Three Regular Tessellations/Hexagons

Theorem

Let $s \in \R_{>0}$ be the side length of the regular hexagons.

For all $x, y \in \Z$, let the center $Q_{x,y}$ of each regular hexagon have Cartesian coordinates:

$Q_{x,y} = s \tuple {x, \sqrt 3 y + \dfrac {\sqrt 3 \paren {3 - m} } 4}$

where:

$m = \begin{cases} 1 & : \textrm {for $x$ even} \\ -1 & : \textrm{for $x$ odd} \end{cases}$

Also, each of the six equilateral triangles has side length $s$, and $Q_{x,y}$ as one of its vertices.

Set:

$i := 3 x$

$j := 2 y + \dfrac {1-m} 2$

The hexagon center $Q_{x,y}$ corresponds to the six vertices:

$A_{i+1, j}, A_{i+1, j+1}, B_{i-1,j}, B_{i-1,j+1}, C_{i,j}, C_{i,j+1} $

as they are labelled in the proof of Three Regular Tessellations:Triangles.

$\triangle A_{i+1, j} B_{i+1, j} C_{i+1, j}$

$\triangle A_{i+1, j+1} B_{i+1, j+1} C_{i+1, j+1} $

$\triangle A_{i-1, j} B_{i-1, j} C_{i-1, j} $

$\triangle A_{i-1, j+1} B_{i-1, j+1} C_{i-1, j+1} $

$\triangle A_{i, j} B_{i, j} C_{i, j}$

$\triangle A_{i, j+1} B_{i, j+1} C_{i, j+1} $

as they are labelled in the proof of Three Regular Tessellations:Triangles.

For all $x' \in \Z$, there exist exactly one $x \in \Z$ and one $k \in \set {-1, 0, 1}$ such that:

$x' = 3 x + k$

For all $y' \in \Z$, there exist exactly one $y \in \Z$ and one $k \in \set {0, 1}$ such that:

$y' = 2 y + k$

which shows that each $\triangle A_{x', y'} B_{x', y'} C_{x', y'}$ is part of exactly one of the hexagons.

$\blacksquare$