# Question 5

Define an $$n$$–colouring of the plane to be a function $f : \mathbb{R}^2 \rightarrow \left\{c_1, \ldots, c_n\right\}.$ In other words, each point in the plane is coloured one of n colours. For which $$n \in \mathbb{N}$$ does there exist an $$n$$–colouring of the plane with the property that no two points in $$\mathbb{R}^2$$ at distance 1 apart have the same colour?

## Solution by Chums

Firstly, if we can produce an $$n$$-colouring of the plane with this property, we can produce an $$(n+1)$$-colouring the same way, by excluding the $$(n+1)$$-st colour.

We will consider polygons circumscribed inside circles of diameter slightly less than 1 (say $$0.9$$). This means that any two points inside the polygons will be at most $$0.9$$ from each other, and so we can safely colour them one colour.

The following tiling of squares proves we can do it with 9 colours:

The following tiling of hexagons gets the number down to 7:

Working from the other end to rule out low numbers, one colour is not enough (consider two points distance 1 apart). Two colours are also not enough: consider an equilateral triangle, where no two vertices can be coloured the same.

Three colours are also not enough: it is easy to check that the following arrangement of ten points requires four colours.

Hence it is possible for seven, but not possible for three.