# Right triangle with edge lengths equal to regular unit polygon edge lengths

This question came up naturally recently from a blog post of John Baez. There is an observation of Euclid that edges of a pentagon, hexagon, and decagon inscribed in a unit circle form the edges of a right triangle:

Euclid used this to construct the icosahedron (see Baez's description for more details). I observed that this follows from a dual description in terms of the dual pythagorean theorem (see also Greg Egan's description with more details).

There are 4 right triangles associated to the 5 Platonic solids with the property that the edge lengths are equal to the edges of regular polygons inscribed in a unit circle:

Question: Are there any other right triangles with edge lengths equal to the lengths of edges of regular polygons inscribed in a unit circle?

Note that a regular polygon is allowed to be immersed, that is a star (or polygram as Sam Nead points out in a comment).

Generalizing the algebraic analysis in Baez's post, this may be formulated in terms of a diophantine equation involving cyclotomic numbers, but I'll leave this formulation up to the reader.

As observed in the blog post, there is a dual description stating that there are 3 regular polygons with edge lengths 1, such that the circumradii of these polygons form the altitudes of a right triangle (which satisfy the dual pythagorean theorem).

Addendum: As Will Sawin points out in the comments, there is also the Galois conjugate hexagon, pentagram, decagram. In fact, one obtains examples associated to all of the Kepler-Poinsot polyhedra.

The pentagram, hexagon, decagram is associated to the great dodecahedron and great icosahedron. The pentagon, hexagon, decagon is associated to the small stellated dodecahedron, and the triangle, decagon, decagram to the great stellated dodecahedron, if I've figured correctly.

Postscript: It appears that a transformation of this question was answered by Gordan in 1877. Gordan showed that the only solutions to $1+\cos(\phi_1)+\cos(\phi_2)+\cos(\phi_3)=0$ with $0<\phi_i<\pi$ and $\phi$ a rational multiple of $\pi$ are permutations of $(\frac23\pi,\frac23\pi,\frac12\pi)$ and $(\frac23\pi,\frac25\pi,\frac45\pi)$ (see. p. 109 of Coxeter). This equation is equivalent to the equation in Sawin's answer, $\zeta_1+\zeta_1^{-1}+\zeta_2+\zeta_2^{-1}-\zeta_3-\zeta_3^{-1}=2$ where $\zeta_1=e^{\phi_1 i}, \zeta_2=e^{\phi_2 i}, \zeta_3=e^{(\pi-\phi_3)i}$ (the degenerate solutions in Gordan's equation are eliminated by assuming $0<\phi<\pi$). The polygon going through the midpoints of edges is similar to the Petrie polygon, which is the projection of the sequence of edges, and lies in the Coxeter plane. The duality in orthogonal triangles associated to each of the regular polyhedra then follows from the symmetry of these equations, but doesn't seem to be observed directly in Coxeter's book.

In fact, if $h$ is the degree of the polygon, then we see a further symmetry of the solutions by letting $\frac{1}{r}+\frac{1}{h}=\frac12$. We see that $\{p,q,r\}$ satisfies equation 6.71 on p.108 of Coxeter: $$\cos^2(\pi/p)+\cos^2(\pi/q)+\cos^2(\pi/r)=1.$$

The solutions are $\{3,3,4\}$ and $\{3,5,\frac52\}$.

On p. 109 (see also regular polyhedron), one sees that permuting the two solutions to this equation, one obtains 3 and 6 polyhedra respectively, indicating a further symmetry not apparent from the right triangle equation.

• Like Atiyah's devil, I have stripped the geometry from your figures, replacing them with triples of numbers, as follows: $(3,4,6)$, $(4,6,6)$, $(3,5,10)$, $(3,10,10)$. I find these easier to think about. Jan 6, 2014 at 20:51
• If stars count as polygons, then one triple missing from the list is triangle, five-pointed star, 10-pointed star. One can obtain it just by a Galois conjugate of the original one. Jan 6, 2014 at 20:54
• @WillSawin: yes, I agree that the problem could be algebraized to obtain a solution, but the motivation came from classical geometric considerations. Also, I think the Galois conjugate example could also be thought of geometrically by the same construction applied to a Kepler-Poinsot polyhedron. en.wikipedia.org/wiki/Kepler-Poinsot_polyhedra Jan 6, 2014 at 21:18
• Immersed polygons appear to be called called polygrams. mathworld.wolfram.com/Polygram.html - Pentagram is a standard one, but octagram was new to me! Jan 6, 2014 at 21:29
• (10,6,5) ;(6,6,4);(4,3,6);(6,3,2);(l,m,n) where l and m sided regular polygons exist on the base and height and the n sided regular polygon exists on thy hypotenuse.- Some examples by solving $Sin^2(\pi/l) + Sin^2(\pi/m )=Sin^2(\pi/n)$ on a spread sheet.
– ARi
Jan 17, 2014 at 11:49

