Distance between Points in Regular Hexagon
Theorem
Let $H$ be a regular hexagon embedded in the Euclidean plane $\R^2$.
Let $s \in \R_{>0}$ be the side length of $H$.
Let $\mathbf x, \mathbf y \in \R^2$ such that $\mathbf x$ and $\mathbf y$ lie in the interior of $H$, or on the circumference of $H$.
Then:
- $\map d {\mathbf x, \mathbf y } \le 2s$
where $\map d {\mathbf x, \mathbf y }$ denotes the Euclidean distance between $\mathbf x$ and $\mathbf y$.
Proof
From Regular Polygon is Cyclic, it follows that $H$ can be inscribed in a circle with center $\mathbf c$.
The circumcircle intersects all vertices of $H$.
From Regular Hexagon is composed of Equilateral Triangles, it follows that the side length $s$ is equal to the distance from $\mathbf c$ to any vertex of $H$.
It follows that the radius of the circumcircle is equal to $s$.
Hence:
\(\ds \map d {\mathbf x, \mathbf y}\) | \(\le\) | \(\ds \map d {\mathbf x, \mathbf c} + \map d {\mathbf c, \mathbf y}\) | by the Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds s + s\) | as $\mathbf x, \mathbf y$ lie in $H$, they also lie in the circumcircle | |||||||||||
\(\ds \) | \(=\) | \(\ds 2s\) |
$\blacksquare$