TITLE: Hexagon packing in a circle
QUESTION [25 upvotes]: Suppose I want to pack hexagons in a circle, as on the drawing below (red indicates "packed" hexagons).  I am wondering what is known about this problem. Specifically, I am interested in an approximation to how many fit (given the radius of the circle and the length of the side of the hexagon) and the error in such approximation.  Any ideas?

REPLY [2 votes]: Posting the trivial estimates so that we can measure progress.
Assume that the side of the hexagon has length $1$, and that the radius of the circle is $r$. The area of a single hexagon is then $6\cdot\sqrt3/4=3\sqrt3/2$.
Let $N$ be the number of hexagons inside the circle. Their total area is less than that of the circle, so we get the inequality
$$
\frac{3\sqrt3 N}2<\pi r^2.
$$
On the other hand the hexagons cover the concentric circle of radius $r-2$ completely. Therefore we get another inequality
$$
\frac{3\sqrt3 N}2>\pi (r-2)^2.
$$
The midway point gives then the ballpark estimate of
$$
N\approx\frac{2\sqrt3\pi(r-1)^2}9
$$
together with the bound on the absolute value of the error
$$
|\Delta N|<\frac{4\sqrt3\pi r}9. 
$$
The link by Chris Culter suggests that it may be possible to get an error term of order $r^{2/3}$. The problem considered there is about the number of lattice points, so it is not exactly equivalent.