Talk:Cubohemioctahedron

Diagram doesn't match table?

I think there is a problem in the diagram. The various article that mention this polyhedron all agree with the table in this article which shows:

• 10 faces - 6 squares and 4 hexagons: 6{4}+4{6}
• A uniform vertex configuration of: 4.6.4/3.6

Other mathematical websites agree.

However if you look at the diagram you will see two triangular faces - bottom foreground and top right. So the number (and type) of faces doesn't match and the vertex configuration is wrong - currently 6.4.3.6.4 assuming that the pattern of triangles and concavities is repeated - give or take a retrograde square. (I can never work out which should be retrograde!)

I think that the correct figure would be a modified Cuboctahedron, with all 8 triangular faces replaced by tetrahedron shaped concavities. (OK these are really the various intersecting hexagons). This agrees my understanding of other websites - for example http://www.korthalsaltes.com/cubohemioctahedron.html

Therefore the two triangular faces should be replaced by tetrahedron shaped concavities.

Does this make sense? If not, I'll try again. If I'm wrong please explain it to me - I'm studying polyhedra at the moment in a maths course, and would like to make sure I've understood vertex configurations and the like.

Many thanks Cje 08:13, 17 April 2006 (UTC)

From each vertex there's two hexagons that pass through the center of the model. There's no triangles. The hexagons are intersecting. That's why you see triangles. It's the same as looking at a pentagram which has intersecting edges, but there's no vertices at the intersections - a pentagram has 5 vertices and 5 edges, just like a pentagon, even though there's 5 nonvertex intersection points as well. Does that help? Tom Ruen 20:41, 17 April 2006 (UTC)

Well, yes and no! We're in complete agreement on how the Cuboctahedron is constructed: and that is reassuring for me. My point is that the figure doesn't make the construction clear.

If, like me, one is new to nonconvex polyhedra it is very easy to look at the yellow coloring between the three vertices at the bottom f the picture and see this as a triangular face. Either some shading here is missing, or the figure needs to be rotated a little? Two of us from the same course independently looked at the figure and misinterpreted it in the same way. It was only when we each built a model using the cutout at the korthalsaltes site that we worked it out. Before that we couldn't see how the vertex configuration matched the picture.

Alternatively, I might take a crack at a textual description based on your comments above.

What do you think Cje 08:44, 18 April 2006 (UTC)

One trick for interpreting the pictures as provided is that "real" vertices are marked by little spheres and real edges by cylinders, while intersecting faces have no special marks. Probably best there should be a link for "nonconvex" that could explain this in general, so not need to be repeated for each of uniform nonconvex polyhedra. No time for me now, so glad if you'd like to try adding something here, or in general. Tom Ruen 00:56, 19 April 2006 (UTC)

A general "nonconvex polygons" sounds like a good idea. Certainly the paragraph you've added to the article helps new polygon students, like myself. I'll be glad to have a go - but it won't be immediately. I'll leave you a note when there is something worth looking at. Alternatively if you get there first, then leave me a note and I'll take a look at it. Cje

Hmm, yes, they do look like triangles. (The lighting gives a subtle cue that they're not, BTW: if they were really triangles they should appear darker like the ones on the cuboctahedron pic, so it implies that they're hexagons and at a different angle. But I don't think we should force readers to puzzle that out.) My concern about rotating the picture is that then it will not be at the same viewpoint as the other octahedral-symmetry polyhedra. Double sharp (talk) 08:19, 5 January 2015 (UTC)