Looks like you're right, BMAD. I haven't taken my 9-long worm example far enough to see that the adversary cannot continue to maintain that pattern indefinitely as long as surgeon does the right thing.
1) draw 3 additional lines parallel to each cevian passing through the remaining vertices of 2 triangles.
2) show that 6 other shaded triangles are mirror images of the green triangle and therefore have the same area
3) show that every part of a shaded triangle that lies outside of the big triangle has an equal unshaded part inside the big triangle. Midpoints of big triangle's sides are points of symmetry.
@bonanova: You're right that if the wind direction has both head/tail component and cross component relative to the trajectory then analyzing only one or the other will result in correct conclusion that wind adds to travel time. However, the only crucial fact that I was trying to show in my previous post was that there exist closed paths (i.e. straight line A->B and back) and wind directions (i.e. perpendicular to the path) for which ignoring the cross wind component will result in incorrect conclusion - same time with and without wind. In other words, in some cases, the entire time penalty is swept under the rug.
I think your simplified analysis still doesn't address the cross wind correctly. Let's say that the closed path is from point A to point B and back in a straight line. It will take longer with the constant cross wind than without any wind.