## 7. Why is the regular hexagon special?

So now we look at one of the regions of type II. We will show that the ratio of the area inside the disc to that of the whole type II region is at most equal to the ratio in the special case when the central angle is 60o. This can be done by an explicit calculation, and the claim then reduces to the observation that the function sin(x) / x is monotonic decreasing in the range between 0 and /2.

But it can also be demonstrated by a purely geometric argument. The figure on the left above shows the special case. In the figure on the right the animation generates the other cases. As the central angle decreases we also generate the image of the left disc under the unique linear transformation which acts by scaling vertically and horizontally, transforming the original (60o) triangle to the narrower isosceles one. This is indicated in the figure above. Linear transformations preserve ratios of areas. The animation then shows how the largest ratio of circular sector to triangle ocurs in a type II sector when the central angle is 60o.

In summary, the density will achieve the maximum possible only when there are no regions of the first or third type and when the central angle of each of the regions of type II is exactly 60o. These criteria uniquely determine the cells in a hexagonal packing. The same proof explains the argument about the 6 discs surrounding a single one, since it shows that the hexagon circumscribing a disc is the Voronoi cell of smallest area containing it.

The analogous argument fails in 3D, since the smallest cell surrounding a sphere is known to be a regular dodecahedron, a shape which cannot partition all of space. In particular, as Kepler himself well knew and well described, this is not the same as the cell in the densest layout of spheres throughout all of space. It is this disparity between local and global behaviour that offers a serious obstacle to a simple proof of Kepler's conjecture in 3D.

@2000 American Mathematical Society