its such a simple idea that im sure its already been done, i know that a hexaminx is nothing like a megaminx, but since a megaminx is made of pentagons is there any twisty puzzle mage of hexagons or even decagons?

Correct me if I'm wrong, but I thought a hexaminx was made out of a megaminx. Called a hexaminx because it has 6 sides..

I think I know what you're thinking though. It's look somewhat like a honeycomb as a web.


I think, assuming it's be a face turning puzzle, all faces would look like this:

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I have no idea how many faces it'd have, though. I think 14?

There can't be one just made of regular hexagons I think. There isn't a Platonic solid with hexagonal faces, or even decagons.

It may be possible with non-regular hexagons but with regular hexagons it is impossible.

Sorry Retr0, not even with 14 sides.

Alex

No polyhedron consists of only hexagonal faces...at least as far as I know. With only hexagonal faces, the geometry just doesn't line up. You have to include some other types in there...most commonly pentagons to make a truncated icosahedron.

Oh, darn. I hoped I was offering a kicka$$ contribution..

So the platonic solids are - Tetrahedron, cube, dodecahedron, octahedron and icosahedron?

the other people were right when they posted in this thread. Tony Fishers hexaminx was made from a megaminx, there is no platonic solid that consists only of hexagons, and a puzzle can be made using hexagons and other shapes.

this is a picture of lee Tutt's Tuttminx, which is composed of hexagonal and pentagonal sides. this is most likely the closest possible outcome to the puzzle you imagined.

It is easy to see this cannot be done if the faces are regular hexagons. If you have three regular hexagons touching at a vertex (and pairwise along the corresponding edges) you will only get a flat shape. You also cannot have less (1 is nonsense and 2 would only glue one atop another) or more (if you tried putting together 4 or more you would get negative curvature and the pieces would never close, a bit like building a hyperbolic plane out of 7 equilateral triangles per vertex).

Here's an argument that does not need regularity of the faces:
Assume there are F faces. Each hexagonal face has 6 edges and every edge is shared by 2 faces, so there are E = 6F/2 = 3F edges. By Euler polyhedral formula we have 2 = V - E + F = V - 3F + F = V - 2F or V = 2F + 2.

Assuming that n faces meet at each vertex we have V = 6F/n vertices. So 6F/n = 2F + 2. We have 3F = nF + n or n(F+1) = 3F. Both sides are integers. F and F+1 are pairwise prime, so F+1 divides 3. The only positive integer with this property is F=2. Our puzzle would have only 2 faces.

Anyone wants to write this out for decagons?

The classification of regular and irregular polygons with regular faces is already complete since 1969 (Johnson, Zalgaller). We have the 5 Platonic solids, there are Archimedean Solids (including two infinite families of N-sided prisms and anti-prisms) and irregular 'Johnson Solids'. The list including pictures can be found on http://mathworld.wolfram.com/JohnsonSolid.html. Maybe Oskar can create a Johson Twisty?

One other remark: be careful with using the Euler formula: it is NOT valid for the void cube!

No problem, just use V - E + F = -8 for the void cube