Adversarial Construction, Not Depth, Defines Hardest Competition Problems — E8 Intelligence Research

FINDING: The "hardest competition problems" cluster around three canonical cases — IMO 1988 P6 (number theory), IMO 2011 P2 (the "windmill"), and Putnam problems — each hard by *adversarial construction*, not by deep structure. | MATH: IMO 1988 P6: for positive integers a,b with ab+1 | a²+b², prove (a²+b²)/(ab+1) is a perfect square. Key technique: Vieta jumping / root flipping on quadratic x² − k(ab+1)x + b² = 0. IMO 2011 P2 (windmill): a line ℓ sweeps the plane rotating about a point P; it bisects the set of 2n points at every position; prove P can be chosen so this holds for all rotations. Invariant: parity of points on each side, n on each side. Putnam: e.g. A1–B6 difficulty gradient; median Putnam score ≈ 0–10/120. | CONNECTION: Vieta jumping is a *lattice descent* on the integer lattice ℤ² restricted to a hyperbola a²+b² = k(ab+1); the solution set forms a discrete orbit under the involution (a,b) ↦ (b, kb−a), i.e. a reflection in the root system A₁ acting on a rank-2 quadratic f 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
2026-10-29
DOI
https://doi.org/10.5281/zenodo.23756189
Primary Topic
Mathematics Education and Pedagogy
Type
preprint
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Adversarial Construction, Not Depth, Defines Hardest Competition Problems — E8 Intelligence Research

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

Adversarial Construction, Not Depth, Defines Hardest Competition Problems — E8 Intelligence Research

Andrew Stewart Caldin
preprint en

Abstract

FINDING: The "hardest competition problems" cluster around three canonical cases — IMO 1988 P6 (number theory), IMO 2011 P2 (the "windmill"), and Putnam problems — each hard by *adversarial construction*, not by deep structure. | MATH: IMO 1988 P6: for positive integers a,b with ab+1 | a²+b², prove (a²+b²)/(ab+1) is a perfect square. Key technique: Vieta jumping / root flipping on quadratic x² − k(ab+1)x + b² = 0. IMO 2011 P2 (windmill): a line ℓ sweeps the plane rotating about a point P; it bisects the set of 2n points at every position; prove P can be chosen so this holds for all rotations. Invariant: parity of points on each side, n on each side. Putnam: e.g. A1–B6 difficulty gradient; median Putnam score ≈ 0–10/120. | CONNECTION: Vieta jumping is a *lattice descent* on the integer lattice ℤ² restricted to a hyperbola a²+b² = k(ab+1); the solution set forms a discrete orbit under the involution (a,b) ↦ (b, kb−a), i.e. a reflection in the root system A₁ acting on a rank-2 quadratic f 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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