Moduli of Right-Angled Hyperbolic Pentagons

A surprise appearance of the golden ratio in hyperbolic geometry.

There are no right angled quadrilaterals in the hyperbolic plane. The angles of a hyperbolic $n$-gon must total less than $(n-2)\pi$ — the deficit is exactly its area — and four right angles come to $2\pi$ on the nose. Five right angles come to $\frac{5\pi}{2}$, comfortably under $3\pi$, leaving an area of $\frac{\pi}{2}$ to spare. So the pentagon is the first polygon that can be right angled at every corner.

Existence of at least one is easy: we can construct a regular hyperbolic pentagon (with all sides the same) by constructing an appropriate triangle, and repeating.

But just like the square isn’t the only right angled quadrilateral in Euclidean geometry, neither is this the only right angled pentagon in the hyperbolic plane. The goal of this note is to completely classify right angled hyperbolic pentagons. In particular, we prove

Thus the moduli space of right angled pentagons is parameterized by

$\mathcal{P}\cong\bigl\{\,a,b>0\ \big|\ \sinh a\,\sinh b>1\,\bigr\}$

though it is worth being precise about what a point of this region is. It is not quite a pentagon: it is a pentagon together with a choice of which adjacent pair is $(a,b)$. There are five corners to pick, and two ways to order the pair at each, so ten choices in all. What the region parameterizes is the marked moduli space.

Those ten relabelings act on the region itself. Moving the marking to the next corner sends $(a,b)$ to $(b,c)$ — a map of $\mathcal{P}$ to itself of order five — and reading a corner the other way sends $(a,b)$ to $(b,a)$. The two together generate the dihedral group $D_5$, and the moduli space of pentagons up to isometry (reflections allowed) is the quotient

$\mathcal{M}\cong\mathcal{P}/D_5$

A pentagon with no symmetry has ten distinct markings, and so ten points of $\mathcal{P}$ lying above it. Symmetric ones have fewer: a mirror-symmetric pentagon has five, and exactly one point of the region is fixed by the whole group. Being fixed forces all five sides equal, so that point is the regular pentagon — and there is precisely one of it.

The Proof

The main idea is simple to state. Draw sides of length $a,b$ perpendicular to each other, then draw common perpendiculars at the endpoints. These either intersect, or they don’t. If they do, no pentagon. If they do not, then there’s a unique shortest common perpendicular. This fifth side forms a right angled pentagon with the others.

To work out the details, we need to find a way to tell when exactly the two sides we extend orthogonally from the segments of length $a,b$ remain disjoint. And to be really quantitative, when they are ultraparallel we should like to know the length of their common perpendicular. The idea is to bring in the trigonometry of hyperbolic triangles, by drawing the geodesic through the endpoints of segments $a$ and $b$. This subdivides each of the right angles there into pairs, $\alpha,\bar{\alpha}$ opposite $a$ and $\beta,\bar{\beta}$ opposite $b$. Call the length of this segment $X$.

The Law of Cosines

The chord cuts the picture into two pieces. One of them is a right triangle, so we have access to many trigonometric relations between $a,b,\bar{\alpha},\bar{\beta}$ and $X$. The other piece is still a mystery: maybe it’s a triangle (finite or ideal) or maybe it’s infinite in area (if the geodesics are ultraparallel). There’s a nice trick here - in the triangle case, one is naturally tempted to employ the law of cosines: but in fact the hyperbolic law of cosines[1^] works in both cases. First, for the triangle - we can compute the angle $\chi$ opposite $X$ via

[1^]: There are actually two hyperbolic laws of cosines. The other one is a direct analog of the Euclidean case; and the one we utilize here has no analog in Euclidean geometry as it allows one to calculate a side length in terms of only information about angles!

$\cos\chi = -\cos\alpha\,\cos\beta+\sin\alpha\,\sin\beta\,\cosh X$

But in the ultraparallel case, the length $d$ of the common perpendicular satisfies the same relation:

$\cosh d = -\cos\alpha\,\cos\beta+\sin\alpha\,\sin\beta\,\cosh X$

Thus, given any leg of length $X$ with angles $\alpha,\beta$ at the endpoints we can compute the common right hand side of this formula, and fully understand the situation:

• If it is less than or equal to[2^] $1$, then the resulting shape is a triangle, and its new angle is the $\arccos$ of the quantity.
• If it is greater than $1$, then it forms a quadrilateral whose new side is the $\operatorname{arccosh}$ of the quantity.

[2^]: When it equals precisely 1, the triangle has an ideal vertex opposite $X$ so the angle is zero. The two sides are then asymptotic — they neither meet nor admit a common perpendicular, and it is exactly the knife edge between the two cases.

