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Please, consider a line segment $AB$ in the Poincaré disk model. Now, consider the set $S$ of all point $P$ in the disk such that the angle $\angle APB$ is constant.

Question: is $S$ a known curve?


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Found something. See this related question, perhaps strongly related. mathoverflow.net/questions/24307/… Note that, in the upper half plane model, if you take your points $A,B$ as being $0, \infty$ along the imaginary axis, you and they get the same answer: the locus is a slanted ray beginning at $0.$ –  Will Jagy Jul 31 '10 at 18:27
Anyway, here is a link to the first page of the article that answers the other question (constant area instead of constant opposite angle) springerlink.com/content/h031vf8c7ydfaxrf –  Will Jagy Jul 31 '10 at 21:56
The next easiest, fixing $A = i$ and $B = \infty$ along the imaginary axis, forcing a right angle at $P = x + i y$ with $ x \geq 0$ gives the hyperbola $ y = \sqrt{1 + x^2}.$ So the answer to your question is at least slightly different from that of question 24307. If the answer with $ A = i$ and $ B = \lambda i$ with finite real $ \lambda > 0$ is a conic section other than a circular arc, well, I do not know the meaning of those in the upper half plane model, but I bet there are books that say. –  Will Jagy Aug 1 '10 at 1:31
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/… –  j.c. Jan 25 '12 at 10:33
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2 Answers 2

In the Klein model, one may see that this is also a circle. Consider a line segment with one point on the center of the disk. One side of the triangle goes through the center. Then orthogonal lines to a line through the center are also orthogonal in the hyperbolic metric, e.g. since they are preserved by reflection. So one sees that a circle is traced out which goes through the origin. If you'd rather center the curve at the origin, then it will be an ellipse, since hyperbolic isometries of the Klein model are projective transformations.

To convert to the Poincare model, take a hemisphere sitting over the disk, and project vertically. The projection of the circle is given by the intersection of a cylinder over the circle with the upper hemisphere. This upper hemisphere is conformally equivalent to the Poincare model, e.g. by inversion through a sphere centered at the south pole of the lower hemisphere. I haven't computed the curve this traces out though.

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This appears to be Theorem 26.1 in the book by Richter-Gebert (the reference pointed out by jc). For angles other than 90 degrees, he points out that the curve is quartic. dl.dropbox.com/u/8592391/Ch26RichterGebert.pdf –  Ian Agol Jan 25 '12 at 17:41
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I imagine the answer to this problem is known, but I don't have a reference for it; if someone has one, I would be interested to see it. Here is my calculation in the upper half-plane.

Following on from Will Jagy's comment, for a fixed angle $\theta$ at $P = x + iy$ and with the other two vertices in the upper half-plane placed at $i$ and $\infty$, the locus $S$ for the vertex $P$ is $$y^2 = 1 + x^2 - 2xy\cot{\theta}.$$ We can use this to extend to the case where the two fixed vertices are at $i$ and $e^h i$, where $h$ is the distance between the two points $A$ and $B$. Suppose $P$ makes the fixed angle $\theta$ with these two vertices, and denote by $\psi$ the angle at $P$ of the ideal triangle with vertices at $e^h i$ and $\infty$. We then have the two equations $$y^2 = e^{2h} + x^2 - 2xy\cot{\psi}$$ $$y^2 = 1 + x^2 - 2xy\cot{(\psi + \theta)}.$$ Using trig identities to remove $\psi$, from this we get $$y^4 - (e^{2h} + 1)y^2 + x^4 + (e^{2h} + 1)x^2 + 2xy(xy + \cot{\theta} - e^{2h}\cot{\theta}) = e^{2h}.$$ I am not personally aware of any particular significance of this locus, though I'd be interested to hear if there is one.

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