Dotty Hexagons

A friend ¹, 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.

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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 $a{n}^2+bn+c$.

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 $1+6\times (\frac{1}{2}n(n-1))$
• … 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?

Footnotes:

1. Thanks to Eli for the suggestion!