Ratio of circumradius to inradius of a regular tetrahedron The circumradius of a set in $\mathbb{R}^n$ is the radius of the smallest sphere enclosing the set.  Similarly, the inradius is the radius of the largest sphere fitting inside a set. 

Question:  What is the ratio $\frac{\text{circumradius}}{\text{inradius}}$ for a regular tetrahedron?


This is one of my all time favorite questions because it admits at least three very different solutions, two of which fit anyone's definition of "elegant".  I'm posting my favorite solution, but I am keen to see what solutions others can come up with. 
 A: Assuming the $\text{centroid}\equiv\text{incenter}\equiv\text{circumcenter}$ of the regular tetrahedron $ABCD$ lies at the origin, i.e. $A+B+C+D=0$, the ratio $\frac{R}{r}$ equals $\frac{OA}{OA'}$ with $A'$ being the centroid of $BCD$. Hence
$$ \frac{R}{r} = \frac{3\|A\|}{\|B+C+D\|} =\frac{3\|A\|}{\|A\|} = 3 $$
with a straightforward generalization to the regular $n$-simplex.
This can be shown also by embedding $ABCD$ in $\mathbb{R}^4$ via $A\mapsto(1,0,0,0),B\mapsto(0,1,0,0),$ $C\mapsto(0,0,1,0),$ $D\mapsto(0,0,0,1)$. Metric computations in this framework are utterly simple.
A: Answer: $3$
Note first that the circumradius $R$ is the distance from the center of the tetrahedron to a vertex, while the inradius $r$ is the distance from the center of the tetrahedron to the center of a face.  You can break a tetrahedron into four smaller tetrahedra by connecting each face to the center.  The height of each smaller tetrahedron is $r$, while the height of the original tetrahedron is $r+R$.  Because they share a base, the ratio of the volume of the big and small tetrahedra is the ratio of their heights, $\frac{r+R}{r} = 1+\frac{R}{r}$.  But four little tetrahedra fill up the big one, so this ratio $=4$.  Hence $\frac{R}{r} = 4-1 = 3$.

The same argument can be generalized to the regular simplex in any number of dimensions.  The ratio of circumradius to inradius for a regular $n$-simplex is $n$.  
A: My second answer.
First, the answer to the analogous question in $\mathbb{R}^2$: "what is the ratio $\frac{\text{circumradius}}{\text{inradius}}$ of an equilateral triangle?"
To visualize this, stack up 6 smaller equilateral triangles, vertex to vertex, inside the triangle:

The inradius $r$ is the smaller triangle altitude, and the circumradius $R$ is two of those, so the ratio $R/r$ is 2/1 = 2.
The situation in 3d is analogous: stack up 20 smaller regular tetrahedra inside the original one, vertex to vertex, in four layers.

The inradius $r$ is one layer height (i.e. one smaller-tetrahedron altitude), and the circumradius $R$ is three layer heights, so the ratio $R/r$ is 3/1 = 3.
A: Referring to the diagram in Yly's answer, we see that r is the
the center's altitude.  Altitude of an average is average of the altitudes (this can be seen by choosing a cartesian coordinate system such that the altitude direction is one of the coordinate axes); therefore r = average of the 4 vertices' altitudes = (apex altitude + 0 + 0 + 0)/4 = (apex altitude)/4.
The circumradius R, on the other hand, is (apex altitude) - (center altitude) = (apex altitude) - (apex altitude)/4 = 3/4 apex altitude.
So their ratio R/r is (3/4 apex altitude) / (1/4 apex altitude) = 3.
