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Given a hyperbolic surface of constant curvature $K=-1/a^2$ embedded in $\mathbb{R}^3$, is there a known formula for the length of the edges of a regular polygon?

I know that the Gauss–Bonnet theorem tells us that the area of an $n$-gon is given by: $$A=(n-2)\pi-\sum_{i=1}^n \alpha_i = n(\pi-\alpha) - 2\pi$$ But i'm not sure how one would go about calculating the edge length without explicitly specifying the surface. Is an explicit formula even possible?

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  • $\begingroup$ Does it matter that the surface is embedded in $\mathbb R^3$? Your questions seem so be only intrinsic properties of the surface. $\endgroup$ Nov 30 '15 at 12:45
  • $\begingroup$ @HenningMakholm - No, but since I'm new to the topic, I wasn't 100% that the length was an intrinsic property. $\endgroup$
    – nbubis
    Nov 30 '15 at 15:21
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Suppose $r$ is the distance one of the vertices of the polygon to its center. Draw the perpendicular bisector of the edge; by symmetry it will pass through the center of the polygon too. This creates a right triangle where

  • The central angle is $\pi/n$.
  • The hypotenuse is $r$.
  • The opposite leg is half the edge length.

Applying the appropriate rule for a right hyperbolic triangle we get $$ \sin \frac{\pi}{n} = \frac{\sinh(x/2a)}{\sinh(r/a)} $$ where $x$ is your sought edge length.

This is easily solved to get $$ x = 2a \sinh^{-1} ( \sin(\pi/n) \sinh(r/a) ) $$


If instead of the radius you know the interior angle $\theta$ at each corner of the polygon, you would use a different rule and get $$ \cosh(x/2a) = \frac{\cos(\pi/n)}{\sin(\theta/2)} $$ giving $$ x = 2a \cosh^{-1}\left( \frac{\cos(\pi/n)}{\sin(\theta/2)} \right) $$

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  • $\begingroup$ Just realized I could use the law of hyperbolic cosines, but I get a slightly different result, which I'll post below. $\endgroup$
    – nbubis
    Nov 30 '15 at 13:48
  • $\begingroup$ Henning - Do you know why we get different results? $\endgroup$
    – nbubis
    Nov 30 '15 at 14:02
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    $\begingroup$ @nbubis: Our results are not really different -- you can convert mine into yours by the double-angle formula for hyperbolic consines: $\cosh(2x)=2\cosh^2(x)-1$, and then $\cos^2\frac\pi n = \frac12+\frac12\cos\frac{2\pi}{n}$ and $\cos^2+\sin^2=1$. $\endgroup$ Nov 30 '15 at 15:21
  • $\begingroup$ Indeed. Thanks! $\endgroup$
    – nbubis
    Nov 30 '15 at 15:29
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Law of cosines to the rescue:

If we know that the angle at each vertex is $\theta$, and the edge length to be found is $x$, then:

$$\cos(2\pi/n)=-\cos^2(\theta/2)+\sin^2(\theta/2)\cosh(x/a)$$ Or: $$x = a\cosh^{-1}\left(\frac{\cos(2\pi/n)+\cos^2(\theta/2)}{\sin^2(\theta/2)}\right)$$

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