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?

• This is related to Fibonacci hashing. There is a proof by Knuth that a irrational angle increment will exhibit at most three different gap sizes, and will successively divide one of the largest gaps. This can be used to proove the above proposition. The Reference is: D. E. Knuth, “The Art of Computer Programming, Vol 3.” Palo Alto, California: Addison-Wesley, 1973, p. 543. Commented Oct 2, 2015 at 20:11
• In the end you can proof it based on D. Knuths work (see Appendix B of my paper ) Commented Jan 14, 2021 at 16:12

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.

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.

• Thank you for the very usefull answer - I wish I had this 10 years ago :-) In the end I published a "sloppy" proof based on Knuths work from 1973 (see Appendix B of my paper ) Commented Jan 14, 2021 at 16:06