Last revision Wednesday 28 August 1996
An interesting excercise is to determine the exradius r of the tetrahedron in terms of the length
of an edge; that is, the radii of the circumscribed and inscribed spheres,
respectively. This is done with the help of basic geometry and the Pythagorean
Theorem. The main observation here is to note that the lengths _{t}OA
and OD are equal, since O is the center of the tetrahedron.
What we wish to determine, then, are the lengths R =
_{t}OD
and r = _{t}OF. Set the length of an edge of the
tetrahedron to be 1. Then AE = 1/2. Note triangle AEF is
a 30-60-90 right triangle, since angle CAB = 60 and F, being
the foot of the perpendicular dropped from center O, is the center
of equilateral triangle ABC. Thus AF is twice EF,
and AE is Sqrt[3] times EF. Hence EF = 1/(2 Sqrt[3])
and AF = 1/Sqrt[3]. Since AF^{2} + DF^{2}
= AD^{2} = 1, DF^{2} = 1 - 1/3 = 2/3. Using
the observation that OA = OD, we have |

It follows that the inradius is simply DF - OD = Sqrt[2/3]
- Sqrt[6]/4 = 1/(2 Sqrt[6]). |