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I have a question about the tesselation of the upper half plane via Ford Circles. Wikipedia says

By interpreting the upper half of the complex plane as a model of the hyperbolic plane (the Poincaré half-plane model) Ford circles can also be interpreted as a tiling of the hyperbolic plane by horocycles.

As far as I understand, the tiling is done by the hyperbolic triangles we get from the Ford Circles and not by the circles itself, is that right? And why do I need horocycles here? Can't I just say that the tiling is done by triangles, whose corners lie on the boundary $\partial \mathbb{H}$ (i.e., are ideal points)?

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  • $\begingroup$ this is in Conway's little book maa.org/press/maa-reviews/the-sensual-quadratic-form There are two tilings, one is the circles which are horocycles in the upper half plane. The other: for each pair of tangent horocycles in the first thing, draw the geodesic tangent to both at the point where they meet. Continue this geodesic only until it bumps into another geodesic. $\endgroup$ – Will Jagy Apr 18 '16 at 17:43
  • $\begingroup$ Why do you assume that the tiling is by triangles not circles? I was about to argue against this assumption, but then I realized that you'd have gaps between the circles; something you usually wouldn't have in a tiling. If you fill those gaps with circles, those are no longer horocycles. Is that the motivation behind your assumption as well? $\endgroup$ – MvG Apr 19 '16 at 1:41
  • $\begingroup$ @Will but how can the circles define a tiling when there are gaps between them? And for the other tiling: Don't you mean that we draw the geodesics orthogonal to the meeting point rather than tangent to it? $\endgroup$ – guest Apr 19 '16 at 11:47
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I think there are mistakes in the wikipedia article

It is not about horocycles but about apeirogons ( https://en.wikipedia.org/wiki/Apeirogon#Apeirogons_in_hyperbolic_plane ,infinite sided polygons) and it is not about the geodesics that intersect where the horocycles meet but about the geodesic that is tangent where the horocycles meet. (the formula given for the geodesic with intersection points is not the geodesic that you need).

From this tangent geodesic take the segment that contains the tangent point and that is between the points where they meet other geodesics

the polygon by connecting these tangent segments form an apeirogon.

in hyperbolic geometry apeirogons have a side length s that depends on the angle between two consequtive sides

$$\Pi(\frac{1}{2} \alpha) = 2s $$

($\Pi$ is the angle of parallelism function, https://en.wikipedia.org/wiki/Angle_of_parallelism )

In the particular Ford circle case:

the apeirogon is the circumscribed apreigon of the horocycle represented by the Ford circle (so the Ford circle is the inscribed horocycle of the apeirogons)

Hopes this helps

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  • $\begingroup$ (I am posting this comment below both answers, since they both helped me a lot) As far as I understand know, we have the following: 1) A tesselation of $\mathbb{H}$ by triangles (whose corners are ideal points), defined through orthogonal geodescis 2) A tesselation of $\mathbb{H}$ by apeirogons, defined through tangent geodescis 3) NO tesselation by Ford circles (i.e. horocycles), since that would leave gaps. Is that correct? $\endgroup$ – guest Apr 21 '16 at 9:22
  • $\begingroup$ as i see it:you have not done 1) yet (but it is possible) and indeed No tesselation hy horocycles, but you can tesselate with apeirogons. see also my new question math.stackexchange.com/questions/1752547/… that i think proofs that apeirogons are not closed curves (ps my question is about an inscribed apeirogon , you use a circumscribed apeirogon, don't mix them up) i rewrote en.wikipedia.org/wiki/Ford_circle $\endgroup$ – Willemien Apr 21 '16 at 10:11
  • $\begingroup$ Willemien, I should emphasize that the tree diagram is a specific example of circumscribing the Ford circles, in that each edge of the infinite tree is tangent to two Ford circles at the point where they are tangent. The construction is a simple outcome using well-known concepts, but i was unable to find any online picture of it. See my answer and, for that matter, the book maths.ed.ac.uk/~aar/papers/conwaysens.pdf as a pdf $\endgroup$ – Will Jagy Apr 21 '16 at 17:40
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It seems the diagram that the OP has in mind is the most popular one, for example http://www-bcf.usc.edu/~fbonahon/STML49/FareyFord.html

John Horton Conway emphasizes a different picture that he calls the "topograph." Geodesic segments are added tangent to each pair of tangent Ford circles. These segments meet up with other such segments. the result is a countably infinite tree, each vertex has valence 3. For that matter, Conway first produces the tree and then says ""...have inscribed circles" which turn out to be the Ford circles. See Conway http://www.maa.org/press/maa-reviews/the-sensual-quadratic-form and Stillwell http://www.springer.com/us/book/9780387955872

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  • $\begingroup$ (I am posting this comment below both answers, since they both helped me a lot) As far as I understand know, we have the following: 1) A tesselation of $\mathbb{H}$ by triangles (whose corners are ideal points), defined through orthogonal geodescis 2) A tesselation of $\mathbb{H}$ by apeirogons, defined through tangent geodescis 3) NO tesselation by Ford circles (i.e. horocycles), since that would leave gaps. Is that correct? $\endgroup$ – guest Apr 21 '16 at 9:22
  • $\begingroup$ @guest almost correct. The apeirogons are each inscribed in a Ford circle. Conway's tree, um, branches circumscribe the Ford circle. $\endgroup$ – Will Jagy Apr 21 '16 at 17:25

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