Googolhedron

Urgh.  Oh, hang on - the resulting shape doesn't even have to approximate a sphere (don't the FiveRegularPolyhedra also have a requirement to be convex?), so just take a tetrahedron, and replace one of the faces with another tetrahedron etc. until you have the right number of facing - to get the last few, replace a tetra that borders more than one existing face.  ProofByHandWaving?, but I think you see what I mean.  --Vitenka

Hmmm. The web site seems to suggest that the shape does approximate a sphere. But using your method, I can conceptualise a shape with 10^100 - 1 faces, which is a multiple of 3.

You might be able to make a geodesic sphere out of a googol equilateral triangles, I'm not sure yet - there are 3 classes of geodesic sphere, and one of them can only do geospheres of 20n^2 facets, where n is any positive integer.  It's a cheat anyway though, because some of the facets will be degenerate, in which case you might as well equally subdivide the faces of a tetrahedron until you got the right number.  I'm not sure about the other two classes of geosphere--Mjb67

Or, use http://www.ifor.math.ethz.ch/staff/fukuda/polyfaq/node12.html to figure out the minimum number of vertices for a polyhedron of 1-googol 2-dimensional faces (i.e. facets) then use http://www.math.niu.edu/~rusin/known-math/95/sphere.faq to place that many vertices equally-distributed around a sphere... hmm - I'm hoping that by having the constraint that 'all facets are similar' but not the constraint that 'all vertices are part of the same number of facets' that it's then possible to join up the neighbours with 1 googol similar triangles... no, it probably doesn't work does it.--Mjb67

Ah, Bakabentham - It's easy, but it doesn't look like a sphere.  It's a bipyramid - take two (googol/2)-based pyramids and stick them together base to base.  It therefore has uniform faces.--Mjb67