## November 11, 2009

### This Week’s Finds in Mathematical Physics (Week 283)

#### Posted by John Baez In week283 of This Week’s Finds, see galaxies in visible, infrared, and ultraviolet light: Read the mystery of who discovered the icosahedron: Theaetetus or the ancient Scots. Read about Lieven le Bruyn’s detective work which helped solve this case! And learn an amazing fact, proved in the last book of Euclid’s Elements, which relates regular polygons with 5, 6, and 10 sides.

The Milky Way is big: But even this big picture is just a low-resolution version of an ultra-high-resolution image created by Axel Mellinger, suitable for use at planetariums.

Posted at November 11, 2009 7:05 AM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2108

### Re: This Week’s Finds in Mathematical Physics (Week 283)

There are a couple of nice pictures at the bottom of the webpage. Notice the Faces and Points alternating symmetry.

Solid …………….Dodecahedron…………Icosahedron
Face shape ………..Pentagon………………Triangle
Faces ………………..12 ……………………..20
Points ………………..20 ……………………..12
Edges ………………..30 ……………………..30

Posted by: Stephen Harris on November 11, 2009 11:57 AM | Permalink | Reply to this

### kids, hands-on, Geometry class; Re: This Week’s Finds in Mathematical Physics (Week 283)

I have a DVD showing me guiding a classroom of Latino high school students from a bad neighborhood, as they are each building the Platonic solids from paper, cutting, folding, gluing, and reporting by building this table on the whiteboard, and spontaneously discussing the duality and self-duality. Within an hour, they were heading towards Euler’s Polyhedral Formula. I tried this again in another predominantly 2nd generation working class Latino neighborhood classroom, for 2 hours, with some Archimedean solids also constructed, and they came closer to the elusive polynomial…

In my Master Teacher’s A.P. Calculus class, I showed some 3-D shapes that violated Euler’s Polyhedral Formula – two tetrahedra sharing a vertex, two cubes sharing an edge, a cube with a square hole cut through it between opposite faces. I asked WHY the formula was violated? Collectively they gave several correct equivalent answers.

Posted by: Jonathan Vos Post on November 12, 2009 1:25 AM | Permalink | Reply to this

### Re: kids, hands-on, Geometry class; Re: This Week’s Finds in Mathematical Physics (Week 283)

Your post is very redeeming and passed by my scant knowledge. The website I linked to did have the option to print out a patterned piece of paper that could be used to construct either the dodecahedron or icosahedron. Of the five Platonic solids, I think only the tetrahedron is self-dual.
I felt a hint of optimism after reading your post; it’s hard to believe how the decline in high school graduates was allowed to happen. Fibonacci lives!

Posted by: Stephen Harris on November 12, 2009 2:49 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Any idea whether Platonic solids were independently discovered outside Europe?

Posted by: Mark Meckes on November 11, 2009 3:05 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Good question!

Of course there are two questions here: whether particular Platonic solids were discovered, and whether the idea of the Platonic solids was discovered. I imagine various different cultures did or did not discover cubes, tetrahedra, etcetera — but I’ve never heard of any culture not influenced by the Greeks becoming interested in the idea of solids made of regular polygons with the same number meeting at each vertex. As I mentioned, in Greece the ‘discovery of the octahedron’ probably consisted of the discovery that there was something interesting about the octahedron.

My wife has pointed out that Chinese mathematicians did not have a word for ‘triangle’ until very late! They knew and used a version of the Pythagorean theorem long before having a word for triangle. They had a word meaning roughly ‘hypotenuse of a right triangle’, but no word for triangle!

This could have made it more difficult to formalize concepts like ‘regular polygon’ or ‘Platonic solid’.

• Lisa Raphals, A ‘Chinese Eratosthenes’ reconsidered: Chinese and Greek calculations and categories, East Asian Science, Technology and Medicine 19 (2002), 10–61.
Posted by: John Baez on November 11, 2009 5:18 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

read an article recently on research on how words effect perception - e.g. inlanguages where nouns can be masculine, feminine or neuter

will see if I can locate the article

Posted by: jim stasheff on November 11, 2009 7:27 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

The “fact we should have learned in high school, but probably never did”, that $cos(2\pi/5) = \phi/2$, appears in the Project Mathematics! videos produced by Tom Apostol and company at Caltech. So, while we didn’t see it in school proper, a few of us got to learn about it on the NASA channel.

Posted by: Blake Stacey on November 11, 2009 11:10 PM | Permalink | Reply to this

### Caltech/Tom Apostol update; Re: This Week’s Finds in Mathematical Physics (Week 283)

Sometime after he became Professor Emeritus and stopped teaching, Tom Apostol shut his Project Mathematics! stand-alone office, and moved back to a Math Department office at Caltech, where he comes several times a week. Snailmail and email about Project Mathematics! are still responded to. And he is writing more Math, publishing more, and winning more prizes for that than ever. Great man!

Posted by: Jonathan Vos Post on November 12, 2009 1:40 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

My dad points out that Alex Mellinger's zoomable map of the Milky Way has his name subtly and repeatedly embedded in it.

Posted by: Toby Bartels on November 12, 2009 5:01 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

And Jim Dolan points out that, despite all this repetition, I managed to get his name wrong!

Posted by: Toby Bartels on November 12, 2009 5:08 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I hope someone works on the puzzle at the end of this Week’s Finds. Part of the puzzle involves making nice pictures of things, but part of the puzzle involves proving things.

I’ve just expanded the puzzle a bit!

Posted by: John Baez on November 12, 2009 9:17 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

“I could explain it easily with lots of pictures” is a little off-putting. It sounds like it’ll need quite a bit of working out.

Any further hints? In those two triangles we have hypoteneuses equal and right angles. Are we to show one of the other angle pairs are equal?

Posted by: David Corfield on November 15, 2009 7:22 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Thanks for having the guts to respond, David! It’s useful to hear that my remark was off-putting. I was feeling very sad that nobody wanted to talk about this stuff… except the number theorist Kevin Buzzard, who wrote:

Actually, I thought that was the best bit ;-) I learned at school the fact that you’re presenting here about the right triangle (with the crappy algebraic proof). I’d never seen the icosahedral proof though!

It wasn’t supposed to be off-putting. I was just trying to say that all this stuff is vastly easier to draw or see than to explain in words. When you see what I’m saying, it’s all perfectly obvious and delightful.

If you take a picture of an icosahedron, like this: you see there’s:

• 1 vertex on top,
• and then an “upper pentagon” made of the 5 vertices next to the top one,
• and then a “lower pentagon” made of the 5 vertices below those,
• and then 1 vertex on bottom.

Next, pick a vertex from the upper pentagon: call this A. Pick a vertex as close as possible from the lower pentagon: call this B. A is not directly above B. Drop a vertical line down from A until it hits the horizontal plane on which B lies. Call the resulting point C.

Draw the triangle ABC. You can just draw it with a magic marker on your computer screen… or print out the picture and draw on that.

Next: do you see why ABC is a right triangle?

If so, it’s easy to see that

$AC^2 = AB^2 + BC^2$

And then comes the magic part:

• the length AB equals the edge of a pentagon inscribed in a circle;
• the length AC equals the edge of a hexagon inscribed in a circle;
• the length BC equals the edge of a decagon inscribed in a circle.

Different circles, but of the same radius! What’s this radius? Take all 5 vertices of the “upper pentagon”. These lie on a circle, and this circle has the right radius.

Why is this stuff true?

First of all: do you see that the length AB equals the edge of a pentagon inscribed in a circle? This is pretty easy.

Do you see that BC equals the edge of a decagon inscribed in a circle of the same radius? This is slightly harder, but not very hard.

So the hard part — and the fun part — is seeing that AC equals the length of a hexagon inscribed in a circle of the same radius!

We can talk about that later. I just want to see if anyone is following me so far…

Posted by: John Baez on November 15, 2009 8:19 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Maybe those who have solved it don’t want to give it away for those who haven’t?

Posted by: John Armstrong on November 15, 2009 11:34 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I spent a few hours trying to find a geometric proof, with no luck at all. I didn’t think my failure was worth a comment, though!

If John wants to disclose his proof, I’m happy to try to illustrate it.

The main sticking point for me was trying to do something that depended on the “pentagon-ness” of the pentagons. Many of the constructions you can perform with a set-up like this can be mimicked on generalisations of the icosahedron where the pentagons are replaced by $n$-gons (i.e. glue two pyramids with $n$-gon bases to the two $n$-gon faces of a suitable antiprism). But obviously the identity only holds when $n=5$, so all such constructions are useless! I tried to incorporate the construction of the pentagon (e.g. as shown in Wikipedia) into the process, but I just kept ending up with various line segments that I knew where congruent, but whose congruence I could only prove by resorting to algebra.

Posted by: Greg Egan on November 16, 2009 2:25 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

O P T C D F S V

I don’t want to clutter this thread up with red herrings, but for the sake of illustration here’s the start of one approach I took that didn’t work.

