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(a)

First, suppose B intersects K. We know that . Let denote the midparallel of B. At the very least, B intersects K at exactly one point, call it C. Then at C, B is its tangent line. Clearly , since the breadth is a, as shown below.

Thus, , so interects .

Next, suppose interects . Using a similar argument, we see that B then intersects K.

(b)

These are parallel sets, where the outer parallel curve of K has distance a/2 from K. From Consequence #1 in Problem11, that the perimeter K is given by . Because these are parallel sets, the perimeter of K is given by . So, .

(c)

We know that . The greatest possible length of the perimeter must be less than or equal to the circumference of a circle whose radius is a / 2. So . It follows that is non-negative.

Problem 17