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I want to draw an (edit)dodecahedron with the vertex in the following coordinates:

 v={{-0.982247, 0, 0.187592}, {0.982247, 
 0, -0.187592}, {-0.303531, -0.934172, 0.187592}, {-0.303531, 
  0.934172, 0.187592}, {0.794654, -0.57735, 0.187592}, {0.794654, 
 0.57735, 0.187592}, {-0.187592, -0.57735, 0.794654}, {-0.187592, 
 0.57735, 0.794654}, {-0.491123, -0.356822, -0.794654}, {-0.491123, 
 0.356822, -0.794654}, {0.491123, -0.356822, 0.794654}, {0.491123, 
 0.356822, 0.794654}, {0.607062, 
 0, -0.794654}, {-0.794654, -0.57735, -0.187592}, {-0.794654, 
 0.57735, -0.187592}, {-0.607062, 0, 
 0.794654}, {0.187592, -0.57735, -0.794654}, {0.187592, 
 0.57735, -0.794654}, {0.303531, -0.934172, -0.187592}, {0.303531, 
 0.934172, -0.187592}}

The way I know to do it is with GraphicsComplex Graphics3D[{Thick, GraphicsComplex[v, Line[i]]}]

Think is that I don't know which vertex are conected by lines.

I can draw the coordinates by the comand:

e1 = Graphics3D[Point[v]]

enter image description here

And now I would know which vertex to join by rotating that figure if I had some way to get the coordinates from the plot or label the points as '1', '2', '3'

Some thoughts, in this line or any other?

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  • $\begingroup$ Nearest[v] might help. So might PolyhedronData["Icosahedron"] together with FindGeometricTransform. If you figure it out, feel free to answer your own question. $\endgroup$ – Michael E2 Feb 20 '15 at 14:08
  • $\begingroup$ Thanks I solved it for this case, cause the ordering of the vertex was the same as Short[i = PolyhedronData["Icosahedron", "FaceIndices"]] then Graphics3D[{Thick, GraphicsComplex[v, Line[i]]}] saved my day. Other way I suppose you have to figure out how to rearrage the list of vertex $\endgroup$ – Popeye Feb 20 '15 at 14:31
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PolyhedronData["Icosahedron", "VertexCount"] == Length@v

(* False*)

Yours isn't an Icosahedron

However, it's a Dodecahedron instead:

f = Nearest@v  
Graphics3D[Line /@ Flatten[MapThread[List, {f[#, 4], ConstantArray[#, 4]}] & /@ v, {2}]]

Mathematica graphics

V10

BoundaryMeshRegion[
 ConvexHullMesh[v],
 MeshCellStyle -> {0 -> {PointSize@.02, Blue}, 1 -> None, 2 -> Opacity@.7}]

enter image description here

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  • $\begingroup$ Feel free to use this BoundaryMeshRegion[ ConvexHullMesh[v], MeshCellStyle -> {0 -> PointSize@.02, 1 -> None, 2 -> Opacity@.2}] because I'm still not sure what is OP goal ;) $\endgroup$ – Kuba Feb 20 '15 at 15:09
  • $\begingroup$ @Kuba I saw your answer, but V9 here :( Thanks! $\endgroup$ – Dr. belisarius Feb 20 '15 at 15:13
  • $\begingroup$ @Kuba Thanks for the edit $\endgroup$ – Dr. belisarius Feb 20 '15 at 15:17
  • $\begingroup$ I'm not able to apply BoundaryMeshRegion to ConvexHullMesh $\endgroup$ – Popeye Feb 27 '15 at 16:12
  • $\begingroup$ @user10712 Welcome! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Dr. belisarius Feb 27 '15 at 16:19

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