Using only the vertices of a regular icosahedron as the corners, how many equilateral triangles can you make?

What if you could only use the vertices of a regular dodecahedron?

(In reply to re: part 2 -- spoiler by Tristan)

When you say "both times" I assume you mean for the smaller and larger triangles in the dodecahedral case.  I see my miscount for the larger triangles, but not for the smaller, as follows:

Suppose we place one face of the dodecahedron on top and one on bottom. Label the vertices of the top face A, B, C, D, E. The edges going down and outward from the top lead to vertices to be labeled F, G, H, I, and J respectively.  Let the vertex joining edges from F and G (downward and outward again) be labeled K, and then in turn, on that level L, M, N and O. Below and inward from K, L, M, N and O are P, Q, R, S and T.

These are the smaller triangles I find:

Around A: FBE
Around B: GCA
Around C: HDB
Around D: ICC
Around E: JAD

Around F: OKA
Around G: KLB
Around H: LMC
Around I: MND
Around J: NOE

Around K: FPG
Around L: GQH
Around M: HRI
Around N: ISJ
Around O: JTF

Around P: TKQ
Around Q: PLR
Around R: QMS
Around S: RNT
Around T: SOP

which agrees with the 20 I previously reported.

For the larger triangles, I see there are two per vertex rather than the one previously reported:

Around A: DOG CJK
Around B: EKH DFL
Around C: ALI EGM
Around D: BMJ AHN
Around E: CNF BIO

etc.

This becomes 20 + 40 = 60 for the dodecahedron.

Posted by Charlie on 2005-08-27 20:30:31