Given a regular tetrahedron of edge length $a$, how do I prove that the circumradius of the tetrahedron is equal to $\frac{\sqrt 6}{4}a$?
[Math] Finding the circumradius of a regular tetrahedron
geometryplatonic-solidssolid-geometry
Related Solutions
One method is to divide the tetrahedron into 8 identical sections or two sets of 4 identical sections and then to find the smallest bounding cube as the sections are randomly rotated. For 8 cubes, I haven't been able to beat 1/2 yet.
Covering an edge-2 tetrahedron with 4 unit cubes is easy (link to code). With some twisting and optimization, the tetrahedron can be covered with 4 cubes with edge length 0.969975. This can likely be improved by allowing the cubes to interact.
Note that the question for how best to cover a triangle with two squares wasn't solved correctly until 2009. Instead of jumping to 8 cubes, it might be best to settle 2 cubes, 3 cubes, and 4 cubes first.
For a tetrahedron with edge length 2 ...
1 cube of edge length 1.41421356237309504
2 cubes of edge length 1.41421356237309504 (??)
3 cubes of edge length 1.3233 (below)
4 cubes of edge length 0.969975 (above)
5 cubes of edge length .848528 (below)
Here's a picture for 3 cubes. The tetrahedron has a base triangle ABC. One cube covers points A, D, mid-AB, mid-AC, and the center of the base. But there is likely improvement by shifting the points mid-AB, mid-AC.
Five cubes seems to have a simple solution.
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.
Best Answer
In Elements XIII, 13 Euclid proves that the square of the diameter of the circumsphere is one and a half times the square of the side of the tetrahedron.
So$$d^2=\frac{3}{2}a^2$$Then if $r$ is the radius$$4r^2=\frac{3}{2}a^2$$making$$r^2=\frac{3}{8}a^2=\frac{6}{16}a^2$$and$$r=\frac{\sqrt 6}{4}a$$