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Truncated Octahedral Conjecture

The truncated octahedral conjecture asserts that the Archimedean truncated octahedron uniquely minimizes surface area among three-dimensional parallelohedra of fixed volume, up to similarity. It was proposed by Bezdek (2006, Conjecture 7.5) and proved by Hales and Song (2026) and independently by Cesaroni and Novaga (2026).

Every three-dimensional parallelohedron of volume V and surface area S satisfies

S>=(3(1+2sqrt(3)))/(4^(2/3))V^(2/3),

with equality iff it is similar to the Archimedean truncated octahedron. The class consists of convex polyhedra that tile space by translations. This restriction distinguishes the result from the general equal-volume foam problem underlying Kelvin's conjecture.

One inequality in the Hales and Song (2026) proof is established by an exact computer-assisted certificate. Hales and Song (2026) report substantial ChatGPT assistance in proof discovery and writing the certificate-verification code. Cesaroni and Novaga (2026) also report ChatGPT assistance in exploring and checking proof arguments.

See also

Kelvin's Conjecture, Parallelohedron, Truncated Octahedron

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References

Bezdek, K. "Sphere Packings Revisited." Europ. J. Combin. 27, 864-883, 2006. https://doi.org/10.1016/j.ejc.2005.05.001.Cesaroni, A. and Novaga, M. "The Truncated Octahedron Minimizes Surface Area Among Parallelohedra of Equal Volume." 2 Sep 2026. https://arxiv.org/abs/2609.02384.Hales, T. and Song, L. "The Truncated Octahedral Conjecture." 1 Sep 2026. https://arxiv.org/abs/2609.00997.

Cite this as:

Weisstein, Eric W. "Truncated Octahedral Conjecture." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/TruncatedOctahedralConjecture.html

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