# The shortest circle is a hexagon

Let ${\|\cdot\|}$ be some norm on ${{\mathbb R}^2}$. The norm induces a metric, and the metric yields a notion of curve length: the supremum of sums of distances over partitions. The unit circle ${C=\{x\in \mathbb R^2\colon \|x\|=1\}}$ is a closed curve; how small can its length be under the norm?

For the Euclidean norm, the length of unit circle is ${2\pi\approx 6.28}$. But it can be less than that: if ${C}$ is a regular hexagon, its length is exactly ${6}$. Indeed, each of the sides of ${C}$ is a unit vector with respect to the norm defined by ${C}$, being a parallel translate of a vector connecting the center to a vertex.

To show that ${6}$ cannot be beaten, suppose that ${C}$ is the unit circle for some norm. Fix a point ${p\in C}$. Draw the circle ${\{x\colon \|x-p\|=1\}}$; it will cross ${C}$ at some point ${q}$. The points ${p,q,q-p, -p, -q, p-q}$ are vertices of a hexagon inscribed in ${C}$. Since every side of the hexagon has length ${1}$, the length of ${C}$ is at least ${6}$.
It takes more effort to prove that the regular hexagon and its affine images, are the only unit circles of length ${6}$; a proof can be found in Geometry of Spheres in Normed Spaces by Juan Jorge Schäffer.