We construct the pentagon as follows: draw a diameter of the green circle through F and the centre O, then construct a perp. bisector, OP, of that diameter. Bisect OF to find the point C, then construct D such that CD=CP. The length of each side of the pentagon is then PD, which we can transfer to the first actual side, PT.

We can then put a perpendicular bisector through PT to find PS as one side of a decagon.

Now what the identity is telling us is that PS is congruent to OD, because OD is one side of the right triangle OPD, OP is the radius of the circle, and PD is the same length as the pentagon side. Somehow, PS=OD must be implicit in the construction here. But I don’t know how to show it without using any algebra!

So I’m looking forward to JB revealing what we can do with an icosahedron that does work …

Posted by: Greg Egan on November 16, 2009 3:05 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Greg wrote:

If John wants to disclose his proof, I’m happy to try to illustrate it.

Okay, great. Especially in the new improved version of This Week’s Finds, I actually tried as hard as I could to disclose the proof without writing something long and confusing. The proof is pretty clear using pictures, and I didn’t want to make it seem scary by giving a long complicated explanation in words.

So, for starters, it would be great if you could draw the points that I called A, B and C back here. It should then be obvious that they form a right triangle.

Next, I claim that:

• The segment AB is the edge of a regular pentagon formed by 5 vertices of the icosahedron. These 5 vertices lie on a circle of some radius $R$.
• The segment BC is the edge of a regular decagon inscribed in some other circle of radius $R$. This circle contains 5 other vertices in the dodecahedron: those of the ‘upper pentagon’.

Pictures illustrating these two points should make them obvious too!

So the only hard part is that:

• The length of the segment AC equals the edge of a regular hexagon inscribed in a circle of radius $R$.

But as I pointed out in week283, the hexagon is — sadly — a red herring. Or so it seems to me! I can’t find a useful hexagon here. So instead, note that the edge of a regular hexagon inscribed in a circle of radius $R$ has length $R$. It’s thus enough to show:

• The length of the segment AC equals $R$.

You may know all this, since it was in week283 before I added any extra clues. But anyway…

To show this fact, it seems helpful to note that the triangle ABC is congruent to another triangle I call $A B C$.

To form this other triangle, let $A$ be any vertex on the upper pentagon, and let $B$ be the top vertex of the icosahedron. Drop a vertical line from $B$ until it hits the plane of the upper pentagon; call the point where it hits $C$.

I think we’ll be able to see that ABC is congruent to $A B C$. Given this, we’re done once we check that:

• The length of the segment $A C$ equals $R$.

And this is really obvious, since $A C$ is a radius of the circle containing the 5 vertices of the upper pentagon!

By the way, none of these ideas are new. I got the idea of the triangle $A B C$ from Mueller’s book Philosophy of Mathematics and Deductive Structure in Euclid’s Elements. In a footnote, he mentions that two other authors, Dijksterhuis (in 1929) and Neuenschwander (in 1975), claimed it’s ‘intuitively evident’ that ABC is congruent to $A B C$.

This annoyed me at first: it didn’t seem intuitively evident at all! After all, the two triangles are situated in the icosahedron in quite different ways. But then all of a sudden something popped, and it became intuitively evident.

Now it’s been almost a week, so it’s not intuitively evident anymore. But if you draw a picture, I think everyone will be able to see it, or at least argue about it.

Also by the way: while Mueller’s book is really great, the picture of an icosahedron he uses to illustrate this discussion is hideously distorted! It’s Fig. 7.10 on page 257. It’s useful but not geometrically accurate.

Posted by: John Baez on November 16, 2009 3:55 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283) OK, so why is the triangle $A B C$ congruent to the triangle $A B' C'$? It’s obvious that they’re both right triangles, and that $A B$ and $A B'$ are congruent, both being equal to a “pentagon side” aka “icosahedron edge”. But how do we show that either $B C$ is congruent to $B' C'$ or $A C$ is congruent to $A C'$?

Sorry to use primes rather than italics to distinguish the vertices of the second triangle, but I think that makes the distinction a lot easier to follow.

(And sorry this isn’t SVG, but it takes a lot of work to massage Mathematica’s 3D figures into well-behaved SVG. If we get the whole proof sorted out, I can always re-do the graphics later.)

Posted by: Greg Egan on November 16, 2009 4:15 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Gee, that was quick! Great picture!

I won’t try to say why $A B C$ is congruent to $A B' C'$. First of all, I’m having some trouble! Second, someone else may enjoy taking a crack at it.

But I’ll say this. I don’t think the argument should proceed by showing that $A C$ is congruent to $A C'$. It’s true… but it’s basically what this fiendish argument was designed to prove.

Remember, $A C'$ is the edge length of a regular hexagon inscribed in your blue circle. So, we need to show $A C'^2$ is the square of the edge of the inscribed pentagon ($A B^2$) plus the square of the edge of the inscribed decagon ($B C^2$). By your picture and the Pythagorean theorem, it’s enough to show $A C = A C'$.

But a slick way to show this would be to find some other method of proving that $A B C$ is congruent to $A' B' C'$.

Posted by: John Baez on November 16, 2009 4:38 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

To make your fourth paragraph correct you need to change ‘plus’ to ‘minus’, and did you mean to write $B C = B C'$ rather than $A C = A C'$, or were you just reiterating something from earlier?

Posted by: David Corfield on November 16, 2009 8:53 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Right, ‘plus’ should be ‘minus’.

… and did you mean to write $B C = B C'$ rather than $A C = A C'$, or were you just reiterating something from earlier?

I was just reiterating the whole argument from the beginning, explaining why we wanted to prove $A C = A C'$ — and thus, why I want these two right triangles to be congruent, since that’s one way to prove this.

But my point was that if our proof that the triangles are congruent involves first showing $A C = A C'$, we can just skip this proof and use the fact that $A C = A C'$!

Posted by: John Baez on November 16, 2009 6:07 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283) This picture is purely to satisfy my curiosity as to how things look if we apply the rotational isometry of the icosahedron that leaves the vertex $A$ fixed and takes $B$ into $B'$. This rotation maps $C$ to $C''$, and takes the blue circle into the red circle.

(I don’t see how this gets us anywhere useful, but it’s the kind of obvious thing that if you don’t try, you’ll just keep wondering about until you do …)

Posted by: Greg Egan on November 16, 2009 11:48 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Well, it looks like you’ve got two sides of a hexagon between CA and AC”…

Posted by: John Armstrong on November 16, 2009 12:41 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Well, it looks like you’ve got two sides of a hexagon between CA and AC”…

No, the angle between them isn’t correct (I’m cheating and using coordinates, but it’s a fast way to check).

Posted by: Greg Egan on November 16, 2009 1:00 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m glad you folks are trying different things. I think I’d just fooled myself into thinking I could prove the congruence of the triangles $A B C$ and $A' B' C'$, due to a mixture of wishful thinking and inaccurate mental visualization.

Of course we can always ‘cheat’ and see how someone solved this problem around 300 BC! Not the congruence of these triangles, but the result for which this is supposed to be a lemma:

Book XIII Proposition 10: If an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.

The first figure looks a lot like Greg’s first attempt! I believe the historians think Euclid thought about the icosahedron but proved this result using only 2d methods. Mueller calls this 2d proof a tour de force.

Posted by: John Baez on November 16, 2009 6:20 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Ah, here’s something a little bit interesting about the diagram above: the angle between the planes of the triangles $A B' C'$ and $A B' C''$ seems to be exactly $\frac{\pi}{2}$. (But don’t confuse this with the angle $C' B' C''$, which is an irrational multiple of $\pi$.)

Now, this is something I haven’t proved according to the rules we’re playing by; I just gleaned it by cheating and doing calculations with the coordinates of the points. But maybe someone else can see if this “conjecture” is provable by Euclidean methods, and/or helpful to our ultimate goal.

Posted by: Greg Egan on November 16, 2009 12:52 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I just gleaned it by cheating and doing calculations with the coordinates of the points. But maybe someone else can see if this “conjecture” is provable by Euclidean methods

It's certainly provable by Euclidean methods, since we can use those methods to set up a coordinate system. But we want a reasonably simple proof! (^_^)

Posted by: Toby Bartels on November 17, 2009 5:45 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m not sure what you mean. Euclidean methods are distinctly weaker than coordinate calculations. You can use coordinates to trisect an angle, for instance.

Posted by: John Armstrong on November 17, 2009 7:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

You can use coordinates to trisect an angle, for instance.

Did Euclid consider it unprovable that for every angle there exists an angle of one third measure? H'm, maybe he did; he probably would never have written about it …

Posted by: Toby Bartels on November 17, 2009 7:33 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283) Here’s another picture, just to make it easier to talk about some things.

The point $D$ is such that $D C'$ and $D C''$ are altitudes of the triangles, when their hypotenuses are taken as their bases, i.e. by choice of $D$, the angles $B' D C'$ and $B' D C''$ are right angles.

I know for sure – but haven’t proved in Euclidean fashion – that $C' D C''$ is a right angle.

Clearly it would be enough to meet our goal if we could prove that $D C'$ is congruent to $D C''$. But of course we’re not yet allowed to use the right angle at $C' D C''$ to help in a proof of that.

