John Conway has given an explanation how a regular Hecatohedron (polyhedron with 100 faces) looks like: http://www.ics.uci.edu/~eppstein/junkyard/hecatohedron.html

Conway explains:

Here's a hecatohedron with full tetrahedral symmetry: Form the "16-reticulated cube", by dividing each face of a cube into 18 smaller "square" faces in the obvious way, giving a 96-hedron. Then tetrahedrally truncate this. I suppose I'd call it the "semi-trivalently-truncated 16-reticulated cube" !

Unfortunatly I was not able to find a way how implement this into Mathematica. Can somebody give me some more hints or explanations? Thanks!

  • $\begingroup$ To get a better idea how it looks like:korthalsaltes.com/model.php?name_en=faceted+sphericons $\endgroup$
    – eldo
    May 24, 2014 at 23:02
  • $\begingroup$ @eldo: That's a different 100-faced polyhedron than the one described in the question. $\endgroup$
    – user484
    May 24, 2014 at 23:15
  • $\begingroup$ This? $\endgroup$
    – wxffles
    May 26, 2014 at 2:12

1 Answer 1


Oh hey, another chance to use my recent post. Define myRegionPlot3D from the linked answer, then do

 Max[x + y + z, x - y - z, -x + y - z, -x - y + z] <= 2.75, (* tetrahedral truncation *)
 {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, 
 Mesh -> 3] (* divide each square face into 4x4 squares *)

enter image description here

Obviously there are ways to draw this polyhedron without all that machinery, but I had the function lying around after which it was just one more line of code. :)

  • $\begingroup$ I'm surprised. The 6 square-faces are divided into 4x4 faces, but this should also change the angles between the faces. Here, you have an angle of 180° between all of your 4x4 faces, and 90° between the 6 square-faces. I thought it might be a some (non-regular/non-symmetric) generalisations of a Icosahedron? Something like a 100-sided dice? $\endgroup$ May 24, 2014 at 23:34
  • $\begingroup$ @Mario: Conway says to divide "each face of a cube into 16 smaller "square" faces in the obvious way", and this was the most obvious way I could think of. :) Perhaps it would be better to make the configuration nonflat by moving vertices outward, but that requires an arbitrary choice of perturbation and the smaller faces would not remain square. The plotted figure is at least topologically/combinatorially correct, but I can delete the answer if you feel it is not the desired solution. $\endgroup$
    – user484
    May 24, 2014 at 23:53
  • $\begingroup$ I think you are right, it's exactly what Conway wrote. I was searching for something like a 100-sided dice (and thought Conway described it), so there was a misunderstanding on my side. So thanks for this nice solution. :) $\endgroup$ May 25, 2014 at 7:06

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