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→Hilbert's construction: asymptotically -> limiting, since the default choice for Wikipedia is "limiting parallel". (I personally prefer "horoparallel", but all that matters is consistency.) |
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{{short description|Theorem in hyperbolic geometry}}
[[File:Ultraparallel.png|thumb|200px|right|[[Poincaré disc
In [[hyperbolic geometry]], two lines are said to be '''ultraparallel''' if they do not intersect and are not [[limiting parallel]].
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==Hilbert's construction==
Let {{mvar|r}} and {{mvar|s}} be two ultraparallel lines.
From any two distinct points {{mvar|A}} and {{mvar|C}} on s draw {{mvar|AB}} and {{mvar|CB'}} perpendicular to {{mvar|r}} with {{mvar|B}} and {{mvar|B'}} on {{mvar|r}}.
If it happens that AB = CB', then the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the [[Saccheri quadrilateral]] ACB'B).
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:<math>x \to \left [ (c-a)^{-1} - (b-a)^{-1} \right ]^{-1} x</math>
Then <math>a</math> stays at <math>\infty</math>, <math>b \to 0</math>, <math>c \to 1</math>, <math>d \to z</math> (say). The unique semicircle, with center at the origin, perpendicular to the one on <math>1z</math> must have a radius tangent to the radius of the other. The right triangle formed by the abscissa and the perpendicular radii has [[hypotenuse]] of length <math>\begin{matrix} \frac{1}{2} \end{matrix} (z+1)</math>. Since <math>\begin{matrix} \frac{1}{2} \end{matrix} (z-1)</math> is the radius of the semicircle on <math>1z</math>, the common perpendicular sought has radius-square
:<math>\frac{1}{4} \left [ (z+1)^2 - (z-1)^2 \right ] = z.</math>
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* The [[Pole and polar|poles]] of these two lines are the respective intersections of the [[tangent line]]s to the boundary [[circle]] at the endpoints of the chords.
* Lines ''perpendicular'' to line ''l'' are modeled by chords whose extension passes through the pole of ''l''.
* Hence we draw the unique line between the poles of the two given lines, and intersect it with the boundary circle
If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter it is also perpendicular in the Beltrami-Klein model, and so we draw a line through the pole of the other line intersecting the diameter at right angles to get the common perpendicular.
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* If both chords are diameters, they intersect.(at the center of the boundary circle)
* If only one of the chords is a diameter, the other chord projects orthogonally down to a section of the first chord contained in its interior, and a line from the pole orthogonal to the diameter intersects both the diameter and the chord.
* If both lines are not diameters, then we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the [[unit circle]] inscribed within it.{{how|date=August 2015}} The poles are opposite vertices of this quadrilateral, and the chords are lines drawn between adjacent sides of the vertex, across opposite corners. Since the quadrilateral is convex,{{why|date=August 2015}} the line between the poles intersects both of the chords drawn across the corners, and the segment of the line between the chords defines the required chord perpendicular to the two other chords.
<!-- ??? "then we may extend the tangents drawn from each pole to produce a [[quadrilateral]] with the unit circle inscribed within it " this is not always the case
Alternatively, we can construct the common perpendicular of the ultraparallel lines as follows: the ultraparallel lines in Beltrami-Klein model are two non-intersecting chords. But they actually intersect outside the circle. The polar of the intersecting point is the desired common perpendicular.<ref>W. Thurston, ''Three-Dimensional Geometry and Topology'', page 72</ref>
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