One more thing: the expected length of the intersection of a solid cube of side s with a random line that intersects it is (2/3)s. Proof: Just as in Matt F.'s answer, the mean cross-sectional length of K (where one integrates over the Grassmannian of affine lines) is the volume of K divided by the mean projected area of K (where the latter is defined by integrating over the linear Grassmannian of planes). We learn from https://math.stackexchange.com/questions/1365486/expected-value-of-cube-proj... that the mean projected area of a cube is (3/2) s^2, so the mean cross-sectional length is (s^3) / ((3/2) s^2) = (2/3)s. Jim Propp On Thu, Jul 25, 2019 at 5:06 PM James Propp <jamespropp@gmail.com> wrote:
It occurs to me that if nobody asks me to explain the 2/3 result for a month or two, and then someone *does* ask, I might have trouble reconstructing the argument, so I'll do it here, even if the next person to read this is my future self.
As the page https://mathoverflow.net/questions/335293/mean-cross-sectional-area explains (see part (1) of Matt F.'s answer), the mean cross-sectional area of K (where one integrates over the affine Grassmannian) is the volume of K divided by the mean width of K (where the latter is defined by integrating over the linear Grassmannian). Here I should mention that some authors rescale the latter quantity by a factor of 2 and still call it the "mean width", but I'm not doing that rescaling here.
So, what is the mean width of a cube? For that matter, what is the mean width of an a-by-b-by-c rectangular parallelepiped?
Before computing that, let's compute the (three-dimensional) mean width of a needle of length 1. Instead of fixing the needle and varying the direction in which we are measuring its width, fix that direction (specifically, let it be the z-direction) and let the needle vary. Put one end of the needle at (0,0,0) and let the other end vary over a sphere of radius 1; we are asking for the average value of |z-0|. Equivalently, let the non-origin end of the needle vary over the upper hemisphere of radius 1; then we are asking for the average value of z. But by the Archimedes hat-box theorem, the average value of z over a hemispherical shell equals the average value of z over a cylindrical shell. Specifically, it is the average of z over the cylinder consisting of points (x,y,z) with x^2+y^2=1 and 0 leq z leq 1. But by vertical symmetry this is 1/2.
It follows that the mean width of a needle of length L is L/2.
Since the rectangular parallelepiped is the Minkowski sum of an interval of length a, an interval of length b, and an interval of length c, and since mean-width is additive with respect to Minkowski addition, the mean width of the rectangular parallelepiped equals a/2 + b/2 + c/2. (We can also see this directly by noting that the width of a cube in the direction v is equal to |ai . v| + |bj . v| + |ck . v| where i, j, and k are the unit vectors in the x, y, and z directions; when we average over v and apply linearity of expectation, we see that this equals the sum of the mean widths of the three needles.)
This shows that the mean width of the a-by-b-by-c box is (a+b+c)/2, or 3s/2 when a=b=c=s.
Finally, returning to Matt F.'s answer, we see that the mean cross-sectional area of a cube of side-length s must be s^3 (the volume of the cube) divided by 3s/2 (the mean width of the cube), or 2(s^2)/3.
Jim
On Thu, Jul 25, 2019 at 1:59 PM James Propp <jamespropp@gmail.com> wrote:
You may remember from a couple of years ago the math-fun discussion of what one gets when one intersects a cube with a random plane that intersects it nontrivially (Subject: Random slice of a cube).
Thanks to communal effort, we learned that the expected number of sides is exactly 4, and that this remains true even if one preselects the orientation of the plane.
In preparing to give a public talk about this result (see https://www.cityguideny.com/event/MoMath--National-Museum-of-Mathematics--20...), I realized that some members of the audience may ask "What about the expected area of the intersection? What about the expected perimeter?"
The answer to the first question is nice (2/3 times the square of the side-length; ask me if you want to know why). I don't know the answer to the second question, but maybe some of you can help me figure it out (e.g., maybe Keith Lynch can dust off his code and use it to estimate expected perimeter). And while we're at it, numerical corroboration of the answer to the first question would be reassuring.
For these two questions, it does NOT suffice to focus on one particular orientation of the cutting plane; you have to average over all of them. The measure I'm using is the unique rotationally-invariant and translationally-invariant continuous measure on the affine Grassmannian (the set of all planes in R^3); it's not a probability measure, but it assigns finite measure to the set of all planes passing through a fixed cube, so by renormalizing you can get a probability measure.
By the way, one of the themes of my talk is the collaborative process by which math is done (in contrast to the myth of the isolated genius), so the sequence of events that brought these results to light is going to be integral to my talk. If you have any concerns that I might acknowledge your contribution and for some reason you don't want me to, please let me know!
Jim Propp