The maximum area of the convex hull of a polyhex

Abstract

A polyhex is an edge-connected set of n cells of the regular hexagonal tiling, where each cell has area one. We prove that the convex hull of a polyhex has area at most (1/6)*ceiling(n^2 + 14n/3), and we show that some polyhex reaches this bound for every n. This proves a conjecture of Kurz from 2008, which asked for the weaker bound (1/6)*floor(n^2 + 14n/3 + 1). The two bounds differ exactly when 3 divides n. We checked the upper bound in the Lean 4 proof assistant with the Mathlib library. We also report a computation over all polyhexes with at most 12 cells, which shows that for these sizes only one shape reaches the maximum, up to rotation and reflection.

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BibTeX RIS

BibTeX

@misc{gaur2026maximum,
  title = {{The maximum area of the convex hull of a polyhex}},
  author = {Pragyaan Gaur},
  howpublished = {arXiv},
  year = {2026},
  month = sep,
  doi = {10.48550/arxiv.2609.30310},
  eprint = {2609.30310},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2609.30310},
  note = {CC BY 4.0},
}

RIS

TY  - UNPB
TI  - The maximum area of the convex hull of a polyhex
AU  - Pragyaan Gaur
PY  - 2026
DA  - 2026/09/23
PB  - arXiv
DO  - 10.48550/arxiv.2609.30310
UR  - https://arxiv.org/abs/2609.30310
AB  - A polyhex is an edge-connected set of n cells of the regular hexagonal tiling, where each cell has area one. We prove that the convex hull of a polyhex has area at most (1/6)*ceiling(n^2 + 14n/3), and we show that some polyhex reaches this bound for every n. This proves a conjecture of Kurz from 2008, which asked for the weaker bound (1/6)*floor(n^2 + 14n/3 + 1). The two bounds differ exactly when 3 divides n. We checked the upper bound in the Lean 4 proof assistant with the Mathlib library. We also report a computation over all polyhexes with at most 12 cells, which shows that for these sizes only one shape reaches the maximum, up to rotation and reflection.
ER  -

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