---
tag: 3JST9B967R
kind: statement
role: theorem
name: Honeycomb theorem
msc: [52C20, 49Q10, 05B45]
uses: [GKJK293DXS, 01KJJ9A400]
about: [GKJK293DXS]
sources:
  - url: https://en.wikipedia.org/wiki/Honeycomb_theorem
same_as:
  wikidata: Q3527261
  wikipedia: Honeycomb theorem
---

## Statement

Let $\Gamma \subseteq \R^2$ be a locally finite graph made of smooth curves, such that $\R^2 \setminus \Gamma$ has infinitely many bounded connected components, all of area $1$.
Let $C$ be the union of these bounded components.
For $r > 0$ let $B_r$ be the disc of radius $r$ about the origin.
Then
$$\limsup_{r \to \infty} \frac{\operatorname{perimeter}(C \cap B_r)}{\operatorname{area}(C \cap B_r)} \ge \sqrt[4]{12},$$
and the regular hexagonal tiling by hexagons of area $1$ attains equality.

## Remarks

Thomas Hales proved the theorem in 1999.
The proof reduces to a single inequality, the hexagonal isoperimetric inequality.
It bounds the perimeter of each cell from below, in terms of its area and of how far the cell is from a hexagon.
Curving an edge outward helps one cell and hurts its neighbour, and the inequality controls this trade.
Before Hales, László Fejes Tóth had proved the result under the hypothesis that all cells are convex.
Check of the constant: a [[regular hexagon|01KJJ9A400]] of side $s$ has area $\frac{3\sqrt{3}}{2} s^2$, so area $1$ means $s^2 = \frac{2}{3\sqrt{3}}$.
Each edge is shared by two cells, so the length per cell is $3s$, and $(3s)^2 = \frac{6}{\sqrt{3}} = 2\sqrt{3} = \sqrt{12}$, so $3s = \sqrt[4]{12} \approx 1.861$.
For comparison, a [[tiling|GKJK293DXS]] by unit squares has length $2$ per cell, which is larger.

## Explanations

### L1 human

Bees build their combs from hexagonal cells.
The theorem says that hexagons are the best choice.
Among all ways to divide the plane into regions of equal area, the hexagonal grid uses the least total length of walls.
This was believed since antiquity and was proved only at the end of the twentieth century.

### L2 human

The theorem is the two-dimensional case of the problem of partitions of least interface area.
The three-dimensional analogue, the Kelvin problem, is open.
Allowing curved edges is what makes the problem hard, since a cell can gain by bending its walls into its neighbours.