User:Anghel/Sandbox

Proof
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 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}$.

From Regular Hexagon is composed of Equilateral Triangles, it follows that each hexagon can be triangulated into six congruent equilateral triangles.

Also, each of the six equilateral triangles has $Q_{x,y}$ as a vertex.

By Three Regular Tessellations:Triangles, there exists a regular tessellation of equilateral triangles with side length s.

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.

The six triangles in the triangulation of the hexagon correspond to the triangles of the regular tessellation:


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

As the triangles form a regular tessellation of the plane, it follows that the hexagons also form a regular tessellation of the plane. qed}}