Posted by: Greg Egan on November 16, 2009 1:28 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

As Thom is reputed to have said:
Very easy to see, very hard to prove.

Posted by: jim stasheff on November 16, 2009 1:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m glad you folks are trying different things. I’m afraid I’d just fooled myself into thinking I could prove the congruence of the triangles $A B C$ and $A' B' C'$, due to a mixture of wishful thinking and inaccurate visualization.

Of course we can always ‘cheat’ and see how someone solved this problem around 300 BC! Not the congruence of these triangles, alas, but the result for which that could be a lemma:

Book XIII Proposition 10: If an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.

The figure illustrating Euclid’s proof looks a lot like Greg’s first attempt! I believe historians think this Proposition was discovered by contemplating the icosahedron, but Euclid proved it using only 2d methods, which were considered more rigorous.

Mueller calls Euclid’s 2d proof a tour de force.

Posted by: John Baez on November 16, 2009 7:02 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

A B C D E G K M F H L N 108 ° 36 ° 18 ° 18 ° 18 ° 54 ° 18 ° 54 ° 72 ° 18 ° 72 ° 54 ° 72 °

SPOILER WARNING! If you want to figure out the two-dimensional proof for yourself, DO NOT read on.

Here’s a version of the proof Euclid gave, adapted from the version JB cited. Rather than proving that various angles here are identical, I’ve just written in the (easily established) numerical values; there’s nothing tricky here, so we might as well take them as given.

Triangle $A B F$ is similar to triangle $B F N$. So $\frac{A B}{B F}=\frac{B F}{F N}=\frac{B F}{B N}$, with the last equality true because the triangles are isosceles with $F N = B N$. Thus ${B F}^2={A B}\cdot{B N}$

Triangle $B A K$ is similar to triangle $K A N$. So $\frac{B A}{A K}=\frac{K A}{A N}$. Thus ${A K}^2={A B}\cdot{A N}$.

Adding our two results, we have: ${B F}^2+{A K}^2={A B}\cdot(A N + B N)={A B}^2$.

$B F$ is our radius, $A K$ is a decagon side, and $A B$ is a pentagon side. Well done Euclid.

Posted by: Greg Egan on November 17, 2009 1:04 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

So what does it look like on the icosahedron?

Posted by: John Armstrong on November 17, 2009 1:57 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

So what does it look like on the icosahedron?

I’m not sure what you’re asking. We could use this 2-dimensional result to take the thing we hoped to prove as given, which then tells us $A B C$ is congruent to $A B' C'$, and that the distance between the two pentagons in the icosahedron (aka the height of a pentagonal antiprism) is equal to the radius of those pentagons.

But as far as I can see, it gets us no closer to proving that congruence of triangles directly on the icosahedron. I’m growing a bit sceptical that there’s really any kind of “short cut” proof involving the icosahedron. Has anyone actually seen such a proof – or have various people just declared that the congruence of the triangles is “intuitively obvious”?

Posted by: Greg Egan on November 17, 2009 3:47 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Great picture and explanation of Euclid’s tour de force, Greg!

Greg wrote:

I’m growing a bit sceptical that there’s really any kind of “short cut” proof involving the icosahedron. Has anyone actually seen such a proof – or have various people just declared that the congruence of the triangles is “intuitively obvious”?

I haven’t seen such a proof. What’s unquestionable is that Euclid used Prop. XIII.10 as part of his construction of the icosahedron in Prop. XIII.16. This makes it quite plausible that Euclid was thinking about the icosahedron when proving Prop. XIII.10. But I don’t see anyone claiming — except me, for while — that the icosahedron gave a quick and easy way to prove Prop. XIII.10, instead of just ‘see it’ or realize that it was an interesting issue. I was hoping that I could ‘see it’ in some way that was more convincing that just ‘seeing it looks approximately true’, yet not easy to formalize using Euclid’s methods.

Mueller, who seems very careful, says that Eva Sachs, in her important Die fünf platonischen Körper said an accurately drawn figure could let someone guess that $AC = AC’$. And that’s believable.

He also says Dijksterhuis and Neuenschwander claimed it’s ‘intuitively evident’ that ABC is congruent to ABC. But since I haven’t read their papers, I don’t know what that’s supposed to mean, exactly.

I’m sad to say that I haven’t gone through Euclid’s proof of Prop. XIII.16. This would probably be the next step, if I had the energy. Unfortunately I’m a bit distracted by other fun stuff!

Posted by: John Baez on November 18, 2009 3:31 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

OK, if the hypothesis is, roughly, that Euclid guessed that $A C = A C'$ on the icosahedron, and that this motivated him to prove $P^2=D^2+H^2$ – enabling him to rigorously construct the icosahedron and rigorously prove what he’d intially just guessed – then that sounds entirely plausible.

Everything else was a bit of a wild goose chase, but at least the scenery was interesting.

Posted by: Greg Egan on November 18, 2009 4:49 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m sorry if it was a wild goose chase. But I still have hopes that there’s a nice proof using the icosahedron, not necessarily using Euclidean methods. Given how ingenious Euclid’s 2d proof was, a good 3d proof might take a few mathematician-weeks to discover. Having the puzzle out there in cyberspace for everyone to see increases the chance that if there’s any insight to be had, someone will have it.

By the way, Euclid’s construction of the icosahedron in Prop. XIII.16 actually mentions a hexagon (as well as a pentagon and decagon). But I don’t understand it.

Posted by: John Baez on November 18, 2009 5:32 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

O L M N P r r a b s s t 36 ° 36 ° 72 ° 72 °

After ungraciously calling JB’s puzzle “a wild goose chase”, I should add that it’s enjoyably addictive nonetheless. And here’s my best shot at a concise proof that the distance between the pentagons in an icosahedron equals their radius.

In the figure above, $L N = 2 s$ is one side of a pentagon inscribed in a circle of radius $r$. Now, Euclid certainly knew that for an isosceles “Golden Triangle” such as $P O N$, the ratio of the long sides to the base was $\Phi\approx 1.618$, where I’m using the “big $\Phi$” that satisfies $\Phi-1=\frac{1}{\Phi}$. So we have $t=\frac{r}{\Phi}$.

We can express $s^2$ two ways, from the two right triangles that $M N$ belongs to: $s^2 = t^2-b^2=r^2-a^2$.

This gives us $s^2 = \frac{r^2}{\Phi^2}-b^2=r^2-(r-b)^2$, and we can easily solve the second equality to get $b=\frac{r}{2\Phi^2}$.

Now, the height $h$ that separates the two pentagons in the icosahedron arises from an equilateral face of the icosahedron – with an edge length of $2 s$ and hence an altitude whose square is $3 s^2$ – being tilted in such a way that its horizontal displacement is just our $b$: the difference between the radius of the circle and the distance from the centre to the midpoint of a pentagon edge.

So we have $h^2=3s^2-b^2=3\frac{r^2}{\Phi^2}-4b^2=r^2 \frac{3\Phi^2-1}{\Phi^4}$.

We can factor $\Phi ^4-3 \Phi ^2+1=(\Phi ^2-\Phi -1) (\Phi ^2+\Phi -1)$, and the first factor is zero by the defining equation of $\Phi$, which means the fraction is $1$ and $h^2=r^2$.

Posted by: Greg Egan on November 18, 2009 3:30 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

In the diagram here we can get $b$ even faster just by noting that the angle at $P N M$ is 18°, so $b$ is the base of half a golden triangle with side $t=\frac{r}{\Phi}$, hence $b=\frac{t}{2\Phi}=\frac{r}{2\Phi^2}$.

It’s also very easy to get from here to $P^2=D^2+R^2$, if we use an identity that’s easily derived from the defining equation for $\Phi$.

$\Phi-1=\frac{1}{\Phi}$

$\Phi^2-\Phi=1$

$\Phi=\Phi^2-1$

$\frac{1}{\Phi^2}=(\Phi-1)^2=\Phi^2-2\Phi+1=\Phi^2-2(\Phi^2-1)+1=3-\Phi^2$

This gives us the soon-to-be-useful formula:

$4b^2=\frac{t^2}{\Phi^2}=t^2(3-\Phi^2)$

Now, the pentagon side is $P=2s$ and the decagon side is $D=t$, so, using $s^2=t^2-b^2$, we have:

$P^2-D^2=4s^2-t^2=4t^2-4b^2-t^2=3t^2-4b^2=\Phi^2t^2=r^2$

Posted by: Greg Egan on November 20, 2009 5:05 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Greg wrote:

Now, Euclid certainly knew that for an isosceles “Golden Triangle” such as PON, the ratio of the long sides to the base was $\Phi \approx 1.618$,…

Yes, this is Prop. XIII.9, which comes right before the one we’re struggling with!

It’s phrased a bit obscurely, at least to modern eyes:

If the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.

‘Extreme and mean ratio’ means ‘golden ratio’. And the side of the hexagon is just the radius of a circle. So this proposition is equivalent to: if a regular decagon is inscribed in a circle of radius $r$, its sides have length $r/\Phi$. And this is equivalent to what you’re saying.

