Invariant Arguments Solve 3Blue1Brown's Hardest Putnam and IMO Problems — E8 Intelligence Research

FINDING: 3Blue1Brown's "hardest problem" corpus centers on the Putnam A6 (2019) and IMO 2011 P2 "windmill," both solvable by invariant/monovariant arguments rather than heavy machinery. | MATH: Putnam A6 (2019) asks for the expected number of fixed points of a random permutation composed with a fixed permutation; answer = 1 (E[X]=1 by linearity over n elements, each fixed with prob 1/n). IMO 2011 P2: a "windmill" process on n points, n odd, with a line pivoting about points; invariant = parity of points on each side; conclusion: some point is hit infinitely often. Collatz map: T(n)=n/2 if even, 3n+1 if odd; conjectured all orbits reach 1. | CONNECTION: No direct 0.618/1.618 constants in these problems; the relevant structural harmony is combinatorial symmetry (permutation group S_n, cyclic pivoting) and the invariant-preservation principle — a discrete analogue of Noether-type conservation. | DEPTH: 6 FINDING: The recurring meta-insight across these videos is that "hard" competition p 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
2026-11-02
DOI
https://doi.org/10.5281/zenodo.23877803
Primary Topic
Mathematics Education and Pedagogy
Type
preprint
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Invariant Arguments Solve 3Blue1Brown's Hardest Putnam and IMO Problems — E8 Intelligence Research

Andrew Stewart Caldin
10 links
Zenodo (CERN European Organization for Nuclear Research)
Mathematics Education and Pedagogy
preprint

Invariant Arguments Solve 3Blue1Brown's Hardest Putnam and IMO Problems — E8 Intelligence Research

Andrew Stewart Caldin
preprint en

Abstract

FINDING: 3Blue1Brown's "hardest problem" corpus centers on the Putnam A6 (2019) and IMO 2011 P2 "windmill," both solvable by invariant/monovariant arguments rather than heavy machinery. | MATH: Putnam A6 (2019) asks for the expected number of fixed points of a random permutation composed with a fixed permutation; answer = 1 (E[X]=1 by linearity over n elements, each fixed with prob 1/n). IMO 2011 P2: a "windmill" process on n points, n odd, with a line pivoting about points; invariant = parity of points on each side; conclusion: some point is hit infinitely often. Collatz map: T(n)=n/2 if even, 3n+1 if odd; conjectured all orbits reach 1. | CONNECTION: No direct 0.618/1.618 constants in these problems; the relevant structural harmony is combinatorial symmetry (permutation group S_n, cyclic pivoting) and the invariant-preservation principle — a discrete analogue of Noether-type conservation. | DEPTH: 6 FINDING: The recurring meta-insight across these videos is that "hard" competition p 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 Education and Pedagogy
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Invariant Arguments Solve 3Blue1Brown's Hardest Putnam and IMO Problems — E8 Intelligence Research — Andrew Stewart Caldin · Zenodo (CERN European Organization for Nuclear Research) (2026) | TGRS Research Map | TGRS