Will a path between $(x, y)$ and $(-x, -y)$ always intersect a 90 degree rotated copy? Suppose we have a path between two points $(x, y)$ and $(-x, -y)$. If we rotate it by 90 degrees around the origin, will the copy intersect the original? (You can add any number of assumptions to avoid pathological cases.)
It seems obvious that it does (if you play with a construction you can see how changing the curve in one place to avoid an intersection with the copy makes it intersect in another place) but I cannot come up with a way to show it. My guess is there is a simple observation I am missing.

Background: This is a technical step in a demonstration that polyominoes that cover two opposite corners of their hull can only tile a rectangle in a certain way. This step comes in to show that if you rotate a polyomino 90 degrees and it does not overlap the original, it also does not overlap the original when you rotate it 180 degrees.)
Edit: After getting a feel for it from the posted answers, I realize it's much more technical than I thought and probably not appropriate for what I wanted to use it for. 
Edit: Actually, this answer given to a version of this question works well for my purposes. Also, see this more general question.

This proof assumes there is no "backtracking" or radial movement (i.e. every line through the origin intersects the curve in exactly one point, with the obvious exception of the line through $(x, y)$ and $(-x, -y)$, which intersects the curve twice). Also, the curve is continuous and goes counterclockwise around the origin from $(x, y)$ to $(-x, -y)$. 
Take the line through the origin and $(-y, x)$ ($90^\circ$ counterclockwise rotated copy of $(x, y)$). If the curve intersects the line at $(-y, x)$ we are done. If not, it intersects the line either outside or inside $(-y, x)$. Let's say inside (like your yellow, green and orange examples). Then the $90^\circ$ rotated curve goes on the inside of $(-x, -y)$.
This means that if we take a continuous "sweep" of lines through the origin, starting with the one through $(-y, x)$ and ending with the one through $(-x, -y)$, then the original curve goes from being on the inside of the rotated curve to being on the outside. By the intermediate value theorem they must intersect.
