0

I want to tile the sphere as many face as possible. And I want every face be the same size and shape. Is it possible to generate more than 100 or 1000 faces of regular convex polyhedron?

Thaina
  • 702

2 Answers2

2

If you want every vertex to touch the same number of faces...then the answer is no, by an argument involving Euler characteristics, which you can find in most elementary topology books. (It is, for example, Exercise 8.1 in w. Massey's "Algebraic Topology, An Introduction".)

If you're willing to have varying vertex degree, then a standard solution is to start from an icosahedron (20 triangular faces), and subdivide each triangular face.

A particularly simple form of subdivision is to add a vertex at the middle of each edge of your face, and connect these three new vertices to each other, which effectively divides your triangle into 4 smaller ones. You can then repeat the process for each smaller triangle, and so on. After $n$ stages of subdivision of all triangles, you will have $20 \cdot 4^{n}$ triangular faces in your polyhedron. And if you project each vertex and edge radially onto the sphere, you'll get a very fine reticulation of the sphere, which is probably what you're after.

John Hughes
  • 102,125
  • That approach yields triangles in the same plane, so this doesn't exactly add new faces. – Ed Pegg Aug 24 '15 at 16:24
  • I have use that approach once. It turn out that when I project each subdivided point into sphere. It then lose the regular property of triangle. Each triangle became difference angle – Thaina Aug 24 '15 at 17:16
  • 1
    Yes, that's correct. Ed Pegg's example shows that if you're willing to abondon "vertex regularity" (as you must, if you want more than 20 triangles) you can get arbitrarily many congruent triangles...but if you want them to be nearly equilateral, you're going to be out of luck (essentially by the same Euler characteristic argument: nearly equilateral means that they have to meet 6 or 5 at a vertex, and that poses unsolvable constraints). My non-congruent-shapes construction is one that works pretty well in many applications, although I admit that it doesn't satisfy your exact requirements. – John Hughes Aug 24 '15 at 17:22
2

You might be thinking of an isohedron. With the disdyakis triacontahedron 120 triangles is possible.

For more than that, chose a long narrow isosceles triangle. A central band will have an equal number of up and down triangles. Pyramids of that same triangle will be above and below it. It might be possible to work in more bands. But this isn't an isohedron. The following is a figure made of 200 identical triangles.

200 identical triangles

Ed Pegg
  • 22,188