Convex polyhedron

[Zapp]

this is more a proof than a puzzle, but who cares, right?

What is the minimum number of triangular sides needed to form a convex polyhedron with more sides than corners. prove it.

I couldn't do it myself and I have no solution, so we just have to trust in the power of math.

also:

Spoiler! :

My hunch would be 3, maybe 2, but I have no way to prove it nor do I know how to approach this problem properly

[xander]

Polyeder? Is that anything like a polyhedron?

xander

[Zapp]

Polyeder? Is that anything like a polyhedron?

xander

whoops, sorry. polyeder is the german word for polyhedron

[xander]

Ah. In that case, I am a bit confused. In your spoiler, you state that you suspect that the minimum is either 2 or 3, but to the best of my knowledge, it is impossible to create a polyhedron with less than four faces (i.e. a tetrahedron is about as simple as things get). Must all the sides be triangular?

I suspect that the answer is 8 (i.e. an octahedron does the job), and that there is a proof utilizing the Euler characteristic, but I'm not sure that I have properly understood the question.

xander

[Zapp]

I did not say that the sides are exclusively triangular. you can use any 2 dimensional object as a side, as long as the polyhedron is convex and has more sides than corners. the problem only asks what the minimum number of triangular sides is necessary to fulfill those conditions.
example: http://upload.wikimedia.org/wikipedia/c ... hedron.jpg
it fulfills the requirements with 8 triangles.

[Gary]

From Zapp's comment (Wed Jun 15, 2011 6:14 pm ) I get 0.

Spoiler! :

A cube has 6 sides, none triangular, and 8 corners.

[Zapp]

From Zapp's comment (Wed Jun 15, 2011 6:14 pm ) I get 0.

Spoiler! :

A cube has 6 sides, none triangular, and 8 corners.

it has to have more sides than corners...