In the 3dimensional hyperbolic space there are given a plane $\mathcal{P}$ and four distinct lines $a_1, a_2, r_1, r_2$ in such positions that $a_1$ and $a_2$ are perpendicular to $\mathcal{P}$, $r_1$ is coplanar with $a_1, r_2$ is coplanar with $a_2$, finally $r_1$ and $r_2$ intersect $\mathcal{P}$ at the same angle. Rotate $r_1$ around $a_1$ and rotate $r_2$ around $a_2$; denote by $\mathcal{S}_1$ and $\mathcal{S}_2$ the two surfaces of revolution they sweep out. Show that the common points of $\mathcal{S}_1$ and $\mathcal{S}_2$ lie in a plane.

2$\begingroup$ Where does the question arise from? $\endgroup$– YCorAug 6, 2022 at 17:17

1$\begingroup$ It's KöMaL A.538. $\endgroup$– TaDAug 9, 2022 at 7:20

$\begingroup$ Here is a link to the problem: komal.hu/feladat?a=feladat&f=A538&l=en $\endgroup$– Sam NeadAug 11, 2022 at 10:54
1 Answer
Let $L$ be the geodesic segment in $P$ with one endpoint at $a_1 \cap P$ and the other endpoint at $a_2 \cap P$. Let $b$ be the midpoint of $L$. We define a plane $Q$ by requiring it to contain $b$ and by requiring it to be perpendicular to $L$ (and thus to $P$). Thus reflection in $Q$ exchanges $a_1$ and $a_2$. It is now an exercise to prove that reflection in $Q$ exchanges the “cones” $S_1$ and $S_2$. Thus $S_1 \cap S_2$ lies in the plane $Q$.
Here I am assuming that the lines $r_1$ and $r_2$ actually meet $P$ — the question is slightly unclear on this detail.

$\begingroup$ If the reflection in $Q$ exchanges $a_1$ and $a_2$, $Q$ is the perpendicular bisection plane of the common perpendicular to $a_1$ and $a_2$. But if so, the $S_1$ is not sure to be the reflection of $S_2$ to $Q$. Because $r_1$ is not sure to be the reflection of $r_2$. $\endgroup$– TaDAug 11, 2022 at 8:59


$\begingroup$ You are correct that $r_1$ and $r_2$ need not be reflections of each other. But some rotation of $r_2$ is a reflection of $r_1$, because of the angle condition. And that suffices. $\endgroup$– Sam NeadAug 11, 2022 at 10:52

1$\begingroup$ yeah, I think I have understood it. By the way, is the proof the same in the euclidean case? Because the $a_1\cap r_1$ isn't the reflection of $a_2\cap r_2$, the intersection won't be on the "perpendicular bisection plane" $Q$. (Maybe it will be a little left or right to the central?) $\endgroup$– TaDAug 11, 2022 at 15:05

1$\begingroup$ It is, the "determine the position (or distance) by angle" is right only in hyperbolic or spherical cases. $\endgroup$– TaDAug 11, 2022 at 15:07