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The diameter of a subset $X$ of $\mathbb{R}^n$ is defined as $\sup\{|x-y|:x,y\in X\}$.

What is the smallest radius $r(d,n)$ such that any subset $X$ of diameter $d$ in $\mathbb{R}^n$ is contained in a ball of radius $r(d,n)$? What are the $X$ that realize this bound? I know that $r(d,n)\leq d$. The equilateral triangle gives $r(d,2)\geq d/\sqrt(3)$ and I think we have equality here but I don't know how to prove it.

Selrach Dunbar
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Jiu
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  • I think this question is really good. It's easy to understand, but surprisingly difficult to solve. – painday Apr 16 '18 at 07:43
  • Indeed, the equilateral triangle (with side length $l$) in $\mathbb{R}^2$ has diameter $d=l$ and can be inscribed in a circle of radius $\frac{l}{2}\cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{1} = \frac{d}{\sqrt{3}}$. – Selrach Dunbar Jul 21 '19 at 19:56

1 Answers1

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Answer: $r(d,n) = d\sqrt{\frac{n}{2(n+1)}}$.

This is known as Jung's theorem (see also this question). The extremal case is the regular $n$-simplex in each dimension.

Daniele Tampieri
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Yly
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  • Could you calrify something for me please? Let $K$ be the ball of diameter $1$ centered at the origin. I do not think one can find a ball of radius $1/sqrt{3}$ that covers $K$. How does this not contradict the theorem? – Ovi Jul 21 '19 at 20:24
  • @Ovi It is possible: you only need a ball of radius $1/2$ (it has diameter $1$), whereas $1/\sqrt(3) \approx 0.577$. – Sambo Jul 21 '19 at 20:35
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    @Sambo Ohh I did not see the word "radius" in the statement of the theorem lol, I assumed it said "diameter" – Ovi Jul 21 '19 at 20:39