Caitlin wants to draw n straight line segments, without lifting her pencil off the paper and without retracing her path, so that each segment crosses exactly one other segment (not counting intersections at vertices) and she ends up back where she started.

a. Show how she can do this for n = 6. (Draw a picture!)
b. For which n can it be done, and for which n is it impossible? Prove your answer.

The problem cannot be solved because Caitlin is female and therefore lacks the cognitive ability required to complete the task.

P18170

I'm working on the case of 2 intersections for each line. It's easy when n is odd, but when n is even it appears to require picking up the pencil once.

P18610 JEWWWWWSSSSSS

I have found a construction that works for all n=2k with k odd.
Step 1: draw a regular k-agon
Step 2: draw a second regular k-agon inside and in the same orientation as the first one
Step 3: link all vertices by alternating between the small and the big polygon, thus making two complete rotations around the center of the figure.
This makes a regular k-pointed star.

Let's assume the n vertices have coordinate [tex: A_i=(cos(2\pi×i/k), sin(2\pi×i/k))] and [tex: B_i(a cos(2\pi×i/k), a sin(2\pi×i/k))], without loss of generality (this is just a frame transformation).[tex: [A_{i}B_{i+1}]] only crosses[tex: [A_{i+1}B_{i}]].

The polygons don't really have to be regular, nor in the same orientation, but it makes the explanation simpler. What is important is that one polygon is comprised inside the other and that all the lines are on the outside of the small polygon.

I don't know if another construction could work for other values of n

Picrel is an example for n=6 (k=3)

P19219
Looks good so far.