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Mathematics LibreTexts

13.1: Angle of parallelism

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    23661
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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 \(\pi/2\).)

    截屏2021-02-24 下午2.07.43.png

    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

    截屏2021-02-24 下午2.24.54.png

    "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

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

    截屏2021-02-24 下午2.16.20.png

    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.