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

\(\ds \)

\(=\)

\(\ds 2s\)

$\blacksquare$