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 $x+y$ even } \\ -1 & \textrm{for $x+y$ odd } \end{cases}$.

From Regular Hexagon is composed of Equilateral Triangles,

Let each triangle be defined as $\triangle A_{x,y} B_{x,y} C_{x,y}$, where the coordinates of its vertices are:

We show that $\triangle A_{x,y} B_{x,y} C_{x,y}$ is an equilateral triangle.

Translate the plane by the translation mapping $\tau: \R^2 \to \R^2$, defined by:


 * $\map \tau {\mathbf p} = \mathbf p - P_{x,y}$

so the translation of the triangle center $\map \tau { P_{x,y} }$ is equal to the origin $\tuple {0,0}$.

From Translation Mapping is Isometry, it follows that this does not change the length of the sides of $\triangle A_{x,y} B_{x,y} C_{x,y}$.

By the Distance Formula, the length of each side is:

From Triangle Side-Side-Side Equality, it follows that equilateral triangles with the same length of their sides are congruent.

By direct calculations, it follows that each vertex of $\triangle A_{x,y} B_{x,y} C_{x,y}$ is identical to five other vertices:


 * $A_{x,y} = A_{x,y-m} = B_{x-2,y} = B_{x-2,y-m} = C_{x-1,y} = C_{x-1,y-m}$
 * $B_{x,y} = B_{x,y-m} = A_{x+2,y} = A_{x-2,y-m} = C_{x+1,y} = C_{x+1,y-m}$
 * $C_{x,y} = C_{x,y+m} = A_{x+1,y} = A_{x+1,y+m} = B_{x-1,y} = B_{x-1,y+m}$

which shows that each side of $\triangle A_{x,y} B_{x,y} C_{x,y}$ is identical to one other side:


 * $A_{x,y} B_{x,y} = A_{x,y-m} B_{x,y-m}$
 * $B_{x,y} C_{x,y} = C_{x+1,y} A_{x+1,y}$
 * $C_{x,y} A_{x,y} = B_{x-1,y} C_{x-1,y}$

Hence, the triangles $\sequence {\triangle A_{x,y} B_{x,y} C_{x,y} }_{x \in \Z, y \in \Z}$ form a regular tesselation of the plane.

qed}}