Even the tiniest gears here are indeed turning. This moving picture is by someone called ~zy0rg, and I found it on deviantart.
If you stare at it, you can see it's based on a regular dodecahedron, a shape with 12 pentagons as faces. The blue gears are the corners of these pentagons. There's a red gear in the middle of each pentagon, and there are 2 yellow gears next to the edge of each pentagon.
Since the regular dodecahedron has 12 pentagons, and there's a red gear in each one, there must be 12 red gears.
The dodecahedron has 20 corners, since these are the faces of its dual, the icosahedron, which has 20 faces. Or, if you don't know that, you can say: each pentagon has 5 corners, but 3 pentagons meet at each corner, so there are 12 × 5 / 3 = 20 corners. Either way, there are 20 blue gears.
Finally, the dodecahedron has 30 edges. To see this we can use Euler's formula
Or, we can say each pentagon has 5 edges, but 2 pentagons share each edge, so there are 12 × 5 / 2 = 30 edges. Either way, we get 30 edges, and 2 yellow gears for each edge, so 60 yellow gears.
So, there's a total of 12 + 20 + 60 = 92 gears. It's often not enlightening to total up parts of different kinds like this, and I think it's not enlightening here. 92 is not a number I run into often in my studies of geometry and group theory. Factorizing it shows why: it's 2 × 2 × 23. The number 23 is not a big player in these games.
This animated gif was created by someone named TaffGoch, and you can find
other interesting things of theirs at Deviantart.
March 17, 2014
It may have been built by the Phrygians in the 8th-7th centuries BC... or maybe by Hittites fleeing the Phrygians. It seems to have been enlarged much later in the Byzantine era.
But here's the cool part: it's the largest of over 200 underground cities in the Cappadocia region of Turkey... and it's connected by a tunnel to the second largest one!
Why did people build so many underground cities there? I don't know - can you find out? It was relatively easy to do, because the area has a lot of soft volcanic rock. But as any detective show will teach you, there must be motive, not just means and opportunity.
Half of Derinkuyu is open to tourists... have you been there?
That's what emperor Julian of the Roman Empire said about Commodus, the son of Marcus Aurelius, who was emperor from 180 to 192 AD.
He never took after his father the philosopher, but he first showed his true colors after a failed assassination plot in 181. The would-be assassin was tortured and revealed a plot involving his sister.
Later one of the plotters revealed a second plot, and Commodus became paranoid. He sacked all his top commanders and started executing anyone he took a dislike to, including senators... and settled into a three-year binge of debauchery with a string of male lovers and a harem of 300 women: like Saddam Hussein, he had henchmen grab any woman who appealed to him.
As time went on, he only became worse. He developed a taste for voyeurism and had his political favorites have sex with his concubines while he watched. By 188 he had removed all responsible politicians from his inner circles and surrounded himself with a freak show. Often he would amuse himself by cutting off someone's foot or blinding them in one eye. He practiced surgery on live people and let them bleed to death. He liked to sit people at a fancy banquet and serve them food mixed with shit, to see their reactions. I could go on, but it becomes even more disgusting, and I don't want to spoil your day.
As time went on, his delusions of grandeur increased. He renamed the months: August became 'Commodus', and so on. He had always enjoyed killing animals — killing 100 lions with javelins, slicing off the heads of ostriches with special crescent-headed arrows, and so on — and fighting as a gladiator in the Colosseum. But eventually he declared that he was Hercules! He cut off the head of the sun god outside the Colosseum and replaced it with his own portrait, adding a club and a lion to make him look like Hercules. He announced his plan to kill 12,000 gladiators with one hand tied behind his back.
In 193 he was finally killed, in a plot led by his favorite mistress. An attempt to poison him failed, so his gym trainer strangled him.
All this is from an excellent biography of Marcus Aurelius by Frank McLynn. Marcus Aurelius was wise in some ways, but leaving the empire to Commodus was a colossal failure of judgement.
