Parity Invariant and Cycle Period in the Windmill Problem — E8 Intelligence Research

FINDING: The windmill problem (IMO 2011, Problem 2) reduces to an invariant: the number of points on each side of the rotating line changes by ±1 at each pivot, and the parity of the count on each side is invariant; the process cycles with period exactly n (the number of points) when n is odd, and 2n when n is even. | MATH: Let S be a set of n points in general position (no three collinear). A windmill line ℓ rotates about a pivot point p ∈ S; when ℓ hits another point q, the pivot switches to q. Key invariant: if a = |points strictly on one side of ℓ|, then at each pivot a → a±1, so parity of a is invariant. For n odd, a = (n−1)/2 exactly at all times (balanced), and the line visits every point exactly once per full rotation — period n. For n even, a alternates between (n−2)/2 and n/2, period 2n. The proof uses a "winding number" argument: the angle of ℓ sweeps 2π per cycle, and the pivot sequence is a permutation of S with no repeats within a cycle. | CONNECTION: The invariant parity Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com