Parity Invariant in the Windmill Process for IMO 2011 Problem 2 — E8 Intelligence Research

FINDING: IMO 2011 Problem 2 (Windmill) is a combinatorial geometry problem of exceptional difficulty, requiring a parity/invariant argument about points in general position and a rotating line. MATH: - Given \(2n+1\) points in general position (no three collinear, no two with same x-coordinate). - A "windmill" process: choose a line through one point \(P\), rotate it clockwise about \(P\) until it hits another point \(Q\); then pivot about \(Q\) and continue. - Key invariant: at any time, the line divides the remaining points into two sets of equal size (each of size \(n\)). - The process visits each point infinitely often; the proof uses the fact that the number of points on each side of the line is invariant mod 2, and the total number of points is odd. - No explicit constants or ratios appear. CONNECTION: - The invariant (equal partition) is a combinatorial symmetry reminiscent of balanced configurations in root systems (e.g., \(A₂ₙ\) root system has \(2n+1\) points Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com