Problem 22: The Intersection of Convex Sets

Before you begin this problem, review the mixed areas of Minkowski as well as Problems 5 and 7.

I. Motions of a Convex Set

To continue our discussion about motions, let's consider the motions of convex sets. Let be a convex set with length and area . The set can be defined by a single fixed point and the angle that the vector forms with the x'-axis.

Suppose we have a fixed convex set of length and area in the plane whose origin is given by . Let's find all the positions of the convex set so that and intersect. The set is the one that "moves", thus we want to find rather than just . We have and

Let and be the support functions at the origins and with respect to the parallel axes x and x' for and , respectively. Fix and translate so that only intersects at a single point. Then must intersect at only one point on its boundary. As is translated, the point will trace the boundary of a convex set formed by the following support function, outlined in black in the animation. This equation holds since the new set formed is a mixed convex set of and . (See Problems 5, and 7).

Let denote the convex set formed by reflecting the set in the point . This is the same as reflecting over both the x-axis and the y-axis. Since these are rigid motions, and are congruent. The support function for is . Recall that the mixed area of a Minkowski set , whose support function is , is given by , where is the mixed area of the sets and . It follows that the measure of the set of translations of such that is equal to . Again from Problem 7, if we use the fact that , we see that the measure of possible positions of is

II. Examples

a) If is a line segment, then , where is the length of .

b) If is a circle of radius R, and we choose to be the center of the circle, then and . This makes sense, as it is the equation for area of an outer parallel set.

III. Convex Sets Contained in Other Convex Sets

Is it possible to find the measure of all sets congruent to that are entirely contained in ? The answer to this problem is very complex. We can simplify the solution if we place certain restrictions on both sets. Suppose that the boundaries of and are continuous. Let denote the greatest radius of curvature for and let denote the smallest radius of curvature for and suppose . As we did before, if we trace the path of , we see that the path outlines a convex set defined by the support function This is true since radius of curvature for this set is . (See Problem 7).
The area of this type of convex set is given by

After integrating over all rotations, we have

To summarize, under the given conditions of curvature, the measure of sets congruent to that are entirely contained in is given by the equation above.

PROBLEM 22:

Let the curvature conditions from above for and hold. Suppose that is a circle of radius whose center is . Find the measure of the sets congruent to that are entirely contained in .

**Hint: Consider the path of .

Solution

References: Santalo