0
$\begingroup$

Given that $u=(1,0)$, $v=(0,1)$, $w=(0,-1)$, find the hyperbolic area of the circle inscribed in the triply asymptotic triangle $[u,v,w]$.

My approach (I'm using the Poincaré disc model):

Set up a system of three equations based on the equal hyperbolic distances between the point (say, $a$) of the incentre of the hyperbolic triangle:

$$r=d_H(a,v-w)$$ $$r=d_H(a,w-u)$$ $$r=d_H(a, v-u)$$

(Is this correct?)

Upon solving the system, get $a=\left(\frac23, 0\right)$, so that $$r=\cosh^{-1}\left(\frac{3\sqrt{10}}{5}+1\right)$$

Do you think this is correct?

$\endgroup$
2
$\begingroup$

If $u-w$ means the geodesic from $u$ to $w$ then yes, $a$ is the point that is equidistant from those three geodesics. But I don't know how you set up and solve that system of equations.

My approach: any three points on a circle can be moved to other three points by a Möbius transformation. So, replace the vertices by $(-1/2, \pm\sqrt{3}/2)$ and $(1,0)$, that is the vertices of an equilateral triangle. Then the center of inscribed circle is $(0,0)$, by symmetry. To find its radius, consider that the geodesic between $(-1/2, \pm\sqrt{3}/2)$ is an arc of a circle with its center on the negative real axis. As this circle is perpendicular to the unit circle, we get a right triangle with $60$ and $30$ degrees, where the shorter leg is the radius of the unit circle. So the hypotenuse is $2$, the center is $(-2,0)$, and the other leg is $\sqrt{3}$. It follows that the Euclidean distance from this geodesic to $(0,0)$ is $2-\sqrt{3}$. Hence, the hyperbolic distance is $$ r = 2\tanh^{-1}(2-\sqrt{3})$$ (Depending on your normalization of the metric.) This is the hyperbolic radius of the inscribed disk. Then there's a formula for its area.

$\endgroup$
  • $\begingroup$ Good approach but it would be even easier with the klein disk model :) $\endgroup$ – Willemien Jul 24 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.