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In Euclidean geometry Thales' Theorem says that if you view a diameter of a circle from any point on the perimeter it occupies exactly $90$ degrees in your field of view.

More generally for any segment $s$ and any angle $\theta$ the set of points from which $s$ occupies exactly an angle of $\theta$ in the field of view is the union of two circle arcs. This follows from the Inscribed Angle Theorem.

In hyperbolic geometry Thales' theorem is false. In fact, there is a distance $d > 0$ such that if you are at distance $d$ from a geodesic then the whole thing occupies exactly $90$ degrees in your field of view. Therefore on a circle of radius $d$ or more there are points from which a diameter occupies less than $90$ degrees.

At least in the case of a geodesic the set of points from which it occupies $90$ degrees is a nice and well known curve (i.e. an equidistant: a curve at a fixed distance from a geodesic). In fact the curve is even an Euclidean circle arc in either the Poincaré disk or the upper half-plane model.

My question is the following: Given a finite segment $s$ in the hyperbolic plane, what does the set of points from which the segment occupies exactly $90$ degrees look like? Is it a well known curve? Has it appeared or been used in relation to other questions?

Also one can ask the same questions for a general angle $\theta$.

Edit: Just realized this is has been asked before on MO. Sorry.

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    $\begingroup$ This question could be related: mathoverflow.net/questions/33977/… $\endgroup$ Commented Jan 25, 2012 at 1:08
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    $\begingroup$ As I mentioned in an answer to the other question, if one positions the geodesic segment in the Klein model to have one endpoint at the origin, then the curve is a circle for which the segment is a diameter. $\endgroup$
    – Ian Agol
    Commented Jan 25, 2012 at 4:38
  • $\begingroup$ I added this comment to the other question, but since I saw your question first, I'll repeat it here: Google books pulled up the following page from Richter-Gebert's recent book "Perspectives on Projective Geometry", though I can't access most of the discussion: books.google.com/… $\endgroup$
    – j.c.
    Commented Jan 25, 2012 at 10:35
  • $\begingroup$ @Agol Very nice! Thanks! Do you know of something that works for angles other than $90$ degrees? $\endgroup$ Commented Jan 25, 2012 at 17:16
  • $\begingroup$ I voted to close this because strictly speaking it is a duplicate; however I much prefer the background and motivation given in the text of this question. +1 $\endgroup$
    – j.c.
    Commented Feb 24, 2014 at 21:52

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