Math Quiz 9

[Jon Stratis]

I finally came up with 0.28967 as my final answer. Work below, colored white so people can ignore it until they've done their own work on the problem Note: I don't remember my calculus very well anymore, so I had to use wolfram alpha to get me (hopefully) close enough!

To solve this (assuming it's correct), I needed a couple of components and assumptions/assertions.

Lets start with a simple case - the average length of line segments contained within a N length line. A good description of this can be found here: http://math.stackexchange.com/questi...ints-on-a-line

What that boils down to, is that if I draw a line such that it intersects two edges of the square, taking the length of that line divided by 3 gives me the average length of every line segment that sits on that line. This little trick lets us GREATLY simplify the calculations, as we can use it to assume we're only looking at lines that intersect two walls of the square, and effectively ignore snort lines that don't intersect the lines of the square.

So, with that in hand, we can now figure out the length of every possible line that intersects two lines of the square. This can be broken down into two categories: lines that intersect adjacent edges of the square, and lines that intersect non-adjacent edges. Due to the symmetry of a square, we know that the first group can be simplified into 4 repetitions (top/right, right/bottom, bottom/left, left/top) and the second into 2 repetitions (top/bottom, left/right). This will come in handy later.

So, lets look at the case where you have adjacent sides. You really have two variables here, X and Y, each being an independent number between 0 and 1 representing their location on one of the sides, and the line that stretches between them. The length of that line, as defined by the Pythagorean Theorem is sqrt(x^2+y^2)... but remember we want that length/3 to get the average length of line segments along it. And remember that we have an infinite number of points between 0 and 1 for both x and y... calculus! So, written in a form wolfram alpha will recognize, integrating over x and y gives us:

integrate integrate (sqrt(x^2+y^2)/3) dx dy from 0 to 1 from 0 to 1

or 0.255065.

We can do the same for the case of non adjacent sides. Here the equation is a little trickier, but ultimately the line length is sqrt((x-y)^2+1). Dividing by 3 and integrating gives us:

integrate integrate (sqrt((x-y)^2+1)/3) dx dy from 0 to 1 from 0 to 1

or 0.358879.

Keep in mind that the average of an integral is that integral times 1/(b-a) - in this case, b-a is 1, so we don't need to do anything else.

So, lets get these two numbers together. Remember, we have 4 sets of the adjacent sides, and 2 sets of the non-adjacent sides. And since those sides all integrated over the same values, we should just be able to average the sets, right? So, averaging those in proportion gives us:

(4*0.255065 + 2*0.358879)/6

or 0.28967.