Hyperbolic Lunes

Section 8.7 Hyperbolic Lunes

Can you find the area of a hyperbolic triangle using some sort of “hyperbolic lune”?

Lunes were introduced in Section 8.6 as segments of spheres. Start at a point, and go off in two (non-collinear) directions. Since all lines intersect twice on the sphere, the region between the two lines is bounded; a lune is one of the regions so determined.

Let's try the same thing in the Poincaré disk. Start at a point, and go off in two (non-collinear) directions. But now the lines go off to infinity!

Mind you, we know where they go – they reach the circle that bounds the Poincaré disk. Points on this circle are called ideal points; they are not part of the Poincaré disk, but are added to it. Ideal points can be added to any model of hyperbolic geometry; they represent the intersection points of “barely parallel” lines, namely the limiting case of a line through a given point parallel to a given line in which the angle of parallelism (see Section 5.3) is actually achieved.

So let's make an infinite triangle by connecting the dots, as shown in Figure 8.7.1. Any such triangle, which we will also refer to as a hyperbolic lune, has three infinite sides. Only one of these sides is a hyperbolic line; the other two are hyperbolic rays. And two of its angles are zero! We use the remaining angle, at the vertex within the Poincaré disk, to label the hyperbolic lune.

What if all three vertices are ideal points? Such triangles, in which all three angles are zero, are called ideal triangles; two such triangles are shown in Figure 8.7.2. Both infinite and ideal triangles can be thought of as limits to infinity of ordinary triangles. It should therefore come as no surprise that SAS congruence still holds for these triangles, so long as the angle used for hyperbolic lunes is the angle in the interior of the disk. Thus, two lunes with the same (nonzero) angle are congruent, and all ideal triangles are congruent to each other. ¹

However, SSS congruence fails, otherwise all hyperbolic lunes would be congruent! Where does the proof fail that shows that SSS follows from SAS?

We are now ready to try to use hyperbolic lunes to determine the area of a triangle. Denote the area of a hyperbolic lune with (nonzero) angle \(\alpha\) by \(A(\alpha)\text{,}\) and the area of the ideal triangle as \(A_I=A(0)\text{.}\)

Claim 8.7.3. Step 1: Supplementary Hyperbolic Lunes.

Two supplementary lunes combine to make an ideal triangle, that is ²

\begin{equation} A(\alpha)+A(\pi-\alpha)=A_I.\label{lsupp}\tag{8.7.1} \end{equation}

We use radians instead of degrees for convenience.

Proof.

See Figure 8.7.4.

Figure 8.7.4. Supplementary hyperbolic lunes.

Claim 8.7.5. Step 2: Addition of Hyperbolic Lunes.

Hyperbolic lunes can be “added” in the sense that

\begin{equation} A(\pi-\alpha)+A(\pi-\beta)=A\bigl(\pi-(\alpha+\beta)\bigr) .\label{ladd}\tag{8.7.2} \end{equation}

Proof.

See Figure 8.7.6. The angles \(\alpha\) and \(\beta\) are below point \(A\text{,}\) separated by the vertical line. So the lune at the bottom has angle \(\alpha+\beta\text{,}\) and the two lunes above it have angles \(\pi-\alpha\) and \(\pi-\beta\text{,}\) respectively. From the figure, we therefore have

\begin{equation} A(\alpha+\beta)+A(\pi-\alpha)+A(\pi-\beta) = A_I\tag{8.7.3} \end{equation}

and using Claim 8.7.3 leads immediately to (8.7.2).

Figure 8.7.6. Combining hyperbolic lunes.

Claim 8.7.5 says that the function \(f(\alpha)=A(\pi-\alpha)\) must be linear, and furthermore that \(f(0)=0\text{!}\) Thus, \(A(\pi-\alpha) = k\alpha\text{,}\) or equivalently

\begin{equation} A(\alpha) = k(\pi-\alpha)\tag{8.7.4} \end{equation}

for some constant \(k\text{.}\) Furthermore, using Claim 8.7.3, we also have

\begin{equation} A_I = k\pi .\tag{8.7.5} \end{equation}

We are finally ready to use our lunes to determine the area of a hyperbolic triangle. Consider the triangle shown in Figure 8.7.7, together with three exterior hyperbolic lunes. Denoting the angles at \(A\text{,}\) \(B\text{,}\) \(C\) by \(\alpha\text{,}\) \(\beta\text{,}\) \(\gamma\text{,}\) respectively, the three lunes have the supplementary angles \(\pi-\alpha\text{,}\) \(\pi-\beta\text{,}\) \(\pi-\gamma\text{,}\) respectively, so from the figure we see that

\begin{equation} A_T + A(\pi-\alpha) + A(\pi-\beta) + A(\pi-\gamma) = A_I\tag{8.7.6} \end{equation}

where \(A_T\) denotes the area of the triangle. Using Claim 8.7.5, we obtain the expected expression for \(A_T\text{,}\) namely

\begin{equation} A_T = k (\pi - \alpha - \beta - \gamma)\label{hsum}\tag{8.7.7} \end{equation}

thus using lunes to show that hyperbolic area is proportional to the defect.

Figure 8.7.7. A hyperbolic triangle and its lunes combine to make an ideal triangle.

A noteworthy feature of this argument is that we have \(shown\) that \(k\lt\infty\text{,}\) since (8.7.7) is clearly finite. Thus, hyperbolic lunes and ideal triangles have finite area!

How can a triangle with infinite sides have finite area? Easy! Consider painting an infinite fence:

A fence is built along the \(x\)-axis for \(x\ge1\text{,}\) with height given by \(ke^{-x}\) for some constant \(k\text{.}\) What is the area of the fence? How much paint is needed to paint it?