Ultraparallel theorem: Difference between revisions

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{{short description|Theorem in hyperbolic geometry}}

In [[hyperbolic geometry]], the '''ultraparallel theorem''' states that every pair of [[ultraparallel line]]s (lines that are not intersecting and not [[limiting parallel]]) has a unique common [[perpendicular]] hyperbolic line.

[[File:Ultraparallel.png|thumb|200px|right|[[Poincaré disc]]: The pink line is ultraparallel to the blue line and the green lines are [[limiting parallel]] to the blue line.]]

In [[hyperbolic geometry]], two lines are said to be '''ultraparallel''' if they do not intersect and are not [[limiting parallel]].

In [[hyperbolic geometry]], theThe '''ultraparallel theorem''' states that every pair of [[(distinct) ultraparallel line]]s (lines that are not intersecting and not [[limiting parallel]]) has a unique common [[perpendicular]] (a hyperbolic line which is perpendicular to both lines).

==Hilberts construction==

==HilbertsHilbert's construction==

Let r and s be two non-intersecting lines.

Let {{mvar|r}} and {{mvar|s}} be two non-intersectingultraparallel lines.

From any two points A and C on s draw AB and CB' perpendicular to r. (B and B' on r)

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' the desired common perpendicular joins the midpoints AC and BB'(by the symmetry of the isocleses birectangle ACB'B ).

.

If not, suppose AB < CB'.

Take A' on CB' so that A'B' = AB.

Trough A' draw a line s', making the same angle with A'B' that s makes with AB.

Then s meets s' in an ordinary point D.

Take a point D' on ray AC so that AD' = A'D.

If it happens that AB = CB', then the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the isocleses[[Saccheri birectanglequadrilateral]] ACB'B ).

Then the perpendicular bisector of DD' is also perpendicular to r.<ref>{{cite book|last1=coxeter|title=non euclidean geometry|isbn=978-0-88385-522-5|pages=190-192}}</ref>

If not, we may suppose AB < CB' without loss of generality. Let E be a point on the line s on the opposite side of A from C. Take A' on CB' so that A'B' = AB. Through A' draw a line s' (A'E') on the side closer to E, so that the angle B'A'E' is the same as angle BAE. Then s' meets s in an ordinary point D'. Construct a point D on ray AE so that AD = A'D'.

Then D' ≠ D. They are the same distance from r and both lie on s. So the perpendicular bisector of DDD'D (a segment of s) is also perpendicular to r.<ref>{{cite book|last1=coxeterH. S. M. Coxeter|titleauthor1-link=nonH. S. M. Coxeter|title=Non-euclidean geometryGeometry|date=17 September 1998|isbn=978-0-88385-522-5|pages=190-192190–192}}</ref>

(If r and s were asymptotically parallel rather than ultraparallel, this construction would fail because s' would not meet s. Rather s' would be limiting parallel to both s and r.)

==Proof in the Poincaré half-plane model ==

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Let

:<math>a < b < c < d</math>

:<math>a < b < c < d</math>

be four distinct points on the [[abscissa]] of the [[Cartesian plane]]. Let <math>p</math> and <math>q</math> be [[semicircle]]s above the abscissa with diameters <math>ab</math> and <math>cd</math> respectively. Then in the [[Poincaré half-plane model]] HP, <math>p</math> and <math>q</math> represent ultraparallel lines.

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Compose the following two [[hyperbolic motion]]s:

:<math>x \to x-a\,</math>

:<math>\mbox{inversion in the unit semicircle.}\,</math>

Then <math>a \to \infty</math>, <math>\quad b \to (b-a)^{-1},\quad c \to (c-a)^{-1},\quad d \to (d-a)^{-1}.</math>

Now continue with these two hyperbolic motions:

:<math>x \to x-(b-a)^{-1}\,</math>

:<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>

The four hyperbolic motions that produced <math>z</math> above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius <math>\sqrt{z}</math> to yield the unique hyperbolic line perpendicular to both ultraparallels <math>p</math> and <math>q</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 ; the chord of intersection will be the desired common perpendicular of the ultraparallel lines.

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.{{how}}

<!-- ??? "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 , they not always form a quadrilateral, nor is the quadrilateral always convex see also http://math.stackexchange.com/q/1382739/88985 -->

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>

<!-- ??? "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 , they not always form a quadrilateral -->

==References==

{{reflistReflist}}

* [[Karol Borsuk]] & [[Wanda Szmielew]] (1960) ''Foundations of Geometry'', page 291.

[[Category:Articles containing proofs]]