# 3-dimensional hyperbolic space [closed]

In the 3-dimensional 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.

• Where does the question arise from?
– YCor
Aug 6, 2022 at 17:17
• It's KöMaL A.538.
Aug 9, 2022 at 7:20
• Here is a link to the problem: komal.hu/feladat?a=feladat&f=A538&l=en Aug 11, 2022 at 10:54

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.

• 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 $r_2$.
• 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. Aug 11, 2022 at 10:52
• 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?)