There is a degenerate case that occurs if a diameter counts as an inscribed digon. In this case, since any triangle inscribed in the circle with one edge a diameter is a right triangle, we can build a right triangle out of any edge of any polygram, an adjacent diameter, and a third edge, which is always an edge of another diagram.

In addition to the polyhedral ones listed, these are the only examples.

The Diophantine equation involving cyclotomic numbers that Ian Agol refers to is $(\zeta_1-1)(\zeta_1^{-1}-1) + (\zeta_2-1)(\zeta_2^{-1}-1)= (\zeta_3-1)(\zeta_3^{-1}-1)$ for $\zeta_1,\zeta_2,\zeta_3$ three roots of unity.

Equivalently:

$\zeta_1 + \zeta_1^{-1} + \zeta_2 + \zeta_2^{-1} - \zeta_3 - \zeta_3^{-1} = 2$

This is an additive identity in algebraic numbers. We can take the trace of each algebraic number in the identity and it will remain an identity. To normalize the trace, we can divide by the degree of the field extension generated by the number. This is especially easy for this equation, because $\zeta+\zeta^{-1}$ is already one step towards taking the trace. So we have:

$tr( \zeta_1)+ tr(\zeta_2) - tr(\zeta_3) = 1$

with the traces appropriately normalized.

Now if $\zeta$ is a primitive $N$th root of unity, the trace of $\zeta_i$ is $\mu(N)/ \varphi(N)$, where $\mu$ is the Mobius function and $\varphi$ is Euler's totient function. So if $a,b,c$ are the orders of the $\zeta_i$, we have:

$$\frac{\mu(a)}{\varphi(a)} + \frac{\mu(b)}{\varphi(b)} - \frac{\mu(c)}{\varphi(c)}=1$$

We can view this as a case of the plane tessellation equation:

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

except $x,y$, and $z$ can be negative numbers or $\infty$. All the solutions to this equation are:

$$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}, \hspace{10pt}\frac{1}{2}+\frac{1}{4}+\frac{1}{4},\hspace{10pt} \frac{1}{2}+\frac{1}{3}+\frac{1}{6}, \hspace{10pt}\frac{1}{2}+\frac{1}{2}+\frac{1}{\infty},\hspace{10pt} \frac{1}{1}+ \frac{1}{y}+\frac{1}{-y}$$

(Proof: One of $1/x$,$1/y$,$1/z$ must be at least $1/3$, so the smallest positive of the $x,y,z$ must be $1$,$2$, or $3$. If it is $3$, the other two must be $3$ as well. If it is $2$, the next-smallest positive one can be $2$,$3$, or $4$. If it is $1$, the next-smallest positive one can be anything. Either way, the last one is determined.)

$\varphi(n)$ is never $3$, so we can eliminate those cases.

If $tr(\zeta_i)=\pm 1$ then $\zeta_i=\pm 1$. Obviously $\zeta_i=1$ does not count, but $\zeta_i=-1$ is the diameter case, which may or may not count.

We are left with the cases of $\frac{1}{4}+\frac{1}{4} + \frac{1}{2}$ and $\frac{1}{2}+\frac{1}{2}+0$. Each one can be split into two cases, depending on which term comes from $\zeta_3$:

$\frac{1}{4}+\frac{1}{4} - \frac{-1}{2}$: $a=10$, $b=10$, $c=3$. By measuring, we see that one shape must be a decagon and one must be a decagram.

$\frac{1}{4}+ \frac{1}{2} - \frac{-1}{4}$: $a=10$, $b=6$, $c=5$. By measuring, we see that either there is a pentagon and a decagon, or a pentagram and a decagram.

