3Blue1Brown's "Hardest Problem": Putnam and IMO Windmill Extremal Arguments — E8 Intelligence Research

FINDING: 3Blue1Brown's "hardest problem" content centers on the Putnam exam and the 2011 IMO "windmill" problem (P2), both combinatorial/geometric extremal arguments rather than constant-driven results. | MATH: Windmill problem: given n points in general position, a line ℓ through one point P rotates about P; when it hits another point Q, pivot to Q. Invariant: the line always bisects the point set (⌊n/2⌋ on each side), so each point is visited infinitely often. No closed-form constant; the core object is a rotating line with a parity/halving invariant. | CONNECTION: The windmill's halving invariant is a 1:1 (n/2 : n/2) partition — a bisection symmetry, not a golden-ratio or base-60 structure. No 0.618/1.618/2.618, no crystallographic root system appears in the sourced problems. | DEPTH: 4 FINDING: The 100 Prisoners riddle resolves via a permutation-cycle argument, yielding a survival probability ≈ 0.31183 for n=100. | MATH: P(success) = 1 − Σ_{k=m+1}^{n} 1/k, with m = n/2. For n=100: Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com