Relating to the Triangle

We want an expression in terms of $a$ and $b$, but right now the law of cosines is in terms of $\bar{\alpha},\bar{\beta}$ and $X$. First, as $\alpha+\bar{\alpha}=\pi/2$ and $\beta+\bar{\beta}=\pi/2$ the trigonometric functions of either of these determine the other. This observation, together with the trigonometry of right triangles implies

$$\begin{aligned} \cos\alpha &= \sin\bar\alpha = \frac{\sinh a}{\sinh X}, &\qquad \cos\beta &= \sin\bar\beta = \frac{\sinh b}{\sinh X},\\[8pt] \sin\alpha &= \cos\bar\alpha = \frac{\tanh b}{\tanh X}, &\qquad \sin\beta &= \cos\bar\beta = \frac{\tanh a}{\tanh X}. \end{aligned}$$

And, of course, the pythagorean identity for the sides:

$\cosh X = \cosh a\,\cosh b$

The Calculation

Starting with the mystery quantity determined by the law of cosines, we substitute terms computed from the triangle above. Watch what happens to $X$: it enters both terms and then leaves again, and the answer depends only on $a$ and $b$.

$$\begin{aligned} ?&=-\cos\alpha\,\cos\beta+\sin\alpha\,\sin\beta\,\cosh X\\[6pt] &=-\frac{\sinh a}{\sinh X}\cdot\frac{\sinh b}{\sinh X} +\frac{\tanh b}{\tanh X}\cdot\frac{\tanh a}{\tanh X}\cosh X \end{aligned}$$

Expanding the definitions of $\tanh a$, $\tanh b$ in the second term and using the pythagorean identity for $\cosh X$:

$$\begin{aligned} \frac{\tanh a}{\tanh X}\cdot\frac{\tanh b}{\tanh X}\cosh X &=\frac{1}{\tanh^2 X}\left(\frac{\sinh a}{\cosh a}\cdot\frac{\sinh b}{\cosh b}\right)\cosh X\\[6pt] &=\frac{1}{\tanh^2 X}\left(\frac{\sinh a}{\cosh a}\cdot\frac{\sinh b}{\cosh b}\right)\cosh a\,\cosh b\\[6pt] &=\frac{\sinh a\,\sinh b}{\tanh^2 X} \end{aligned}$$

Plugging this back in:

$$\begin{aligned} ?&=-\frac{\sinh a\,\sinh b}{\sinh^2 X}+\frac{\sinh a\,\sinh b}{\tanh^2 X}\\[6pt] &=\sinh a\,\sinh b\left(\frac{1}{\tanh^2 X}-\frac{1}{\sinh^2 X}\right)\\[6pt] &=\sinh a\,\sinh b\left(\coth^2 X-\operatorname{csch}^2 X\right)\\[6pt] &=\sinh a\,\sinh b \end{aligned}$$

where the final line uses $\coth^2 X-\operatorname{csch}^2 X=1$, the cosecant–cotangent identity. After all that cancellation we see that our mystery quantity is just $\sinh a\,\sinh b$, and when this quantity is $\leq 1$ the edges intersect (perhaps ideally) and there is no possible hyperbolic pentagon constructible from the configuration. But when $\sinh a\,\sinh b>1$, the sides are ultraparallel and there is a unique geodesic intersecting each orthogonally. That is, there is a unique hyperbolic pentagon with all right angles, proving our theorem.

Description of All Sides

With a precise characterization of all right angled pentagons in hand, we turn to the quantitative problem of determining their side lengths. Label the sides $a,b,\ldots, e$ clockwise around the pentagon as below:

First, the law of cosines determines $d$ in terms $a$ and $b$. This identity turns out to be incredibly useful well beyond the study of pentagons, so we box it off:

Applying this rule to other pairs of sides, we see that $\cosh a = \sinh c\,\sinh d$, and as $d$ and $a$ are known, this gives $c$. Likewise, $\sinh b\,\sinh c = \cosh e$ and since we already know $b$ and $c$ this fixes $e$. Unpacking these give explicit descriptions in terms of $a,b$:

Appendix: The Regular Pentagon

One point of $\mathcal{P}$ is fixed by the whole $D_5$ action, and it is worth measuring: the regular pentagon, where all five sides agree. Call this common side length $e$. Then by the trigonometry of pentagons we see

$\sinh e\,\sinh e = \cosh e$

And as $\sinh^2 e =\cosh^2 e-1$, the quantity $u=\cosh e$ satisfies the quadratic polynomial $u^2-1=u$, or

$u^2=u+1$

The defining equation for the golden ratio, which has $\varphi=\frac{1+\sqrt{5}}{2}$ as its unique positive root.