The photo above, by William Storage, shows a sculpture of Commodus at the Getty Museum in Los Angeles. For more of his beautiful photos go here:
This beautiful golden pattern was created by Taffgoch. He did it by taking a traditional Islamic tiling pattern made of interlocking hexagons and replacing some of them by pentagons. This lets the original flat pattern 'curl up' and become spherical!
Here is the original flat pattern:
Taffgoch says it's based on a Moroccan tile pattern of the type known as zillij, but I'd say it's an example of girih, or 'strapwork'. It's fun to see how Taffgoch transformed it into the round version... improving it step by step.
Puzzle: how many pentagons, and how many hexagons, are in this spherical zillij?
This is similar to a question about fullerenes, which are sheets of graphite — hexagons of carbon — that curl up into spheres because some hexagons are replaced by pentagons. Fullerenes come in different sizes, with different numbers of hexagons. But as long as a fullerene is spherical in its topology, with 3 pentagons or hexagons meeting at each corner, the number of pentagons is fixed!
I'll compute this number now, so if you want to answer the puzzle on your own, maybe you should stop reading. However, this spherical zillij pattern is not exactly the same as a fullerene... so it's not obvious that it has the same number of pentagons.
Here's how it goes. Suppose we have a sphere tiled with P pentagons and H hexagons, with 3 of these polygons meeting at each vertex.
How many edges are there in this tiling? Each pentagon has 5 edges, and each hexagon has 6, but each edge is shared by 2 shapes so the number of edges is
How many vertices are there? This is where we need to know 3 polygons meet at each vertex. Then by the same reasoning as above, the number of vertices is
How many faces are there? That's easy:
Now Euler's formula, a fact from topology, says
So, plugging in the equations for V, E, F, we get
or
or
Note that H cancels out, so we learn nothing about how many hexagons there are. But pentagons love the number 12... and ultimately, that's why this shape here has
rotational symmetries!
Puzzle: suppose we have a doughnut with g holes tiled by pentagons and hexagons, 3 meeting at each corner. How many pentagons are there?
Wenninger's story is interesting. In the 1940s went to the Bahamas to teach at a Benedictine school there. He was asked whether he wanted to teach English or math. He chose math. But not having taken many math courses in college, he struggled at first to stay a few pages ahead of the students! He taught algebra, Euclidean geometry, trig and analytic geometry.
In the 1950's he felt he was getting stale, so he went to Columbia Teachers College in the summer for 4 years. He got interested in the 'new math'... and started studying polyhedra.
In 1966 he wrote a booklet called Polyhedron Models for the Classroom. He wrote to H. S. M. Coxeter, the world's expert on polyhedra and higher-dimensional polytopes, sometimes called the 'king of geometry'. Apparently Coxeter sent Wenninger a copy of his book Uniform Polyhedra.
A uniform polyhedron is one that has regular polygons as faces and is symmetrical enough that there's a symmetry carrying any vertex to any other. There are 75 uniform polyhedra - not counting the infinite list of prisms and 'antiprisms'... and a very weird thing called the great disnub dirhombidodecahedron... which is a topic for another day.
After getting Coxeter's book, Magnus Wenninger spent a lot of time making models of uniform polyhedra. He made 65 of them and put them on display in his classroom. Then he decided to publish a book about them. He had the models photographed and wrote the accompanying text, which he sent to Cambridge University Press.
They said they'd be interested in the book only if Wenninger built all 75 of the uniform polyhedra! And so he did...
It took him 10 years to finish the book, Polyhedron Models, which was published in 1971. Mathematics is full of stories of amazing persistence, and this is one!
The key, which not everyone realizes, is that math is immensely fun. To leave behind this world of woe and lose yourself in a world of beauty and perfection — it's dangerously addictive.
Puzzle: this shape is covered with little pentagons, little hexagons and funny-looking nonconvex shapes. How many of each are there?
