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While I like the Poincaré disk for aesthetic reasons, sometimes the half-plane model is easier, and I'd say this is such a case. Consider the ideal triangle with vertices at $(\pm1,0)$ and at infinity. It's boundary is the upper half of the unit circle, together with two vertical tangents to that circle. The horocycles incident with the point at infinity are ...

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What you are asking for is the theory of hyperbolic 2-dimensional orbifolds. It's a very big theory. If you do not add additional hypotheses, it becomes somewhat unreasonable to expect a good description of the theory. Even if you add enough hypotheses to tame the question, it still requires a lot of mathematics to even describe the classification. Let me ...

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The triangle with angles $\frac \pi \beta$, $\frac \pi \beta$ and $\frac \pi \gamma$ is an isosceles triangle. Tracing the height to the uneven side, two triangles with angles $\frac \pi 2$, $\frac \pi \beta$ and $\frac \pi {2 \gamma}$ are obtained. And the assertion follows.

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The fixed circle has centre $(4,4)$ and radius $6$. The locus of the centre of the circle is actually two part-parabolas. When the circle lies above the $x$ axis, the focus of the parabola is at the centre of the circle $(4,4)$ and the directrix is the line $y=-6$ This is because the distance from the centre of the variable circle to the point $(4,4)$ is ...

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Yes, this is true. The proof is not specific to hyperbolic metric: one can argue the same way about the Euclidean metric on the plane, or in higher dimensions. Step 1: Compose $f$ with a Möbius transformation that sends $f(0)$ to $0$. This reduces the problem to the case $f(0)=0$. Step 2: Since $f$ is an isometry, every circle around the origin is ...

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You can rediscover the geometry if you assume that the isometries of the geometry are given by some homographies. The group of orientation-preserving homographies who leave $\Bbb P^1(\Bbb R)$ invariant is $G = PGL_2^+(\Bbb R) = GL_2^+(\Bbb R) / \Bbb R^*I_2$ (the $+$ indicates we only keep matrices with a positive determinant). Its Lie algebra is $A = ... 1 This is not a proper answer, is as vague as also perceived in the question but hopefully impacts what you are attempting to say. At any rate I do hope it survives down-voting. Take a circular disk and let its boundary length increase indefinitely. The surface area in the neighborhood of boundary increases and spreads to the center by deformation ... 1 A regular pattern in the hyperbolic plane will appear self-similar and fractal in common models of the hyperbolic plane. You'd see copies of the same theme repeated at different scales. So there is at least some relation here. To make things less vague, you'd first have to define the term “fractal”. On this topic, Wikipedia currently writes (text by ... 1 The answer is (d) : a parabola made of the union of 4 parabolic arcs belonging to parabolas$P_1$and$P_2$with a common focus$O(4,4)$and resp. directrices with equation$y=-6$for the arcs of parabola above the$x$axis and$y=6$for the arcs of parabola below the$x$axis. Recall about the focal def. of a parabola: it is the locus of points at equal ... 1 It's the subgroup$\langle xyx,y \rangle$of$\langle x,y \mid x^2,y^\beta,(xy)^{2\gamma} \rangle$. The index of this subgroup$2$, so checking that the subgroup has the presentation$\langle z,y \mid z^\beta,y^\beta,(xy)^\gamma \rangle$(with$z=xyx$) is routine. 1$\$\frac{\alpha z+\beta}{\overline\beta z+\overline\alpha}\cdot\frac{\overline\alpha\overline z+\overline\beta}{\beta \overline z+\alpha}=\frac{|\alpha|^2|z|^2+\alpha\overline\beta z+\overline\alpha\beta\overline z+|\beta|^2}{|\beta\overline z +\alpha|^2}=\frac{|\alpha|^2|z|^2+\alpha\overline\beta z+\overline\alpha\beta\overline z+|\beta|^2}{|\beta\overline z ...

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