# Dotty Hexagons

A friend ((Thanks to Eli for the suggestion!)), justifiably proudly, shared on Facebook that he’d worked out the number of dots he’d need to draw a hexagon with $n$ dots on each edge. I thought it was a nice puzzle!

Before we go anywhere, I’ll clear up what exactly we mean by a hexagon with $n$ dots per side. Here’s a hexagon with $n = 3$:

* * * * * * * * * * * * * * * * * * *

… which has 19 dots. Can you come up with a general expression for the number of dots? I came up with three methods, including the same one that my friend did. Spoilers below the line.

### Option 1: mechanically

When $n=1$, the number of dots is 1.

When $n=2$, the number of dots is 7.

When $n=3$, the number of dots is 19.

We’re looking at an area, so it’s almost certainly a quadratic sequence of the form $an^2 + bn + c$.

This gives three equations:

- $a + b + c = 1$ [1]
- $4a + 2b + c = 7$ [2]
- $9a + 3b + c = 19$ [3]

Subtracting [1] from [2] gives $3a + b = 6$

Subtracting [2] from [3] gives $5a + b = 12$

It’s not too hard to see that $a = 3$ and $b = -3$; looking back at [1] gives $c = 1$.

Therefore the number of dots is $3n^2 - 3n + 1$.

### Option 2: triangles

This is a way that made me happy to think about: each hexagon is made of six triangles with a base of $n-1$ and and extra dot in the middle. To make a triangle with base $n-1$, you need $\frac{1}{2}n(n-1)$ dots, so in all we need $3n(n-1) + 1$, which is $3n^2 - 3n + 1$ as before.

### Option 3: building up

This is my friend’s observation:

- You start with a dot
- You add six dots for the second hexagon
- You add twelve dots for the third hexagon
- You add 18 for the fourth
- … and so on
- So each hexagon has one dot, plus 6 times the sum of the numbers from 1 to $n-1$
- … which is $1 + 6 \times \br{\frac{1}{2}n(n-1)}$
- … or $3n^2 - 3n + 1$.

That *sum* works out to be the same as for option 2, but the intuition behind it is interestingly different.

Do you have any alternative approaches?