Last revision Wednesday 28 August 1996
Octahedron
(f = 8, e = 12, v = 6)
Like the cube, the octahedron's ex- and inradius are very easy to determine.
As usual, we let the side of the octahedron have length 1. Thus AE
= 1/2. Now, it is clear that ABCD is a square (from symmetry). Then
O
is the center of this square, and is the foot of the perpendicular dropped
from vertex F. Thus OE = EA, and OA = OB
= OF. So OA = Sqrt[2]/2 = 1/Sqrt[2], which is the exradius
Ro.
To find the inradius, not shown here, we construct the perpendicular from
point O to the center of equilateral triangle
ABF at G.
AGO being right, GO2 +
AG2 =
OA2. Since AG2 = EG2
+ EA2 = 4/3 EA2,
ro2
= GO2 = OA2 - 4/3 EA2
= 1/2 - (4/3)(1/4) = 1/6, and ro = 1/Sqrt[6]. |