Pentagon-Hexagon-Decagon Identity

Icosahedron Illustrating Pentagon-Hexagon-Decagon Identity - Greg Egan

Icosahedron Illustrating Pentagon-Hexagon-Decagon Identity – Greg Egan

Suppose we inscribe a regular pentagon, a regular decagon, and a regular hexagon in circles of the same radius. If we denote the respective edge lengths of these polygons by $P$, $D$ and $H$, then these lengths satisfy the identity

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

This means that the edges of a pentagon, decagon and hexagon of identical radii can fit together to form a right triangle!

Euclid stated this beautiful but mysterious identity as Proposition 10 of Book XIII of the Elements. This is the last book of the Elements, the one which deals with properties of the Platonic solids. He used Proposition 10 as part of his construction of the regular icosahedron in Proposition 13.

This has led some historians to suggest that the pentagon-decagon-hexagon identity was first discovered in the course of research on the icosahedron. The idea is this. If you hold an icosahedron so that one vertex is on top and one is on bottom, you’ll see that its vertices are arranged in 4 horizontal layers. From top to bottom, these are:

  • 1 vertex on top
  • 5 vertices forming a pentagon: the "upper pentagon"
  • 5 vertices forming a pentagon: the "lower pentagon"
  • 1 vertex on bottom

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 $C$:

It is easy to check that $ABC$ is a right triangle. If we apply the Pythagorean theorem to this triangle we get the equation

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

But to see this, we need to check that:

  • 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? The five vertices of the lower pentagon lie on the circle shown in blue. This circle has the right radius.

Using this idea, it’s easy to see that the length $AB$ equals the edge of a pentagon inscribed in a circle. It’s also easy to see that $BC$ equals the edge of a decagon inscribed in a circle of the same radius. The hard part is showing that $AC$ equals the edge of a hexagon inscribed in a circle of the same radius… or in other words, the radius of that circle! (The hexagon appears to be a red herring.)

To prove this, it suffices to show the following marvelous fact: the distance between the "upper pentagon" and the "lower pentagon" equals the radius of the circle containing the vertices of the upper pentagon!

Can you prove this?

In Ian Mueller’s book Philosophy of Mathematics and Deductive Structure in Euclid’s Elements, he suggested various ideas the Greeks could have had about this. Today’s image shows one. Let’s look at it again:

Icosahedron Illustrating Pentagon-Hexagon-Decagon Identity - Greg Egan

The trick is to construct a new right triangle $AB’C’$. Here $B’$ is the top vertex, and $C’$ is where a line going straight down from $B’$ hits the plane containing the upper pentagon.

Remember, we’re trying to show the distance between the upper pentagon and lower pentagon
equals the radius of the circle containing the vertices of the upper pentagon.

But that’s equivalent to showing that $AC’$ is congruent to $AC$.

To do this, it suffices to show that the right triangles $ABC$ and $AB’C’$ are congruent! Can you do it?

In the references to Mueller’s book, he says the historians Dijksterhuis (in 1929) and Neuenschwander (in 1975) claimed this is “intuitively evident”. He also notes that Eva Sachs, in her book Die Fünf Platonischen Körper, suggested an accurately drawn figure could let someone guess that the distance between the two pentagons equals the radius of either one. But that’s not a proof.

You can see a proof due to Greg Egan here:

• Pentagon-hexagon-decagon identity: Proof using the icosahedron, nLab.

Egan put some other proofs of the pentagon-hexagon-decagon identity here:

Pentagon-hexagon-decagon identity, nLab.

Also see:

• John Baez, This Week’s Finds in Mathematical Physics, Week 283, and discussion on the n-Category Café.

• Eva Sachs, Die Fünf Platonischen Körper, zur Geschichte der Mathematik und der Elementenlehre Platons und der Pythagoreer, Berlin, Weidmann, 1917, pp. 102–104. See page 102–103 here and page 104 here.

• Ian Mueller, Philosophy of Mathematics and Deductive Structure in Euclid’s Elements, MIT Press, Cambridge Massachusetts, 1981, pp. 257–258 and references therein.

Of course, we can prove also prove the pentagon-hexagon-decagon identity using algebra. Start with a unit circle. If we inscribe a regular hexagon in it, then clearly

$$ H = 1 $$

So we just need to compute $P$ and $D$. If we think of the unit circle as living in the complex plane, then the solutions of

$$ z^5 = 1 $$

are the corners of a regular pentagon. So let’s solve this equation. We’ve got

$$ 0 = z^5 – 1 = (z – 1)(z^4 + z^3 + z^2 + z + 1) $$

so ignoring the dull solution $z = 1$, we must solve

$$ z^4 + z^3 + z^2 + z + 1 = 0$$

This says that the center of mass of the pentagon’s corners lies right in the middle of the pentagon.

