Looking for a proof that the diameter of the smallest bounding circle is less than or equal to $\frac{2}{\sqrt{3}}$ times the diameter of the set

Question

This came up while I was attempting to solve an old journal problem. It's not the easiest result to search for so I figured I would ask.

Let $E$ be a subset of $\mathbb{R}^2$, then the diameter of $E$ is

$$\operatorname{diam} E = \sup \{d(p,q) \mid p,q \in E\}$$

where $d$ is the Euclidean metric on $\mathbb{R}^2$. As this answer states, the diameter of $C$, the smallest disc capable of covering $E$, is less than or equal to $\frac{2}{\sqrt{3}} \operatorname{diam} E$. The answer also claims that this inequality is "not straightforward to prove." Where can I find a proof of this?

Answer 1

Jung's theorem states that if $K$ is a compact subset of $R^n$ and $$ d=\max_{p,q\in K}\|p-q\|_2, $$ there exist a closed ball with radius $$ r\leq d\sqrt{\frac{n}{2(n+1)}}$$ that contains $K$, and the equality is achieved only by the regular $n$-simplex.

The proof relies on a convexity argument and the use of Helly's theorem.

Answer 2

Assume without loss of generality that $E$ is compact. Consider the smallest circumscribed circle $C$ of $E$. To fix ideas, assume that $C$ has unit radius (the problem is scale-invariant).

Form the intersection $T=C\cap E$. Then $T$ has the property that it is not contained in any semicircle of $C$ (otherwise by first variation $C$ would not be the smallest circle).

Either $T$ consists of a pair of opposite points of $C$, in which case the result is immediate, or one can choose three points $P,Q,R\in T$ such that the set $\{P,Q,R\}$ does not lie in any semicircle of $C$. The shortest arcs then satisfy $$PQ+QR+RP=2\pi.$$ Therefore one of the three arcs is at least $\frac{2\pi}{3}$. This is equivalent to the answer you want.