By the way: readers will note that Greg is using $\Phi = 1.618...$ while I was using $\phi = 1/\Phi = 0.618...$. They’re both useful and they’ve both been called the golden ratio. But you’ll notice that Euclid doesn’t use the term ‘golden ratio’ — he uses ‘extreme and mean ratio’.

Indeed, if anyone out there thinks that ancient Greeks ran around in togas philosophizing about the “golden ratio” and calling it “$\Phi$”, they’re wrong. This number was named $\Phi$ after the Greek sculptor Phidias only in 1914, in a book called The Curves of Life by the artist Theodore Cook. And, it was Cook who first started calling $1.618...$ the golden ratio. Before him, $0.618...$ was called the golden ratio! Cook dubbed this number “$\phi$”, the lower-case baby brother of $\Phi$.

In fact, the whole “golden” terminology can only be traced back to 1826, when it showed up in a footnote to a book by one Martin Ohm, brother of Georg Ohm, the guy with the law about resistors. Before then, a lot of people called $\phi$ the “Divine Proportion”. And the guy who started that was Luca Pacioli, a pal of Leonardo da Vinci who translated Euclid’s Elements. In 1509, Pacioli published a 3-volume text entitled Divina Proportione, advertising the virtues of this number. Some people think da Vinci used the divine proportion in the composition of his paintings. If so, perhaps he got the idea from Pacioli. But nobody is sure.

Maybe Pacioli is to blame for the modern fascination with the golden ratio; it seems hard to trace it back to Greece. These days you can buy books and magazines about “Elliot Wave Theory”, a method for making money on the stock market using patterns related to the golden number. Or, if you’re more spiritually inclined, you can go to workshops on “Sacred Geometry” featuring talks about the healing powers of the golden ratio. But Greek texts seem remarkably quiet about this number.

The first recorded hint of it is Prop. II.11 in Euclid’s Elements. It also shows up elsewhere in Euclid, especially Prop. VI.30, where the task is “to cut a given finite straight line in extreme and mean ratio”, meaning a ratio $A:B$ such that

$A:B::(A+B):A$

i.e., “$A$ is to $B$ as $A+B$ is to $A$”.

(Useless erudition recycled from week203.)

Posted by: John Baez on November 18, 2009 4:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

this is just a wild guess, but a possible candidate for your sought hexagon may be here:
call what you called upper and lower pentagonal circles pentagonal opposites. through your point A there are two more pentagonal circles besides the upper pentagonal circle. draw a perpendicular from A to the pentagonal opposite of each of the remaining two pentagonal circles through A. the two perpendiculars give two possible sides of a hexagon with the right radius. let A’ be the point on the lower pentagon, which is in your image icosahedron.jpg below A (i.e. in the back). do the same construction as before. one has now four possible sides of a hexagon. the big question is now wether the connecting segments between the ends of the two twosegmentpolygonallines have the right length. as said its just a wild guess.

Posted by: nobody on November 18, 2009 8:31 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Nobody, all the perpendicular lines joining opposite pentagons are parallel to some “axis” of the icosahedron, i.e. the lines joining pairs of opposite vertices. But no two axes lie at 60° or 120° to each other, as they would have to if any subset of the perpendiculars were the edges of a regular hexagon.

If you normalise all the vertex coordinates to a length of 1, each vertex $A$ has dot products with the others of either $\frac{1}{\sqrt{5}}$ (five times, for the pentagon that forms the base of a pentagonal pyramid with $A$ at the apex), $-\frac{1}{\sqrt{5}}$ (five times, for the pentagon opposite the first) or $-1$ (for the vertex directly opposite $A$).

Posted by: Greg Egan on November 18, 2009 9:54 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

in my above coment 2 of the vertices (A and A’) of the “virtual” (and probably nonexisting) hexagon would also be vertices of the icosahedron the other 4 would be vertices on the respective pentagonal decagons. it would be interesting to know what is the angle between two triangles of the icosahedron, which meet in a vertex A but which are not adjacent?

i have no 3-D applet that works. with an applet one could may be check immediately. in the drawings i cant see wether the two two-segments-lines lie in the same plain (meanwhile I have rather the impression they are not). if one could turn the object then one could see this probably at first sight.

Posted by: nobody on November 19, 2009 3:27 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Nobody, the construction you’ve described can’t be part of a regular hexagon. The line segments you have meeting at $A$ meet at an angle of $cos^{-1}(-\frac{1}{\sqrt{5}})\approx 116^{\circ}$, not $120^{\circ}$. Also, the distance between your opposite vertices, $A$ and $A'$, is the diameter of the sphere in which the icosahedron is inscribed, not the diameter of the circle in which each pentagon is inscribed.

Posted by: Greg Egan on November 19, 2009 6:46 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

“A and A’, is the diameter of the sphere in which the icosahedron is inscribed”

oops. true. i should have seen this. :(

Posted by: nobody on November 19, 2009 9:38 AM | Permalink | Reply to this

### Thanks also to my subcontractor da Vinci…; Re: This Week’s Finds in Mathematical Physics (Week 283)

Not looking at my notes from when I taught Math to Art grad students, Pacioli got da Vinci to draw the illustrations for him!

Leonardo da Vinci (1452-1519) was the quintessential renaissance man: artist, mathematician, scientist, and engineer. He was a great lover of geometry, and devoted much time to it starting in his early forties. His most outstanding polyhedral accomplishment is the illustrations for Luca Pacioli’s 1509 book The Divine Proportion. At right is one of the illustrations from that book. The term Ycocedron Abscisus in the title plaque means truncated icosahedron, and the term Vacuus refers to the fact that the faces are hollow. (The drawings are beautifully hand colored like this in the Ambrosiana manuscript, reprinted by Officina Bodoni, 1956, and also by Silvana Editoriale, 1982.)

These are the first illustrations of polyhedra ever in the form of “solid edges.” The solidity of the edges lets one easily see which edges belong to the front and which to the back, unlike simple line drawings where the front and back surfaces may be visually confused. Yet the hollow faces allow one to see through to the structure of the rear surface. This is a brilliant new form of geometric illustration, one worthy of Leonardo’s genius for insightful graphic display of information. However, it is not clear whether Leonardo invented this new form or whether he was simply drawing from “life” a series of wooden models with solid edges which Pacioli designed. If Pacioli designed these models, then he deserves the credit for this new “solid edge” idea, but it is likely that Leonardo designed them.

Posted by: Jonathan Vos Post on November 20, 2009 7:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

BTW, the fact that the height of the pentagonal pyramid sitting on top of one of the pentagons (the length $B' C'$ in the diagram back here) is a decagon’s side length is about as easy to prove directly from the same calculations as the ones that show the distance between the pentagons. In this case, we want the altitude of an icosahedron’s face to be tilted so that it spans a distance $a$ in the diagram a couple of posts back, so if we call the height of the pyramid $H$ and use $s^2=r^2-a^2$:

$H^2=3s^2-a^2=3r^2-4a^2=r^2(3-4(1-\frac{1}{2\Phi^2})^2)=r^2(\frac{-\Phi ^4+4 \Phi ^2-1}{\Phi ^4})$. The numerator minus $\Phi^2$ is $-\Phi ^4+3 \Phi ^2-1=-(\Phi ^2-\Phi -1) (\Phi ^2+\Phi -1)=0$, so the numerator itself equals $\Phi^2$.

So $H^2=t^2$, where $t=\frac{r}{\Phi}$ is a decagon’s side length.

Posted by: Greg Egan on November 18, 2009 10:21 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

O L M N P r r b s s t 36 ° 36 ° 72 ° 18 ° 54 ° r = t Φ t 2 b=t/Φ 2 b=t/Φ r = t Φ 2 b=t/Φ t

By the Golden Triangle rule for isosceles triangles with angles of 36°, triangle $N O P$ gives us $r=t\Phi$. From the same rule for triangle $N M P$ glued to its mirror image in the line $N M$ we have $2b=t/\Phi$.

From the right triangle $N M P$ we have $s^2=t^2-b^2$, hence $(2s)^2=4t^2-4b^2$.

The whole large multi-coloured rectangle has an area of $(2s)^2$, because the green square, the light blue rectangle, and each of the two red rectangles all have areas of $t^2$, but then we need to subtract the light red square where the two red rectangles overlap, and that has area $4b^2$.

Now if we subtract a single $t^2$, the light blue rectangle, then we’re left with a square of side length $r$, thanks to the fact that $\Phi=1+1/\Phi$. So $(2s)^2-t^2=r^2$, which is the pentagon-decagon-hexagon identity.

Posted by: Greg Egan on November 21, 2009 1:33 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Wow! A proof Euclid could have found! But this is the proof — the one that really explains why the result is true!

Posted by: John Baez on November 21, 2009 5:17 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

But this is the proof — the one that really explains why the result is true!

The proof from the book?

