Unusual Polyhedron

You pointed in the correct family. From youR link,... it looks close to this

tornado, should count up the sides,... and as well as the shapes,... and one may be able to extrapolate a name to do a search.

The sides were counted in the OP.

In wonder who thought that was OT, when it was entirely on topic. Maybe somebody who can't even count to two.

You need to stop taking something so petty, so personally Just look at being marked OT was made by someone that may has an axe to grind, so the OT Mark has no value.

Uh-huh. I wondered who that might be.

Wondered huh?,... I was go to offer to counter it with a GA to put your wondering mind to rest... but I see it’s gone now,... feel better?,... or are you still wondering?

I don't think it conforms to most of the various definitions of semiregular. Not all vertices have the same sequence of polygons around them. One could call it an icosahexahedron, but that doesn't seem specific enough. I have in mind a joke name to be shared later.

A lot of semiregular polyhedrons are constructed by truncating (chopping off the pointy ends) of another polyhedron.

Hexagonal Trapezohedron (Hexagonal bipyramid with one pyramid rotated 30 deg

Cutting off the top and bottom forms the hexagonal faces 30 degrees offset and 12 pentagonal faces.

If the angle of the original trapezohedron is just right (i.e. wider or taller), you get regular pentagons as shown on the left.

But this is not your polygon. With a different angle, you can get equilateral triangles on the sides of the hexagons (which is part of the original trapezohedron) and carving off the 12 vertices around the outside, you get squares (diamonds) touching the vertices of the hexagons. On the right is shown one diamond attached to a triangle.

So, I believe your polyhedron could be a truncated hexagonal trapezohedron.

Intriguing. I haven't yet satisfied myself that one can get both equilateral triangles and squares that way. If it does work, your name suggestion is perfect.

Oops, stop the presses. The octagon version doesn't work, after all. I think it's back to only the hexagon version.

Thanks for posting this. It's been fun playing with it.

I've been trying to justify buying a 3D printer, and I think I've found a use - making a polygon collection.

Still looking for a name?

What about Bob?

or Bill, or George, or anything but Sue...

Haaaa.. great reference.

This would make an interesting gambling game. Based on the structure, calculate the odds of the various faces coming up on a roll of the polyhedron. The hexes would be high, the squares low, and the triangles very low. From the photo, I'm not sure that the squares or triangles would come up parallel to the board (I think not), but there would always be a single uppermost face no matter how it landed. Or the game is played above a mirror and you read the bottom face. Odds of a pair of triangles coming up would be really low.

jhammond

It stands on any of its faces. It is hollow paper; I don't know if being solid would change that.

It's symmetrical, so whatever face it is resting on, the center of gravity should be above the center of that face and it would be stable.

As far as figuring the probabilities of each of the faces, a good approximation would be to figure the solid angle (in steradians) of each surface as seen from the center (center of gravity). If you can calculate the coordinates of each vertex, the solid angles of a hexagon, square, and triangle can be figured out.

This calculation would be exact if the polyhedron were laid down with a random orientation. Maybe if it were tumbling and spinning, the rotational inertia might vary the statistics if it bounces on a vertex. The vertices are not equidistant from the center.

If you want a good die, I guess it's safest to stick with the Platonic solids.

Some others have identical faces all around, too. Bipyramids, trapezohedra, rhombic dodecahedron, rhombic triacontahedron.

You are right, there is no need for the sides to be regular polygons for a randomly thrown die to have equal probabilities. Rhombus or kite-shaped surfaces would suffice.

There is another factor. I understand that a skillful thrower can control a regular cubic die to some extent.

https://www.casinocenter.com/controlling-the-dice-is-it-fact-or-fiction/

It might be even easier to cheat by controlling the throw if the die is not symmetrically shaped. For example, a pentagonal bipyramid is flattened and one side or the other could be favored by the throw. Just a thought...

http://mathworld.wolfram.com/PentagonalDipyramid.html

It's an 18-sided, semi-regular polyhedron, consisting of 10 squares and 8 equilateral triangles, with all sides being equal, so that would make it qualify as a semi-regular octodecahedron.

It's a little embarassing to take two trys to build the correct model, and get the correct 12 squares, 12 equilateral triangles, and the proper two hexagons (or would that be 12 more equilateral triangles?), for a semi-regular, 26-sided, polyhedron.

(My original model was of the same genre, but only had one square on the top, and one more on the bottom, as a sort of ''little brother'' to the OP polygon?)

