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.

Hexagon as unit disk
Hexagon as unit disk

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.

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