Proof: number of intersection points of borders of a convex polygon and its translate never greater than 2 How can I prove the observation that the borders of a convex polygon $P$ with no parallel sides in $\mathbb{R}^2$ and a translate of $P$ by any vector $t\neq 0 \in \mathbb{R}^2$ that is not parallel to any side of $P$ intersect in no more than two points?  I've used this argument in a proof and thought it was obvious, but now that I'm rechecking everything I've written I struggle to come up with a formal proof. 
Any idea would be appreciated :).
 A: That is not true in general (e.g. allow parallel sides):
Take a unit square and translate its along $\vec{v} = (1,\frac{1}{2})$. This vector is neither parallel nor perpendicular to any of the sides, but the intersection contains the whole segment $\big((1,\frac{1}{2}),(1,1)\big)$.
If you do not allow parallel sides, then here is a sketch:


*

*Arrange the polygon such that the translation vector is vertical.

*Because of the assumptions, no side of the polygon is vertical; for that reason, there is exactly one left-most point $W$ and one right-most point $E$.

*Split the polygon border into the "upper side" and "lower side" (the split is at the left- and right-most points).

*The intersection of the two convex polygons is also a convex polygon with no side vertical; in particular it also has its left-most and right-most points.

*Prove that the intersection of borders happens exactly at the two above.

*Find a translation vector with the same direction and magnitude such that there is exactly one point of intersection of the two copies of the polygon. Name the upper side point $N$ and lower side point $S$.

*Prove that exactly one intersection happens between $WS$ and $NW$, and exactly one intersection happens between $SE$ and $EN$.


I hope this helps $\ddot\smile$
