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One for the mathematicians

 
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There was a question recently about dice which got me thinking about how to construct an N-sided die - ie, a polyhedron whose faces are polygons of equal size and shape, but not necessarily regular. Obviously, the Platonic solids provide solutions for 4, 6, 8, 12 and 20; but what about other numbers like 5, 7 and 9?

I've seen solutions that involve sticking two "crenellated cones" together to make a dice with "kite"-like faces, but many of these simply repeat numbers on each side to provide the spread required, which to my mind is cheating.

One manufacturing possibility that occurred to me was to construct a sphere, mark N equidistant points on its surface, and then grind down the surface at each mark to create a flat circular "face" centred on each mark until their edges touch.

So question 2 is: is it possible to create 5 (or 7 or 9) equidistant points on the surface of a sphere? It seems like it ought to be possible, but I have absolutely no idea how I'd go about it.

Any suggestions?

Winston
 
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1) I don't think there's a way to construct an N-sided polyhedron in general, using only one shape and size of polygon.

By no means is this any kind of proof, but just Google 5-sided die, and all you will find are either curved or discontinuous faces.

2) I believe a tetrahedron is the only three-dimensional body in which all points are equidistant. If you grind the points down until the newly created surfaces all touch each other, you have a new tetrahedron.

 
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I don't think the issue is the mathematics -- although the math is hard too.

I think this issue is fairness. Or to be more exact, you need to be able to easily prove that the dice is fair. And you need to be able to prove it to the cynic gamer, who believes that everything has a conspiracy...

Using a die with sides that are not flat, with sides that are not exactly the same shape, with sides that have complex relationships with other sides, or with weird features, is just *not* going to be accepted.

Using a 10 sided die, where each number is represented twice (once with and touching each poles, and you know what I mean) may work. Using a 6 sided die, where one side represents a "roll again" may work too.

Henry
 
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Stephan van Hulst wrote:By no means is this any kind of proof, but just Google 5-sided die, and all you will find are either curved or discontinuous faces.


I did, and what you generally get is a 10-sided die with faces as I described and numbers duplicated.

However, if you start with a sphere, it seems to me that you ought to be able to grind (or "slice") 5 (or 7 or 9) circular faces of equal size and uniformly distant from each other, such that when you "roll" it, it has an equal chance of landing on one of them. Obviously, the placement of the numbers to indicate which face it landed on will also be interesting.

But then my question is: how do you place the 5 (or 7 or 9) "face centres" on the surface of a sphere?

Winston
 
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Henry Wong wrote:Using a 6 sided die, where one side represents a "roll again" may work too.


Yeah, but that's cheating too.

It just seemed like an interesting problem to me...

Winston
 
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How about a stick? whose cross section is a pentagon? ... and also put tapered edges (to a point) on the two ends of the stick, so that it can never land that way?

Henry
 
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If you create a pyramid-like shape, with base a regular N-side (ie a tetrahedron when N = 3, and a pyramid when N = 4), and you glue a stick to the center of the base, then you can use the whole thing as a toll. The bottom facing side is what counts. Is this also cheating? (if so, why?).

To place N points evenly distributed on a sphere, well, there is no general solution, but there are many articles on this subject. For instance this one

 
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Henry Wong wrote:How about a stick? whose cross section is a pentagon?


Excellent suggestion! You've obviously played "pencil cricket".

However, I'm still interested in a "polyhedral" or "cropped sphere" solution - if there is one.

Winston
 
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Piet Souris wrote:To place N points evenly distributed on a sphere, well, there is no general solution, but there are many articles on this subject. For instance this one


Aha! Thanks Piet.

I had an idea you might come up with something....

Winston
 
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Winston Gutkowski wrote:However, if you start with a sphere, it seems to me that you ought to be able to grind (or "slice") 5 (or 7 or 9) circular faces of equal size and uniformly distant from each other, such that when you "roll" it, it has an equal chance of landing on one of them


I think you can do this, but dice constructed this way that aren't platonic solids, necessarily have a spherical dome at the opposite side of a flat face. That means that each number would have to be printed on a dome and not on a face.

If you keep grinding until there are no domes, your dice will have no volume. Either way, in general I think a polyhedron is impossible.
 
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Stephan van Hulst wrote:I think you can do this, but dice constructed this way that aren't platonic solids.


But there are only 5 of those aren't there? What I was more interested to know was whether there was a Platonic-like solution for 5, 7 or 9 (or any other value of n) faces that uses any convex polygon of uniform shape and size.

Winston
 
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I'm not sure, but I guess that depends on what you mean by "Platonic-like".

I was commenting on your method of constructing such a body, which I believe can not result in a polyhedron unless it's platonic.
 
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Surely somebody got his PhD for doing that work, some time in the 15th century? Is there any way to grind 7 flats onto a sphere in a symmetrical distribution? I suspect not.
 
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Stephan van Hulst wrote:I'm not sure, but I guess that depends on what you mean by "Platonic-like".


I'm not sure either, but I'd say that it has to have the characteristics of a "die":
1. All faces must be the same shape and size, but not necessarily regular polygons.
2. All face surfaces must be flat.
3. It must be the closest analog to a sphere for the number of faces it has - ie, if you make it from a sphere, it's the shape that removes the least amount of material.

Winston
 
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Campbell Ritchie wrote:Surely somebody got his PhD for doing that work, some time in the 15th century? Is there any way to grind 7 flats onto a sphere in a symmetrical distribution? I suspect not.


Well, Piet's link suggests that there is; but the maths appears to be quite complicated (still going through it). Or did you mean 7 flats with no "domes" in between? I'd say that's my "polyhedral" question.

Winston
 
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That article seems to suggest that only the five Platonic solids are strictly evenly distributed. You can make a seven‑sided die with the electrostatic repulsion technique, for example, but how are you going to persuade a punter it is ‑fair‑. It doesn't matter what size the flats are, as long as they are large enough for the die to lie still and all the same size.
 
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