# Wax Mathematical Solution

The following is a solution to Puzzle #10 – Wax Mathematical. Take a look if you haven’t already.

Question #1

We’re first asked to find an expression for $c(n)$, the number of cells in a hive with $n$ layers around the nucleus. If no divine inspiration is forthcoming, it’s best to just write out a few terms, and see if you can spot a pattern.

From the diagram,

$c(0)=1$

$c(1)=1+6$

$c(2)=1+6+12$

$c(3)=1+6+12+18$

The first term persists, since the hive nucleus is always present. For every layer that is added, some multiple of six cells joins the hive. By inspection, the general pattern must be

$\displaystyle c(n)=1+6\sum_{i=0}^n i$

The sum of the integers from one to $n$ is a known function of $n$:

$\displaystyle \sum_{i=0}^n i =\frac{1}{2}n(n+1)$

Expanding this out gives the solution

$c(n)=3n^2+3n+1$

Question #2

Next we calculate $g(n)$, the number of gaps in the hive. Note that we’re only counting those formed between three adjacent cells, and not the bumps at the edge of the hive.

The diagram shows that

$g(0)=0=6\times 0$

$g(1)=6=6\times 1$

$g(2)=24=6\times 4$

$g(3)=54=6\times9$

Then in general,

$g(n)=6n^2$

Question #3

The next task is to find the area of the gap between three adjacent circles. The key to this problem is that the centres of the three circles are connected by an equilateral triangle. The diagram below should make it clearer how to find the solution.

The area of the triangle is half the product of its base length and its height. The base of the triangle has width $2r$. Its height $h$ can be found from Pythagoras’ theorem. Splitting the equilateral triangle in half gives two right-angled triangles with side lengths $r$ and $h$. So

$r^2+h^2=(2r)^2$

$h=\sqrt{3}r$

So the area of the equilateral triangle connecting the three adjacent circles is

$\displaystyle \frac{1}{2}\sqrt{3}r\cdot2r=\sqrt{3}r^2$

We can also evaluate the area of the region where the circles and their connecting triangle overlap. This region consists of three circular sectors. Each sector comprises one sixth of a circle. Then the sectors combined comprise one half of a circle. Hence their collective area is

$\displaystyle \frac{1}{2}\pi r^2$

This means the area of the gap must be

$\displaystyle A=\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2$

Question #4

Here, we bring together all we’ve learnt so far.

The entire hive consists of both the cells and the gaps. The area of the cells $C(n)$ is the product of $c(n)$, the number of cells, and the area of a single cell. It is given that the radius of each cell is $r$, so

$C(n)=\pi r^2 c(n)$

$C(n)=(3n^2+3n+1)\pi r^2$

The area of the gaps $G(n)$ is given by

$\displaystyle G(n)=\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2 g(n)$

$\displaystyle G(n)=6n^2\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2$

Then the fraction $f$ of the hive that the gaps constitute is

$\displaystyle f(n)=\frac{G(n)}{G(n)+C(n)}$

$\displaystyle f(n)=\frac{6n^2\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2}{6n^2\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2+(3n^2+3n+1)\pi r^2}$

The question stipulates that we take the limit in which $n$ becomes unboundedly large – this corresponds to a hive that fills the entire plane. When $n$ goes to infinity, terms of order $n$ squared dominate overwhelmingly, so terms of order $n$ and below can be ignored. Hence

$\displaystyle f(\infty)\to\frac{6n^2\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2}{6n^2\Big(\sqrt{3}-\frac{\pi}{2}\Big)r^2+3n^2 \pi r^2}$

Simplification yields

$\displaystyle f(\infty)=\frac{2\Big(\sqrt{3}-\frac{\pi}{2}\Big)}{2\Big(\sqrt{3}-\frac{\pi}{2}\Big)+\pi}$

$\displaystyle f(\infty)=\frac{2\sqrt{3}-\pi}{2\sqrt{3}}$

$\displaystyle f(\infty)=1-\frac{\pi}{2\sqrt{3}}$

Numerically, this comes out to be

$f(\infty)\equiv 9.310\%$

to three decimal places. So for a hive made up of closely packed circles, just under one tenth of the available space will be wasted.