Home > Combinatorics, IMO, Problem Solving > IMO 2011 Problem 2

## IMO 2011 Problem 2

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

IMO 2011 Problem 2

Solution: This problem was very nice, creative and inspiring. Something a bit new which looks very nice. I will present a solution I found on AoPS, which I consider very interesting.

For any point $P$, there is a line which cuts the remaining points in two groups, which differ with at most $1$ point from one another. This kind of a line works for the problem. Why? Imagine how such a line creates the windmill: it touches one point, then one side gains one point, and then this side loses the pivot, leaving the situation as before, i.e. with sides of the line differing by at most one point.

Pick now one point and a line which has the above property. At a given moment, the windmill will be parallel to this line, and by uniqueness, will coincide with this line. This reasoning shows that any point will be pivot in a $2\pi$ rotation of the windmill.

Categories: Combinatorics, IMO, Problem Solving Tags:
1. August 3, 2011 at 8:46 am

Well, in case if the number of points is even you should also fix which side of the plain contains an extra point, otherwise the line with said property wouldn’t be really unique.

• August 4, 2011 at 5:31 am

Yes, that is true. Thank you. 🙂