Dodecahedron

The regular dodecahedron is the Platonic Solid ($P_4$) composed of 20 Vertices, 30 Edges, and 12 Pentagonal Faces. It is given by the symbol $12\{5\}$, the Schläfli Symbol $\{5,3\}$. It is also Uniform Polyhedron $U_{23}$ and has Wythoff Symbol $3\,\vert\,2\,5$. The dodecahedron has the Icosahedral Group $I_h$ of symmetries.

A Plane Perpendicular to a $C_3$ axis of a dodecahedron cuts the solid in a regular Hexagonal Cross-Section (Holden 1991, p. 27). A Plane Perpendicular to a $C_5$ axis of a dodecahedron cuts the solid in a regular Decagonal Cross-Section (Holden 1991, p. 24).

The Dual Polyhedron of the dodecahedron is the Icosahedron.

When the dodecahedron with edge length $\sqrt{10-2\sqrt{5}}$ is oriented with two opposite faces parallel to the $xy$-Plane, the vertices of the top and bottom faces lie at $z=\pm(\phi+1)$ and the other Vertices lie at $z=\pm(\phi-1)$, where $\phi$ is the Golden Ratio. The explicit coordinates are

$$\pm \left({2\cos({\textstyle{2\over 5}}\pi i), 2\sin({\textstyle{2\over 5}}\pi i), \phi+1}\right)$$ (1)

$$\pm \left({2\phi\cos({\textstyle{2\over 5}}\pi i), 2\phi\sin({\textstyle{2\over 5}}\pi i), \phi-1}\right)$$ (2)

with $i=0$, 1, ..., 4, where $\phi$ is the Golden Ratio. Explicitly, these coordinates are

$${\bf x}_{10}^\pm = \pm(2, 0, {\textstyle{1\over 2}}(3+\sqrt{5}))$$ (3)

$${\bf x}_{11}^\pm = \pm({\textstyle{1\over 2}}(\sqrt{5}-1), {\textstyle{1\over 2}}\sqrt{10+2\sqrt{5}}, {\textstyle{1\over 2}}(3+\sqrt{5}))$$ (4)

$${\bf x}_{12}^\pm = \pm(-{\textstyle{1\over 2}}(1+\sqrt{5}), {\textstyle{1\over 2}}\sqrt{10-2\sqrt{5}}, {\textstyle{1\over 2}}(3+\sqrt{5}))$$ (5)

$${\bf x}_{13}^\pm = \pm(-{\textstyle{1\over 2}}(1+\sqrt{5}), -{\textstyle{1\over 2}}\sqrt{10-2\sqrt{5}}, {\textstyle{1\over 2}}(3+\sqrt{5}))$$ (6)

$${\bf x}_{14}^\pm = \pm({\textstyle{1\over 2}}(\sqrt{5}-1), -{\textstyle{1\over 2}}\sqrt{10+2\sqrt{5}}, {\textstyle{1\over 2}}(3+\sqrt{5}))$$ (7)

$${\bf x}_{20}^\pm = \pm(1+\sqrt{5}, 0, {\textstyle{1\over 2}}(\sqrt{5}-1))$$ (8)

$${\bf x}_{21}^\pm = \pm(1, \sqrt{5+2\sqrt{5}}, {\textstyle{1\over 2}}(\sqrt{5}-1))$$ (9)

$${\bf x}_{22}^\pm = \pm(-{\textstyle{1\over 2}}(3+\sqrt{5}), {\textstyle{1\over 2}}\sqrt{10+2\sqrt{5}}, {\textstyle{1\over 2}}(\sqrt{5}-1))$$ (10)

$${\bf x}_{23}^\pm = \pm(-{\textstyle{1\over 2}}(3+\sqrt{5}), -{\textstyle{1\over 2}}\sqrt{10+2\sqrt{5}}, {\textstyle{1\over 2}}(\sqrt{5}-1))$$ (11)

$${\bf x}_{24}^\pm = \pm(1, -\sqrt{5+2\sqrt{5}}, {\textstyle{1\over 2}}(\sqrt{5}-1)),$$ (12)

where ${\bf x}_{1i}^+$ are the top vertices, ${\bf x}_{2i}^+$ are the vertices above the mid-plane, ${\bf x}_{2i}^-$ are the vertices below the mid-plane, and ${\bf x}_{2i}^-$ are the bottom vertices. The Vertices of a dodecahedron can be given in a simple form for a dodecahedron of side length $a=\sqrt{5}-1$ by (0, $\pm\phi^{-1}$, $\pm\phi$), ($\pm\phi$, 0, $\pm\phi^{-1}$), ($\pm\phi^{-1}$, $\pm\phi$, 0), and ($\pm 1$, $\pm 1$, $\pm 1$).

For a dodecahedron of unit edge length $a=1$, the Circumradius $R'$ and Inradius $r'$ of a Pentagonal Face are

$$R' = {\textstyle{1\over 10}}\sqrt{50+10\sqrt{5}}$$ (13)

$$r' = {\textstyle{1\over 10}}\sqrt{25+10\sqrt{5}}.$$ (14)

The Sagitta $x$ is then given by

$$x\equiv R'-r'={\textstyle{1\over 10}}\sqrt{125-10\sqrt{5}}.$$ (15)

Now consider the following figure.

