Measurements on a unit circle Given a unit circle and a point $P_0$ on it, we call $i$-step ($i=1, 2, \ldots$) taking a point $P_i$ on the circle such that the arc lenght $P_{i-1}P_i=n\in\mathbb{N}$ (we use the convention of taking measurements anticlockwise).


*

*Prove/disprove that after a finite numbers of steps there aren't $i, j$ such that $P_iP_j>\alpha$, for every $\alpha\in(0, 2\pi)$.

*Fixed $\alpha$, determine the first $k$-step after that the previous statement is true (if it's true!). If it's not possible to determine exactly such $k$, give upper and lower bounds on it.
Edit: $n$ and $\alpha$ are fixed.
 A: This is a consequence of the equidistribution theorem. Your step is a fraction $n/2\pi$ of the circle, which is irrational, so the $P_i$ are uniformly distributed. Specifically for any $\alpha>0$ there is a $N(\alpha)$ that guarantees that after $N(\alpha)$ steps some $P_i$ falls in the arc $[\frac{i\alpha}{2},\frac{(i+1)\alpha}{2}]$ for every $0\le i < 4\pi/\alpha$ guaranteeing the maximum distance is at most $\alpha$.
The number of steps needed for a given $\alpha$ is related to the behavior discrepancy
$$
D(k) = \sup_{A}\left| \frac{\#\{P_i \in A\}}{k}-\frac{|A|}{2\pi}\right|
$$
where the $\sup$ is over all arcs $A$, $|A|$ is the arc length and $\#\{P_i\in A\}$ is the number of the first $k$ points that land in $A$. A theorem of Schmidt says that there are arbitrarily large $k$ such that $D(k)\ge \log k/25 k$, so in general you will need on the order of $\frac{1}{\alpha}\log \frac{1}{\alpha}$ steps.
I don't have a simple upper bound, but here are some references that deal with this problem:


*

*Ramshaw: On the discrepancy of the sequence formed by the multiples of an irrational number

*Drmota & Tichy: Sequences, Discrepancies and Applications

*Kuipers & Niederreiter: Uniform Distribution of Sequences

