Animation Of Cutting A Dodecahedron From A Cube?

Here are detailed instructions for cutting a tree stump into a regular dodecahedron. The essential play a joke on is to make a template for the dodecahedron's dihedral bending. It should be possible to utilise like techniques to make an icosahedron.

Yous might try practicing on fruit first: it'south cheap and easy to cut.

Addendum: There are like shooting fish in a barrel tricks for the tetrahedron and octahedron if y'all're starting with a cube. To get the octahedron, you only slice all the corners off: Mark the centers of 3 adjacent faces, cutting in the plane that contains these three points, and y'all have cut one confront of an octahedron. To get the tetrahedron, color the vertices of the cube in alternating colors, say red and blackness. And then on each face draw the diagonal from ane red vertex to the other. These diagonals will form the edges of the tetrahedron; cut in planes that each comprise three reddish vertices. To get the cube, cut nothing.

Half dozen of the thirty icosahedron edges and all twelve of its vertices lie on faces of the cube. Each icosahedron vertex lies near one cube edge and at that place is clearly the same number of icosahedron vertices as cube edges. $\dots$ Finally, from the model, find that it is easy to list the $(x,y,z)$ coordinates of the twelve vertices:

$(\pm 1 , \pm \phi, 0)$,

$(0, \pm ane, \pm \phi)$, and

$(\pm \phi, 0, \pm i)$, where $\phi$ is the gilded ratio.

from here.

You lot tin can make a dodecahedron equally follows:

Start with an octahedron $\mathcal O$ with edges of length $1$. You can colour its faces inblack and white in such a way that no two faces of the aforementioned color share an edge, and then you tin orient each edge of $\mathcal O$ so that when you move along it, with your head pointing towards the outside of $\mathcal O$, you accept white on your left. Now pick a number $\theta\in[0,ane]$ and on each edge of the octahedron mark the indicate which is at distance $\theta$ from the starting vertex of that edge (co-ordinate tothe orientations we stock-still)

If $\theta\in(0,1)$, then this construction gives us 12 points, one one each edge. If you call back about this a fiddling flake y'all'll see that for some value of $\theta$ these twelve points are the vertices of a dodecahedron: this is a simple consequence of the intermediate value theorem from calculus. One can explicitely compute this, with some work, and information technology turns out that we need $\theta$ to be the inverse of the gold ratio.

The following Mathematica code gives an interative thingie to run across this construction:

          vertex = Flatten[    Map[NestList[RotateLeft, #, 2] & ,  {{1, 0, 0}, {-i, 0, 0}}, {ane}],     1]; n = Length[vertex]; edges = Select[Subsets[vertex, {ii}],     EuclideanDistance[#[[1]], #[[two]]] == Sqrt[two] &]; faces = Select[Subsets[vertex, {three}],     Complement[Subsets[#, {ii}], edges] == {} &]; red = FindVertexCover[    Graph[Rule @@@       Select[Subsets[faces, {ii}],        Length[Intersection[#[[1]], #[[2]]]] == 2 &]]]; orientedRedFaces = Map[Role[f,      Partition[      If[Dot[Cross[f[[2]] - f[[1]], f[[three]] - f[[2]]], Total[f]] < 0,        f[[{1, 3, 2}]], f], two, 1, 1]     ], reddish]; orientedBlueFaces = Map[Function[f,      Sectionalization[      If[Dot[Cross[f[[two]] - f[[1]], f[[iii]] - f[[2]]], Total[f]] < 0,        f[[{ane, three, 2}]], f], 2, 1, 1]     ], Complement[faces, reddish]]; orientedEdges = Flatten[orientedRedFaces, 1]; Manipulate[  Graphics3D[{    Line /@ orientedEdges,    {PointSize[0.02],      Point[t #[[1]] + (1 - t) #[[two]]] & /@ orientedEdges},    FaceForm[Red],    Polygon /@      Map[t #[[i]] + (one - t) #[[2]] &, orientedRedFaces, {ii}],    Polygon /@      Map[(1 - t) #[[1]] + t #[[ii]] &, orientedBlueFaces, {ii}]    }, Boxed -> Faux],  {{t, i/GoldenRatio, "\[Theta]"}, 0, 1}  ]

This produces images such as

enter image description here

merely the interesting thing is really the interactivity which allows you to play with it and see how the motion-picture show changes when you change $\theta$.

Now of course this does not reply your question, because I synthetic a dodecahedron from an octahedron, and you wanted an icosahedron from a cube! But ane can dualize this constructing, taking advantage of the fact that the dual of a cube is an octahedron, and the dual of an icosahedron is a dodecahedron. I'll get out all that fun to y'all.

Source: https://newbedev.com/how-to-cut-a-cube-into-an-icosahedron

Posted by: sochahaphe1951.blogspot.com