The golden angle

Conventions. Write $\operatorname{fd}(u,v)=\{v-u\}\in[0,1)$ for the forward distance from $u$ to $v$ on the circle. The gap after $x_k$ is $\operatorname{fd}(x_k,x_{k'})$, where $x_{k'}$ is the first point met going forward. Since $\alpha$ is irrational, $x_j=x_l$ only if $j=l$.

Small N. For $N=1$ there is one gap, of length 1. For $N=2$ the gaps are $\{\alpha\}$ and $1-\{\alpha\}$. Assume $N\ge3$.

Set-up. Let $a$ minimise and $b$ maximise $\{k\alpha\}$ over $k\in\{1,\dots,N-1\}$ (both unique). Put $\delta_1=\{a\alpha\}$ and $\delta_2=1-\{b\alpha\}$. Since $N-1\ge2$, $a\ne b$ and $0<\delta_1<\{b\alpha\}<1$, so $\delta_1,\delta_2>0$ and $\delta_1+\delta_2=1-(\{b\alpha\}-\{a\alpha\})<1$.

Basic bounds. For $j\ne l$ in $\{0,\dots,N-1\}$ and $m=|l-j|\in\{1,\dots,N-1\}$:

• if $l>j$, $\operatorname{fd}(x_j,x_l)=\{m\alpha\}\ge\delta_1$, with equality iff $m=a$;
• if $l<j$, $\operatorname{fd}(x_j,x_l)=1-\{m\alpha\}\ge1-\{b\alpha\}=\delta_2$, with equality iff $m=b$.

Claim. The gap after $x_k$ is $\delta_1$ if $k+a\le N-1$; $\delta_2$ if $k\ge b$ and $k+a\ge N$; and $\delta_1+\delta_2$ if $k+a\ge N$ and $k\le b-1$. These cases cover every $k$, so every gap lies in $\{\delta_1,\delta_2,\delta_1+\delta_2\}$, which proves the theorem.

Case 1: $k+a\le N-1$. Then $\operatorname{fd}(x_k,x_{k+a})=\delta_1$. Suppose some $x_l$ lies strictly inside that arc. If $l>k$, then $\operatorname{fd}(x_k,x_l)\ge\delta_1$, a contradiction. If $l<k$, then $k+a-l\in\{1,\dots,N-1\}$, so $\operatorname{fd}(x_l,x_{k+a})\ge\delta_1$, again a contradiction.

Case 2: $k\ge b$ and $k+a\ge N$. Then $\operatorname{fd}(x_k,x_{k-b})=1-\{b\alpha\}=\delta_2$. If some $x_l$ lies strictly inside: for $l<k$, $\operatorname{fd}(x_k,x_l)\ge\delta_2$; for $l>k$, $l-(k-b)\in\{1,\dots,N-1\}$, so $\operatorname{fd}(x_l,x_{k-b})\ge\delta_2$. Either way, a contradiction.

Case 3: $k+a\ge N$ and $k\le b-1$. Let $c=k+a-b$. Then $1\le N-b\le c\le a-1\le N-2$, and $c\ne k$ since $a\ne b$. Its forward distance from $x_k$ is $\{(a-b)\alpha\}=\{\delta_1+\delta_2-1\}=\delta_1+\delta_2$. Suppose $x_l$ lies strictly inside, so $0<\operatorname{fd}(x_k,x_l)<\delta_1+\delta_2$.

• If $l>k$, let $m=l-k$. Then $1\le m\le N-1-k\le a-1$, so $\delta_1<\{m\alpha\}<\delta_1+\delta_2$. Then $\{(a-m)\alpha\}=1-(\{m\alpha\}-\delta_1)>1-\delta_2=\{b\alpha\}$, with $a-m\in\{1,\dots,N-1\}$, contradicting the maximality of $\{b\alpha\}$.
• If $l<k$, let $m=k-l$. Then $1\le m\le b-1$, so $\delta_2<1-\{m\alpha\}<\delta_1+\delta_2$, that is $\{b\alpha\}-\delta_1<\{m\alpha\}<\{b\alpha\}$. Then $\{(b-m)\alpha\}=\{b\alpha\}-\{m\alpha\}\in(0,\delta_1)$, with $b-m\in\{1,\dots,N-1\}$, contradicting the minimality of $\delta_1$.

So the successor of $x_k$ is $x_c$, at distance $\delta_1+\delta_2$. If $k+a\le N-1$ we are in Case 1; otherwise either $k\ge b$ (Case 2) or $k\le b-1$ (Case 3). ∎