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  "body": "\n## Problem\n\nAmong all partitions of $\\R^3$ into cells of equal volume, find one whose interfaces have the least [[surface area|0S29Y4ZW0Y]] per cell.\n\nTo make the question precise, fix the cell volume to $1$.\nThen ask for the least value of the quantity\n$$\\limsup_{R \\to \\infty} \\frac{\\mathcal{H}^2(\\Sigma \\cap B_R)}{\\abs{B_R}}.$$\nHere $\\Sigma$ is the union of the cell boundaries, $\\mathcal{H}^2$ is area, and $B_R$ is the ball of radius $R$ about the origin.\n\n## Remarks\n\nKelvin proposed as the answer a partition into congruent cells shaped like truncated octahedra, arranged as in the bitruncated cubic [[honeycomb|HTJ03RFKFH]].\nHis cells are not exactly polyhedra: their hexagonal faces are slightly curved, so that they satisfy [[Plateau's laws|381RB634WS]].\nIn 1993 Denis Weaire and Robert Phelan found a partition into two kinds of cells of equal volume with about $0.3\\%$ less area than Kelvin's.\nSo Kelvin's proposal is not optimal.\nThe problem is open: no partition is known to be optimal, and it is not known whether the Weaire–Phelan structure is.\nThe two-dimensional analogue is solved: by the [[honeycomb theorem|3JST9B967R]] of Thomas Hales, the regular hexagonal tiling is optimal in the plane.\n\n## Explanations\n\n### L1 human\n\nHow should one divide space into bubbles of the same size, so that the soap film needed is as small as possible?\nKelvin suggested a pattern of identical fourteen-faced cells.\nA century later, a better foam was found with two kinds of cells, and nobody yet knows the best one.\n",
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      "first_stated": "1887",
      "posed_by": "William Thomson (Lord Kelvin)"
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