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# 13.1: Angle of parallelism

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Let $$P$$ be a point off an h-line $$\ell$$. Drop a perpendicular $$(PQ)_h$$ from $$P$$ to $$\ell$$; let $$Q$$ be its foot point. Let $$\phi$$ be the smallest value such that the h-line $$(PZ)_h$$ with $$|\measuredangle_h Q P Z|=\phi$$ does not intersect $$\ell$$.

The value $$\phi$$ is called the angle of parallelism of $$P$$ to $$\ell$$. Clearly, $$\phi$$ depends only on the h-distance $$s=PQ_h$$. Further, $$\phi(s) \to \pi/2$$ as $$s \to 0$$, and $$\phi(s) \to 0$$ as $$s \to \infty$$. (In the Euclidean geometry, the angle of parallelism is identically equal to $$\pi/2$$.)

If $$\ell$$, $$P$$, and $$Z$$ are as above, then the h-line $$m=(PZ)_h$$ is called asymptotically parallel to $$\ell$$. In other words, two h-lines are asymptotically parallel if they share one ideal point. (In hyperbolic geometry, the term parallel lines is often used for asymptotically parallel lines; we do not follow this convention.)

Given $$P \not\in \ell$$, there are exactly two asymptotically parallel lines thru $$P$$ to $$\ell$$; the remaining parallel lines are called ultra parallel.

On the diagram, the two solid h-lines passing thru $$P$$ are asymptotically parallel to $$\ell$$; the dashed h-line is ultra parallel to $$\ell$$.

Exercise $$\PageIndex{1}$$

Show that two distinct h-lines $$\ell$$ and $$m$$ are ultraparallel if and only if they have a common perpendicular; that is, there is an $$h$$-line $$n$$ such that $$n \perp \ell$$ and $$n \perp m$$.

Hint

"only-if" part. Suppose $$\ell$$ and $$m$$ are ultraparallel; that is, they do not intersect and have distinct ideal points. Denote the ideal points by $$A, B, C$$, and $$D$$; we may assume that they appear on the absolute in the same order. In this case the h-line with ideal points $$A$$ and $$C$$ intersects the h-line with ideal points $$B$$ and $$D$$. Denote by $$O$$ their point of intersection.

By Lemma 12.3.1, we can assume that $$O$$ is the center of absolute. Note that $$\ell$$ is the reflection of $$m$$ across $$O$$ in the Euclidean sense. Drop an h-perpendicular $$n$$ from $$O$$ to $$\ell$$, and show that $$n \perp m$$.

"If" part. Suppose $$n$$ is a common perpendicular. Denote by $$L$$ and $$M$$ its points of intersection with $$\ell$$ and $$m$$ respectively. By Lemma 12.3.1, we can assume that the center of absolute $$O$$ is the h-midpoint of $$L$$ and $$M$$. Note that in this case $$\ell$$ is the reflection of m across $$O$$ in the Euclidean sense. It follows that the ideal points of the h-lines $$\ell$$ and $$m$$ are symmetric to each other. Therefore, if one pair of them coincides, then so is the other pair. By Exercise 12.1.1, $$\ell = m$$, which contradicts the assumption $$\ell \ne m$$.

Proposition $$\PageIndex{1}$$

Let $$Q$$ be the foot point of $$P$$ on h-line $$\ell$$. Then

$$PQ_h=\dfrac{1}{2} \cdot \ln \dfrac{1 + \cos \phi}{1 - \cos \phi},$$

where $$\phi$$ is the angle of parallelism of $$P$$ to $$\ell$$.

In particular, if $$P \notin \ell$$ and $$\beta =|\measuredangle_h XPY|$$ for some points $$X,Y\in\ell$$, then

$$PQ_h < \dfrac{1}{2} \cdot \ln \dfrac{1+\cos \dfrac{\beta}{2}}{1- \cos \dfrac{\beta}{2}}.$$

Proof

Applying a motion of the h-plane if necessary, we may assume $$P$$ is the center of the absolute. Then the h-lines thru $$P$$ are the intersections of Euclidean lines with the h-plane.

Let $$A$$ and $$B$$ denote the ideal points of $$\ell$$. Without loss of generality, we may assume that $$\angle APB$$ is positive. In this case

$$\phi=\measuredangle QPB=\measuredangle APQ=\dfrac{1}{2} \cdot\measuredangle APB.$$

Let $$Z$$ be the center of the circle $$\Gamma$$ containing the h-line $$\ell$$. Set $$X$$ to be the point of intersection of the Euclidean segment $$[AB]$$ and the line $$(PQ)$$.

Note that, $$PX = \cos \phi$$. Therefore, by Lemma 12.3.2,

$$PX_h=\ln \dfrac{1+\cos\phi}{1-\cos\phi}.$$

Note that both angles $$PBZ$$ and $$BXZ$$ are right. Since the angle $$PZB$$ is shared, $$\triangle ZBX \sim \triangle ZPB$$. In particular,

$$ZX \cdot ZP = ZB^2;\]$$

that is, $$X$$ is the inverse of $$P$$ in $$\Gamma$$.

The inversion in $$\Gamma$$ is the reflection of the h-plane across $$\ell$$. Therefore

$$\begin{array} {rcl} {PQ_h} & = & {QX_h =} \\ {} & = & {\dfrac{1}{2} \cdot PX_h =} \\ {} & = & {\dfrac{1}{2} \cdot \ln \dfrac{1 + \cos \phi}{1 - \cos \phi}.} \end{array}$$

The last statement follows since $$\phi > \dfrac{\beta}{2}$$ and the function

$$\phi \mapsto \dfrac{1}{2} \cdot \ln \dfrac{1+\cos\phi}{1-\cos\phi}$$

is decreasing in the interval $$(0,\dfrac{\pi}{2}]$$.

Exercise $$\PageIndex{2}$$

Let $$ABC$$ be an equilateral h-triangle with side $$100$$. Show that

$$|\measuredangle_h ABC|<\frac1{10\ 000\ 000\ 000}.$$

Hint

By triangle inequality, the h-distance from $$B$$ to $$(AC)_h$$ is at least 50. It remains to estimate $$|\measuredangle_h ABC| using Proposition \(\PageIndex{1}$$. The inequalities $$\cos \phi \le 1 - \dfrac{1}{10} \cdot \phi^2$$ for $$|\phi| < \dfrac{\pi}{2}$$ and $$e^3 > 10$$ should help to finish the proof.