Math 2.0

Convex and discrete geometry Theorem

Honeycomb theorem

Builds on TessellationRegular polygon

Statement

Let Γ⊆ℝ2\Gamma \subseteq \R^2 be a locally finite graph made of smooth curves, such that ℝ2∖Γ\R^2 \setminus \Gamma has infinitely many bounded connected components, all of area 11. Let CC be the union of these bounded components. For r>0r > 0 let BrB_r be the disc of radius rr about the origin. Then lim sup⁡r→∞perimeter⁡(C∩Br)area⁡(C∩Br)≥124,\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 11 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 of side ss has area 332s2\frac{3\sqrt{3}}{2} s^2, so area 11 means s2=233s^2 = \frac{2}{3\sqrt{3}}. Each edge is shared by two cells, so the length per cell is 3s3s, and (3s)2=63=23=12(3s)^2 = \frac{6}{\sqrt{3}} = 2\sqrt{3} = \sqrt{12}, so 3s=124≈1.8613s = \sqrt[4]{12} \approx 1.861. For comparison, a tiling by unit squares has length 22 per cell, which is larger.

Going deeper

For a student

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.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Honeycomb_theorem
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