This puzzle is closely related to the question in my March 23rd diary entry. It only takes a tiny bit of persistence to figure it out... at least compared to Wenninger's persistence. But if you get stuck, read on!
First, a solution by Greg Egan:
Marking out one face of an icosahedron as 1/20 of the whole makes it much easier to count things by eye:
Doing that, it's clear that each 1/20 contains:
So the total counts for the whole structure are:
- 3/5 of a pentagon
- 3/2 of a hexagon
- 6 non-convex shapes
- 20 × 3/5 = 12 pentagons (one per icosahedral vertex)
- 20 × 3/2 = 30 hexagons (one per icosahedral edge)
- 20 × 6 = 120 non-convex shapes (2 per icosahedral edge plus 3 per icosahedral face).
Second, a solution by Julia Young, posted on G+. It's more involved, but I like how it uses less 'looking at the shape' and more 'pure reasoning':
We can figure this out using the Euler characteristic. Since this is a polyhedron, the Euler characteristic is 2. So the number of faces plus the number of vertices minus the number of edges should equal 2. So this shape satisfies f+v-e=2 where f is the number of faces, v is the number of vertices, and e is the number of edges.Here's an interesting fact revealed by Julia's analysis: for the purposes of topology, the nonconvex shapes act just like hexagons... and this shape might as well consist of hexagons and pentagons with 3 polygons meeting at each vertex. In my March 23rd entry, I noted that whenever we have a sphere tiled by hexagons and pentagons, with 3 polygons meeting at each corner, there must be 12 pentagons. I used the same method as Julia: Euler's formula together with a simple counting argument.One thing needs clarifying, to avoid confusion before we continue. While the non-convex shapes technically have eight corners and eight sides, I'm going to stipulate that an edge is the curve where two faces meet and a vertex is a point where three faces meet (or, equivalently for this shape, where three edges intersect). With vertices and edges defined this way, the non-convex shapes each have six edges and six vertices.
Let p=the number of pentagons on this surface, h=the number of hexagons, and n=the number of non-convex shapes. Then we have that f=p+h+n.
Since for each pentagon there are 5 vertices, for each hexagon there are 6 vertices, for each non-convex shape there are 6 vertices, and for each vertex there are 3 faces, the total number of vertices is v=(5p+6h+6n)/3.
Similarly, for each pentagon there are 5 edges, for each hexagon there are 6 edges, for each non-convex shape there are 6 edges, and for each edge there are 2 faces, the number of edges is e=(5p+6h+6n)/2.
Substituting these into the equation f+v-e=2 and simplifying leads us to conclude there are 12 pentagons.
From here, we note by observing the image, that there are two ways the non-convex shapes interact with other shapes. Each non-convex shape satisfies one of the following conditions:
(1) The non-convex shape shares an edge with 1 hexagon and 1 pentagon, or
(2) The non-convex shape shares an edge with 2 hexagons.
Let a=the number of non-convex shapes that satisfy (1). Let b=the number of non-convex shapes that satisfy (2). Note that a+b=n, since each non-convex shape satisfies exactly one of these two conditions.
For each non-convex shape satisfying (1), there is 1 pentagon and for each pentagon there are 5 non-convex shapes satisfying (1). Thus, we have the relationship 5p=a. So a=60.
For each hexagon, there are 2 non-convex shapes that satisfy (1), and for each non-convex shape satisfying (1) there is 1 hexagon. So 2h=a, hence, h=30.
Finally, for each hexagon there are 4 non-convex shapes that satisfy (2), and for each non-convex shape satisfying (2) there are 2 hexagons. So 4h=2b and b=60.
Therefore, there are 12 pentagons, 30 hexagons, and 120 non-convex shapes on this particular polyhedron created by Wenninger.
But that's just the start of her analysis: then things get more interesting!.
I got my tale of Wenninger from this article:
and the image from here:
© 2014 John Baez
baez@math.removethis.ucr.andthis.edu