User:Anghel/Sandbox

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

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


 * $r \tuple{ \dfrac 3 2 x, \sqrt 3 y + \dfrac{ \sqrt 3} 2 m}$

where $m = \begin{cases} 0 & \textrm{for even $x$} \\ 1 & \textrm{for odd $x$} \end{cases}$.

If we label the vertices of each hexagon as $V_1, V_2, \ldots, V_6$ in anticlockwise direction with $V_1$ as the lowest left vertex, their coordinates are:

We show that $V_1, \ldots, V_6$ define the vertices of aregular hexagon.

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

As $V_2 = V_4 - r \tuple { 0, \sqrt 3 }$ and $V_1 = V_4 - r \tuple { 0, \sqrt 3 }$, it follows by Translation Mapping is Isometry that


 * $V_4 V_5 = V_1 V_2 = r $

As $V_3 = V_5 - r \tuple { - \dfrac 3 2, \dfrac {\sqrt 3} 2 }$ and $V_6 = V_2 - r \tuple { - \dfrac 3 2, \dfrac {\sqrt 3} 2}$, it follows by Translation Mapping is Isometry that


 * $V_5 V_6 = V_2 V_3 = r $

As $V_6 = V_4 - r \tuple { \dfrac 3 2, \dfrac {\sqrt 3} 2 }$ and $V_1 = V_3 - r \tuple {\dfrac 3 2, \dfrac {\sqrt 3} 2 }$, it follows by Translation Mapping is Isometry that


 * $V_6 V_1 = V_3 V_4 = r $

By definition of dot product, the

From Angle Between Vectors in Terms of Dot Product, it follows that:

qed}}