Lemma 3

Let bi be the closest point of B from a vertex a of A.  If µ is the moving direction (clockwise or counterclockwise) from bi to bi+1 then, for a complete cycle through all vertices of A, µ changes no more than twice.

Proof :

Let's first illustrate this lemma with an example :

As shown above, moving from b1 to b2 follows a counterclockwise path on the boundary of B, while b3 to b5 implies a clockwise traversal.  Because no movement occured from b2 to b3, b3 is called a stationary point.

So, suppose now we go along a chain monotone in some direction d, from a point an to an+1 :

If we put by hypothesis that bn is the closest point of B from an, where can be located bn+1 ? We already proved (lemma 1b) that line L is a supporting line of polygon B, which is located below L. So bn+1 can only be below L.

However, bn+1 can not be below L at the right of P : any point in that area is further from an+1 than bn. If there is no point of B on the left side of P, then bn+1 could only be at the same place than bn. Except for this case, bn+1 can not be neither on P, because any point on P below L is also further from an+1 than bn.

So bn+1 is necessarily below L, but most important, at the left of P, that is in the same direction d that we went from an to an+1.  We conclude that moving along a direction d on a monotone chain of ai's make bi's move in the same direction.  In the worst case, some bi may be at the same place than bi-1 (stationary point), but this doesn't change our conclusion.

Proving lemma 3 is now straightforward :  travelling around a convex polygon is like moving on two monotone chains of opposite directions. Because bi's will move in the same direction than a i's (or maybe not move at all), then the moving direction of bi's can not change more than twice.


Obviously, lemma 3 doesn't say that µ will change twice : for instance, if the starting vertex of the figure above would have been the leftmost or rightmost point of A, then only one change of direction would have had occured. Similarly, we can easily find polygons A and B such that there would be no change in µ, because bi's would move only in one direction, or not move at all.