Posted by: Toby Bartels on November 21, 2009 7:52 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

It really puzzled me that Euclid proved the Golden Triangle result, then didn’t use it for this! Maybe he thought it was “purer” somehow to exhibit a proof that didn’t need it as a lemma. His proof in the Elements is really dazzling, but it made me feel like he’d pulled a coin out from behind my ear.

I bet someone (maybe in the 15th century or later, if the Greeks really weren’t that into $\Phi$), has done a whole lot of diagrammatic versions of identities obeyed by $\Phi$, including the one I used for this proof, which corresponds to $\Phi^2+\frac{1}{\Phi^2}=3$ … categorifying all the equations that follow from $\Phi-1=\frac{1}{\Phi}$.

Posted by: Greg Egan on November 21, 2009 11:13 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Greg wrote:

It really puzzled me that Euclid proved the Golden Triangle result, then didn’t use it for this!

Mueller sketches a version of Euclid’s proof of Prop. XIII.10 (the pentagon-decagon-hexagon identity). His sketch uses the fact that $F A K$ is a golden triangle. But then he says:

Euclid’s reasoning differs from that just described in one significant respect. He does not use the fact that $F A K$ is an isosceles triangle with vertex angle of 36°, and he shows no sense of the relationship of XIII.9 to material proved in connection with the inscription of the regular pentagon.

Later:

In this case [the proof of Prop. XIII.9] Euclid’s failure to exploit a connection with work already done in book IV may be due to a failure to recognize the direct connection between the regular decagon and the isosceles triangle with vertex angle of 36°. This failure may, of course, be taken as further evidence for Euclid’s slavish following of independent sources. However, it is also possible to suppose that Euclid chose to prove the minimum number of results needed for the inscription of the regular pentagon and accordingly did not put that material in a form directly usable in book XIII. In any case, there is always a danger in historical interpretation in mistaking the obvious to us for the eternally obvious.

I think that there’s a lot of debate over to what extent Euclid was a ‘mere systematizer’ of work that had been done before. We know incredibly little about him!

Of course, even if he was a ‘mere systematizer’, his accomplishments may have been huge! Developing the axiomatic method, for example.

Posted by: John Baez on November 22, 2009 3:38 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Having invoked “the Golden Triangle rule” a few times, I should probably offer a proof of it. This is incredibly simple; I’ve just adapted the proof of Euclid’s Prop. XIII.9 that JB cited earlier.

E A B C D r t r r 72 ° 36 ° 72 ° 36 ° 36 ° 108 °

Triangle $B D E$ is similar to triangle $B E C$. So:

$\frac{B D}{B E}=\frac{B E}{B C}$

$\frac{r+t}{r}=\frac{r}{t}$

$1+\frac{t}{r}=\frac{r}{t}$

But the definition of the Golden Ratio is that $\Phi$ is the (positive) solution of:

$1+\frac{1}{\Phi}=\Phi$

So $\frac{r}{t}=\Phi$, as is the side-to-base ratio for any isosceles triangle with angles 72°-36°-72°.

Posted by: Greg Egan on November 21, 2009 7:59 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

1 Φ Φ 1 A B’ –A –B’ C’ P Q R O

Suppose we construct an icosahedron based on three orthogonal “duads” (see John’s page “Some Thoughts on the Number Six”). These are rectangles in the golden ratio, with vertices of the icosahedron at their corners. In the diagram above, I’ve projected everything into the plane of one duad, which has the vertices $A$, $B'$, $-A$ and $-B'$ as its corners. Where two vertices are projected to the same point in the diagram, I’ve drawn a circle around the point.

It’s obvious that all vertices will be equidistant from the centre of symmetry, and not too hard to prove that all edge lengths arising from this construction are equal. (Here I’ve normalised the edge length to 1.) So it’s not difficult to see that we really do have an icosahedron here.

The right triangle $O R P$ is similar to the right triangle $C' A B'$. So $\frac{O R}{O P}=\Phi=\frac{C' A}{C' B'}$. But $C' A$ is just the radius of our circle in which the pentagon is inscribed, so this tells us, by the Golden Triangle rule, that $C' B'$ is a decagon edge length.

And of course $A B'$ is a pentagon edge length, so we’ve proved the pentagon-decagon-hexagon identity.

Posted by: Greg Egan on November 22, 2009 2:40 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Here’s another view of the “duads” mentioned here, making it a bit clearer how they fit into the icosahedron.

Posted by: Greg Egan on November 22, 2009 11:56 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

This view shows off the regular hexagon hiding in the icosahedron. I keep finding myself wanting to use this to prove Prop. XIII.10. But its edge length equals that of the obvious regular pentagons in the icosahedron. So, instead of a regular pentagon and regular hexagon both inscribed in circles of the same radius, nature gives us a regular pentagon and a regular hexagon with the same edge length, together with a regular decagon inscribed in the same circle that the pentagon is inscribed in.

Posted by: John Baez on November 22, 2009 5:17 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

John Baez wrote:

This view shows off the regular hexagon hiding in the icosahedron. I keep finding myself wanting to use this to prove Prop. XIII.10. But its edge length equals that of the obvious regular pentagons in the icosahedron.

I don’t think the edge length of the hexagon here equals the pentagon’s edge length. The edges that form the hexagon aren’t coplanar, and in this view they’re all foreshortened. The triangle in the centre of the view does have the pentagon’s edge length, though, because it’s orthogonal to the direction of the view.

Toby Bartels wrote:

somebody should note that Greg added this stuff to the Lab

Sorry, I got so caught up in that nLab entry that I forgot to link to it!

One thing I added in the nLab entry is a bit more detail, proving that the duad construction really does give a regular icosahedron. It’s not at all hard, and it involves the same identity in $\Phi$ as a lot of the other calculations:

$\Phi^2+\frac{1}{\Phi^2}=3$

That identity is also at the heart of the rectangle-area proof. It all boils down to the fact that if you take a square of length $1$ and two $\Phi \times \frac{1}{\Phi}$ rectangles, the sum of their areas, which is obviously $3$, can also be seen as the sum of two squares, one of side length $\Phi$ and the other of side length $\frac{1}{\Phi}$.

1 1/Φ 1/Φ Φ Φ 1 1/Φ
Posted by: Greg Egan on November 22, 2009 11:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Greg wrote:

I don’t think the edge length of the hexagon here equals the pentagon’s edge length. The edges that form the hexagon aren’t coplanar, and in this view they’re all foreshortened.

You’re right, of course. Posted by: John Baez on November 24, 2009 5:15 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Maybe I missed it, but somebody should note that Greg added this stuff to the Lab: pentagon decagon hexagon identity.

Posted by: Toby Bartels on November 22, 2009 6:05 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

1 Φ Φ 1 A B –A –B C P Q R S O T α α α 90–α

I just noticed, very belatedly, that this projection of the icosahedron makes it almost trivial to see that the distance between the pentagons, $S T$, equals their radius, $A C'$!

No mention of $\Phi$ is needed, let alone any calculations with it; it follows directly just from the fact that $Q R$ is parallel to $A B'$, and that both are orthogonal to $O P$. That’s enough to show that the triangle $A B' C'$ is congruent to triangle $S P T$.

Of course, you still need the calculations with $\Phi$ to prove that this duad construction yields a regular icosahedron. But I think merely realising that the icosahedron’s edges split up into orthogonal pairs gives you the fact that the distance between the pentagons equals their radius.

Posted by: Greg Egan on November 23, 2009 10:10 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Great! This is getting very close to my dream of a ‘quick icosahedral proof’!

Posted by: John Baez on November 24, 2009 2:21 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

If you rotate the icosahedron around the OA axis in the image above by Greg Egan (OA and B’ is assumed to be lying in the viewing plane) by an angle of 18 degrees, then the projection of the rotated point B’ and one of the double points at Q will have the same distance from OA if one uses the same unaltered viewing plane. One can convince oneself of this fact by looking at the pentagon in which QB’ lies. Call the projection of the rotated points Q1 and B1. Likewise one obtains points R1 and -B1. So after this rotation one has almost a hexagon formed by A, B1, R1, -A, -B1, Q1, however the sides AB1 and AQ1 (and their opposites) are still too large. One can now tilt the icosahedron in the plane in which OA lies and which is perpendicular to the viewing plane (thus it doesn’t in particular alter the distance of Q1 and B1, to the projection of OA (this distance should be h=ST*cos18°) in order to obtain a hexagon, which is so to say the outer shadow of the icosahedron. h is just one half of the distance between two overnext neighbours of this hexagon. If you use Johns parametrization (i.e. B’=(g,1,0) etc. then (no guarantee) the edge length of this hexagon should be 4/Sqrt(3)*(sqrt(g^2-1/4)/sqrt(g^2+1).

So if one argues that h is modulo a factor cos30^° the length of an edge of a hexagon which is the hexagon “shadow” of an icosahedron and observes that it is also modulo a factor the radius in a pentagon (again by looking at the icosahedron) then one has some interpretation of pentagon-decagon-hexagon identity in terms of the hexagonal shape of the shadow of the icosahedron.

