TITLE: Does a closed curve exist for which a square cannot intersect it 8 or more times?
QUESTION [20 upvotes]: To phrase my question more clearly: Imagine you have a game with two players, Minnie and Maxime. Minnie starts by defining some closed curve. Then Maxime translates, rotates, and scales a square with the goal to maximize the number of intersections between the edges of the square and the curve. Minnie tries to think of a curve to minimize the number of intersections. Is it possible for Minnie to get a curve for which Maxime cannot transform the square to get at least 8 intersections?
For example Maxime will be able to get 8 intersections with a circle, an infinite amount with a triangle (overlapping straight edges) and 16 with some kind of croissant shaped closed curve.

REPLY [3 votes]: This isn't an answer, but I need to post a picture in order to demonstrate that the crescent idea doesn't work.  See the following:

This handles the case where the upper arc subtends more than $\pi/2$ radians.  And in the "sufficiently flat" case, we have:

which I think is quite interesting since one side has zero intersections but the opposite side makes up for this by having four.  I think this is a key configuration that demonstrates that any non-convex shape admits a line that intersects the boundary at four distinct points.