2016
DOI: 10.1216/rmj-2016-46-3-801
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The geometry of cyclic hyperbolic polygons

Abstract: A hyperbolic polygon is defined to be cyclic, horocyclic, or equidistant if its vertices lie on a metric circle, horocycle, or a component of the equidistant locus to a hyperbolic geodesic, respectively. Convex such n-gons are parametrized by the subspaces of (R + ) n that contain their side length collections, and area and circumcircle or "collar" radius determine symmetric, smooth functions on these spaces. We give formulas for and bounds on the derivatives of these functions, and make some observations on t… Show more

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Cited by 10 publications
(45 citation statements)
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“…If the coefficient ∂ ∂dn D 0 (T i + n (d(t))) + D 0 (T i − n (d(t))) is non-zero then we can solve for d n (t), yielding a first-order ODE in d n (t). We claim this holds at t = 0, ie for the T i ± n (d), and therefore at all possible values of d(t) near d. Given the claim, Picard's theorem on the existence of solutions to first-order ODE implies there is a smooth solution d n (t) for small t (note that smoothness of D 0 is proven in [4]). We will apply results from Section 2 of [3], together with [4, Proposition 2.3], to prove the claim.…”
Section: Increasing Injectivity Radiusmentioning
confidence: 77%
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“…If the coefficient ∂ ∂dn D 0 (T i + n (d(t))) + D 0 (T i − n (d(t))) is non-zero then we can solve for d n (t), yielding a first-order ODE in d n (t). We claim this holds at t = 0, ie for the T i ± n (d), and therefore at all possible values of d(t) near d. Given the claim, Picard's theorem on the existence of solutions to first-order ODE implies there is a smooth solution d n (t) for small t (note that smoothness of D 0 is proven in [4]). We will apply results from Section 2 of [3], together with [4, Proposition 2.3], to prove the claim.…”
Section: Increasing Injectivity Radiusmentioning
confidence: 77%
“…Second, by construction Delaunay cells are cyclic or horocyclic; that is, inscribed in metric circles or horocycles, respectively. In [4] there are calculus formulas describing the derivative of area with respect to side length for such polygons.…”
Section: The Delaunay Tessellationmentioning
confidence: 99%
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“…. , b k−1 ) ∈ [0, ∞) k−1 , since f (b 1 ,...,b k−1 ) (x) → 0 as x → ∞ there is an unbounded interval consisting of possible choices of b k for which equation (4) has no solutions. In the remainder of this section, we will analyze the case of three-holed spheres using an elementary direct method, providing a corrected version of the theorem in this case.…”
Section: The Systole Of Loops On the Three-holed Spherementioning
confidence: 99%
“…Analogously, two horocyclic ideal triangles with the same compact side length are isometric (this is easy to prove bare hands, or cf. [10,Prop. 3.4]), and every one has a reflection exchanging its two vertices in H 2 .…”
Section: Some Examples Showing Sharpnessmentioning
confidence: 99%