A line intersecting segments

Question

There are n parallel segments on a plane, for any three of them, there exist a line crossing all three of them, how to prove there exist a line crossing all n segments?

Thank you very much.

Answer 1

Hint:

• The only case where you would not be able to find such line looks like this (or reflection):

  [------------]
                  [-----------]
   [------------]
  

• Find a line that minimizes maximum segment-direction distance from every segment, that is, minimizes $\max_i \operatorname{dist_{segment\ dir}}(I_i, l)$ where $I_i$ are the segments and $l$ is the line.
• Either the maximum is zero or you can take 3 segments that are causing $\max > 0$ (two on one side, one on the other) and reach contradiction.
• Don't forget to handle edge cases, e.g. all segments being collinear.

I hope this helps $\ddot\smile$

Answer 2

I put a recently found answer to end this question: since the segments are parallel, assume them parallel with y axis, for the i'th segment, it can be represent with its lowest point and highest point $(x_i,y_{i,1}),(x_i,y_{i,2})$. A line, $y=ax+b$ crossing this segment is to say: $$y_{i,2}<ax_i+b<y_{i,2}$$ Now define set $$G_i=\{(a,b)|y_{i,2}<ax_i+b<y_{i,2}\}$$ This represents all (a,b) pairs so that line y=ax+b cross the i'th line. For any 3 segment, there is one crossing line is equivalent as for any 3 sets $$G_i,G_j,G_k$$, they have non-empty intersection.

Since all $G_i$ are convex (each of them are the intersection of two half space), according to Helly's Theorem, all $G_i$ have non-empty intersection. Therefore the problem solved.