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Let K1 and K2 be two bounded sets in the plane with perimeters L1 and L2. Knowing this, we obtain the following definition.
External Cover: The external cover Ce of K1 and K2 is the boundary of the convex hull of . Note that the convex hull of a set of points in the plane is defined as the intersection of all convex sets that contain this set of points. |
If you have a group of convex sets, think of the convex hull as a the region created by a rubber band that is stretched around all of the sets.
You can think of the external cover as a rubber band placed around the two sets, where the length of the rubber band is denoted Le. In the same fashion, and assuming that K1 and K2 do not overlap, let the internal cover Ci be a rubber band placed around the two sets so that it crosses at a point O in between them. We will denote its perimter by Li. These are illustrated in the figure below.
Now we consider the set of lines that intersect either only K1 or only K2. These lines have a total of 6 points where they intersect the curves . Thus, m6 denotes the measure of the set of these lines. The lines that separate the sets K1 and K2 have four points of intersection; namely, two with and two with . Similarly, we will denote the measure of the set of these lines by m4.
Earlier in Problem 11, we saw that the measure of the set of lines that intersect a convex set is equal to the perimeter of that set. Thus, the measure of the set of lines intersecting K1 while not intersecting K2, denoted m6', and the set of lines intersecting K2 and not K1, denoted m6'', are as follows:
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(1) |
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and so the total of m6 is
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(2) |
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Due to the fact that Ce itself is also a closed curve, it follows that
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(3) |
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Again, recall from Problem 11 that there is result in geometric probability that states that integrating the number of times a line intersects a convex curve over all lines that intersect this curve yields twice the perimeter. In other words,
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Applying this idea to the four curves we have considered, we see that
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(5) |
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You don't really have to understand how equation (5) comes from equation (4), just realize that there is a justification for it. Finally, we can determine m4, m6', m6'', and m10 in terms of their respective perimeters. Thus, we have
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| (6) |
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So, we have a measure for the set of lines that (a) intersect both K1 and K2, (b) intersect K1 and not the other K2, and (c) vice versa, and (d) that separate K1 and K2.
Now, suppose K1 and K2 are two bounded, convex, disjoint sets, and suppose that G is a random line that is under the condition of having to intersect the convex hull of K1 and K2. Then we can determine the following probabilities:
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The probability that G intersects K1 and K2 is:
The probability that G intersects K1 without intersecting K2 is:
The probability that G intersects K2 without intersecting K1 is:
The probability that G separates K1 and K2 is:
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