Subfield: Combinatorics


Question: Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another point $Q$ of $\mathcal{S}$. The point $Q$ now takes over as the new pivot. This process continues indefinitely, with the pivot always being a point from $\mathcal{S}$.

Show that for a suitable $P \in \mathcal{S}$ and a suitable starting line $\ell$ containing $P$, the resulting windmill will visit each point of $\mathcal{S}$ as a pivot infinitely often.

Solution:
Give the rotating line an orientation and distinguish its sides as the oranje side and the blue side. Notice that whenever the pivot changes from some point $T$ to another point $U$, after the change, $T$ is on the same side as $U$ was before. Therefore, the number of elements of $\mathcal{S}$ on the oranje side and the number of those on the blue side remain the same throughout the whole process (except for those moments when the line contains two points).
<img_3677>

First consider the case that $|\mathcal{S}|=2 n+1$ is odd. We claim that through any point $T \in \mathcal{S}$, there is a line that has $n$ points on each side. To see this, choose an oriented line through $T$ containing no other point of $\mathcal{S}$ and suppose that it has $n+r$ points on its oranje side. If $r=0$ then we have established the claim, so we may assume that $r \neq 0$. As the line rotates through $180^{\circ}$ around $T$, the number of points of $\mathcal{S}$ on its oranje side changes by 1 whenever the line passes through a point; after $180^{\circ}$, the number of points on the oranje side is $n-r$. Therefore there is an intermediate stage at which the oranje side, and thus also the blue side, contains $n$ points.

Now select the point $P$ arbitrarily, and choose a line through $P$ that has $n$ points of $\mathcal{S}$ on each side to be the initial state of the windmill. We will show that during a rotation over $180^{\circ}$, the line of the windmill visits each point of $\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\mathcal{S}$ and select a line $\boldsymbol{\ell}$ through $T$ that separates $\mathcal{S}$ into equal halves. The point $T$ is the unique point of $\mathcal{S}$ through which a line in this direction can separate the points of $\mathcal{S}$ into equal halves (parallel translation would disturb the balance). Therefore, when the windmill line is parallel to $\ell$, it must be $\ell$ itself, and so pass through $T$.

Next suppose that $|\mathcal{S}|=2 n$. Similarly to the odd case, for every $T \in \mathcal{S}$ there is an oriented



line through $T$ with $n-1$ points on its oranje side and $n$ points on its blue side. Select such an oriented line through an arbitrary $P$ to be the initial state of the windmill.

We will now show that during a rotation over $360^{\circ}$, the line of the windmill visits each point of $\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\mathcal{S}$ and an oriented line $\ell$ through $T$ that separates $\mathcal{S}$ into two subsets with $n-1$ points on its oranje and $n$ points on its blue side. Again, parallel translation would change the numbers of points on the two sides, so when the windmill line is parallel to $\ell$ with the same orientation, the windmill line must pass through $T$.