Given that $l_1$ and $l_2$ are hyperbolic lines and they are ultraparallel, prove that there exists a line perpendicular to both $l_1$ and $l_2$.

My progress: In the half-plane model, I can prove that the situation when $l_1$ is a Euclidean vertical line and $l_2$ is a Euclidean semi-circle. However, I cannot prove the statement when $l_1$ and $l_2$ are both Euclidean circles. I let the radius of the two circles containing $l_1$ and $l_2$ be $r_1$ and $r_2$. I let the radius of the circle containing the perpendicular line be $r$. I let the center of the circles containing $l_1$, $l_2$ and the perpendicular line to be $c_1$, $c_2$ and $c$. Then I have the equation: $$ (c-c_1)^2 = r_1^2+r^2$$ $$ (c-c_2)^2 = r_2^2+r^2$$ I want to show that $r>0$ but I couldn't.

  • $\begingroup$ Consider using an isometry beforehand to arrange for one of the lines to be a vertical Euclidean line, so that you don't have to worry about the situation with two circles. $\endgroup$ – Neal Sep 12 '16 at 1:19
  • $\begingroup$ @Neal Thanks for your comment. Sorry, I have just started to learn hyperbolic geometry. How to convert one of the lines to a vertical Euclidean line? $\endgroup$ – Kenneth.K Sep 12 '16 at 1:45
  • $\begingroup$ Do you know what the isometries of the Poincare upper half plane are? If so, it is not so hard to find one that does the trick. $\endgroup$ – Alfred Yerger Sep 12 '16 at 2:00

The two circles $l_1$ and $l_2$ have a unique line, called the radical axis, with the property that a circle is orthogonal to both $l_1$ and $l_2$ if and only if it is center is on the radical axis. The radical axis is orthogonal to the segment connecting the centers of the two circles $l_1$ and $l_2,$ which in our case is the real line $\mathbb{R} = \partial \mathbb{H}^2$ (the boundary at infinity of the hyperbolic plane). Now, draw the radical axis of $l_1$ and $l_2$ and find it's intersection point $O$ with the the real axis $\mathbb{R} = \partial \mathbb{H}^2$. Draw a line $t$ passing through $O$ and tangent to circle $l_1$ (or you can choose $l_2$ if you prefer, it doesn't matter). Let $T$ be the point of tangency of the line $t$ and $l_1$. Now the circle centered at $O$ and radius $|OT|$ (Euclidean :) ) is orthogonal to both $l_1$ and $l_2$ and its center $O$ is on the real axis. Hence this is the unique geodesic orthogonal to both $l_1$ and $l_2$.


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