*Last revision Wednesday 28 August 1996*
*Dodecahedron*
(*f* = 12, *e* = 30, *v* = 20)
To derive the exradius *R*_{d} of the dodecahedron, we
will use an interesting fact (which we will also prove): there exist six
vertices which coincide with the vertices of a cube. This arises because
four of the threefold axes of the dodecahedron coincide with the four threefold
axes of hte cube, and three of the twofold axes of the dodecahedron coincide
with the three fourfold axes of the cube. We can see this in the figure
at right. This gives us a means of construction. Consider the cube with
face*ABCD*.
Let edge *AB* = 2. Then, as we found for the cube, the exradius is
Sqrt[3]. Thus our goal is to find the length of an edge of the dodecahedron.
To this end, bisect *AB, BC, CD, DA* at *H, J, K, L* and draw
*HK*
and *JL* intersecting at the center of the face *N*. Locate point
*P*
on *JN* such that *P* divides *JN* in mean and extreme ratio;
that is, *JN*/*NP* = phi, the golden ratio. Similarly, locate
*M*
so *MN* = *NP*. Erect perpendiculars *EP* and *FM*
to face *ABCD* such that *MN* = *NP* = *EP* = *FM*.
I claim that *E, F* are vertices of the dodecahedron.
To show this, we draw *RG* in the same manner as for *EP*.
Thus we must demonstrate that *AGBFE* is a planar, regular pentagon.
(By symmetry, the entire dodecahedron is similarly constructed.) First,
we show that *A, B, E, F, G* indeed all lie in the same plane. Clearly,
*A,
B, E, F* are planar (symmetry), and *A, B, G* are planar. Thus
these planes are the same if *GRH* is similar to *HNS*. But this
is obvious, because *GR/RH* = *HN*/*NS* = phi; therefore
*AGBFE*
is planar.
Second, we show that *AE* = *EF*. Note *NJ* = 1 so *NP*
= 1/phi, and
*PJ* = 1 - 1/phi = 1/phi^{2}. Thus by the Pythagorean
Theorem, *AP*^{2} (in yellow) = *PJ*^{2} + *AJ*^{2},
and *AE*^{2} = *AP*^{2} + *EP*^{2},
from which it follows that |

or *AF* = 2.
Therefore *AGBFE* is a regular pentagon, *E, F* are vertices
of the dodecahedron, and the proof is complete. The length of an edge is
2/phi, so for an edge length of 1, the exradius is Sqrt[3]/(2/phi) = Sqrt[3]
phi/2 = Sqrt[3](1+Sqrt[5])/4.
This construction appears in Euclid's *Elements*, but rather than
using algebraic methods to prove the claim as we have done here, he uses
geometric arguments. Another interesting sidenote is that we can inscribe
five cubes in the dodecahedron, thus regaining the fivefold symmetry axes
through each face. Furthermore, since
the tetrahedron can be inscribed in a cube, replacing each cube with a
tetrahedron results in a compound of five tetrahedra. However, the alignment
of the vertices of one tetrahedron and the dodecahedron is such that no
twofold axes of symmetry are preserved; thus there are two distinct ways
to inscribe five tetrahedra in a dodecahedron, and they are mirror images
or *enantiomorphs*. |