Theorem: Three Regular Tessellations/Triangles

Equilateral triangles form a regular tessellation:
:400px

Proof:
Let $s \in \R_{>0}$ be the side length of the equilateral triangles.
For all $x, y \in \Z$, let the center $P_{x, y}$ of each triangle have Cartesian coordinates:
:$P_{x, y} = s \tuple {\dfrac 1 2 x, \dfrac {\sqrt 3} 2 y + \dfrac {3 - m} {4 \sqrt 3} }$
where $m = \begin{cases} 1 & \textrm {for $x + y$ even} \\ -1 & \textrm {for $x + y$ odd} \end{cases}$.
Let each triangle be defined as $\triangle A_{x, y} B_{x, y} C_{x, y}$, where the coordinates of its vertices are:
{{begin-eqn}}
{{eqn | l = A_{x, y}
      | r = P_{x, y} +  s \tuple {-\dfrac 1 2, \dfrac {-m} {2 \sqrt 3} }
}}
{{eqn | l = B_{x, y}
      | r = P_{x, y} +  s \tuple {\dfrac 1 2, \dfrac {-m} {2 \sqrt 3} }
}}
{{eqn | l = C_{x, y}
      | r = P_{x, y} +  s \tuple {0, \dfrac m {\sqrt 3} }
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = A_{x, y} B_{x, y}
      | r = \sqrt {\paren {s \paren {-\dfrac 1 2 - \dfrac 1 2} }^2 + \paren {s \paren {\dfrac {-m} {2 \sqrt 3} + \dfrac m {2 \sqrt 3} } }^2} 
}}
{{eqn | r = \sqrt {s^2}
}}
{{eqn | r = s
      | c = as $s > 0$
}}
{{eqn | l = B_{x, y} C_{x, y}
      | r = \sqrt {\paren {s \paren {\dfrac 1 2 - 0} }^2 + \paren {s \paren {\dfrac {-m} {2 \sqrt 3} - \dfrac m {\sqrt 3} } }^2}
}}
{{eqn | r = \sqrt {\dfrac 1 4 s^2 + \paren {\dfrac {-3 m} {2 \sqrt 3} s}^2}
}}
{{eqn | r = \sqrt {s^2}
      | c = as $m^2 = 1$
}}
{{eqn | r = s
}}
{{eqn | l = C_{x, y} A_{x, y}
      | r = \sqrt {\paren {s \paren {0 + \dfrac 1 2} }^2 + \paren {s \paren {\dfrac m {\sqrt 3} + \dfrac m {2 \sqrt 3} } }^2} 
}}
{{eqn | r = s
      | c = by calculations as for $B_{x, y} C_{x, y}$
}}
{{end-eqn}}
From Triangle Side-Side-Side Congruence, 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 $\sequence {\triangle A_{x, y} B_{x, y} C_{x, y} }_{x \mathop \in \Z, y \mathop \in \Z}$ form a regular tesselation of the plane.
{{qed}}
Category:Three Regular Tessellations
Category:Equilateral Triangles
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