Posted by: stubido on November 24, 2009 12:26 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Stubido, I agree that you can find a projection of the icosahedron such that the shadow is a regular hexagon. The diagram here is an example of that. (Actually that diagram is a tiny bit skewed from the perfect viewing angle, because it wasn’t designed deliberately to illustrate the hexagonal outline.)

I’m not clear, though, what ties this in any simple way to the pentagon-decagon-hexagon identity. The radius of the “shadow” hexagon differs from the radius of the circle in which the pentagon is inscribed; of course with enough work we can calculate the precise factor by which it differs … but I don’t understand what that actually gives us!

Posted by: Greg Egan on November 24, 2009 2:08 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Circle whose radius is icosahedron/pentagon edge length Sphere in which icosahedron is inscribed Circle in which shadow hexagon is inscribed Circle in which pentagon is inscribed

Here’s a projection of an icosahedron as a perfect regular hexagon, with various other things shown for comparison.

I’m not putting this forward as part of any argument; it just seems like a useful thing to be able to refer to, since it can be hard to keep straight what the relative sizes of these various circles are.

Posted by: Greg Egan on November 24, 2009 3:29 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

At first I was shocked that the red circle is smaller than the green one. I thought if you neatly fit an icosahedron in a sphere of glass, and looked at the icosahedron’s shadow — holding it so its shadow is a regular hexagon — that hexagonal shadow would fit tightly in the sphere’s circular shadow!

But I was wrong. It’s the same ‘foreshortening’ effect that tricked me before.

The 6 vertices of the icosahedron whose shadows form the hexagon’s corners don’t lie on a plane. They lie on 2 separate parallel planes: 3 form an equilateral triangle on one plane, 3 form an equilateral triangle on another. So, the true distance between opposite vertices is greater than the distance between their projections.

But only slightly!

Posted by: John Baez on November 24, 2009 5:31 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

The image above is correctly that what you get if you perform my above described rotations.

Greg Egan wrote: “of course with enough work we can calculate the precise factor by which it differs… but I dont understand what that actually gives us!”

What someone gets from what I wrote is I guess completely up to that someone.

It seems to me that the point about all this magic about g and the icosahedron arises basically from the fact that
***********
sin18°=1/(2g)
***********
(or likewise cos18°=1/2*sqrt(1+g^2)).
So e.g. it follows from r sin18° = t/2 that
r=t*g.
similarily the pentagon-hexagon-decagon identity can be restated in terms of trigonometric addition identities, as the pentagon edge length is 2*r*sin(2*18°).

So the rotation (described above) by 18° is crucial. It says that 1/2 sqrt(3) times the edge length of the shadow hexagon is r*cos18^° = r/2*sqrt(1+g^2)(which gives you what I wrote before if you take the above coordinates, where hopefully r= 2*g/(sqrt(g^2+1)). Moreover if one interprets the above pentagons as projections of a pentagon of an icosahedron then this r*cos18° is immediately visible as the projection of a radius of a pentagon.

John Baez wrote:”So, the true distance between opposite vertices is greater than the distance between their projections.”

Yes this is achieved by the above described rotation in the plane in which OA lies and which is perpendicular to the viewing plane.

Posted by: stubido on November 24, 2009 7:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Stubido, John and I are playing a slightly crazy game: we’re trying to see how someone who was a contemporary of Euclid might have reasoned about these things, using methods that resemble the proofs that were used in that time. So although we know the trigonometric definitions of the golden ratio, and all the algebraic consequences that follow from them, we’re trying to find different ways to derive the same results.

Posted by: Greg Egan on November 24, 2009 9:41 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I think we can assemble a chain of reasoning now that makes the pentagon-decagon-hexagon identity seem close to inevitable!

Suppose we take three congruent rectangles, and assemble them symmetrically so that their long edges are all mutually orthogonal, and their short edges are all mutually orthogonal. Like this:

We then join up the corners to make an icosahedron, as in the diagram above. Eight faces of this icosahedron (the ones centred on the diagonals of the eight “octants” into which the space here is divided) will always be congruent and equilateral; that’s guaranteed by the construction, whatever the precise ratio of the side lengths of the rectangles. The other twelve faces will be isosceles triangles, all congruent to each other.

Now, it’s clear that by making the rectangles long and skinny, we can make the isosceles triangles as long and skinny as we like. But what about making them short and squat? How far can we go in that direction?

If we make the three rectangles into squares, as above, the isosceles triangles have two 45° angles. So between this extreme, and the long-and-skinny extreme, it’s clear that we can choose some intermediate degree of skinniness for the rectangles that will turn the 12 isosceles triangles into equilateral triangles.

So, a regular icosahedron must exist, and it will have six edges that form three mutually orthogonal pairs.

Having reached that conclusion, we can now use the construction below, where we project everything into the plane of one of our rectangles.

1 Φ Φ 1 A B –A –B C P Q R S O T α α α 90–α

We can’t yet justify the specific value, $\Phi$, that we’ve written here as the length of the rectangles … but we don’t need to! Without any calculations that depend on $\Phi$, we can see that $Q R$ is parallel to $A B'$, and that both are orthogonal to $O P$. That’s enough to show that the triangle $A B' C'$ is congruent to triangle $S P T$, and hence the distance between the pentagons, $S T$, equals their radius, $A C'$.

Now we can appeal to John’s original construction: $A B$ is a pentagon edge, $B C$ is a decagon edge, and $A C = S T = A C'$ is the radius of the circle in which the pentagon and decagon are inscribed. So the congruent right triangles $A B' C'$ and $A B C$ constitute proofs of the pentagon-decagon-hexagon identity.

And then, as a final bonus, we can simply appeal to Euclid’s golden triangle lemma – which gives us the ratio between the radius and the decagon edge length as $\Phi$ – and work backwards to show that this is also the ratio of length to width for our three orthogonal rectangles. This follows simply from the fact that the right triangle $O R P$ is similar to the right triangle $C' A B'$.

Posted by: Greg Egan on November 25, 2009 2:56 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Wow! This is great!

It’s not a proof Euclid could have done, since it uses the intermediate value theorem to show that somewhere between two extreme cases (in fact the octahedron and the cuboctahedron) there lies a regular icosahedron.

But never mind. What matters much more is that this step is intuitively evident — and easy to formalize using today’s math.

I’ve myself used a similar intermediate value theorem argument to prove the existence of the regular icosahedron. Namely, I take two ‘caps’ formed by 5 equilateral triangles, which will become the top and bottom of our regular icosahedron. I turn them correctly… and then note that at some distance from each other, they must be the top and bottom of a regular icosahedron.

But your argument is better, because it instantly tells us that the vertices of the regular icosahedron lie on 3 congruent orthogonal rectangles!

You’ll see that in week284 I got lazy and simply pointed the reader to your $n$Lab page instead of explaining your arguments. Writing up summaries of those talks here at UCR soaked up all my energy. But now I’m glad I didn’t try to explain your previous arguments in detail… because the latest one is even better!

I hope you copy it over to the $n$Lab.

Posted by: John Baez on November 25, 2009 3:18 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

John wrote:

I’ve myself used a similar intermediate value theorem argument to prove the existence of the regular icosahedron.

Ah, I’d been wondering what your own answer was to “Exercise 1” in Some thoughts on the number 6! I never had a good answer myself, until this argument finally dawned on me this morning.

I hope you copy it over to the nLab.

Done.

BTW, my apologies for ever dismissing this as a “wild goose chase”!

Posted by: Greg Egan on November 25, 2009 4:36 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Indeed, I’d been going to point you towards the answer to “Exercise 1”, when I discovered that I’d never included my answer to this exercise.

Thanks for putting your latest proof on the $n$Lab. Wild geese are great when you actually catch them!

My one criticism of the $n$Lab entry is that you still say “Both of these [proofs] rely on a lemma concerning the ratio of sides in a ‘golden triangle’.” If I understand it correctly, a great feature your latest proof is that it reaches the pentagon-decagon-hexagon identity without needing the golden triangle lemma!

So, I’ll change the wording here a bit. I’ll also add a bit about the historical context of this problem. If I say something wrong, maybe you’ll fix it.

Posted by: John Baez on November 26, 2009 2:46 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Thanks very much for tidying up the nLab entry and adding the historical context. You’re right, of course, that the icosahedron proof doesn’t need the golden triangle lemma to reach the pentagon-decagon-hexagon identity; I still invoke it at the end to go back and show that the duads are in the golden ratio, but of course that’s just a nice bonus, it’s not actually necessary.

Posted by: Greg Egan on November 26, 2009 3:08 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Maybe you need to explicitly say somewhere in the nLab entry that Euclid did not use degrees though he probably knew of their contemporary use in surveying and astronomy.

Thus something like the Golden Triangle Lemma in true Euclidian style would have to start off by constructing a decagon.

I’m not sure how much of your “new” proofs depend on performing arithmetic on degree values which are a decategorification of geometric constructions.

Posted by: RodMcGuire on November 26, 2009 4:53 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m not sure how much of your “new” proofs depend on performing arithmetic on degree values which are a decategorification of geometric constructions.

