Source: IMO-2011-P2
Category: Operations and Strategies

Problem:
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. 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.

Solution:
Let $|\mathcal S| = n+1$. Orient the line $\ell$ in some direction, and color the plane such that its left half is red and its right half is blue. 

The critical observation is that the number of points of $\mathcal{S}$ on the red side of $\ell$ does not change, nor does the number of points on the blue side, except at the brief moments when $\ell$ contains two points. To see this, consider what happens when the rotating line $\ell$ meets a new point $Q \in \mathcal{S}$. Because no three points of $\mathcal{S}$ are collinear, $\ell$ contains exactly two points at this moment: the old pivot $P$ and the new pivot $Q$. As the line continues its clockwise rotation around $Q$, the point $P$ leaves the line and enters the half-plane that $Q$ previously occupied, while $Q$ takes over as the pivot. Consequently, the net number of points on the red side and the blue side of the oriented line $\ell$ remains strictly invariant throughout the entire process.

We can pick the initial configuration of the point $P$ and the line $\ell$ so that there are exactly $\lfloor n/2 \rfloor$ points on the red side and $\lceil n/2 \rceil$ points on the blue side. As the windmill process continues indefinitely, the line $\ell$ eventually completes a full $180^\circ$ rotation. At this point, the line has reversed its orientation, meaning the red and blue sides have effectively switched positions in the plane. 

Because the counts of points on the red and blue sides are invariant, and the red and blue half-planes have swapped regions in the plane, the windmill must have passed through every point of $\mathcal{S}$ during this $180^\circ$ rotation. If a point had not been passed through, it would have remained on the same color side of the oriented line, which corresponds to the opposite side of the unoriented line in the plane—a contradiction. 

Therefore, the line $\ell$ passes through every point of $\mathcal{S}$ during any $180^\circ$ rotation. Since the process continues indefinitely, the line undergoes infinitely many such rotations, and thus the windmill uses each point of $\mathcal{S}$ as a pivot infinitely many times.


Grading guidelines:
(Partial) 1. Introduced an oriented line and a two-coloring of the half-planes, and proved the key invariant that the numbers of points of S on each side of the oriented line stay constant throughout the windmill (using that at each switch exactly two points lie on the line and the old pivot moves to the other side). 2. Explained how to choose the initial pivot P and direction of ℓ so that the invariant side-counts are balanced (e.g. ⌊n/2⌋ on one side and ⌈n/2⌉ on the other), by rotating a line through a fixed point and applying a continuity/step argument. 3. Proved that after a total rotation of 180° the oriented line has reversed, so the two colored half-planes have swapped, and derived at least one nontrivial consequence toward covering all points (e.g. a point cannot stay on the same colored side for the entire 180° rotation). (Almost) 1. Gave the invariant side-count argument and the 180° reversal idea, and concluded that every point must be met/pivoted during each 180° rotation, but left a small gap in justifying why a point not used as a pivot would force an impossible change in the invariant counts. 2. Proved all main steps but did not fully justify existence of the initial configuration with side counts ⌊n/2⌋ and ⌈n/2⌉ (e.g. asserted without proof that such a line through some P exists). 3. Solution is almost complete, but made minor mistakes which are not negligible.