*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
*R*_{o}.
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, *GO*^{2} +
*AG*^{2} =
*OA*^{2}. Since *AG*^{2} = *EG*^{2}
+ *EA*^{2} = 4/3 *EA*^{2},
*r*_{o}^{2}
= *GO*^{2} = *OA*^{2} - 4/3 *EA*^{2}
= 1/2 - (4/3)(1/4) = 1/6, and *r*_{o} = 1/Sqrt[6]. |