Last time I showed what happened if you took a cube and chopped off its corners more and more until you reached its dual: the octahedron. Today let’s do the same thing starting with a dodecahedron!
Just as a cube has 3 squares meeting at each vertex, a dodecahedron has 3 pentagons meeting at each vertex. So, instead of the Coxeter diagram for the cube:
everything today will be based on the Coxeter diagram for the dodecahedron:
The number 5 is much cooler than the number 4, which is, frankly, a bit square. So the shapes we get today look much more sophisticated, at least to my eyes. But the underlying math is very similar: we can use diagrams to keep track of these shapes as we did before.
Dodecahedron: •—5—o—3—o
First we have the dodecahedron, with all pentagons as faces:
I like this shape so much I gave a lecture about it, and you can see the slides here:
• John Baez, Tales of the dodecahedron: from Pythagoras through Plato to Poincaré.
Truncated dodecahedron: •—5—•—3—o
Then we get the truncated dodecahedron, with decagons (10-sided shapes) and triangles as faces:
Icosidodecahedron: o—3—•—3—o
Then, halfway through, we get the aptly named icosidodecahedron, with pentagons and triangles as faces:
Like that other ‘halfway through’ shape the cuboctahedron, every edge of the icosidodecahedron lies on a great circle’s worth of edges.
Truncated icosahedron: o—5—•—3—•
Then we get the truncated icosahedron, with pentagons and hexagons as faces:
This one is so beautiful that a whole sport was developed in its honor!
Icosahedron: o—5—o—3—•
And then finally we get the icosahedron, with triangles as faces:
Again, I like this one so much I gave a talk about it:
• John Baez, Who discovered the icosahedron?
I almost feel like telling you all the stuff that’s in these talks of mine… and if I turn these blog posts into a book, I’ll definitely want to include it all! But there’s a lot of it, and I’m feeling a bit lazy—so why not just go check it out?
Puzzle. Why did Thomas Heath, the great scholar of Greek mathematics, think that Geminus of Rhodes is responsible for the remark in Euclid’s Elements crediting Theatetus with discovering the icosahedron?
Azimuth 
Typo: the faces of the truncated dodecahedron look like decagons and not dodecagons as you state.
Whoops—thanks!
An American football is a prolate spheroid apparently. The originals were just pig bladders (hence pigskins).
From Wikipedia: The familiar 32-panel association football ball design is sometimes referenced to describe the truncated icosahedron Archimedean solid, carbon buckyballs or the root structure of geodesic domes (thought they looked familiar beyond the world of football!)
The air inside of course makes it a sphere and I suppose the panels that are stitched together in that particular manner are both practical and artistic in basis.
I am a sports illiterate, so maybe now I can bore sports lovers as well as being bored by them! teehee.
Yes, I’m all in favor of fighting boredom by being more boring than the people who bore me.
I don’t know why the soccer ball is patterned after a truncated icosahedron, but I’m very glad it is. An American football is a bit too sharp at the ends to be a prolate spheroid:
You get a prolate spheroid where you rotate an ellipse around its long axis. I don’t know if there’s a mathematical specification of the shape of an American football!
John said: Yes, I’m all in favor of fighting boredom by being more boring than the people who bore me.
I like that! I wouldn’t have tried it at school though.
http://en.wikipedia.org/wiki/Football_%28ball%29 says:
“Rugby league is played with a prolate spheroid shaped football which is inflated with air.”
and
“The ball used in rugby union, usually referred to as a rugby ball, is a prolate spheroid essentially elliptical in profile.”
There’s one more thing, John, in terms of the earlier typo pointed out by Ezequiel above: the change was made from dodecagons to decagons, but in the parentheses, it still says 12-sided shapes, instead of 10-sided shapes.
If were less honest I’d say I included these mistakes to check if you’re paying attention.
I’ll fix that. As you can see, it wasn’t a typo—it was a thinko. Thanks!
Hi John, I couldn’t open the link to your talk on the icosahedron.
It’s working for me now. Maybe the U.C. Riverside computer was down for repairs for a while this weekend. Try it again?
