Recall from Problem 11 that the mean length of chords made by random lines intersecting a convex set is given by

Thus, the average length of the chords made by a square of side length l lying in a field of random lines is thus

In order to analyze this general case, however, we must analyze each of its subcases. The first one being

We shall denote the position of points A and B by the distance above either the x-axis or y-axis, depending on which pair of opposing sides of the square the two points are located.

Thus, let's denote the position of A by x and the position of B by y. Since both A and B have equal probablity of being at any point on their side, then we say that is a uniform probability distribution. Thus, the probability that A lies in the interval is simply the length of that interval divided by the length of the side, or . Similarly, the probability that B lies in the interval is . Moreover, the probability that A and B lie in the intervals and simultaneously is . Using the Pythagorean Theorm, we see that

Thus, to find the average length of AB, we need to integrate this length formula over all possible lengths using the probability element from above. Thus, we have

which is a very difficult integral to solve. So to help us, I suggest we consider this transformation:

A picture of the square having gone through this transformation is located above. Once we account for this transformation, our integral now takes the form

Now we can move on to the second subcase.

This is very similar to Subcase #1, except we denote the position of A and B by their respective distances x and y from the vertex that lies between them. Again using the Pathagorean Theorem, we find that the length of the chord from A to B is . Using the same probability element, the average length of the chord AB is given by the integral

Again, solving this integral difficult. Although the region of integration is nice, the integrand is to difficult to work with. Thus, our solution this time will be to convert to polar coordinates (remember them, right?). Consider the following diagram that depicts the possible x-y values that determine the chords (It does not represent the square as in the previous diagram). Notice that once increases to , that the length of the hypotenuse of the triangle changes from to .

Thus, referring to the diagram at the very beginning of our discussion of Subcase #2, we have

The diagram up top is transformed into the diagram on bottom by way of these polar coordinates. The two curved parts of the region are (bottom) and (top). Recall that when integrating in polar coordinates that dxdy becomes . Thus, our integral becomes

Now that we have determined all the subcases, we can determine the overall average length for Case #2 by taking the weighted average. Since only one side of the square out of four is defined as opposite of a particular side, then the probability that two points end up on opposite sides is 1/4; since two sides of four are defined to be adjacent to a particular side, then the probability that two points end up on adjacent sides is 1/2; and similarly, the probability that two points end up on the same side is 1/4. Thus, the weighted average of the lengths we found in our subcases yields the overall average length for Case #2. Thus, we have

In this case, there are two classes of chords that occur, those where A and B are located on opposite sides from one another and those where A and B are located on side adjacent to one another. Without loss of generality, we can assume that the point A is restricted to one side of the square and the angle is restricted to the interval . That is to say that the average length of the chords considered under these conditions is the same as the average length of the chords considered under all conditions, much the same way that (1/2)=(6/12). The probability element the lies in is and the probability element that x lies in the interval is . Thus, the probability element that lies in and that x lies in the interval simultaneously is . By trigonometry, the angle Considering the diagram above, when point B is located on the adjacent to the side on which A lies (remember we are only considering in ), then the length of the chord is and . When point B is located on the side opposite the side on which A lies, then the length of the chord is and . Stating this more formally, we have

Thus, our integral is given by

However, to make the evaulation of this integral easier, we will change the order of integration. In doing so, our integral changes to

The chord AB, that is depicted above, is made up of the two smaller chords AP and PB. The chord AP is contained in the region , while the chords PB is contained in the section . Thus, the average length of AB with respect to is the sum of the average lengths of AP and PB, with respect to . Let P be determined by x and y coordinates of a uniform distribution. Thus, because of symmetry, the average length of AB is twice the average length of AP. Using the same symmetry argument that was used in Case #3, the average length of AP with respect to and x can be determined when is restricted to , and so we have

Recall from Case #3 that probability element with respect to and x was . Similarly, with respect to , x, and y, the probability element is . Thus, the average length of a chord in this case is given by the following integral:

Let us denote the first triple integral by I₁ and the second by I₂. Luckily, it turns out that these two integrals are equal. Below is a diagram of the surface . The region of the box below the surface is the region of integration for I₁ and the region of the box above the surface is the region of integration for I₂. However, we will only be concerned with the lower region as we are going to only evaluate I₁ and then double that answer. To make things easier, we are going to, first, express the surface in terms of x and (that is, ) and, second, change the order of integration to . In order to make this change in the order of integration, it is required that for x and are located in the gray region in the diagram under the statement of Case #3. The gray region can be divided into two subregions:

The region of integration is a volume (why?), and so this volume of integration is divided into two subvolumes, given by:

Thus, our integral becomes

Thus, we have that

Below the diagram is a graph of the surface performed by Maple. The -axis of this graph is upside down so that the surface is easier to view.

In summary