Re: [math-fun] Random slice of a cube
On 2017-10-08 21:16, Keith F. Lynch wrote:
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. ....)
What about choosing three points independently and uniformly on a sphere that encloses the cube, and using the plane defined by those three points, skipping any planes that don't intersect the cube? Would that get you the same answer?
I am not a visual thinker. I'm not even sure what numbers of sides the polygon could have. It can obviously have 3, 4, or 6. I doubt it can have more than 6, but I'm not sure. I'm even less sure whether it can have 5. gosper.org/pentagonal cubeslice.png --rwg
I would ask, not what the average number of sides is, but the proportion of each number of sides. That way I can quickly get any of countless possible means, including the arithmetic mean, geometric mean, harmonic mean, mode, median, root-mean-square, etc. Maybe some of them are more interesting than others. Also, what's the distribution of areas of the intersection? And what are the answers to the same questions for the other four regular solids, and for the sphere? Similarly in other dimensions.
When I was initially thinking about the problem, I wasn't sure there were pentagonal cross-section, but then I realized that there were, by continuity. Once you have a hexagonal cross-section, you can perturb the cutting plane slightly, so that it's in general position. Then, as you translate the plane perpendicular to itself, the moving plane never meets more than one vertex of the cube at a time, so the transition-events all become of the form "one edge shrinks to length zero", with the number of sizes of the polygon decreasing by exactly 1. I like Keith's suggestion that one looks at the probabilities, and not just the mean. It would be interesting if the distribution failed to be unimodal; e.g., if pentagons were rarer than both hexagons and quadrilaterals. Jim Propp On Mon, Oct 9, 2017 at 1:15 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2017-10-08 21:16, Keith F. Lynch wrote:
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. ....)
What about choosing three points independently and uniformly on a sphere that encloses the cube, and using the plane defined by those three points, skipping any planes that don't intersect the cube? Would that get you the same answer?
I am not a visual thinker. I'm not even sure what numbers of sides the polygon could have. It can obviously have 3, 4, or 6. I doubt it can have more than 6, but I'm not sure. I'm even less sure whether it can have 5. gosper.org/pentagonal cubeslice.png --rwg
I would ask, not what the average number of sides is, but the proportion of each number of sides. That way I can quickly get any of countless possible means, including the arithmetic mean, geometric mean, harmonic mean, mode, median, root-mean-square, etc. Maybe some of them are more interesting than others. Also, what's the distribution of areas of the intersection? And what are the answers to the same questions for the other four regular solids, and for the sphere? Similarly in other dimensions.
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