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As we saw in Problem 5, the equation for the support line that is distance p away from the origin at angle is given by:

Assuming that is differentiable, we get the equation: . We can then solve this system of equations for x and y, yielding:

From the first diagram it is easy to see that the point H is described by the coordinates and it follows that HP =

Let's assume that p is of class C_² and from equation (2), we have:

where is the radius of curvature and s is the arc length. From equation (3), it is clearly necessary that .

Now let's find the length of the boundary of the convex set. Again supposing that p is of class C_² , the length is given by the equation:

The perimeter can be expressed in terms of the width function in addition to the support function given above.

In addition to the perimeter, the area of a convex set can also be described in terms of its support function. Suppose that K is broken down into a series of triangles of height p and base ds, so that each one shares a common vertex. We can then use these triangles to evaluated the area. Call this area F.

or

Suppose K is a convex set of constant width, i.e. constant. Then the perimeter of K is .

Let K_r be an outer parallel set as defined in Problem 5 whose boundary is . If K has a support function , then it follows that has a support function at . Let be the perimeter of K_r and let be its area. These equations follow:

The lenght and area of an inner parallel set can be found by substitutin r for -r.

Mixed Convex Sets and Mixed Areas of Minkowski

Suppose we have two bounded convex sets, whose support functions are and respectively. Let's also suppose that . Let . This is the support function for the convex set , which is defined as the mixed convex set of and .

Next, we would like to calculate both the perimeter and the area of these mixed convex sets. Let L be the perimeter of . Naturally,

where and are the perimeters of and .

The area of has the form:

	and

	and