# Simple cases of Jung's Theorem

Jung's theorem states that if $A\subset\mathbb{R}^n$ and $0<d=\operatorname{diam}(A)<\infty$, then there exist a unique closed ball $\bar{B}(x,r)$ of radius $r$ where $$r=\sqrt{\frac{n}{2(n+1)}}d$$that contains A. The book I'm reading (an introductory real analysis book) only mentions this results and asks, as an exercise, to check that the theorem is true when $A$ is an equilateral triangle in the plane or a regular tetrahedron in the space, and that the constant $\sqrt{\frac{n}{2(n+1)}}$ is the best possible .

Now, I can see that in those simple cases, $A$ is indeed contained in a closed ball with center its center of gravity and radius $r$ just described. However, I can't figure out why this radius is the best possible... This is just chapter 1 of the book, so I'm expecting some simple ways to explain this but it's quite challenging.

To summarize, assuming $A$ is an equilateral triangle in the plane or a regular tetrahedron in the space, why is the radius $r$ the best possible to cover $A$? Also, why is such a closed ball with radius $r$ unique? Please enlighten me.

Let $R$ be the set of all real numbers $r>0$ such that the given set $A\subset{\mathbb R}^n$ is contained in some closed ball of radius $r$. Then the number $\rho:=\inf R$ is uniquely determined.
Jung's theorem states that $$\rho\leq\sqrt{{n\over2(n+1)}}\>{\rm diam}(A)\ .$$
Furthermore it can be proved by some compactness arguments that there is actually a ball $B_\rho$ of radius $\rho$ containing $A$, and this ball is uniquely determined.
If $\triangle ABC$ is equilateral with base of lenght $a.$ Then the radius of excircle of $\triangle ABC$ is equal to $$R=\frac{1}{2} \cdot \frac{a}{\cos\frac{\pi}{3}}=\sqrt{\frac{2}{2(2+1)}}\mbox{diam} (\triangle ABC),$$ since the $\mbox{diam} (\triangle ABC)=a.$
The uniqueness of such ball follows fram the fact that a intersection of two distinct balls of radius $r$ can be covered by ball with radius less than $r.$