Proof that golden angle successively divides the largest gap by the golden ratio? The golden angle divides the circumference of a circle by the golden ratio.
"If radial spokes are placed successively into the circle, each spaced by a golden angle increment, then each additional spoke divides the currently largest existing azimuthal gap between two successive spokes."
I have read frequently about this property of the golden angle, and I can see this behaviour by plotting (or calculating) the first few spokes.
Does anybody know a proof for the above mentioned property of the golden angle (preferably citeable)?
Note:
The following answer proves that the golden angle is the irrational angle wich leads to spokes with a maximal distance to its neighbours, but it does not explicitly prove the above mentioned property: Does the golden angle produce maximally distant divisions of a circle?
 A: Conclusions
My analysis shows that adding $F_n$ (the $n$-th Fibonacci number) successive golden angles corresponds to an angle of plus or minus $\theta_n=\frac{2\pi}{\varphi^n}$. Here $\varphi$ denotes the golden ratio and $\theta_0=2\pi$ whereas $\theta_2=\frac{2\pi}{\varphi^2}$ is the renown golden angle. The sequence
$$
\{\theta_n\}=\pi\cdot\{2,\quad\sqrt 5-1,\quad-\sqrt 5+3,\quad2\sqrt 5-4,\quad...\}
$$
is most simple to calculate by subtracting succesive angles like for instance (leaving out the factor $\pi$ for simplicity)
$$
\begin{align}
2-(\sqrt 5-1)&=-\sqrt 5+3\\
(\sqrt 5-1)-(-\sqrt 5+3)&=2\sqrt 5-4\\
&\text{etc.}
\end{align}
$$
At the $(F_n-1)$'th step in the process the circle is divided into $F_n$ parts of size $\theta_{n-2}$ and $\theta_{n-1}$ where the former is the largest angle of the two has $F_{n-1}$ instances and the smaller angle has $F_{n-2}$ instances. From that step on until the $(F_{n+1})$-th step each of the largest angles will be subdivided. So the said statement is correct and can be proven.
Here is a dynamic diagram of it made with GeoGebra:
Link to GeoGebraTube
Analysis
All angles $\theta_n$ have the form $(-1)^n(a_n-b_n\sqrt 5)$ for positive integers $a_n,b_n$. One can check that
$$
\begin{align}
\{a_n\}&=\{2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199,...\}=\{L_n\}\\
\{b_n\}&=\{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,...\}=\{F_n\}
\end{align}
$$
where $L_n$ are the Lucas numbers. The equation
$$
\pm\frac{\theta_n}{\pi}=L_n-F_n\sqrt 5=(-\sqrt 5+3)s-2t
$$
has the unique solution $[s,t]=\left[F_n,\frac{3F_n-L_n}{2}\right]$. It is unique since $s$ has to equal $F_n$ because $\sqrt 5$ is irrational. This is closely connected to Advanced Algebra and the fact that $\mathbb Z[\sqrt 5]=\{a+b\sqrt 5\}$ is a $2$-dimensional vectorspace over $\mathbb Z$ so the coefficients $a,b$ are unique.
Now due to the analysis, the gap between spoke $0$ and spoke $F_n$ has to be $\theta_n$. Here spoke $0$ denotes the radius at angle $0$ - equal to zero golden angles. Refer to my dynamic diagram for visualization of it. By shifting these we see that the gap between spoke $k$ and spoke $(k+F_n)$ must also be $\theta_n$.
Focusing on spoke $0$ we see how the sequence of Fibonacci spoke numbers $0,1,1,2,3,5,...$ fall alternatingly on each side of spoke $0$ at angles plus and minus $\theta_n$ starting with spoke $F_1=1$ dividing into the negative $\theta_1$ and also spoke $F_2=1$ dividing into the positive angle $\theta_2$. Next time spoke $F_3=2$ divides the former greatest angle negative $\theta_1$ by adding negative $\theta_3$ and so on.
After each spoke $F_n$ this pattern around spoke $0$ is copied around spoke $k=1,2,...,(F_{n-1}-1)$. I do not have the time to go into further detail for the moment. The last part here is a bit fluffy and could be elaborated upon.
A: The property that the last point divides one of the largest gaps is true for all irrationals $\alpha$. Please refer to this paper by Tony Ravenstein, specifically proposition 4.2 and the first statement of its proof. (Also note that, the question's situation of dividing a circle's circumference is equivalent to dividing the interval [0,1] by fractional-part of multiples of $\alpha$).
Now, if $\alpha$ is the golden-ratio, it can be proved that the ratio of the new produced gaps is also in golden-ratio. In fact, all of the gaps (there are 3 distinct, due to Three-Gap Theorem) are always in golden-ratio. To prove this, we need to use the relation of the three gap lengths to the continued fraction of $\alpha$. For that, in addition to above paper, we may also refer this (theorem 3) and this (theorems 2 and 4; this one is written by me).
These references tell us about gap-lengths being related to leftmost/rightmost corner-points ordinal. These two corner points (their ordinal) themselves are related to the continued-fraction.
For golden-ratio, these corner points ordinal will come out to be some consecutive fibonacci-numbers (we need to work out from the details in these references). Once we have the gap-lengths expressed in terms of fibonacci-numbers, we can use fibonacci-number's properties to derive that their ratio is golden-ratio itself.
