A line intersecting segments 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.
 A: 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 $\ddot\smile$
A: 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.
