Problem 2

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 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 next meets a point of

S

. This process continues indefinitely. Show that we can choose a point

P

S

and a line going through

P

such that the resulting windmill uses each point of

S

as a pivot infinitely many times.

Step 5 of 6: Each direction has at most one balancing line

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In plain words

Sweeping a line of fixed ordinary direction across all the points, the right-count ticks down by exactly one point at a time, so exactly one vertex on that sweep can be the balancing pivot.

\left\lfloor\frac{n-1}{2}\right\rfloor

Detailed analysis

Fix an ordinary direction $\delta$ and slide a line with direction $\delta$ across the plane; because $\delta$ is ordinary no two vertices tie for the same position along the sweep, so the number of vertices to the right decreases by exactly $1$ every time the sweeping line passes a vertex. Consequently there is exactly one vertex at which this count equals $\left\lfloor\frac{n-1}{2}\right\rfloor$ at the moment of passing, so at most one balancing line has direction $\delta$ .