Circle is the only shape with the smallest maximum arc-chord ratio

Given a nontrivial rectifiable closed plane curve $\gamma\colon\mathbb{R}/\ell\mathbb{Z}\to\mathbb{R}^{2}$ of length $\ell$ parameterized by the arclength, the arc-chord constant of $\gamma$ is defined as

That is, it is the maximum ratio between the arc length and the chord length among all of its segments of length at most $\frac{\ell}{2}$. Note that the curve needs to be simple (i.e., have no self-intersection) in order for its arc-chord constant to be finite.

Obviously, the arc is in general longer than the chord, so $c_{\mathrm{AC}}(\gamma)$ is bounded below by $1$. But if you think about it, than you can immediately see that any closed curve you can imagine would have the arc-chord constant strictly away from $1$, so it is natural to ask what is the minimum value of it.

Intuitively, it sounds plausible to believe that circles are the only closed curves with the smallest arc-chord constant, which is $\frac{\pi}{2}$. This post is about a proof of this fact. The original argument is due to Seok Hyeong Lee.

Suppose that $c_{\mathrm{AC}}(\gamma) \leq \frac{\pi}{2}$. We will show that $\gamma$ must be a circle, which further shows $c_{\mathrm{AC}}(\gamma) = \frac{\pi}{2}$. Think of a new curve $f\colon\mathbb{R}/\ell\mathbb{Z}\to\mathbb{R}^{2}$ given as $f(s) \mathrel{\unicode{x2254}} \gamma(s) - \gamma(s + \ell/2)$. By the assumption $c_{\mathrm{AC}}(\gamma) \leq \frac{\pi}{2}$, for any $s\in\mathbb{R}/\ell\mathbb{Z}$, $f(s)$ cannot lie inside the open disk of radius $\ell/\pi$ centered at the origin.

On the other hand, clearly the total length of $f$ cannot exceed $2\ell$ by its definition. Since $f(s+\ell/2) = -f(s)$ holds for all $s$, the segment \(f\vert_{[0,\ell/2]}\) of $f$ must have the length at least $\ell$ because $f$ cannot pass through the disk of radius $\ell/\pi$.

Although this sounds intuitively very clear, I could not find a proof simpler than the one given below. Consider a new curve $g\colon s\mapsto \frac{\ell}{\pi\vert f(s)\vert}f(s)$, i.e., the projection of $f$ onto the circle of radius $\ell/\pi$. Since $f$ is Lipschitz, Radamacher’s theorem shows that $f$ is almost everywhere differentiable and $f$ is equal to the Lebesgue integral of its derivative. The same is true for $g$ since $f$ is strictly away from the origin. Note that

so

where $\cdot^{\perp}$ denotes the $90^{\circ}$-degree counterclockwise rotation. Hence,

holds for all $s$. In particular, the length of the segment \(g\vert_{[0,\ell/2]}\) must be at most $\ell$, but since $g(0) = -g(\ell/2)$ and any geodesic on the circle of radius $\ell/\pi$ connecting two antipodal points must have the length $\ell$, it follows that the length of \(g\vert_{[0,\ell/2]}\) is precisely $\ell$. Hence, the above inequality is actually an equality (at least for almost every $s$), thus repeating this once more with \(g\vert_{[\ell/2,\ell]}\) shows that

1. $f$ is a constant-speed parameterization with the speed $2$ (which means that $\vert f’\vert = 2$ almost everywhere), and
2. $\vert f\vert = \frac{\ell}{\pi}$ holds almost everywhere.

Now, by definition of $f$, we know

and since both of $\gamma’(s)$ and $\gamma’(s+\ell/2)$ are of size $1$, $\vert f’(s)\vert = 2$ is possible only when $\gamma’(s+\ell/2) = -\gamma’(s)$. In other words,

holds almost everywhere, or everywhere since $\gamma$ is Lipschitz, which implies that $c \mathrel{\unicode{x2254}}\frac{\gamma(s) + \gamma\left(s+\frac{\ell}{2}\right)}{2}$ is a constant. Note that this shows that

holds, thus $\gamma$ is a curve contained in the circle of radius $\frac{\ell}{2\pi}$ centered at $c$. By considering the universal covering of circles, it can be shown that the only such closed curves with finite arc-chord constant (so that without self-intersection) are the regular ones, i.e., the ones making exactly one complete turn without backtracking, in either counterclockwise or clockwise directions. Hence, the proof is done.