3Blue1Brown's Hardest Problems: Putnam and IMO Windmill — E8 Intelligence Research

FINDING: 3Blue1Brown's "hardest problem" content centers on the Putnam (a 6-problem, 12-hour undergraduate exam) and the 2011 IMO Problem 2 ("windmill"), where difficulty arises from combinatorial/geometric structure rather than heavy machinery. | MATH: Putnam scoring: 120 points max, median score frequently 0–10; IMO P2 windmill: a line ℓ rotates about a point P on a plane containing a finite set S of n points, each point hit by ℓ is "flipped" to the opposite side; problem asks whether some P yields infinitely many flips for every S. | CONNECTION: The windmill is a rotational symmetry / reflection (Z₂) action on a point set; the invariant is the count of points per half-plane, preserved under the flip — a discrete symmetry-conservation argument, not a golden-ratio or lattice-harmony result. | DEPTH: 3 FINDING: The 100 Prisoners Riddle has a counterintuitive answer (~31% success via the "follow-the-cycle" strategy), driven by permutation cycle structure. | MATH: For n prisoners, strat Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

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Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2026-10-27
DOI
https://doi.org/10.5281/zenodo.23683679
Primary Topic
Mathematics and Applications
Type
preprint
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3Blue1Brown's Hardest Problems: Putnam and IMO Windmill — E8 Intelligence Research

Andrew Stewart Caldin
12 links
Zenodo (CERN European Organization for Nuclear Research)
Mathematics and Applications
preprint

3Blue1Brown's Hardest Problems: Putnam and IMO Windmill — E8 Intelligence Research

Andrew Stewart Caldin
preprint en

Abstract

FINDING: 3Blue1Brown's "hardest problem" content centers on the Putnam (a 6-problem, 12-hour undergraduate exam) and the 2011 IMO Problem 2 ("windmill"), where difficulty arises from combinatorial/geometric structure rather than heavy machinery. | MATH: Putnam scoring: 120 points max, median score frequently 0–10; IMO P2 windmill: a line ℓ rotates about a point P on a plane containing a finite set S of n points, each point hit by ℓ is "flipped" to the opposite side; problem asks whether some P yields infinitely many flips for every S. | CONNECTION: The windmill is a rotational symmetry / reflection (Z₂) action on a point set; the invariant is the count of points per half-plane, preserved under the flip — a discrete symmetry-conservation argument, not a golden-ratio or lattice-harmony result. | DEPTH: 3 FINDING: The 100 Prisoners Riddle has a counterintuitive answer (~31% success via the "follow-the-cycle" strategy), driven by permutation cycle structure. | MATH: For n prisoners, strat Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

Zenodo (CERN European Organization for Nuclear Research)
Mathematics and Applications
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