Nothing depends on this. Marking up these diagrams with angles in degrees is simply a shorthand for long-winded statements about which angles are equal, or equal to twice some other angle, and so on, in the Euclidean original. There are no “protractor constructions” where angles are created that could not be constructed by compass and straight-edge.

Posted by: Greg Egan on November 26, 2009 5:49 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’m sorry for a premature posting that mostly got you angry.

I mainly wanted to bring up that

1) Euclid seemed fanatical about only using correspondences and not a decategorification into degrees.

2) Decategorification relates this to category theory.

3) The Euclidian constructive arithmetic that one can can perform on degrees is not ordinary arithmetic but is strangely limited - one can divide by 2 but not by 3.

I would guess someone has formalized this strange Euclidian degree arithmetic but with a limited google I can’t find anything, much less a category that handles it.

I like you and applaud your proofs. I was just trying to edge you to the categorical implications.

Posted by: RodMcGuire on November 26, 2009 6:42 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

No offence taken, Rod! I just wanted to set your mind to rest that the degrees were there solely for the reader’s convenience.

Posted by: Greg Egan on November 26, 2009 7:02 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I’ve added a note to the nLab entry that says:

Euclid himself would not have used numerical values like this, and they are not a part of the proof; they’re merely a short-hand that allows the reader to see at a glance the relationships between angles that the original proof established by routine methods.

Posted by: Greg Egan on November 26, 2009 6:09 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Rod wrote:

Thus something like the Golden Triangle Lemma in true Euclidian style would have to start off by constructing a decagon.

You can see Euclid’s proof of the Golden Triangle Lemma here — it’s Prop. 9 of Book XIII. You’ll note he doesn’t actually start by constructing a regular decagon! Instead, he feels free to use the fact that the angle between vertices of this decagon is 1/5 the angle between opposite points on a circle.

I think I’ve seen people question how religiously Euclid adhered to the practice of only talking about things that he’d constructed by straightedge and compass. I seem to recall people saying that he became a bit more relaxed in Book XIII, the book on solid geometry.

How do you draw a 3d shape with a straightedge and compass, after all? And how do the axioms of plane geometry suffice to prove results about 3d shapes?

Perhaps more to the point, Greg did nothing with degree values of angles that couldn’t easily be done without them. They’re just a way of making arguments easier to follow.

As far as I can see, the only argument on the $n$Lab page that couldn’t be transformed into utterly pedantic Euclidean reasoning is Greg’s use of the intermediate value theorem in constructing the icosahedron.

Anyway, I never thought the main point of this game was to follow Euclid’s rules. I thought the point was to find a clear and simple proof of the pentagon-decagon-hexagon identity, one that obtains

$P^2 = D^2 + H^2$

by exhibiting a right triangle with sides $P, D,$ and $H$. And that’s what Greg did.

Posted by: John Baez on November 26, 2009 6:44 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I got ahold of Eva Sachs’ 1917 book Die Fünf Platonischen Körper!

If anyone out there reads German, I’d love a translation of what she says about the icosahedron and the pentagon-hexagon-decagon identity. Click on the pictures for larger views:   I think the first page gets interesting around “Der Satz XIII 10 lautet…” (“Proposition XIII.10 says…”) and more interesting after “Dieser Satz, obwohl planimetrisch, ist durch Betrachtung der ebenen Geometrie nicht leicht zu finden”. (“This proposition, though planar in character, is not so easy to find through considerations of plane geometry.”)

You’ll notice that she focuses our attention on the right triangles ZWQ and QEP, which are the ones Greg Egan considered here: Proving that these are congruent is the key to the pentagon-decagon-hexagon identity!

I’m especially curious about the footnote on page 104, and also the remark further up this page saying “So fand er den Satz XIII 9…” (“This is how he found Proposition XII.9”, namely the golden triangle lemma).

As you can see, I can read German a little, but not very well.

Posted by: John Baez on November 26, 2009 11:05 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

May it be that the construction of the pentagon is stemming from a pythagorean or what is more likely it is stemming from Eudoxos, it cant be attributed to Theaetet, just as the proposition XIII 8 on which it is based on.

In contrast to this propositions XIII 9 and XIII 10 belong to Theaetet. I will try to show that they were found in connection to the construction of the icosahedron.

Proposition XIII 10 reads: If a regular pentagon is inscribed into a circle then the square of a pentagon edge is equal to the square of a regular hexagon edge (so in particular of the radius of that circumscribing circle) plus the square of the edge of the regular decagon, which is inscribed into the same circle.

This proposition, though planar in character, is not so easy to find through considerations of plane geometry. In contrast to this - by looking at the icosahedron - one can guess to some extent that the sides/edges of the three forms comply to this. At the body a rectangular triangle whoose hypotenuse is the edge of the pentagon and whose one cathetus is the edge of the decagon, the other the radius of the pentagon circumscribing circle is appearing twice. At the beautiful figure of Heath (the XIII Books of Eukl. El. III p. 487) which I am allowed to reprint here with the permission of the University Press Cambridge, these are for example the triangles ZWQ and QEP. Similarily from the drawing of the body it could be deduced that WQEV is a square, whose edge is forming the radius of the circumfencing circle of the pentagon QRSTU. If the geometer had proven with regard to the plane figure (XIII 10) that what he could deduce by merely looking at the solids body, he could likewise detect by looking at the icosahedron that the diameter of the sphere, which circumscribes the icosahedron (ZX) is composed of double the edge of the regular decagon (ZW + VX) and the radius of the circle which circumscribes the regular pentagon QRSTU, which fits into the same circle as the regular decagon. This was stemming from the triangles we were just talking of: In triangle ZWQ one had that ZW=VX was the edge of the decagon. In triangle QEP the edge QE=QW=WV was the edge of the hexagon which was inscribed into the same circle as the pentagon and decagon which was the circumscribing circle of the pentagon. Hence in order to construct the icosaeder Theaetet only needed to ask what is the edge of the hexagon (i.e. the radius of the circumscribing circle of the decagon) in relation to the edge of the regular decagon plus the edge of the hexagon, that is he only needed a proposition in which the proportion between the sum of both sides to the hexagon edge was asked for.
In such a way he found proposition XIII 9: “If the edge of the regular hexagon is extended by the edge of a regular decagon, which is inscribed into the same circle, then the resulting line is divided by the golden mean and the bigger section is the edge/side of the hexagon (i.e. the radius of the circumscribing circle of the decagon).
That this proposition is stemming from Theaetet arises also from another consideration. We saw at the construction of the regular pentagon and decagon that Euklid seemed not to know of the proposition: “The decagon side is the bigger section of the by the golden mean divided radius of its circumscribed circle.” This is supported by the formulation of XIII 9. If this proposition of the decagon would have been known, then from this proposition and XIII 5 the proof of XIII 9 would emerge.^1)
___________
1)The proof of the proposition of the relation of the decagon side to the radius of the circumscribing circle could be supplied in reverse order as the exercise in IV 10 was solved. In the figure there are: R\triangle the decagon edge; BA and A\triangle the radii of the circumscribing circles; each of the angles AB^\cdot\triangle = A \triangle^\cdotB double as big as the angle BA^\cdot\triangle, so finally A\Ìamma=B\triangle after construction. According to III 32 it could be shown that B\triangle was the tangent to the circle around the triangle A\Gamma\triangle was. and according to III 37 that B\triangle^2=\GammaA^2=BA*B\Gamma, i.e. B\triangle is the bigger section of the by the golden mean divided radius AB.__________

This is again displaying that XIII 9 was stemming from Theaetet and that it was found in connection to the construction of the icosahedron.

The construction of the five bodies (proposition 13-17) is stemming from Theaetet. Vogt (Bibl.math 3. F. IX p. 47)^1 was rightly so polemizing against Tannery, who (Geom p. 101) attributes the construciton of the five polyhedra to the pythagoreans, while according to him Theaetet had only added the calculation of the proportion of the edges to the radius of the circumscribing sphere: since in the first place the exact construction is relying on this calculation. As it leads to irrational proportions for all five bodies, it couldn’t have existed before Theaetets proof of the irrational or at least before Theodoros discovery, so that the inner context of the historical events is in accordance with the tradition (Suidas), which attributes the construction of the five bodies to Theaetet.

Posted by: translation on November 27, 2009 4:26 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Thanks, anonymous translator, for your kind help! I also thank Urs Schreiber, who emailed me a quick translation.

Let me try polishing it a bit more, based on both your work:

The construction of the pentagon may be due to a Pythagorean, or more likely Eudoxos; it cannot be attributed to Theaetetus, nor can Proposition XIII 8, on which it is based.

In contrast to this, Propositions XIII 9 and XIII 10 belong to Theaetetus. I will try to show that they were found in connection to the construction of the icosahedron.

Proposition XIII 10 reads: “If a regular pentagon is inscribed into a circle then the square of a pentagon edge is equal to the square of a regular hexagon edge (so in particular the radius of the circumscribing circle) plus the square of the edge of the regular decagon, which is inscribed into the same circle.”

