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We divide a plane ($\mathbb{R}^2$) into infinite number of regions each of area equal $1$. We can use only (one-dimensional) curves which may meet at points.

Fix a point $p$ on a plane and consider circles with center at the point $p$ and radius $r$. Let $N(r)$ be the number of regions inside the circle and $L(r)$ the total length of all curves inside this circle.

What is best possible arrangement of curves if we want to minimize $$\lim_{r\rightarrow\infty}\frac{L(r)}{N(r)}.$$

I guess that the curves should be edges of hexagons, i.e. they should form the honeycomb structure. In that case $\lim_{r\rightarrow\infty}\frac{L(r)}{N(r)}=\frac{4}{\sqrt[4]{27}}$ because on average we need to draw $4$ new edges of hexagon to create every new region and the length of the edge of hexagon which has volume $1$ is $1/\sqrt[4]{27}$.

Actually, if we assume that all regions have to be of the same shape then the only possible choices are: triangular, square or hexagonal lattice. It is easy to show that hexagonal lattice is the best. But do the regions really have to have all the same shape?

This is probably a standard problem. In fact it was solved by bees long time ago, but do they find the optimal solution?

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This reminds me of the type of problems in Burago-Burago-Ivanov's "Metric Geometry"; sorry I can't think of a chapter, nor page. –  user99680 Jun 26 at 19:19
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This is the honeycomb conjecture, proven by Hales in 1999.

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Instead of providing only a link to a solution, you should also summarize main ideas, methods and results of the solution you linked to. Besides, links are not guarantied to be valid forever. –  VividD Jul 2 at 23:00
The content of the links is literally the very conjecture in the question and the information that Hales proved it in 1999. Not sure what more you expect me to add. –  Rahul Jul 3 at 0:31
The prove might be found here arxiv.org/abs/math/9906042 –  user72829 Jul 3 at 10:00
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