I am working through Lectures in Discrete Geometry, and one of the questions goes as follows:
Any point set of diameter $1$ in $\mathbb{R}^2$ can be enclosed in a disc of radius $1/\sqrt{3}$. Prove this statement, then find a $r = r(d)$ for which a point set of diameter $1$ in $\mathbb{R}^d$ can be enclosed by a $d$-disc of radius $r$. Prove it.
I know the structure of the general proof, using Helly's Theorem. However, a key step in the proof is noting that any equilateral triangle with side length $1$ will always have a distance of $1/\sqrt{3}$ from its center to its vertices.
The $d$-dimensional generalization should have the exact same structure as the $2$-dimensional proof, but I cannot find the $r$. There is a hint in the back of the book which says:
$\sqrt{2d/(d+1)}$
But I cannot parse the hint. Is this supposed to be $r$? In that case, why is $r(2) = \sqrt{4/3}$, rather than the correct $1/\sqrt{3}$?
I believe that $r$ should be the distance I ask for in the title, but I am having trouble coming up with an argument for a distance dependent only on $d$, and I still don't understand the hint.
There is a nice embedding of the regular $d$-dimensional simplex into $\mathbb{R}^{d+1}$, by taking the vertices to be the unit vectors along every axis, i.e., $(1,0,0,\ldots,0), (0,1,0,\ldots,0), \ldots, (0,0,\ldots,0,1)$.
This tetrahedron has a center of $(a,a,a,\ldots,a)$, where $a=\frac1{d+1}$. It has a side length of $\sqrt{2}$.
Taking the squared differences in each coordinate, we see that the distance from this center to any vertex is $\sqrt{d\cdot a^2+(1-a)^2} = \sqrt{\frac{d+d^2}{(d+1)^2}}=\sqrt{\frac{d}{d+1}}$.
Scaling down by a factor of $\sqrt{2}$ so that we normalize to a unit tetrahedron, our final answer is $\sqrt{\frac{d}{2(d+1)}}$. You can verify that at $d=1$ this is $1/2$, and at $d=2$ is it $1/\sqrt{3}$, as expected.
Perhaps the book misplaced this factor of $2$? I am not sure how they arrived at that value.