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The truncated octahedron minimizes surface area among parallelohedra of equal volume
Abstract
We prove that the regular truncated octahedron uniquely minimizes surface area among all parallelohedra of fixed volume. Equivalently, every three-dimensional parallelohedron P satisfies (ℋ2(∂ P))/|P|2/3 ≥ (3(1+2√3))/42/3, with equality if and only if P is similar to the regular truncated octahedron. Among the non-truncated Fedorov types we prove a stronger sharp bound, attained uniquely by the regular rhombic dodecahedron.
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BibTeX
@misc{cesaroni2026truncated,
title = {{The truncated octahedron minimizes surface area among parallelohedra of equal volume}},
author = {Annalisa Cesaroni and Matteo Novaga},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.02384},
eprint = {2609.02384},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.02384},
}
RIS
TY - UNPB
TI - The truncated octahedron minimizes surface area among parallelohedra of equal volume
AU - Annalisa Cesaroni
AU - Matteo Novaga
PY - 2026
DA - 2026/09/09
PB - arXiv
DO - 10.48550/arxiv.2609.02384
UR - https://arxiv.org/abs/2609.02384
AB - We prove that the regular truncated octahedron uniquely minimizes surface area among all parallelohedra of fixed volume. Equivalently, every three-dimensional parallelohedron P satisfies (ℋ^2(∂ P))/|P|^2/3 ≥ (3(1+2√3))/4^2/3, with equality if and only if P is similar to the regular truncated octahedron. Among the non-truncated Fedorov types we prove a stronger sharp bound, attained uniquely by the regular rhombic dodecahedron.
ER -