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 [quote="Scurra"]Surely there would be a point at which the number wouldn't fit into the available space? So it can't be infinite. Maybe 0.99 of infinite?[/quote]
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Zandor
Posted: Fri Feb 15, 2013 1:18 am    Post subject: 1

Lepton wrote:
Tetrahedron: How many different ways can you number the four corners of a tetrahedron?

2 (if mirrored tetrahedrons are considered different)

Quote:
Cube: Is the cube more stable spinning (quickly) while balanced on a vertex or on a side?

Vertax. If you spin it on a side there is more friction making it unstable

Quote:
Octahedron: How many different ways can you number an octahedron with the digits 1 to 8 such that (a) the sum of opposite sides is 9, (b) the numbers 1-4 are all adjacent to the same vertex, and (c) all the odd numbers are adjacent to the same vertex?

4 (if mirrored octahedrons are considered different)

Quote:
Dodecahedron: Which has more edges, a dodecahedron or an icosahedron?

As these polyhedrons are dual the number of edges is equal (30)

Quote:
Icosahedron: One way to compare the "roundness" of two solids is to find the ratio of the volume to the surface area: ratios closer to 1/3 (a sphere) are "rounder". For an icosahedron and a dodecahedron, each with an edge length of 1, which is more "round"?

Dodecahedron: 0.371
Icosahedron: 0.252
This would give the dodecahedron as an answer, although this definition of roundness is really [insert negative word].
The Potter
Posted: Thu Feb 14, 2013 10:48 pm    Post subject: 0

Cube: Is the cube more stable spinning (quickly) while balanced on a vertex or on a side?

A higher moment of inertia will be more stable. When spinning a block about the center axis, the for the moment of inertia is
(1/12)m(L^2 + D^2)
For spinning about the longest axis it is
(1/6)m(L^2 D^2 + L^2 W^2 + D^2 W^2) / (L^2 + D^2 + W^2)

For both cases L=D=W=s.
Spinning about an axis on a face is (1/6) m s^2
Spinning about the vertex is (1/6) m s^2

It appears there isn't a difference.
groza528
Posted: Thu Feb 14, 2013 12:38 pm    Post subject: -1

For 1) I think it depends on whether we assume that the vertices are indistinguishable (until being labeled, that is.) Because if so then any organic chemist knows immediately that there are only two chiral configurations. If not I guess it would be 4! or 24.
Lepton*
Posted: Thu Feb 14, 2013 6:06 am    Post subject: -2

gftt: I didn't say it was a good measurement! In this case, the interplay between the definition and the edge length is what makes this question interesting/surprising for me.

bgg1996: And there are, of course, limitations from quantum information theory if the tetrahedron is physical. Bah, use 1-4.

Scurra: I'm planning to build a model solar system, and I needed something to situate between the heavenly spheres.
bgg1996
Posted: Thu Feb 14, 2013 12:25 am    Post subject: -3

Scurra wrote:
Surely there would be a point at which the number wouldn't fit into the available space? So it can't be infinite. Maybe 0.99 of infinite?

In Euclidean Geometry, you could write numbers as small as you want or you could make a line at some rational distance from the vertex and pair the distance with an integer so that 1/2 down is 1, 1/3 is 2, 2/3 down is 3, 1/4 down is 4, 3/4 down is 5, 1/5 down is 6, etc.
In the real world, you can't make lines at arbitrary perfectly rational locations, but in the real world, there are no tetrahedrons.
Scurra
Posted: Wed Feb 13, 2013 10:03 pm    Post subject: -4

Surely there would be a point at which the number wouldn't fit into the available space? So it can't be infinite. Maybe 0.99 of infinite?
bgg1996
Posted: Wed Feb 13, 2013 9:14 pm    Post subject: -5

Lepton wrote:
Tetrahedron: How many different ways can you number the four corners of a tetrahedron?

infinity. [1,2,3,4], [1,2,3,5], [1,2,3,6]... ...[1,2,3,197], etc.
Scurra
Posted: Wed Feb 13, 2013 8:27 pm    Post subject: -6

Lepton wrote:
I fulfilled a lifelong dream today by purchasing a set of dice in the shapes of the Platonic solids.
Really? Have you started playing old school D&D or something?
gftt*
Posted: Wed Feb 13, 2013 7:21 pm    Post subject: -7

Lepton wrote:

Icosahedron: One way to compare the "roundness" of two solids is to find the ratio of the volume to the surface area: ratios closer to 1/3 (a sphere) are "rounder".

This is a bad measurement, as it is not constant if you scale the solid. A sphere of radius 1 and a sphere of radius 2 would have different "roundness" factors.

The ratio of surface area cubed to volume squared would be more appropriate.
Lepton
Posted: Wed Feb 13, 2013 6:29 pm    Post subject: -8

I fulfilled a lifelong dream today by purchasing a set of dice in the shapes of the Platonic solids. The following are some simple questions that came up while I was playing with them.

Tetrahedron: How many different ways can you number the four corners of a tetrahedron?

Cube: Is the cube more stable spinning (quickly) while balanced on a vertex or on a side?

Octahedron: How many different ways can you number an octahedron with the digits 1 to 8 such that (a) the sum of opposite sides is 9, (b) the numbers 1-4 are all adjacent to the same vertex, and (c) all the odd numbers are adjacent to the same vertex?

Dodecahedron: Which has more edges, a dodecahedron or an icosahedron?

Icosahedron: One way to compare the "roundness" of two solids is to find the ratio of the volume to the surface area: ratios closer to 1/3 (a sphere) are "rounder". For an icosahedron and a dodecahedron, each with an edge length of 1, which is more "round"?