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Two hyperbolic lines that don't intersect and are not parallel (that is, they are ultra parallel) do have a unique common perpendicular. The common perpendicular intersects the two hyperbolic lines at two points that are the closest points between the two lines. The task is to find the common perpendicular. Since the Poincaré upper half plane model is ...


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A more general depiction of hyperbolic geodesic ultra-parallel lines is available by employing Bi-polar coordinates. https://en.wikipedia.org/wiki/Bipolar_coordinates Wikipedia link above gives the above in its geometrical description but not in the context of presently discussed Poincaré $ \mathbb {H}^2 $ shortest lines/minimum distance between skew lines ...


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Here are some rough hints on how to analyze the convergence to get you started. Since the integrand is continuous on $(d,\infty)$ it suffices to check the integral close to the singularity of the integrand at $x=d$ and in the tail that goes off to $x=\infty$. For $r\gg 1$ we have $\cosh(r) \sim \frac{e^{r}}{2}$ and if $r\gg d$ then $\cosh(r) \gg ...


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no, take $N=4,$ one figure is a "square" with four equal edges and four equal angles at the vertices. The other figure is a rhombus, same four edges but hte vertex angles in two pairs. For example, we could just glue two equilateral triangles together along one edge. Isometries preserve angles too.


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$\pi- \epsilon$. The sum of the angles of a hyperbolic triangle comes out to $\pi - Area(\Delta ABC)$, so by making the area close to zero, your angles will be close to $\pi$ (think of $\epsilon$ as a small constant). On the other hand, by making a very large hyperbolic triangle, the angles approach zero and the sides are nearly parallel. (Try this in a ...


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The area of a spherical triangle is given by $${\rm area}(\triangle)={\rm angle\ sum}-\pi\ ,$$ and for a hyperbolic triangle we have $${\rm area}(\triangle)=\pi-{\rm angle\ sum}(\triangle)\ .$$ Both formulas are a consequence of the famous Gauss-Bonnet theorem. It follows that in a hyperbolic triangle we have $${\rm angle\ sum}(\triangle)=\pi-{\rm ...


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As I understand the research, there are two approaches to making finite hyperbolic planes. The one in my textbook comes from balanced incomplete block designs, where each line (block) has the same number of points on it. These don't lend themselves naturally to ideal (omega) points. The other approach starts from the projective geometry idea. In that ...


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I claim that such a (possibly anticonformal) isometry exists if $N=3$ and the $A_i$ are not on the boundary of the plane, and may or may not otherwise. For $N=3$, consider the case of $A_1 = i$ and $A_2 = ti$ where $0<t<1$. Note that, using the usual metric $A_1 A_2 = -2\ln(t)$. Assume that we are given $A_1 A_3 = R$. Consider the hyperbolic circle, ...


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Transformations of the form $\frac{az+b}{\bar b z + \bar a}$ are assumed to be normalized: $a \bar a + b \bar b = |a|^2 + |b|^2 = 1$. But $f(z)=e^{i\theta} z$ is not normalized. When you normalize it you get $$f(z) = \frac{e^{i\theta/2} z + 0}{0 + e^{-i\theta/2}} $$ and the universe is saved.


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I don't really understand the problem. (1) The tangents are parallel lines only if one of the segments is a diameter. (2) As shown below, there is no quadrilateral if the chords lie on the same side of a diameter: In this case the line connecting the intersection points of the tangent lines will intersect both of the cords. This is because the ...



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