IMO 2011 Problem 2

This one got out with help of various hints, including the figure shown below and the big topic 'invariant'. But still nice to get it proved semi-independently. The figure is from the book Problem-Solving Methods in Combinatorics authored by Pablo Soberón.

Let $S$ be a finite set of at least two points in the plane.  Assume that no three points of $S$ are collinear. A windmill is a process that starts with a line $L$ going through a single point $P\in S$. The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point belonging to $S$. This point, $Q$, takes over as the new pivot, and the line now rotates clockwise about $Q$, until it meets a point of $S$. This process continues indefinitely. Show that we can choose a point $P$ in $S$ and a line $L$ going through $P$ such that the resulting windmill uses each point of $S$ infinitely many times.

Proof:

Imagine the line $L$ has two different sides, colored black and white. Observe that the number of points in $S$ on each side of $L$ never change. Have $L$ cut through the 'center' of points in $S$. Specifically, initialize $L$ such that the number of points in $S$ on both sides of $L$ differ by at most one. Suppose that there is a point $X$ not touched by $L$ at all. Assume at two points in time $L$ is in the position of the red and blue lines in the figure. Hence $X$ must lie between the blue and red lines, and moreover by the first observation the number of points in $S$ below the blue line equals the number of points in $S$ above the red line. Thus the difference of numbers of points on each side of $L$ differ by at least $2$, a contradiction.

$\blacksquare$