$\frac{1}{2} + \frac{1}{2} - 0$: $a=6$, $b=6$. By measuring, we find that the third side must be a square.

$\frac{1}{2} + 0 - \frac{-1}{2}$: $a=6$, $c=3$. By measuring, we find that the second side must be a square.

• A good source for this sort of thing is Conway and Jones, Trigonometric diophantine equations, available at matwbn.icm.edu.pl/ksiazki/aa/aa30/aa3033.pdf (although there is more recent work). Jan 7, 2014 at 1:35
• Ok, thanks. I suppose the diameter of a circle gives a bigon, which could be considered a degenerate polygon. I think I've worked out another approach, which basically reverses the steps in the geometric argument, to show such a triangle leads to a polyhedron. But I'll have to think about what the degenerate case might correspond to, maybe a 2-dimensional polyhedron :) ? Jan 7, 2014 at 4:04
• Yes exactly. I think a two-dimensional polyhedron is a polygon, and I think that's exactly what's going on in the degenerate case. This reminds me a lot of the classification of finite subgroups of $SO(3)$. I'm going to think a little bit about how to prove a full classification from this algebraic perspective. Jan 7, 2014 at 4:34
• Great! That's an elegant approach. Jan 7, 2014 at 18:33

These are the equivalent images for the Kepler-Poinsot polyhedra:

One way to see why one of the polygons (the "green" one) is invariant under taking the dual of the polyhedron is to note that if the faces of the polyhedron are $n$-gons and the vertex figures are $v$-gons, the "yellow" polygon (whose vertices are all the midpoints of all the edges incident on some vertex of the polyhedron) will be a $v$-gon, with each edge subtending an angle of $\frac{2\pi}{v}$, while the "blue" polygon centred on the vertex will be a $\frac{2n}{n-2}$-gon, where each edge subtends an angle of $\frac{(n-2)\pi}{n}=\pi-\frac{2\pi}{n}$ (i.e. the interior angle between the edges of an $n$-gon).

The difference of the squares of the sines of half these angles is:

$$\sin\left(\frac{\pi}{v}\right)^2 - \sin\left(\frac{(n-2)\pi}{2n}\right)^2 = -\cos\left(\pi \left(\frac{1}{n}+\frac{1}{v}\right)\right) \cos\left(\pi \left(\frac{1}{n}-\frac{1}{v}\right)\right)$$

where the RHS is now clearly invariant under an exchange of $n$ and $v$. And of course this quantity is the squared sine of half the angle subtended by each edge of the "green" polygon.

And for completeness, here are some images of the construction in the degenerate case, where the polyhedron becomes a polygon and the "yellow" polygon becomes a digon:

• Cool! There's some kind of duality going on here, where the right triangle has vertices of the hypotenuse a vertex and center of one polyhedron, and vice versa for the other (with the tetrahedron self-dual, corresponding to the isosceles right triangle). Jan 7, 2014 at 18:21
• @IanAgol: I think it's the octahedron, not the tetrahedron, where the associated right triangle is isosceles. In this odd sense of duality, the tetrahedron and the cube are "dual", the dodecahedron and great stellated dodecahedron are "dual", the icosahedron and small stellated dodecahedron are "dual", and the great dodecahedron and great icosahedron are "dual". Jan 8, 2014 at 23:34
• What's required for that duality is that both polyhedra have the same vertex figure, and that it's possible to arrange them so each has a vertex at the other's centre and all midpoints of edges incident on these vertices match up between the two polyhedra. So a cube and tetrahedron have the same triangular vertex figure and the right metric properties to be dual in this sense. Jan 8, 2014 at 23:46
• Good point, I mixed it up. A consultation of your figures makes this apparent. I think there ought to be a geometric answer to this question, which makes the duality obvious. Given 3 regular polygons with unit edge lengths forming the altitudes of a right triangle, one take a fundamental region for each polygon (so for a polygon of degree $d$, a right triangle with angle $2\pi/d$ and an opposite edge of length $\frac12$, so that $2d$ of these make up the polygon). Then you can take the red right triangle, and add on these 3 right triangles meeting along an edge. There are two ways to extend Jan 9, 2014 at 3:37
• this tetrahedron (along the green face) to two orthoschemes, which form the fundamental domains for the two dual polyhedra quotiented by their respective symmetry groups. Jan 9, 2014 at 3:39