Talk:Tetrahedron
|
|
Maybe I'm just spacially challenged, but I can't agree with the following sentence from the article:
A tetrahedron can be embedded inside a cube so that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces.
If you obey "each vertex is a vertex of the cube", then the statement that "each edge is a diagonal of one of the cube's faces" is wrong, maximally 3 edges of the tetrahedron can be diagonals of the cube, the others must lie on the edges of the cube. WDYT? --snoyes 00:17 Feb 26, 2003 (UTC)
Consider vertices at (0, 0, 1), (0, 1, 0), (1, 0, 0) and (1, 1, 1) on the unit cube.
The edges of the tetrahedron made by these points are:
- (0, 0, 1), (0, 1, 0)
- (0, 1, 0), (1, 0, 0)
- (0, 0, 1), (1, 0, 0)
- (0, 0, 1), (1, 1, 1)
- (0, 1, 0), (1, 1, 1)
- (1, 0, 0), (1, 1, 1)
All face diagonals (I think: it's late, I'm tired). The Anome 00:35 Feb 26, 2003 (UTC)
- Yip, I'm just spaTially (and spelling) impaired. Thanks for your time. --snoyes 01:24 Feb 26, 2003 (UTC)
- The buckminster-fuller web site has a great animation that shows this :-) -- Tarquin 11:11 Feb 26, 2003 (UTC) ... and I can't find it on google. I have it on my HD though, email me Snoyes & I'll send it back to you :-)
A cube can be divided into 5 tetrahedra. The central one is a regular tetrahedron whose edges are the diagonals of the cube's faces. The other 4 tetrahedra are not regular with 3 edges diagonals of the cube and the other 3 edges edges of the cube. Thus everything said above is correct. You're just talking about different tetrahedra.