A topic from April

by t0rajir0u, Jun 5, 2007, 5:30 AM

Problem: Show that an equiangular polygon with a prime number of sides, all of which are also of rational length, must also be equilateral.

Proof: The prime condition suggests that geometry is not all there is to this problem; this, along with the rational condition, strongly suggests that some kind of number theory is required. We will begin with a complex number interpretation.

The existence of an equiangular $ n$-gon with side lengths $ a_{0}, a_{1}, ... a_{n-1}$ corresponds to the truth of the equation

$ \sum_{k = 0}^{n-1}a_{k}\omega^{k}= 0$

Where $ \omega = e^{i\frac{2\pi}{n}}$ is a primitive $ n^{th}$ root of unity. To see why this is true, consider the vector that points each vertex of an equiangular $ n$-gon to the next vertex; the angle that each successive vector makes with the $ x$-axis is $ 0,\frac{2\pi}{n},\frac{4\pi}{n}, ...$, and the lengths of these vectors are of course the lengths of the sides.

We are given the additional constraint that all of the side lengths must be rational. Here we recall the concept of minimal polynomials (read that post before tackling the following argument!).

As stated in the above post, the minimal polynomial of a primitive $ n^{th}$ root of unity $ \omega$ over the rationals is

$ \Phi_{n}(x) =\prod_{1\le d\le n, (d, n) = 1}(x-\omega^{d})$

For $ n$ a prime $ p$, this becomes

$ \Phi_{p}(x) = 1+x+x^{2}+...+x^{p-1}$

Recall the original side-length equation. It is a polynomial with rational coefficients of degree $ p-1$ such that $ \omega$ is a root. But since $ \Phi_{p}(x)$ is the minimal polynomial (which is of degree $ p-1$), it follows that $ \Phi_{p}(x)$ divides that polynomial; in other words,

$ \sum_{k = 0}^{p-1}a_{k}\omega^{k}= a_{p-1}\sum_{k = 0}^{p-1}\omega^{k}$

And therefore $ a_{0}= a_{1}= ... = a_{p-1}$; in other words, the polygon is equilateral. QED.


Extension: Describe the equiangular polygons with rational side lengths that are not equilateral.

Solution: Clearly, $ n$ cannot be prime. The first obvious example is that of a rectangle with rational side lengths $ a, b$. The next might be a hexagon with rational side lengths $ a, a, b, a, a, b$. These results can in fact be generalized in an algebraic way.

Again, let an equiangular $ n$-gon have sides $ a_{0}, ... a_{n-1}$. Let $ \omega = e^{i\frac{2\pi}{n}}$ be a primitive $ n^{th}$ root of unity. Again,

$ \sum_{k = 0}^{n-1}a_{k}\omega^{k}= 0$

It's more interesting from here. The cyclotomic polynomial $ \Phi_{n}(x)$ has degree $ \varphi(n) < n-1$ (for composite $ n$), so we have interesting solutions. In particular, the above polynomial can most generally take the form

$ \Phi_{n}(\omega )\sum_{k = 0}^{(n-1)-\varphi(n)}s_{k}\omega^{k}= 0$

Expanding and equating coefficients gives a description of all possible equiangular $ n$-gons. In particular, for $ n = 4$ we have $ \Phi_{4}(\omega) = (\omega^{2}+1)$ and therefore

$ (\omega^{2}+1)(a\omega+b) = a\omega^{3}+b\omega^{2}+a\omega+b$

Precisely the polynomial corresponding to a general rectangle, and for $ n = 6$ we have $ \Phi_{6}(\omega) = (\omega^{2}-\omega+1)$ and therefore

$ (\omega^{2}-\omega+1)(a\omega^{3}+b\omega^{2}+c\omega+d) = a\omega^{5}+(b-a)\omega^{4}+(a-b+c)\omega^{3}+(b-c+d)\omega^{2}+(c-d)\omega+d$

This expression is more complicated because, in fact, the general form of an equiangular hexagon is more complicated. The special case we identified before occurs when $ c = b, a = d$.


Further Extension: Describe all equiangular polygons.

Solution: Surprisingly, this case is more boring. For one thing, the prime restriction disappears when the rational restriction on side lengths is lifted.

When we are not restricted to the rationals, the minimal polynomial of any complex number $ a+bi$ over the reals is merely $ x^{2}-2ax+(a^{2}+b^{2})$ (this result is otherwise known as the conjugate root theorem), so, in other words, the minimal polynomial of any primitive $ n^{th}$ root of unity $ e^{i\frac{2\pi}{n}}$ over the reals is just

$ (x-e^{i\frac{2\pi}{n}})(x-e^{i\frac{2(n-1)\pi}{n}}) = x^{2}-2\cos\frac{2\pi}{n}x+1$

For $ n = 4, 6$ this polynomial corresponds to the minimal polynomial over the rationals because $ \varphi(4) =\varphi(6) = 2$, but of course this is not true generally.

The general description is rather annoying; it is given by the coefficients of

$ \left(\omega^{2}-2\cos\frac{2\pi}{n}\omega+1\right)\sum_{k = 0}^{n-3}s_{k}\omega^{k}= 0$

We see that fixing $ n-2$ sides of an equiangular $ n$-gon uniquely determines the other two. This makes geometric sense; extending the two unknown sides arbitrarily, they coincide at exactly one point, which determines their proper lengths. So this geometric fact is in fact an algebraic consequence of the conjugate root theorem!


The real beauty of these results, in my opinion, is the way in which a purely algebraic result is given a very intuitive geometric interpretation.

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