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Below you see the Rhombicuboctahedron. If you put an additional point in the blue triangle, you make three blue triangles out of one. Now you connect a yellow square with two adjacent small blue triangle and you end up with a blue-yellow hexagon.

$\hskip2.7in$enter image description here

Drawn in the plane this would look like this:

$\hskip2.1in$enter image description here

How is the resulting 3D object called, if it has a name at all...

draks ...
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  • I assume you move the "additional point in the blue triangle" toward the outside so it lays at the intersection of the three plans defined by the three closest yellow squares ? – Evargalo May 29 '17 at 13:21
  • @Evargalo no they are centered in the face and each new point is connected to the vertex of the triangle... – draks ... May 29 '17 at 13:24
  • I can see the images now ... Isn't the second one a Truncated Octahedron ? https://en.wikipedia.org/wiki/Truncated_octahedron – Donald Splutterwit May 29 '17 at 13:25
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    @Donald I don't think so. Based on the Schlegel diagram in the post, there are too many hexagons -- 12 here, versus 8 in the truncated octahedron. EDIT: wait.... are those pentagons in the post? I mistook them for another set of hexagons. – pjs36 May 29 '17 at 13:30
  • @draks: then I don't understand that step: "Now you connect a yellow square with two adjacent small blue triangles and you end up with a blue-yellow hexagon.". I seems to me than the yellow square and the two adjacent small blue triangles are not even in a same plan ? – Evargalo May 29 '17 at 13:33
  • @Evargalo I just figured out how to picture it. The point you place in the triangle will be above the shape; you're really adding a pyramid whose apex is the new point. Anyway, the new point is placed in such a way that it lies in the planes containing the three adjacent squares. I'll try and get on Sage at some point and see if I can supply an actual 3D picture. – pjs36 May 29 '17 at 13:47
  • @pjs36 no pentagons – draks ... May 29 '17 at 14:57
  • @Evargalo I didn't care about them being in the same plane. Do we have to? – draks ... May 29 '17 at 14:58
  • @draks How can you pretend the 'blue-yellow' shape to be an hexagon if it's not planar ? By the way, if the new point was exactly in your blue triangle, the result would be the exact same Rhombicuboctahedron you strated from. I think the new point has to be 'above' the triangle, as pjs36 explained probably more clearly than myself... – Evargalo May 29 '17 at 16:04

2 Answers2

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Here is your solid:

enter image description here

It's part of the family of the chamfered cubes, but I don't think it has a name on its own.

lesath82
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I think this has to be this since it has $12$ hexagons and $6$ squares as requested.

https://en.wikipedia.org/wiki/Chamfer_(geometry)

enter image description here

enter image description here

They call it truncated rhombic dodecahedron on this page.

zwim
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  • This is not exactly the solid the OP talks about, you can see that the exagons are too "fat". If you cut away their sharpest verices, you get rectangles, not squares equal to the purple ones. But it't the correct family. – lesath82 May 29 '17 at 15:12
  • @lesath82 the solid is perfectly fine... – draks ... May 29 '17 at 15:52
  • @draks I guess you are interested in its topology, so the exact shape of faces does not matter. What is important is that it is ok for you – lesath82 May 29 '17 at 16:28
  • @lesath82 the required solid is indeed a truncated rhombic dodecahedron in the general sense. The 4-face vertices have been cut off in this example to give squares, they just need to be cut back a lttle further to give the exact shape described in the OP. I'm not sure there are any conventional rules on how to truncate a rhombic dodecahedron (how far to cut back the 4-face vertices, and indeed whether to cut back the 3-face vertices or not - they have been left intact in this case) but this is most definitely a rhombic dodecahedron which has been truncated in a particular way. – Level River St May 29 '17 at 19:59
  • @Level I'm perfectly fine with that. In the beginning I just thought that the OP needed the solid built following his exact instructions, but this is not the case. In fact there is a smooth transition going from a rhombic dodecahedron at one extreme towards a cube at the other. If you cut slightly the dodecahedron vertices you call it truncated, if you slice away the cube's edges, you call it chamfered. – lesath82 May 29 '17 at 20:27
  • Wow, I didn't thought my post would be subject to such a heated discussion :-) But yes, generally polyhedrons are classified using regular polygons, so here the hexagons have $6$ equal sides, but the drawing by lesath92 resembles more the original picture given by OP, yet the topology is the same, I think this is what matters. I found an animated version of it here -> http://maths.ac-noumea.nc/polyhedr/troncat1.htm – zwim May 29 '17 at 20:40
  • @zwim the hexagons may have equal sides in your example but they are not regular, nor can they be. Where 3 hexagons meet the angle of the hexagon vertex is 109.47 degrees (as in the rhombic dodecahedron) so the other angles are larger. – Level River St May 29 '17 at 20:55