\zeta_n^{n+2}, \sigma^2\left(\zeta_n\right)=\zeta_n^{n+3}$, and so on. If $\zeta_n^n$ produces a shift of $360^{\circ}$ around the complex unit circle, then $\sigma\left(\zeta_n\right), \ldots, \sigma^{n-1}\left(\zeta_n\right)$ simulates a shift of $\frac{360^{\circ}}{n}$, or $\frac{2 \pi}{n}$ radians. Thus all automorphisms of $\mathbb{Q}\left(\zeta_n\right)$ are symmetrical w.r.t its vertices on the complex unit circle. This implies that every respective conjugate of $\zeta_n$ lies exactly $180^{\circ}$ from each other.

Since all regular polygons are symmetrical, the action of each automorphism in Aut$\left[\mathbb{Q}\left(\zeta_n\right) / \mathbb{Q}\right]$ on $\zeta_n$ forms an n-gon on the complex plain.

Now, we know that the first 2 roots of unity, one trivial and the other being primitive, are conjugate pairs of each other. Therefore we must choose two of them, which will be given by $_nC_2=\frac{n(n-1)}{2}$. But, we need to subtract n from it since no diagonals can be formed using adjacent vertices. Thus, $\frac{n(n-1)}{2} - n$ = $\frac{n(n-1)-2n}{2}$. This implies that $\frac{n(n-3)}{2}$ must be the number of diagonals of any n-sided polygon.

In conclusion, by leveraging the properties of polynomial factorization, field theory, and automorphisms, we have derived a concise formula for the number of diagonals in any (n)-sided polygon. The expression $\frac{n(n-3)}{2}$ encapsulates the count of non-adjacent vertex pairs, considering the symmetrical actions of automorphisms on the roots of unity in the complex plane. This result not only provides a mathematical representation of polygon diagonals but also showcases the elegant interplay between algebraic structures and geometric concepts.