Using the Pythagorean Theorem on the figure then gives

$${z_1}^2+m^2 = (R'+r)^2$$ (16)

$${z_2}^2+(m-x)^2 = 1$$ (17)

$$\left({z_1+z_2\over 2}\right)^2+R'^2 = \left({z_1-z_2\over 2}\right)^2+(m+r')^2.$$ (18)

Equation (3) can be written

$$z_1z_2+r^2=(m+r')^2.$$ (19)

Solving (1), (2), and (19) simultaneously gives

$$m = r'={\textstyle{1\over 10}}\sqrt{25+10\sqrt{5}}$$ (20)

$$z_1 = 2r'={\textstyle{1\over 5}}\sqrt{25+10\sqrt{5}}$$ (21)

$$z_2 = R'={\textstyle{1\over 10}}\sqrt{50+10\sqrt{5}}.$$ (22)

The Inradius of the dodecahedron is then given by

$$r={\textstyle{1\over 2}}(z_1+z_2),$$ (23)

so

$$r^2 = {\textstyle{1\over 4}}\left({{\textstyle{1\over 10}}\sqrt{50+10\sqrt{5}}+{\textstyle{1\over 5}}\sqrt{25+10\sqrt{5}}\,}\right)^2$$

$$= {\textstyle{1\over 40}}(25+11\sqrt{5}\,),$$ (24)

and

$$r=\sqrt{25+11\sqrt{5}\over 40}={\textstyle{1\over 20}}\sqrt{250+110\sqrt{5}} = 1.11351\ldots.$$ (25)

Now,

$$R^2 = R'^2+r^2=[{\textstyle{1\over 100}}(50+10\sqrt{5}\,)+{\textstyle{1\over 400}}(250+110\sqrt{5}\,)]$$

$$= {\textstyle{3\over 8}}(3+\sqrt{5}\,),$$ (26)

and the Circumradius is

$$R=a\sqrt{{\textstyle{3\over 8}}(3+\sqrt{5}\,)} = {\textstyle{1\over 4}}(\sqrt{15}+\sqrt{3}\,)=1.40125\ldots.$$ (27)

The Interradius is given by

$$\rho^2 = r'^2+r^2=[{\textstyle{1\over 100}}(25+10\sqrt{5}\,)+{\textstyle{1\over 400}}(250+110\sqrt{5}\,)]$$

$$= {\textstyle{1\over 8}}(7+3\sqrt{5}\,),$$ (28)

so

$$\rho={\textstyle{1\over 4}}(3+\sqrt{5}\,) = 1.30901\ldots.$$ (29)

The Area of a single Face is the Area of a Pentagon,

$$A={\textstyle{1\over 4}}\sqrt{25+10\sqrt{5}}\,.$$ (30)

The Volume of the dodecahedron can be computed by summing the volume of the 12 constituent Pentagonal Pyramids,

$$V = 12({\textstyle{1\over 3}}Ar)$$

$$= 12({\textstyle{1\over 3}})({\textstyle{1\over 4}}\sqrt{25+10\sqrt{5}}\,)({\textstyle{1\over 20}}\sqrt{250+110\sqrt{5}}\,)$$

$$= {\textstyle{1\over 20}}(75+35\sqrt{5}\,)={\textstyle{1\over 4}}(15+7\sqrt{5}\,).$$ (31)

Apollonius showed that the Volume $V$ and Surface Area $A$ of the dodecahedron and its Dual the Icosahedron are related by

$${V_{\rm icosahedron}\over V_{\rm dodecahedron}} = {A_{\rm icosahedron}\over A_{\rm dodecahedron}}$$ (32)

The Hexagonal Scalenohedron is an irregular dodecahedron.

See also Augmented Dodecahedron, Augmented Truncated Dodecahedron, Dodecagon, Dodecahedron-Icosahedron Compound, Elongated Dodecahedron, Great Dodecahedron, Great Stellated Dodecahedron, Hyperbolic Dodecahedron, Icosahedron, Metabiaugmented Dodecahedron, Metabiaugmented Truncated Dodecahedron, Parabiaugmented Dodecahedron, Parabiaugmented Truncated Dodecahedron, Pyritohedron, Rhombic Dodecahedron, Small Stellated Dodecahedron, Triaugmented Dodecahedron, Triaugmented Truncated Dodecahedron, Trigonal Dodecahedron, Trigonometry Values Pi/5, Truncated Dodecahedron

References

Cundy, H. and Rollett, A. Mathematical Models, 3rd ed. Stradbroke, England: Tarquin Pub., 1989.

Davie, T. ``The Dodecahedron.'' http://www.dcs.st-and.ac.uk/~ad/mathrecs/polyhedra/dodecahedron.html.

Holden, A. Shapes, Space, and Symmetry. New York: Dover, 1991.