The n-Category Café

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

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.

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 …

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.

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 …)

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.

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.

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

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.

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.

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 $h^2=r^2$.

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.)

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.

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!

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.

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.

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.

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.