Does the golden angle produce maximally distant divisions of a circle?

I just ran across a video that claimed that the sequence of multiples of the golden angle produces some sort of optimal spacing around a circle for all possible iterations (this is a little hand-wavy, I'm aware).

Of course, any irrational angle has the property that it won't ever repeat and will form a dense set as the limit goes to infinity, but intuitively the golden angle seems to work much better than an irrational very close to $\frac{\pi}{2}$.

I'm looking for a way to more formally state this idea of optimal spacing for any iteration as a means to either prove or disprove that the golden angle provides this property (and is it unique?).

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Watching the video, it seems to me that what's nice about using this angle is as follows: the ratio between the smallest and largest gap between the angles $a_1,\ldots,a_n$, which are the first $n$ iterations produced by the given angle $a$ arranged from least to greatest, is constant. Does that seem right? – Alex Becker Dec 23 '11 at 5:51

Given an arbitrary angle $\theta$ which is not a multiple of $\pi$, to say that $n \theta$ and $m \theta$ are close to each other on the circle is to say that $(n-m)\theta$ is close to $0 \bmod 2\pi$, so to say that the fractional part of $(n-m) \frac{\theta}{2\pi}$ is close to $0$. If you want to avoid this, you want to pick a value of $t = \frac{\theta}{2\pi}$ with the property that, for any integer $n-m$, the fractional part of $(n-m)t$ is never too close to $0$.

Phrased another way, for any integer $q$ we never want $qt$ to be too close to another integer $p$. Phrased yet another way, we never want $t$ to be too close to a rational number $\frac{p}{q}$. So we are looking for irrational numbers which are hard to approximate by rationals.

The closest rational approximations (in a suitable sense) to an irrational number can be read off from truncations of its continued fraction $$t = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + ...}}$$

and in particular are better approximations if the corresponding entries $a_i$ of the continued fraction are large. That means that the best irrational number for the job is the one with the property that each $a_i$ is as small as possible, so $$1 + \frac{1}{1 + \frac{1}{1 + ...}}$$

which turns out to be precisely the golden ratio. (Why? Because the above number $t$ has the property that $t = 1 + \frac{1}{t}$, or $t^2 = t + 1$, so it is either the golden ratio or its conjugate, and it's greater than $1$ so it must be the golden ratio.)

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Very nice answer. +1 – Alex Becker Dec 23 '11 at 5:52
Great answer, thanks. As a side note, $\frac{1}{\phi}$ works just as well since it's the reverse direction on a circle. – cobbal Dec 23 '11 at 6:07
Does anybody know a citable (and more elaborate) version of this proof? – Stiefel Jun 11 '14 at 11:32