Partitioning 5 Rays: Nonempty Intersection

[jjjja]

I need to show the following thing: Given a collection of 5 rays (half-lines) in the plane, show that it can be partitioned into two disjoint sets such that the intersection of the convex hulls of these two sets is nonempty.

 

[Office_Shredder]

Is this a homework question? What have you done so far to try to prove it?

 

[jjjja]

I need to show the following thing: Given a collection of 5 rays (half-lines) in the plane, show that it can be partitioned into two disjoint sets such that the intersection of the convex hulls of these two sets is nonempty.

Hi, I made this post in a hurry and was unaware of any forum etiquette, so I will explain the problem in more detail now. Sorry for the bad intro into the community.

About the problem, I am solving a homework. For one of the tasks, I am proving something and was given the hint to use this variant of Radon's theorem in one of the steps. I have managed to do the whole proof just assuming that this is true, but for the homework to be complete I need to prove the claim as well. However, I don't even have an idea on how to start it. So let me put it as follows:

Let $\mathcal{R}=\{r_1,r_2,r_3,r_4,r_5\}$ be a collection of rays in the plane. Show that there are disjoint and nonempty sets $A$ and $B$ which partition $\mathcal{R}$ such that $conv(A)\cap conv(B)$ contains a ray.

Thanks for any help!

P.S. The website won't let me preview my LaTeX, so I hope everything formats okay when I post. Sorry if anything messes up.

 

[Office_Shredder]

There might be a clever generalizable proof by using radon's theorem directly, but here's something kind of direct and low level. If you only have three rays, under what conditions can you form disjoint sets whose convex hulls intersect?

 

[mfb]

To just have a non-empty intersection you don't even need the rays, five points are sufficient. "contains a ray" is a new requirement in post 3 which makes the problem more interesting. As the convex hull of a ray already contains a ray one of the sets can be a single ray and you can show that there has to be a set of 4 rays that cover the fifth one.