# Planar kelvin problem

What is the minimal possible value of the maximal total side length shared by any two tiles in a tiling of the plane if all tiles have the same area $A$?

$\text{Total side length} = \text{Length-integral of the curve formed by the intersection of two tiles}$

i) Using a finite set of tiles
ii) Using any set of tiles.

This is known for $3D$ as http://mathworld.wolfram.com/KelvinsConjecture.html
I have found no info on the planar case, so it may be trivial, in which case I want to see the proof.

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I fail to see how this has anything to do with elementary set theory... – Asaf Karagila Nov 10 '11 at 18:29
My guess: hexagonal tiling. – Tim Seguine Nov 13 '11 at 15:09

If by maximal sidelength you mean the maximum total sidelength rather than the maximum of any one side, then I suspect that a hexagonal tiling is the optimum, which has maximal side length $\sqrt{\frac{2A}{3\sqrt{3}}}$.
I think you got the maximal side length wrong. If the side of the hexagon is $s$, the area is $A = 3\sqrt{3}s^2/2$, so the maximal sidelength is $\sqrt{2A/3\sqrt{3}}$. – Craig Nov 15 '11 at 3:57