There are three great families of semiregular polyhedra in 3 dimensions: the tetrahedron family shown here, the cube/octahedron family shown in Part 5, and the dodecahedron/icosahedron family shown in Part 6 […]
Dear John Baez,
here we found how an icosahedral group evolves towards the great dodecahedron:
http://quantumcinema.uni-ak.ac.at/site/research/topology-of-lie-algebra/
Is it possible to specify this subgroup on the what we call the
“rosetta path”?
We would be greatful for any hints !
As I understood you call this shape in the center an Epitahedron. Unfortunately it is a bit hard to see in the video, wether all its vertices always end up on vertices of the Icosahedron, so given this I unfortunately find it really hard to help you further than with this. sorry.
But other people here may eventually see more than me in this video, it looks nice.
Eventually the referees of the paper you are submitting to might also give you hints.
Thanks, for your reply :
The epitahedron E+ shares 5 vertices with the icosahedron.
Related findings are published already:
Visualisation of the icosadedral group in a new 3D representation, In: Experience-centered Approach and Visuality in The Education of Mathematics and Physics, Ed.: Jablan Slavik et al., Kaposvar: Kaposvar University, pp.195-197. (2012)
I guess the answer needs somebody who is familiar with
Group Theory AND Geometry like Plato, Galois, F. Klein or Bertram Kostant,- or somebody else strumbling along
THIS blog ?
Oh it looked, as if you wanted to publish it somewhere else. Sorry I have no library account, so I have only your video to look at. It looks as if your Epitahedron has 6 vertices. Is that correct?
Yes seems so.
yes: 6 vertices, 7 faces, and 11 edges; it is one of the 34 topologically distinct convex heptahedra, but golden … we will publish it soon online too.
And sometimes it seems that mathematicians nowadays rather have footballs in their minds instead of Plato’s triangles ;-)
So if I understood correctly your epitahedron is the last of the heptahedra in this Wikipedia list.
This is also hard to see in the video, are the 5 vertices, which touch the icosahedra of your heptahedron always the same? (It looks so, but a video is not so overly useful for studying your geometry. Why don’t you put your 3D model on your website via a 3D viewer?) It looks as if your 5 points are the 5 points you get if you take the five points which are on three perpendicular planes as shown here. I.e. it looks a bit as if you e.g. take the quadrilateral formed by two green and two red points and then one blue point for the tip. Is this correct?
I understand that you want John to answer your question. He is extremely busy. He has a lot of duties, like his students at his university for which he has to care for and so there is not so much time left for all that math orphans. This holds also true for other academic experts. See I have also a question for him ( or some other geometry wizzard), which he hasn’t answered yet.
That is I could of course try to find the answer myself, but this would be rather tedious and timeintensive, especially since I am not familiar with all the dialects spoken in this branch of mathematics and then in the end after a cumbersome and lonesome archeological and linguistical investigation someone would point out to you that this question was answered in a russian paper in the early seventies. (which might still be quite unreadable though).
Thank you for your time to deal with my problems, nad!
In fact the epitahedron is fromed from one pentagon, – a heptahedron which Hilbert apparently had in mind, when he was searching for a 3D version of the Moebius strip;- not exactly the known heptahedron in the list, homomorphous but its edges are in the golden ratio.
Many thanks you for your link, II did it not see as clearly before: The epitahedron(E±)seems to serve for a 3D version of Euklid’s 10th proposition:
1 E± corresponds to the pentagon, 2 E± form a hexagon and
5 E± form a decagon
And with your second question it converges to an interesting solution: No, the perpendicular planes are not important here; the axis of roational axes are corresponding to the planes of the pentagons in an icosahedron which are forming a great dodecahedron;- again in the 3D manner, which seems to prove the above assumption. q.e.d.
Alas, the above question how it all related to the icosahedral groups remains still open…
And thanx for the tip: we’ll work it out on a 3D viewer too.
If you like, you may see more details of the Epitahedron
I could discuss with Roger Penrose:
http://quantumcinema.uni-ak.ac.at/site/expert-talk/sir-roger-penrose/
[…] Part 6: Interpolating between a Platonic solid and its dual. Example: the dodecahedron/icosahedron […]