Looking up “polyhedron” in Wikipedia shows only five polyhedrons . Why is the collection so limited? I would have thought every polygon matches up to a corresponding polyhedron. Is there a simple explanation why this is not the case?
…And, I should ask, what happens if I try joining hexagons, for instance, along their edges? Will I form a surface that will not yield a closed object? 
There are many different kinds of polyhedra, in many varieties. However, there only five Platonic solids (that is, ones whose vertices all have the same number of faces meeting at them, and whose faces are all non-intersecting copies of the same regular polygon (one whose edges all have the same length and whose angles are all equal)).
To start with, let’s note that the same regular polygon can be used to make more than one regular polyhedron. For example, a tetrahedron and an octahedron both have triangular faces. So simply specifying a polygon isn’t enough to say how you intend to turn it into a polyhedron. (It doesn’t tell you how many faces meet at each vertex.)
If you try joining regular hexagons along their edges, you’ll find you won’t be able to wrap them into a closed surface. You’ll only be able to tessellate a flat plane (with three hexagons around each vertex).
And with more than six sides, you won’t even be able to do that: you won’t be able to get any three faces to meet at a vertex at all.
Why is that? Well, consider: as you walk all the way around a polygon, you turn by 360 degrees (a full revolution). So at each vertex in a regular n-gon, the direction you are facing turns by 360/n degrees. This means the interior angle is 180 - 360/n degrees (i.e., 1/2 - 1/n full revolutions).
This gets larger as n gets larger. When n hits 6, this becomes 120 degrees (1/3 of a full revolution).
At every vertex of a polyhedron, at least three faces touch. The total angle around that point will be 360 degrees if they touch in a flat plane, and less than 360 degrees if it’s bent into a corner [for example, in a cube, at each corner, there are three faces with 90 degree interior angles meeting, for a total of 270 degrees].
Thus, the angle at each corner must be less than 120 degrees (1/3 of a full revolution).
So for n > 6, we can’t bring three faces to match at a point; for n = 6, we can tessellate a plane but not bend into a polyhedron. For a polyhedron, we need n < 6.
This means we can only use triangles, squares, or pentagons. With triangles, the interior angle is 60 degrees, and you could use anywhere from 3 to 5 faces meeting at a point before you hit the 360 degree limit. With squares, the interior angle is 90 degrees, and you can only use 3 faces meeting at a point before you hit the limit of 360 degrees. Finally, with pentagons, the interior angle is 108 degrees, and you can only use 3 faces meeting at a point.
So those are the five possibilities.
Hey, thanks. I’ve always kinda wondered about that ever since learning the regular polyhedra way back in fifth grade (I was a nerd, big surprise for a Doper, I know), and that was a very clear and concise explanation.
If you’re willing to use regular pentagons along with your hexagons, you can make one of these.
Sheesh, Indistinguishable, I know Euclid’s Elements has been around for a while, but that’s no excuse for spoiling the ending.
Indistinguishable, many thanks for an exceptionally clear explanation. It got me wondering whether we 3D folk are just lucky that all five of the possible regular polyhedra actually exist. It seemed fortuitous, so I looked up whether the same magic occurs in higher dimensions. No dice. Even with all that extra space, in 5D and higher, there are only 3 regular polytopes, the hypertetrahedron, the hypercube, and the hyperoctohedron. 4D people are lucky to have six platonic solids, the highest number in any dimension greater than 2 (in which there are an infinite number). You no doubt knew all that, but it was intriguing news to me.
“Every polygon has its polyhedron
Just like every night has its dawn
Just like every cowboy sings his sad, sad song
Every polygon has its polyhedron … .”
Well, three dice. The rules for 5-dimensional Dungeons and Dragons must be a lot simpler than ours.
And I seem to recall that there’s a novel Platonic solid that shows up somewhere in the vicinity of 50 dimensions, too (48 dimensions, maybe?), but I can’t remember the details.
Everything I’ve read says there are only the three in more than five dimensions, and that’s more consistent with the general notion that low-dimensional spaces are the weird ones.
Yeah, I just went Googling for it, and I can’t find it either. It’s quite possible that I’m conflating it with some other factoid about a higher-dimensional space, but what I’m remembering is vague enough that it’d be very difficult to figure out what that other factoid is.
Probably something about wacky 7-spheres.
I hope it doesn’t seem quibbling to ask you to insert the word “convex” into your definition. Without the “convex” Wikipedia claims there are nine regular polyhedra:
It’s not quibbling, but it’s covered by my saying “non-intersecting”. (I debated whether “convex” or “non-intersecting” was the cleaner condition to state, but in the end I decided “non-intersecting” was at least the more readily understood condition (even though “convex” is closer to what I actually used in my (presentation of Euclid’s) proof…).)
hold it a sec. you’re talking about two things (correct me if i’m wrong):
as i understand the OP question, the answer should be, yes, because any polygon could be a cross-section of a prismatic crystal with the same number of sides on the long axis but of course must have at least xP +2 planes to enclose the space.
now, a square may be found in cross-section in a cube (6P) a square bi-pyramid/octahedron(8P,) or a four-sided rhombic prism with pyramidal points (12P.)
if xS = xP, then it’s a totally different thing.
(Of course, some people may have definitions of the word “polyhedron” in their head to start with which already assume convexity or non-intersection or other such things.
Speaking of which, as a somewhat interesting related note, the definitional subtleties which kept popping up over the historically tumultuous evolution of the jargon “polyhedron” (Can “polyhedra” be self-intersecting? Can “polyhedra” have internal holes? Etc., etc.) happen to have been a key theme in one of the major books on philosophy of mathematics, Lakatos’s “Proofs and Refutations”.)
Sorry, I’m afraid I don’t quite understand what you’re saying…
the so-called platonic solids may be formed only by triangles, squares or pentagons, exclusive of each other. that’s only 3 possible solutions to one interpretation of the OP’s question “how many types of polygons can form a polyhedron?”
my own interpretation to the question is “does any polygon have a 3D expression?” the answer is yes, all since the polygon is simply a cross-section.
asking “does every polygon have its polyhedron?” is vague to me. triangles, squares, rectangles, trapezoids, even pentagons can bound one side of a polyghedron. or am i using the wrong term? i just use the term “crystal.”
Sure, there are many, many types of polyhedra. I said that at the beginning of my initial post; I didn’t mean to suggest there weren’t any other kinds of polyhedra, if that’s what you’re taking issue with. It’s just that the OP saw a list which suggested there were only five polyhedra. I assumed this was in fact a list of Platonic solids and tried to explain what that was all about.
you did and pretty well too.
It’s possible to make a fair die in a shape other than a regular polyhedron (e.g. the d10, basically two five-sided pyramids joined base-to-base).