Hexahedron madness

[bonanova]

A polyhedron is a solid with plane faces.

It is called convex if each of its faces can be placed flat on a table top.

The simplest convex polyhedron is the tetrahedron, with four triangular faces.

It can have an endless variety of shapes, but the network of its edges is always the same.

Its faces cannot be anything but triangles.

If we increase the number of faces to five, we can sketch two different configurations:

a pyramid with a quadrilateral base, and a triangular prism.

Note that the pyramid has 5 vertices and 8 edges; the prism has 6 vertices and 9 edges.

Those are the easy cases, so let's take it to six.

How many topologically distinct, convex, six-faced solids [hexahedra] can you sketch?

One, of course, is the cube.

Note that changing the square faces of the cube into rectangles or trapezoids does not change its topology.

Also note that truncating the apex of a square pyramid produces the same topology as a cube.

In each case, there are six quadrilaterals sharing an edge with four of the others.

To be distinct, the number of edges or vertices or sides of the faces must differ.

[Guest]

a pentagonal pyramid

Edited September 7, 2009 by nrduren

[Guest]

I think that your suggestion is the only other possibility, nrduren. One must remember that the multiplicity of any vertex (the number of edges which converge to the vertex) can not be greater than five. Equally, no face can more than pentagonal (which the base of your pyramid is.) Also the number of edges must be the number of vertices + 4, from Euler's Theorem. If one suggests a four sided prism, for example, one ends up with a topological cube, which BN has already enumerated. I think the problem might become perhaps too exciting if one were to consider an icosahedron.

Edited September 7, 2009 by jerbil

[Guest]

removing 1 edge from a cube.

[plainglazed]

Two tetrahedra stuck together (sharing a face) would have five vertices and nine edges. Combinging the pyramid and the prism (sharing a four sided face) where the prism is truncated on an angle at each end such that it's triangular faces are in the same plane as two opposing triangular faces of the pyramid. This shape would have eleven edges and seven vertices (may be the same as p_m's answer above). Seems there also might be one with twelve edges and eight vertices. Still thinking.

edit: never was very good with irregular plurals or spelling in general for that matter...

Edited September 7, 2009 by plainglazed

[bonanova]

Shape [edges, vertices]

.

1. Cube [12, 8] - OP
2. Pentagonal pyramid [10, 6] nduren
3. Cube with edge removed [11, 7] psychic_mind
4. "Siamese tetrahedra" [9, 5] plainglazed.
   pg's other suggestion is, as he suspected, the same as #3.

.

There are more.

[plainglazed]

a volume of pentagonal cross-section truncated at an angle such that each of the two resulting pentagonal shaped faces share an edge - eight vertices, twelve edges.

[bonanova]

a volume of pentagonal cross-section truncated at an angle such that each of the two resulting pentagonal shaped faces share an edge - eight vertices, twelve edges.

Yes.

Shall we say two to go. One is pretty; one is weird.

And both will also have the same number of edges and faces as ones already found.

Only the 5 vertices, 9 edges case will be unique.

The remaining cases will differ in the shapes of one or more faces.

[Guest]

Shape [edges, vertices]

.

1. Cube [12, 8] - OP
2. Pentagonal pyramid [10, 6] nduren
3. Cube with edge removed [11, 7] psychic_mind
4. "Siamese tetrahedra" [9, 5] plainglazed.
   pg's other suggestion is, as he suspected, the same as #3.

.

There are more.

I made a tent [11,7]:

tent.bmp

[Guest]

I made a tent [11,7]:

tent.bmp

[11,7] shape?

11-7.bmp

[Guest]

Here's my thoughts and assumptions. I'm not actually able to see the spoilers on my work computer for some reason, so likely all the ones I've found have been found already. First, some analysis:

A topologically distinct, convex, six-faced solid is defined by the number of edges on each of its sides. For example, a cube has six 4 edged sides. Any other convex polyhedron with six 4 edged sides will not be topologically distinct from a cube. (I'm assuming this is by definition. Is it ok to assert this without proof?)

Each edge in the polyhedron will be shared between exactly 2 sides. So if we take each side as separate 2 dimensional shapes and count up the edges, then divide by 2, that is the number of edges in our polyhedron. For example a cube is 6 squares with 4 edges each (24). Divide by 2 and our cube has 12 edges. Since each edge must be shared, the total number of edges of the sides taken separately must be an even number. Also, no side can share more than one edge with another individual side (any more and our sides would lie flat on top of each other).

Each side must have at least 3 edges (of course). Since each edge on a side is shared with exactly one other side and no side can share more than one edge with another, if any or our sides had 6 or more edges, our polyhedron would need to have more than 6 sides. Finally our polyhedron cannot have more than 2 sides with 5 edges: A 5 edged side needs to share an edge every other side in our 6 sided polyhedron. Once we've attached our 3 sides to each other there is no way we can finish forming our polyhedron with only 3 plane sides to work with. (I'm low on time and having trouble proving this without using pictures and common sense...)

Anyways, so if we split our polyhedron into 6 individual sides we will have a combination of triangles, squares and up to two pentagons whose edges add up to an even number. This gives us 7 possibilities:

6 triangles (18/2 = 9 edges)

Connect 2 tetrahedrons by an identical side. (5 vertexes)

1 pentagon and 5 triangles (20/2 = 10 edges)

A pyramid with 5 sides and a 5-sided base. (6 vertexes)

4 squares and 2 triangles (11 edges)

Take a pyramid (4 edged base) and pull out the middle of the base down to look like the roof of a house. (7 vertexes)

6 squares (12 edges)

A cube (6 vertexes)

2 pentagons, 2 squares and 2 triangles (12 edges)

Picture 2 pentagons attached at one edge, opened up like a clam. Draw lines attaching the remaining three vertexes on the top pentagon to the corresponding vertexes on the bottom one. (6 vertexes)

1 pentagon, 2 squares and 3 triangles (11 edges)

Take our clam above, shrink one of the edges opposite the hinge edge of our clam. This vertex is now connected to two lines leading down to the bottom pentagon. The top pentagon is now a square. (7 vertexes)

2 squares and 4 triangles (12 edges)

Now take the bottom pentagon and shrink the edge opposite the hinge edge of our clam but make sure it's not the one directly under the one we shrunk on the top. Shrink it to a point. Now we've got a clam of 2 squares. Each vertex opposite the hinge has edges leading to the other two vertexes above. As long as one of the squares is not actually a perfect square (I guess I've been using the terms square too loosely, I mean quadrilateral) then it should work out (I think. Having trouble visualizing this) (8 vertexes)

Please excuse any errors I've made in my haste but I think you get the idea...

[Guest]

Speaking for myself, I think that an excellent solution. Of course in Britland we use the term "vertices."