This is a unit, folded from a square, based on equilateral triangles. But it does a really nice job making any polyhedron with equilateral triangles for faces. Two units make a tetrahedron. Four make an octahedron. Ten make an icosahedron, which feels really nice! Six units make a snub disphenoid. There's all sorts of things you can make with this unit! (And there's some cool math involved too!)
Hmmm, you don't even need Euler's formula. Since if all the faces are triangles, then E = 3F/2 (3 edges per triangle, 2 faces per edge). And since both E and F are whole numbers, F must be an even number or else you would end up with an extra half an ...
Hmmm, you don't even need Euler's formula. Since if all the faces are triangles, then E = 3F/2 (3 edges per triangle, 2 faces per edge). And since both E and F are whole numbers, F must be an even number or else you would end up with an extra half an edge haha. Wow, a friend of mine likes using a similar dual-triangle unit but I never realized it could be "all-encompassing".
Posted by Daniel Kwan on Thu Jun 14 16:26:03 2007
Yes :)
Yes, Daniel, you can prove that any polyhedron with only triangle faces must have an even number of faces. If I recall, it pops out of Euler's formula V-E+F=2 and some counting tricks.
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(view all 4 comments)Right!
Yep. There ya go.
Posted by Thomas Hull on Sat Jun 16 18:20:47 2007
ooh..
show full show summaryHmmm, you don't even need Euler's formula. Since if all the faces are triangles, then E = 3F/2 (3 edges per triangle, 2 faces per edge). And since both E and F are whole numbers, F must be an even number or else you would end up with an extra half an ...
Hmmm, you don't even need Euler's formula. Since if all the faces are triangles, then E = 3F/2 (3 edges per triangle, 2 faces per edge). And since both E and F are whole numbers, F must be an even number or else you would end up with an extra half an edge haha. Wow, a friend of mine likes using a similar dual-triangle unit but I never realized it could be "all-encompassing".
Posted by Daniel Kwan on Thu Jun 14 16:26:03 2007
Yes :)
Yes, Daniel, you can prove that any polyhedron with only triangle faces must have an even number of faces. If I recall, it pops out of Euler's formula V-E+F=2 and some counting tricks.
Posted by Thomas Hull on Wed Jun 13 22:21:29 2007