[math-fun] Random slice of a cube
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.) Jim Propp
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.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Oops! Allan is right. I should've said, erect the plane and see if *it* intersects the cube. That would give us something equivalent to the highbrow version (which I feel is the "right" one to use). A different "wrong" probability measure to use is, once you've picked P, choose a random plane perpendicular to OP that intersects the cube. The probability measure I have in mind favors certain directions over others, because the cube isn't equally wide in all directions. Thanks, Allan! Jim Propp On Sunday, October 8, 2017, Allan Wechsler <acwacw@gmail.com> wrote:
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 <javascript:;>> 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.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Here is a related question: What is the locus of points Q such that the plane through Q perpendicular to OQ intersects the cube? It obviously contains the cube itself as a subset. For concreteness, assume the cube to be centered at the origin of a coordinate system, and the vertices have all coordinates with absolute value 1. Is it the union of eight spheres, centered at the points halfway between the vertices and the origin, with radius sqrt(3)/2? On Sun, Oct 8, 2017 at 8:24 AM, James Propp <jamespropp@gmail.com> wrote:
Oops! Allan is right. I should've said, erect the plane and see if *it* intersects the cube. That would give us something equivalent to the highbrow version (which I feel is the "right" one to use).
A different "wrong" probability measure to use is, once you've picked P, choose a random plane perpendicular to OP that intersects the cube. The probability measure I have in mind favors certain directions over others, because the cube isn't equally wide in all directions.
Thanks, Allan!
Jim Propp
On Sunday, October 8, 2017, Allan Wechsler <acwacw@gmail.com> wrote:
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 <javascript:;>> 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.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Allan Wechsler -
James Propp