Now, quartic equations can always be solved using radicals, but it’s a lot of work. Luckily, we can solve this one by repeatedly using the quadratic equation! And that’s why the Greeks could construct the regular pentagon using a ruler and compass.

The trick is to rewrite our equation like this:

$$ z^2 + z + 1 + z^{-1} + z^{-2} = 0 $$

and then like this:

$$ (z + z^{-1})^2 \, + \,(z + z^{-1}) \, -\, 1 = 0 $$

If we write

$$z + z^{-1} = x $$

our equation becomes

$$x^2 + x \; – 1 = 0 $$

Solving this, we get two solutions. One of them is the golden ratio

$$ x = \phi = \frac{\sqrt{5} -1}{2} \approx 0.6180339\dots $$

Next we need to solve

$$ z + z^{-1} = \phi $$

This is another quadratic equation:

$$ z^2 – \phi z + 1 = 0 $$

with two conjugate solutions, one being

$$ z = \frac{\phi + \sqrt{\phi^2 – 4}}{2} $$

This is a fifth root of unity in the first quadrant of the complex plane, so we know

$$ z = \exp(2 \pi i/5) = \cos(2\pi/5) + i \sin(2\pi/5) $$

So, we’re getting

$$ \cos(2\pi/5) = \phi/ 2 $$

A fact we should have learned in high school, but probably never did! Now we’re ready to compute $P$, the length of the side of a pentagon inscribed in the unit circle:

$$ P^2 = |1 – z|^2 = (1 – \cos (2\pi/5))^2 + \sin^2 (2\pi/5) =
2 – 2 \cos (2\pi/5) = 2 \; – \phi $$

Next let’s compute $D$, the length of the side of a decagon inscribed in the unit circle! We can mimic the last stage of the above calculation, but with an angle half as big:

$$ D^2 = 2 – 2 \cos(\pi/5) $$

To go further, we can use the half-angle formula:

$$ \cos(\pi/5) = \sqrt{\frac{1 + \cos (2\pi/5)}{2}} = \sqrt{\frac{1}{2} + \frac{\phi}{4}} $$

This gives

$$ D^2 = 2 \; – \sqrt{2 + \phi} $$

But we can simplify this a bit more. As any lover of the golden ratio should know,

$$ 2 + \phi \approx 2.6180339\dots $$

is the square of

$$ 1 + \phi \approx 1.6180339\dots $$

So we really have

$$ D^2 = 1 – \phi $$

And now we’re done! We see that the pentagon-hexagon-decagon identity

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

simply says:

$$ 2 – \phi = 1 + (1 – \phi) $$


Visual Insight is a place to share striking images that help explain advanced topics in mathematics. I’m always looking for truly beautiful images, so if you know about one, please drop a comment here and let me know!

18 comments on “Pentagon-Hexagon-Decagon Identity

  1. Ian Agol on said:

    Here’s another proof using the icosahedron:

    Take an icosahedron whose edge lengths are 2. Consider a plane going through the center of the icosahedron, and cutting the edges it crosses through in half (so e.g. the plane midway between the two planes drawn in your diagram). This cuts the icosahedron in a regular decagon with edge lengths 1. So the distance from the center of the icosahedron to an edge is $1/D$. If we take a plane cutting off a corner of the icosahedron, so that it intersects the adjacent faces at the midpoints, then this cuts off a regular pentagon with edge lengths 1, so the distance from this vertex to a midpoint of an edge is $1/P$. But we have a right triangle with vertices the center of the icosahedron, a vertex, and an adjacent edge midpoint. The altitudes of this right triangle are therefore $1/P$, $1/H$ and $1/D$ from the above discussion (one altitude is half an edge, one is the radius of the midscribed sphere going through the midpoints of the edges, and one is the radius of the circumcircle of the corner pentagon). This is equivalent to $P^2=H^2+D^2$ by some simple geometry.

    • johnbaez on said:

      Cool—I’ll have to think about that a bit!

      By the way, simple LaTeX works in comments on this blog; just put dollar signs around math expressions as usual. I tested this out by adding dollar signs to your comment, just for fun.

    • Greg Egan on said:

      That’s ingenious!

      I especially like the use of the Reciprocal Pythagorean Theorem (which I’ve taken to calling the Dual Pythagorean Theorem), the most unjustly neglected result in elementary geometry. That theorem is the reason we can find, say, the overall spatial frequency of a wave from individual spatial frequencies measured along coordinate axes, or the overall gradient of a planar patch of land from its gradient in the north-south and east-west directions, as square roots of sums of squares.

      • johnbaez on said:

        I was going to bring this to your attention, because I learned the Dual Pythagorean Theorem from your Orthogonal trilogy, I guess the first volume.

