13.1: Angle of parallelism
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 \(\ell\) .
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
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"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 \(\ell \ne m\).
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
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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
is decreasing in the interval \((0,\dfrac{\pi}{2}]\) .
Let \(ABC\) be an equilateral h-triangle with side \(100\) . Show that
\(|\measuredangle_h ABC|<\frac1{10\ 000\ 000\ 000}.\)
- Hint
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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.