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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, 2015 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, 2015 at 14:31

1 Answer 1

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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 -> {[email protected], Blue}, 1 -> None, 2 -> [email protected]}]

enter image description here

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  • $\begingroup$ Feel free to use this BoundaryMeshRegion[ ConvexHullMesh[v], MeshCellStyle -> {0 -> [email protected], 1 -> None, 2 -> [email protected]}] because I'm still not sure what is OP goal ;) $\endgroup$
    – Kuba
    Feb 20, 2015 at 15:09
  • $\begingroup$ @Kuba I saw your answer, but V9 here :( Thanks! $\endgroup$ Feb 20, 2015 at 15:13
  • $\begingroup$ @Kuba Thanks for the edit $\endgroup$ Feb 20, 2015 at 15:17
  • $\begingroup$ I'm not able to apply BoundaryMeshRegion to ConvexHullMesh $\endgroup$
    – Popeye
    Feb 27, 2015 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! $\endgroup$ Feb 27, 2015 at 16:19

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