Your lowbrow procedure excludes some planes that do intersect the cube. Pick a point Q just "above" the center of one of the cube's faces, (so it would have been excluded by your procedure) and erect the plane through Q perpendicular to OQ. Now tilt the plane just a little toward one of the four vertices of the chosen face, until the plane nicks that vertex. There is a point Q' in this tilted plane such that the tilted plane is perpendicular to OQ'; I contend that this can be done so that Q' is still outside the cube, and hence the tilted plane would have been excluded even though it intersects the cube. On Sun, Oct 8, 2017 at 7:48 AM, James Propp <jamespropp@gmail.com> wrote:
If I choose a random plane that intersects a cube, creating a polygon, how many sides does the polygon have on average?
(The most lowbrow way to define the probability measure I have in mind when I say "random plane" is to circumscribe a sphere of radius r around the cube with center O. Choose a point uniformly P at random on the sphere, and now choose a point Q uniformly at random on segment OP. If Q is in the cube, erect a plane through Q perpendicular to OP. Otherwise, start over. The most highbrow way is to go via the standard measure on the affine Grassmannian of all planes in 3-space; if we restrict to the set of planes that pass through the cube, we get a finite measure, so we can rescale to get a probability measure.)
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