To make an icosahedron out of Sierpinsky tetrahedrons is difficult because regular tetrahedra can't tile in space.

The dihedral angle of a tetrahedron is ~70.53. So the first step would be to make tetra's with 72'angles (360/5) and to tile them into two sets of five tetrahedra to form the top and bottom of the icosahedron, and then i would have to fill the middle area with 10 tetrahedra.

What is the easiest way to roughly make an icosahedron from tetrahedra? what are the dimensions of the tetrahedra involved, are they all the same? ideally it would be possible to see through all the triangles in a line from one end to the other, is it possible?

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Perhaps I misunderstand, but if the goal is to partition an icosahedron into tetrahedra, why not just connect each face to the centroid?

Incidentally, if each icoshedron edge is $1$ unit long, then the tetrahedron edges incident to the centroid have length $$ \frac{1}{2} \sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)} \approx 0.95 \;. $$ So the tetrahedra are slightly non-regular.

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    $\begingroup$ oh ok i see what you mean, if i had some tetrahedra, i could align them with an icos devided into tetra and resize them. cool idea, i will see how i can resize my sierpinskies using x-y-z axis resizing, a bit difficult. $\endgroup$ – com.prehensible Jun 26 '15 at 19:57
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    $\begingroup$ Thanks! distance from centroid to mid-face = = 0.755761314s. distance from centroid to mid-side = = 0.809016995s. distance from centroid to a vertex = = 0.951056517s. from kjmaclean.com/Geometry/Icosahedron.html $\endgroup$ – com.prehensible Jun 27 '15 at 2:24

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