Parity and Cyclic Invariants in the IMO Windmill Problem — E8 Intelligence Research

FINDING: The 2011 IMO windmill problem reduces to a parity/cyclic argument on point sets, not a deep algebraic structure. | MATH: The core invariant is the number of points on each side of the rotating line; the "windmill" process cycles through all lines determined by pairs of points, and the key lemma is that for any starting line not through a point, the number of points on each side changes by ±1 per step, with a fixed total n. The solution uses a discrete continuity argument: the line rotates continuously, and when it passes through a point, that point switches sides; the invariant is that the line always has exactly ⌊n/2⌋ points on one side (for odd n) or n/2 on each side (for even n) after a full cycle. No explicit equations beyond counting; the proof is combinatorial. | CONNECTION: Weak — the problem's structure resembles a cyclic permutation of the n points (the order in which the line hits them), but there is no golden ratio, no base-60, no crystallographic symmetry. The only Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com