Problem: Let $\alpha$ be an irrational number with $0 < \alpha < 1$,
and draw a circle in the plane whose circumference has length $1$.
Given any integer $n\ge 3$,
define a sequence of points $P_1$, $P_2$, \dots, $P_n$ as follows.
First select any point $P_1$ on the circle, and for $2\le k\le n$ define $P_k$
as the point on the circle for which the length of arc $P_{k-1}P_k$ is $\alpha$,
when travelling counterclockwise around the circle from $P_{k-1}$ to $P_k$.
Suppose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$.
Prove that $a+b\le n$.

Solution: No points coincide since $\alpha$ is irrational.

Assume for contradiction that $n < a+b < 2n$.
Then it follows that \[ \ol{P_n P_{a+b-n}} \parallel \ol{P_a P_b} \]
as shown below.

\begin{center}
\begin{asy}
  size(9cm);
  transform t = shift(3.4,0);
  pair A = dir(170);
  pair B = -conj(A);
  pair C = dir(130);
  pair D = -conj(C);

  draw(unitcircle);
  draw(A--B, red);
  draw(C--D, blue);
  dot("$P_a$", A, A);
  dot("$P_b$", B, B);
  dot("$P_n$", C, C);
  dot("$P_{a+b-n}$", D, D);

  draw(t*unitcircle);
  draw(t*(A--B), red);
  draw(t*(C--D), blue);
  dot("$P_a$", t*A, A);
  dot("$P_b$", t*B, B);
  dot("$P_{a+b-n}$", t*C, C);
  dot("$P_n$", t*D, D);
\end{asy}
\end{center}

This is an obvious contradiction
since then $P_{a+b-n}$ is contained in the arc
$\arc{P_a P_b}$ of the circle through $P_n$.