# A closed path is made up of 11 line segments. Can one line, not containing a vertex of the path, intersect each of its segments?

This problem is taken from the book Mathematical Circles by Dmitri Fomin, et al., translated by Mark Saul and published by the American Mathematical Society. Can anyone describe what the question actually means?

• Start with a simpler problem. Replace 11 with 3. Can one line, not containing any vertex of a triangle, intersect all three of its sides? (A "side" is a closed, bounded line segment - not the whole line, extending beyond the vertices.) If not, why not? Can you use the same argument for 11 instead of 3? Aug 28, 2016 at 2:09

Can anyone describe what the question actually means?

For example, here is a closed path consisting of six line segments where a line can pass through all of them avoiding any vertex (endpoints of the line segments). The question asks if it is possible to do this with 11 line segments.

Alas, I didn't read the question too well. The OP was not asking for a proof, but a proof is here in the hidden portion.

The answer is no. Suppose contrary that a line $L$ exists with the required property. This line cuts the plane into two open half plane. Note that two consecutive vertices cannot belong in the same half plane. Do you see a contradiction now?

• Odd cycles in a bipartite graph? Dec 31, 2016 at 17:56