MathLabs

Problem 2

Let SS be a finite set of at least two points in the plane. Assume that no three points of SS are collinear. A windmill is a process that starts with a line going through a single point P∈SP\in S. The line rotates clockwise about the pivot PP until the first time that the line meets some other point belonging to SS. This point, QQ, takes over as the new pivot, and the line now rotates clockwise about QQ, until it next meets a point of SS. This process continues indefinitely. Show that we can choose a point PP in SS and a line going through PP such that the resulting windmill uses each point of SS as a pivot infinitely many times.
Step 3 of 6: The right-count stays balanced forever
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In plain words

Whenever the pivot changes, the old and new pivot simply trade sides, so the right-count never actually changes throughout the whole windmill.

⌊n−12⌋\left\lfloor\frac{n-1}{2}\right\rfloor
Detailed analysis

Start the windmill from the ordinary balancing line built in the previous step. At each pivot change the line sweeps past exactly one vertex, and the old pivot and the new pivot simply swap sides while every other vertex stays on the side it was already on; so the right-count of the line is exactly the same immediately before and immediately after every pivot change. Since it began at ⌊n−12⌋\left\lfloor\frac{n-1}{2}\right\rfloor, every ordinary line that ever occurs during this infinite windmill again has exactly ⌊n−12⌋\left\lfloor\frac{n-1}{2}\right\rfloor vertices to its right, i.e. it is again a balancing line.