Recently, a reader of this blog contacted me about a polyhedron he wished to model. His name is Dave Smith, and he had already done much of the work involved, but needed help finishing off his project. Here’s the picture he e-mailed me.
The visible faces are regular pentagons — four of them. The invisible faces are isosceles trapezoids, in two “bowtie” pairs which share their shortest edges with those of their reflections. I e-mailed Smith, and told him the truth: I didn’t have a clue how to make this in Stella 4d, the program I use to make the rotating polyhedra on this blog (including the one below). I also told him I wasn’t giving up — merely enlisting help with his puzzle.
And, with that, I went to Facebook, posting the image above, along with an explanation, and request for help finishing it. This may not be what most people think of when they consider Facebook, but I have deliberately sought out experts there in many fields, including geometry, to make the social-networking site useful in unusual ways, such as getting help with geometrical puzzles I can’t solve alone. Three geometricians with skills which exceed mine (Wendy Krieger, Tom Ruen, and Robert Webb, who wrote Stella 4d) began discussing the figure. One of them, Tom Ruen, sent me .stel files (That’s what Stella 4d uses) for multiple figures, getting closer each time. With the last such virtual model Tom sent me, I was able to “tweak” it to get the pentagons regular.
This eight-faced figure has two edge lengths, the shorter appearing only twice, as the shared, shorter base within each “bowtie” pair of isosceles trapezoids — and these two edge lengths are in the golden ratio. A type of octahedron, it also has an interesting form of symmetry — it reminds me of pyritohedral symmetry, but is not; the features seen in pyritohedral symmetry in relation to the x-, y-, and z-axes of coordinate space only show up here in relation to two of these three axes. This symmetry-form is called dihedral symmetry.
And it only took five people to figure all of this out!
RobertLovesPi.net 
Neat!
Leslie
LikeLiked by 1 person
Re Smith polyhedra – If you replace the pentagons with five equilateral triangles on each, then you end up with a 24 sided figure (20 triangles, 4 isosceles trapezoids). I also made up an attractive 32 sided polyhedron using card, made up of 24 trapezoids and 8 equilateral triangles. However, I’m putting it down as a ‘near miss’ due to the faces not lying completely flat.
I’d like to thank everybody that contributed in some way as I could not have got this far without your help – Dave S.
LikeLiked by 1 person
Using a recent snapshot of the Antiprism polyhedron software
http://www.antiprism.com/snapshots/
The model can be made and viewed with this command
off_util -D f0,2,9,3,6,11 -x V dod | off_trans -y full | to_nfold 2 | minmax -a u -s 40 | conv_hull | antiview
As an explanation, the first part (off_util) takes a dodecahedron and deletes the faces around opposite vertices, leaving a belt of six pentagons. The next part (off_trans) aligns the belt on its main symmetry axis (needed for the next part). The next part (to_nfold) makes a similarly connected model but reduces the 3-fold axis to a 2-fold axis. The next part (minmax) makes the resulting four pentagons regular. The next part (conv_hull) makes the convex hull in order to create the missing faces. The final command displays the model in a 3D viewer.
LikeLike
Thanks for your input. That is an interesting way of creating the model – not that I fully understand it mind! I have a couple of other polyhedrons that utilise the trapezoid if you are interested.
LikeLiked by 1 person
There are two combination /rubik’s type puzzles based on this geometry. The Quirky Gem 1 which was first solved by the inventor. Also the Pentajumble by a different inventor, which I solved last year. In these puzzles only the pentagons rotate. Also the puzzles are classified as jumbling-only. Jumbling is a complex mathematical property that will have to be explained at a later time. The Pentajumble was the hardest puzzle that I ever solved and I’ve solved many of the various rubik’s type puzzles since 2003.
LikeLiked by 1 person