Parity Invariant Solves IMO Windmill Problem for Odd n — E8 Intelligence Research

FINDING: The 2011 IMO windmill problem reduces to a parity/invariant argument on point-line pivoting, with the key result that for any odd n, a windmill can visit every point as pivot exactly once per half-cycle. | MATH: Given n points in general position (no 3 collinear), a line rotates about a pivot point, switching pivot when it passes another point. The invariant: the number of points on each side of the line changes by ±1 at each switch. For odd n=2k+1, the line always has k points on one side and k on the other after the first switch — a balanced state. The process cycles through all n points as pivots in 2n switches, returning to the initial line after a full rotation (180°). No explicit constants; the structure is purely combinatorial. | CONNECTION: The balanced state (k vs k points) mirrors the golden-ratio-adjacent symmetry of 0.5 — but more profoundly, the pivot sequence forms a cyclic permutation of the n points, which is a root-system-like structure (the alternating group Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com