This proposition, though planar in character, is not so easy to find through considerations of plane geometry. In contrast to this, by looking at the icosahedron, one can more or less guess that the sides of the three forms obey this equation.

On this body we see two appearances of a right triangle whose hypotenuse is the edge of the pentagon, whose one side is the edge of the decagon, and whose other side is the radius of the circle in which the pentagon is inscribed.

In the beautiful figure of Heath (Euclid’s Elements vol. 3 page 487) which I am allowed to reprint here with the permission of the Cambridge University Press, these are for example the triangles ZWQ and QEP.

Similarily one may see from the figure that WQEV is a square whose sides form the radius of the circle in which the regular pentagon QRSTU sits.

Once the geometer had proven for the plane figure (XIII 10) what he could deduce by merely looking at this 3-dimensional figure, he could likewise realize from regarding the icosahedron that the diameter ZX of the sphere circumscribing this body is composed of twice the side of the regular decagon (ZW+VX) and the diameter (VW) of the outer circle of the regular pentagon QRSTU, that is enscribed inside the same circle as the regular decagon. This followed from the triangles that we just talked about: In triangle ZWQ one had that ZW=VX was the edge of the decagon. In triangle QEP the edge QE=QW=WV was the edge of the hexagon which was inscribed into the same circle as the pentagon and decagon.

Hence in order to construct the icosahedron, Theaetetus only needed to ask what is the edge of the hexagon (i.e. the radius of the circumscribing circle of the decagon) in relation to the edge of the regular decagon plus the edge of the hexagon. That is, he only needed a proposition in which the proportion between the sum of both sides to the hexagon edge was asked for. In such a way he found proposition XIII 9: “If the edge of the regular hexagon is extended by the edge of a regular decagon, which is inscribed into the same circle, then the resulting line is divided by the golden mean and the bigger section is the edge/side of the hexagon (i.e. the radius of the circumscribing circle of the decagon).

That this proposition comes from Theaetetus also follows from another consideration. We saw at the construction of the regular pentagon and decagon that Euclid seemed not to know of the proposition: “The decagon side is the bigger section of the radius of its circumscribed circle, when this radius is divided according to the golden ratio.” This is supported by the formulation of XIII 9. If this proposition of the decagon would have been known, then from this proposition and XIII 5 the proof of XIII 9 would emerge.

[Footnote: The proof of the proposition concerning the relation of the decagon side to the radius of the circumscribing circle could be supplied in reverse order, as the exercise in IV 10 was solved. In the figure there are: $R\Delta$ the decagon edge; BA and A$\Delta$ the radii of the circumscribing circles; each of the angles $A\dot{\mathrm{B}} \Delta$ = A$\dot\Delta \cdot$ B are twice as big as the angle B$\dot{\mathrm{A}}\Delta$, so finally A$\Gamma$=B$\Delta$ after construction. According to III 32 it could be shown that B$\Delta$ was the tangent to the circle around the triangle A$\Gamma\Delta$. And according to III 37, B$\Delta^2=\Gamma\mathrm{A}^2=BA \times \mathrm{B}\Gamma$, i.e. $B\Delta$ is the bigger section of the radius AB divided according to the golden ratio.]

This again shows that XIII 9 stemmed from Theaetetus, and that it was found in connection to the construction of the icosahedron.

The construction of the five bodies (Propositions 13-17) is also due to Theaetetus. Vogt (Bibl.math 3. F. IX p. 47) was thus correct in his polemics against Tannery, who (Geom p. 101) attributes the construction of the five polyhedra to the Pythagoreans, while according to him Theaetetus had only added the calculation of the proportion of the edges to the radius of the circumscribing sphere. In the first place, the exact construction relies on this calculation. As it leads to irrational proportions for all five bodies, it could not have existed before Theaetetus proof of the irrationality, or at least before Theodoros’ discovery, so that the inner context of the historical events is in accordance with the tradition (Suidas), which attributes the construction of the five bodies to Theaetetus.

Posted by: John Baez on November 27, 2009 8:00 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

For people who don’t like using the Intermediate Value Theorem, here’s a purely geometric construction of the regular icosahedron.

We set up three congruent orthogonal rectangles, each in the golden ratio, and use their twelve corners as the 12 vertices of the icosahedron, like this:

It’s obvious that 6 of the edges here are congruent, because they’re all equal to the short side of our rectangle. For simplicity, we’ll set that short side to 1, so the long side is $\Phi$.

It’s also obvious that all 24 other edges are congruent to each other. So all we have to do is prove that one of them has a length of 1.

1 Φ Φ 1 A B –A –B C P Q R S O T α α α 90–α 1/(2Φ) Φ/2 1/2

If we take an edge that projects onto $A Q$ in the diagram above as our example of the other edge length, we can easily see that the square of $A Q$ itself is $\frac{3}{4}$: the light green square has area $\frac{1}{4}$, and each red triangle has area $\frac{1}{2}\times \frac{\Phi}{2} \times \frac{1}{2 \Phi} = \frac{1}{8}$. By construction, both vertices that project onto $Q$ are a distance of $\frac{1}{2}$ from the plane, so the total length of either edge that projects to $A Q$ is $1$. So all $30$ edges are now shown to have a length of $1$.

Posted by: Greg Egan on November 27, 2009 3:48 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I found a nicer way to avoid using the intermediate value theorem.

If we project the generic version of our icosahedron onto the plane of one of its rectangles, we get this:

1 x x 1 A B P Q R S O

In order to make $A P R$ into a straight line – which amounts to making the pentagon of vertices that project onto $A P R$ into a planar figure – we need to satisfy:

$\frac{x+1}{x}=x$

This is just the defining equation for the golden ratio, $\Phi$.

Next, we note that the golden triangle lemma can be used to establish that in the figure below, the larger of the two golden triangles here associated with a regular pentagon is $\Phi$ times taller than the smaller one.

36° 72° 72° 72° 72° 36° 36° 36° 108° 108° 36° 36°

Now, by choosing $x=\Phi$, as well as making the pentagon that projects onto $A P R$ into a planar figure, we’ve actually made every set of five vertices that are neighbours to any given vertex in the icosahedron lie in a plane. For example, the two pentagons that project onto $A B R S Q$ must also be planar. But since they’re planar, their projection here will retain the ratio between the perpendicular distance from $A B$ to the vertex that projects to $S$, which is $\frac{x+1}{2}$, and the perpendicular distance to the vertices that project to $R$ and $Q$, which is $\frac{x}{2}$. Our choice of $x=\Phi$ makes that ratio, in the projection, $\frac{\Phi+1}{\Phi}=\Phi$.

But we’ve just established that the ratio is $\Phi$ for a regular pentagon, and the pentagons projecting to $A B R S Q$ are already so symmetrical (with at least four congruent edges, and mirror symmetry in the perpendicular bisector of the fifth edge, $A B$) that if they share that ratio, they too must be regular. This means that the two separate sets of congruent edges in the generic icosahedron are actually the same size, and the icosahedron itself is regular.

Posted by: Greg Egan on December 1, 2009 12:03 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

I wrote:

the pentagons projecting to $A B R S Q$ are already so symmetrical (with at least four congruent edges, and mirror symmetry in the perpendicular bisector of the fifth edge, $A B$) that if they share that ratio, they too must be regular.

That’s not quite correct! The pentagon with the constraints I list has two degrees of of freedom: the ratio of lengths of the fifth edge to the others, and one free angle between edges.

In order to prove that it’s regular, we need one more constraint as well as the height ratio: we need the ratio of its width to the fourth edge (i.e. $\frac{Q R}{A B}$) to match that of a regular pentagon. And of course it does; this ratio is $\Phi$ as well.

Posted by: Greg Egan on December 1, 2009 12:59 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 283)

Here’s a variation on the previous 2-dimensional proof that displays right triangles that exhibit the pentagon-decagon-hexagon identity.

2 b 2 b 2 b t t 2 b r t t 2 b r 2 b 2 b

The diagram above is similar to one used in a common proof of Pythagoras’s theorem, but we’ll use it to establish the value of the square of the hypotenuse of a right triangle with sides $r$ and $t$, where $r$ is the hexagon edge and $t$ is the decagon edge. The length $b$ is the distance from the midpoint of a pentagon edge to the decagon vertex that lies between the pentagon edge’s two vertices (picture in this post). The golden triangle lemma plus the definition of the golden ratio give us $t+2b=r$, and Pythagoras’s theorem gives us $s^2=t^2-b^2$, where $s$ is half a pentagon edge.

The tilted square in the centre has a hypotenuse as its side. Its area is $4b^2$, from the central pink square, plus $4t^2$, from the four multi-coloured $t\times t$ squares arranged around the central one, minus four times the area of the dark green triangle that needs to be excluded (the contributions from the yellow triangles inside and outside the region cancel out).

The dark green triangle has both base and height of $2b$, so each one has an area of $2b^2$, and the total area of the tilted square is thus $4t^2-4b^2=4(t^2-b^2)=(2s)^2$, the square of a pentagon edge.

Posted by: Greg Egan on November 28, 2009 1:58 AM | Permalink | Reply to this

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