Can there exist $3$, $4$ and $5$-faceted shapes with congruent flat sides in $\mathbb{R}^3$? I rose this question in my discrete math class (the unit on probability) not too long ago. For instance, a two-sided shape (like a coin) can be one with any geometrical shape as its "side," such as a circle (as with a coin) or a square, or any other connected shape in $\mathbb{R}^2$, and in tossing this shape there is a $1/2$ "probability" it will land on one side as opposed to another. A cube (like a dice) is a shape with $6$ sides, all congruent flat squares, and upon tossing this object into the air, there is a $1/6$ chance of landing on any given side.
Now suppose I wished to toss a $3$-sided object such that all facets of this shape are congruent, as in the penny or dice, resulting in a $1/3$ chance of landing on any given side.
Can there exist solids in $\mathbb{R}^3$ (or elsewhere) with $3,4$ and $5$ flat facets, such that all facets are congruent? Naturally these shapes may exist with curved sides, but I'm more concerned with solids of flat facets. I'm certain one of $3$ sides cannot exist, but perhaps some of you can enlighten me on how I might prove these cases?
What about a solid with $n$-faces? Perhaps this is related to packing problems (and hence I will add that tag)?
 A: One of the dies you want you can make. The others you can fake.
The tetrahedron is the first platonic solid. It has four triangular faces.
The platonic solids give you fakes for probabilities of 1/3 (label the faces of a cube with 1, 2, and 3 each twice) and 1/5 (label the faces of an icosahedron with 1 through 5, each four times).
You're right that a real 3 is impossible. I'm pretty sure 5 is too. Maybe I'll come back to this with proofs, if noone else provides them.
For some larger values of n, check out the wikipedia page for deltahedra:
http://en.wikipedia.org/wiki/Deltahedron
A: First, the solution in $R^3$:
In any number of dimensions $d$ the simplest polytope that can exist, or simplex, has $d+1$ vertices. This is easy to prove; the $0$-simplex has (indeed is) 1 vertex and the $d+1$-simplex has 1 vertex more than the $(d)$-simplex.
Thus, the simplest in 3D must have four vertices, and the only solution is the tetrahedron. Any isosceles triangle will do, it does not have to be regular. Four isosceles triangles make an elongated tetrahedron, or disphenoid.
Thus, stepping back one, three flat faces are not possible.
Five is a bit more subtle. The edges of the face polygons meet in pairs, so if you count up the total edges of all the face polygons it will be twice the number of edges in the polyhedron. So there must be an even number of odd-numbered polygons. Five is an odd number, so at least one face must be even-sided; 2 is the degenerate digon, 6 has too many sides to meet only four others, so it would have to be a quadrilateral. Erecting more qudrilaterals on each side does not close the figure, you need a sixth one to top it off. So no, just five congruent faces is not possible.
For even numbers of faces $n$, for $n\ge3$ the $n/2$ bipyramids and trapezoherda provide solutions. Note that the cube is the trapezohedron for $n=3$
I am not sure about odd values of $n$, but I suspect there may be none.
You also ask about "elsewhere" than $R^3$. The hemicube has just three square faces and is a tiling of the projective plane, just as convex polyhedra are tilings of the sphere. Similarly we would be able to construct congruent-faced tilings from the hemi-pentagonal bipyramd and trapezohedron, each with five faces. Toroidal polyhedra may be constructed as tilings or dissections of a rectangle, which is then rolled up and glued to form a torus. Clearly, for $n\ge3$ a stack of $n$ parallel slices will give you an $n$-faced solution. One can also have fun in hyperbolic spaces, but I do not know them well enough to give examples.
A: What you are describing are the platonic solids (link to Wikipedia).