    • Ian Agol on said:

      Incidentally, from this perspective, one obtains right-triangles associated to the tetrahedron and octahedron as well. For the tetrahedron, one has a square, hexagon, and triangle. For the octahedron, one has two hexagons and a square. So these are relatively boring!

      • Ian Agol on said:

        Actually, I just realized there’s also right triangles associated to the cube and dodecahedron too. The cube is the same as its dual, the octahedron (triangle, square, hexagon). To the dodedahedron, one gets a triangle, decagon, and decagon-star, i.e. a 10-pointed star that wraps 3 times around.

      • Greg Egan on said:

        It’s incredibly cool that this can be generalised!

        I suppose you get the “large” polygon (analogous to the decagon in the icosahedral case) by taking a plane that passes through the centre of the polyhedron and the midpoints of two edges. When the polyhedron has equilateral triangles as faces, you can pick any two edges of the same face, and the polygon will then have edges whose length is half the polyhedron’s edge length. This gives a decagon for an icosahedron, a square for a tetrahedron, and a hexagon for an octahedron.

        But when the polyhedron’s faces aren’t triangular, what’s the rule? For example, for a cube you could choose midpoints of two adjoining edges, which would give you a hexagon from the slicing plane, or two opposite edges, which would give you a square. But in both cases, the edges of the polygon are no longer half the length of the polyhedron’s edge. Is that OK — do we just accept that as the nature of the result?

        • Ian Agol on said:

          @Greg Egan: Yes, in each of these cases, one has 3 regular polygons which share an edge which connects midpoints of adjacent edges of the regular polygon face of the polyhedron. There is the polygon on the plane going through the center, the plane cutting off a corner, and the plane of the face. Then there is a right triangle cutting the three planes in its altitudes (with vertices on a vertex, an edge midpoint, and the center of the polyhedron), to which one applies the dual Pythagorean theorem.

    • Greg Egan on said:

      @Ian Agol: Thanks!

      I’ve put a picture of all 5 cases here:

      http://gregegan.net/images/GPHD.png

      The construction is: choose a vertex $V$ of the polyhedron, and edges $E_1$ and $E_2$ that are incident on $V$. Let $M_1$ and $M_2$ be the midpoints of those edges. We then have three regular polygons with $M_1 M_2$ as one of their edges:

      The green polygon, whose centre is $C$, the centre of the polyhedron.

      The yellow polygon, whose vertices are all the midpoints of all the edges that are incident on $V$, so it lies in a plane that truncates that vertex.

      The blue polygon, whose centre is $V$ and which lies in the plane of the face of the polyhedron containing $V$, $E_1$ and $E_2$.

      The red triangle is a right triangle whose vertices are $V$, $C$ and $M_1$, with two perpendicular sides whose lengths are equal to the radii of the green and blue polygons, and whose altitude with the hypotenuse as base is the radius of the yellow polygon.

    • Greg Egan on said:

      @Ian Agol: I’ve taken the liberty of writing up your proof here:

      http://nlab.mathforge.org/nlab/show/pentagon+decagon+hexagon+identity#IcosahedronDual

      If there’s anything incorrect, or anything you wish to improve, this page is editable.

      • John Baez on said:

        Great – I’d been planning to do that myself, but not relishing the prospect—not because this proof isn’t cool, but because I don’t understand it yet, and I’m busy doing other stuff. It’s good to have lots of information about this identity in one place. When I get around to understanding this new proof, I may blog about it here!

      • Ian Agol on said:

        Cool, I’ll have a look! I also posted something on Google+ in order to advertise it (I used your image with credit). I’ll add a link to your write-up.

        thanks,
        Ian
        https://plus.google.com/109869432137613080522/posts/33crmNFaov6

      • Ian Agol on said:

        I added a question on mathoverflow, whether there are other right triangles whose edges are the lengths of edges of regular polygrams inscribed in a unit circle. Note that one other example comes by Galois conjugation, and may be interpreted in the same way from Kepler-Poinsot polyhedra. http://mathoverflow.net/q/153761/1345

    • Greg Egan on said:

      Thanks for all the new information, Ian. This just keeps getting better!

      I made an image of the Kepler-Poinsot polyhedra:

      http://gregegan.net/images/GPHD2.png

  2. John Miller on said:

    Says 2 plus phi = 2.618…

    ???

    • johnbaez on said:

      Could you be a bit less laconic in your question? I’ll try to read your mind. There are two numbers called the golden ratio: “little phi”

      φ ≈ 0.6180339…

      and its reciprocal “big Phi”:

      Φ ≈ 1.6180339…

      In this article I’m only using φ, which I defined to be (-1 + √5)/2.