• 10:35 AM

I'll take a stab at it!

If it isn't possible to extend the edges of the tetragonal anti-wedge to make a pyramid, that is, if the edges are skew, my whole theory is pointless. So read this with that caution in mind! I'm guessing that it is possible, but I have nothing to back that idea up.

I just see that you have no responses and I'm trying to help in the few ways that I can.

So, this is just a WILD guess:

The tetragonal anti-wedge is like two tetrahedra each with one 90 degree dihedral angle placed abut so that they form an pyramid with a 4-sided base that is not a square. You then truncate each tetrahedra separately forming two triangular faces opposite the 4-sided base.

You could then minimize the surface to volume ratio for the pyramid with a 4-sided base, then truncate the smallest possible triangular faces (thus decreasing the volume as little as possible, and increasing the surface area as little as possible)

If you can call the pyramid with the the 4-sided base a "degenerate tetragonal anti-wedge" (with two triangular sides of area 0) that could be even better, although in some contexts it might be illegal.

Of course, you would need to find out how to maximize the volume of the pyramid shape...

I'd love to know what the answer to this questions is when you find it! Please let us know.

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Re: I'll take a stab at it!

I just thought this through again and I've got it all wrong! I'll let you know if I come up with anything.

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Re: I'll take a stab at it!

Thanks for your efforts nonetheless! I wrote to George Hart, who hosts the Pavillion of Polyhedreality to see if he is feeling generous enough to give some insight or a solution. Maybe I will appeal to the owners of a few other websites to see if I can get anything else. Before I posted here I did attempt to see if there were any online resources, but this seems to be a pretty obscure problem.

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The problem is not quite right yet. Here is the answer to the current problem, but it is silly:

Take any particular version of the polyhedron you want with finite nonzero edge lengths, vertex angles, surface area, and volume.

Now double it in size (take the coordinates of the polyhedron and double them component-wise for instance).

The vertex angles remain unchanged, the edge lengths have increased by a factor of 2, the surface area has increased by a factor of 4, and the volume by a factor of 8. The surface-area-to-volume ratio has now been multiplied by 4/8 = 1/2, so it is much smaller.

Just repeat this until you are satisfied the ratio is low enough.

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So how would you reword the problem to state the less trivial question I originally intended to ask?

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It's a little unclear how to reword it, that's why I didn't post an attempt.

For simple shapes like polygons you can say "for a given perimeter and number of sides, what is the largest area one can get?"

For your question it is not entirely clear if one should put a bound on:

1) The maximum side length

2) The total side lengths ("perimeter" is a little weird)

3) The surface area

Something along these lines is called geometric analysis and is totally not my thing so I'm not sure what a good reference/intro would be. Your question is very similar to the famous "isoperimetric inequalities" which basically says that for a polygon you should try to make it a circle, and for a regular, convex polyhedron you should try to make it a sphere.

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How about this:

For a given surface area of a tetragonal antiwedge, what combination of edge lengths and vertice angles will yield the lowest surface area to volume ratio?

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