Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1998
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1998

[Date Index] [Thread Index] [Author Index]

Search the Archive

some trivia about Geometry`Polytopes`



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




  • Prev by Date: Newbie needs help with 3D plot
  • Next by Date: Re: Q--PolynomialReduce
  • Prev by thread: HTML export
  • Next by thread: How to get the solution area of inequation?