If the 12 squares of are taken off the OP solid, but still adjacently connected, and then, laid out flat, the squence of squares would be seen connected left, right, left, right, etc., and connectable as a circular ''ring'' of squares

That could suggest that the OP solid could be more effectively viewed as a single section of an endless conduit, with hexagonal (openings) on each of the two sides of each section, and not just a single, solid, multi-hedron.

• Demi-hexatriahedron?
• Tri-hexahedron?
• Bi-nonahedron?

Takeyourpickahedron...

You might enjoy this site...

https://polyhedra.tessera.li/

You can modify and rotate polyhedra.

That's a good source. I also found the Johnson solids as a Wikipedia subheading. But so far I haven't seen "my" polyhedron anywhere than on my desk/shelf, where it has been for 50 years.

Back then, I thought I had found a disproof if its existence. But so many years have intervened that I became unsure. That is why the thread title says polyhedron/ would-be polyhedron.

Your truncated 6-trapezohedron suggestion made me look again, and for a brief while it looked possible. And seductive! Who wouldn't like to create a publishable article on a whole new family of polyhedra?

Alas, not so. I have found a new proof different from whatever my long-forgotten proof was. I'm still figuring out how to describe it easily. My AutoCAD is on another off-line computer, and this computer doesn't have any graphics programs that I am familiar with.

In a subsequent post I will try to provide a simple explanation.

The simplest analysis I have found so far is to consider the square touching the bottom surface a bit to the right of center. Make that touching point be the origin (0,0), with the other corners being (1,0), (0,1) and (1,1) (Assuming you are looking perpendicularly at this square.)

Rotate the image ~50º clockwise. Now you see a square above tilted 45º away and back, and a square to the right also tilted 45º away and back. In projection, both of these squares become rectangles of dimensions √2/2 by 1.

Label the relevant points thus (CDGF being the first square):
A B
C D E
F G H

In projection, AB = CD = FG = CF = DG = EH =1; AC = BD = DG = EH = √2/2.
On the actual tilted squares, AD = DH = √2. However, AH = √2 (1 + √2/2) = √2 + 1.

By law of cosines, AH² = AD² + DH² - 2 AD AH cos θ; solving numerically gives θ ≈ 117.2º. However, for this whole construction to work, θ must be 120º. (Because the AD and DH square diagonals are sides of a hexagon.)

This discrepancy is small enough to escape detection in the model.

OK, you got me sucked in!

I started by constructing a hexagon and on each edge an equilateral triangle. Only the vertices are shown.

Each triangle is made up of two blue vertices "*" from the hexagon and a red vertex "+". I rotate the red vertices about the axis defined by the two blue vertices by an angle phi.

Now consider the angle formed from each blue dot to the two closest red dots above it. As the rotation angle phi increases, this angle gets smaller. At some point, this angle equals 90 degrees, and it forms the corner of the square.

I computed the dot product of the vectors between a blue vertex and the nearest two red vertices. At 90 degrees, the dot product is zero.

phi = 43.393724991 degrees (approximately).

That's half your polygon, the other half fits on top.

I think I'm agreeing with you, this is a pseudosemiregular polyhedron. (That could be why there seems to be no matching picture on the internet.)

Using Octave, I constructed a hexagon and on each side an equilateral triangle. I rotated the top vertices up an angle phi = 43.393724991 degrees so that the angle between each bottom vertex and the two above were exactly 90 degrees (see #31). This located the second tier of vertices.

Having 3 corners of the square faces permitted locating the 4th corner in the third tier.

Finally, the top hexagon was placed exactly, rotated 30 degrees from the bottom hexagon. The angle was measured from one vertex of the top tier (the hexagon) to the two neighboring vertices in the third tier below which would determine the angle of the square's top corner.

Bottom line: The second and third tiers were constructed from the bottom tier. The top tier was constructed from the bottom tier. Measuring the top corner angle of a square face revealed a discrepancy. The computed angle for the top square corner was 91.027 degrees. (Angle is computed as arc cosine of the dot-product of vectors between vertices.)

Bottom tier hexagon, Blue "*"

Second tier vertices, red "+"

Third tier vertices, green "o"

Top tier hexagon, orange "x"

What it boils down to is that the clamshells don't quite zip together--but close.

I've very much enjoyed revisiting this.

The joke name I gave it is poly-phony.

If I make it out aright the polyhedron has 26 faces, which makes it an eikosiexihedron (Greek eikosi = 20, exi = 6)