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The truncated octahedron minimizes surface area among parallelohedra of equal volume
- Published
- Source
- OpenAlex
- Published in
- arXiv (Cornell University)
- DOI
- 10.48550/arxiv.2609.02384
- Original
- Read it at the publisher
- Topics
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 \[ \frac{\mathcal H^2(\partial P)}{|P|^{2/3}} \ge \frac{3(1+2\sqrt3)}{4^{2/3}}, \] 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.
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