```in the package Geometry`Polytopes`, we can find the vertices of the
regular polyhedrons.
Nice for use in programs that deal with Dodecahedrons, Icosahedrons and
the like.

For those who would like these vertices in exact (as opposed to floating
point) numbers,
I propose (after debugging?) the use (inclusion?, substitution?) of the
following:

Vertices[Tetrahedron]//InputForm

{{0, 0, Sqrt[3]}, {0, 2*Sqrt[2/3], -(1/Sqrt[3])}, {-Sqrt[2], -Sqrt[2/3],
-(1/ Sqrt[3])}, {Sqrt[2], -Sqrt[2/3], -(1/Sqrt[3])}}

Vertices[Octahedron]//InputForm

{{0, 0, Sqrt[2]}, {Sqrt[2], 0, 0}, {0, Sqrt[2], 0}, {0, 0, -Sqrt[2]},
{-Sqrt[ 2], 0, 0}, {0, -Sqrt[2], 0}}

Vertices[Cube]//InputForm

{{1/Sqrt[2], 1/Sqrt[2], 1/Sqrt[2]}, {-(1/Sqrt[2]), 1/Sqrt[2],
1/Sqrt[2]}, {-( 1/Sqrt[2]), -(1/Sqrt[2]), 1/Sqrt[2]},
{1/Sqrt[2], -(1/Sqrt[2]), 1/Sqrt[2]}, {-(1/Sqrt[2]), -(1/Sqrt[2]),
-(1/Sqrt[ 2])}, {1/Sqrt[2], -(1/Sqrt[2]), -(1/Sqrt[2])},
{1/Sqrt[2], 1/Sqrt[2], -(1/Sqrt[2])}, {-(1/Sqrt[2]), 1/Sqrt[2],
-(1/Sqrt[ 2])}}

Vertices[Dodecahedron]//InputForm

{{Sqrt[(5 - Sqrt[5])/10], (3 - Sqrt[5])/2, Sqrt[(5 + Sqrt[5])/10]},
{-Sqrt[5/2 - 11/(2*Sqrt[5])], (-1 + Sqrt[5])/2, Sqrt[(5 +
Sqrt[5])/10]},
{-2*Sqrt[1 - 2/Sqrt[5]], 0, Sqrt[(5 + Sqrt[5])/10]}, {-Sqrt[5/2 -
11/(2* Sqrt[5])], (1 - Sqrt[5])/2, Sqrt[(5 + Sqrt[5])/10]},
{Sqrt[(5 - Sqrt[5])/10], (-3 + Sqrt[5])/2, Sqrt[(5 + Sqrt[5])/10]},
{Sqrt[(5 + Sqrt[5])/10], (-1 + Sqrt[5])/2, Sqrt[5/2 -
11/(2*Sqrt[5])]},
{-Sqrt[1 - 2/Sqrt[5]], 1, Sqrt[5/2 - 11/(2*Sqrt[5])]}, {-Sqrt[2 -
2/Sqrt[ 5]], 0, Sqrt[5/2 - 11/(2*Sqrt[5])]},
{-Sqrt[1 - 2/Sqrt[5]], -1, Sqrt[5/2 - 11/(2*Sqrt[5])]},
{Sqrt[(5 + Sqrt[5])/10], (1 - Sqrt[5])/2, Sqrt[5/2 - 11/(2*Sqrt[5])]},

{Sqrt[1 - 2/Sqrt[5]], 1, -Sqrt[5/2 - 11/(2*Sqrt[5])]},
{-Sqrt[(5 + Sqrt[5])/10], (-1 + Sqrt[5])/2, -Sqrt[5/2 -
11/(2*Sqrt[5])]},
{-Sqrt[(5 + Sqrt[5])/10], (1 - Sqrt[5])/2, -Sqrt[5/2 -
11/(2*Sqrt[5])]},
{Sqrt[1 - 2/Sqrt[5]], -1, -Sqrt[5/2 - 11/(2*Sqrt[5])]}, {Sqrt[2 -
2/Sqrt[ 5]], 0, -Sqrt[5/2 - 11/(2*Sqrt[5])]},
{Sqrt[5/2 - 11/(2*Sqrt[5])], (-1 + Sqrt[5])/2, -Sqrt[(5 +
Sqrt[5])/10]},
{-Sqrt[(5 - Sqrt[5])/10], (3 - Sqrt[5])/2, -Sqrt[(5 + Sqrt[5])/10]},
{-Sqrt[(5 - Sqrt[5])/10], (-3 + Sqrt[5])/2, -Sqrt[(5 + Sqrt[5])/10]},
{Sqrt[5/2 - 11/(2*Sqrt[5])], (1 - Sqrt[5])/2, -Sqrt[(5 +
Sqrt[5])/10]}, {2* Sqrt[1 - 2/Sqrt[5]], 0, -Sqrt[(5 + Sqrt[5])/10]}}

Vertices[Icosahedron]//InputForm

{{0, 0, Sqrt[10/(5 + Sqrt[5])]}, {2*Sqrt[2/(5 + Sqrt[5])], 0, Sqrt[2/(5
+  Sqrt[5])]},
{(-1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])], 1, Sqrt[2/(5 + Sqrt[5])]},
{-((1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])]), Sqrt[-1 + 10/(5 + Sqrt[5])],
Sqrt[ 2/(5 + Sqrt[5])]},
{-((1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])]), -Sqrt[-1 + 10/(5 + Sqrt[5])],
Sqrt[ 2/(5 + Sqrt[5])]},
{(-1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])], -1, Sqrt[2/(5 + Sqrt[5])]},
{(1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])], Sqrt[-1 + 10/(5 + Sqrt[5])],
-Sqrt[2/( 5 + Sqrt[5])]},
{-((-1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])]), 1, -Sqrt[2/(5 + Sqrt[5])]},
{-2* Sqrt[2/(5 + Sqrt[5])], 0, -Sqrt[2/(5 + Sqrt[5])]},
{-((-1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])]), -1, -Sqrt[2/(5 + Sqrt[5])]},

{(1 + Sqrt[5])/Sqrt[2*(5 + Sqrt[5])], -Sqrt[-1 + 10/(5 + Sqrt[5])],
-Sqrt[ 2/(5 + Sqrt[5])]}, {0, 0, -Sqrt[10/(5 + Sqrt[5])]}}

It's trivial, but applying N[..] to these is faster than the inverse
operation (:-))

wouter.
Dr. Wouter L. J. MEEUSSEN
w.meeussen.vdmcc@vandemoortele.be
eu000949@pophost.eunet.be

```

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