Rotating Line Invariant Proves Equal Partition of Odd Point Sets — E8 Intelligence Research

FINDING: IMO 2011 Problem 2 (Windmill) — combinatorial geometry problem requiring proof that for any finite set of points in general position, there exists a line through a point that, when rotated, passes through exactly one new point at a time, partitioning the set into two equal halves at each step. MATH: No explicit equations; key concept is parity and invariant: number of points on each side of rotating line changes by ±1 at each pivot; initial configuration must have odd total points (2n+1) to achieve equal halves (n on each side) at some rotation. CONNECTION: The rotation symmetry and pivot sequence evoke cyclic group C_{2n+1} action on points; the invariant (difference in counts) is a discrete harmonic function on the circle. No direct golden ratio or base-60 link. DEPTH: 6 — elegant combinatorial invariant, but isolated to contest problem; no universal constant or physical law. FINDING: IMO 2025 Problem 1 (easiest of hardest set) — solution likely involves number theory Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com