Re: [math-fun] Random slice of a cube
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.
My method was to pick a point Q at a distance less than the cube root of 3 from the cube's center, i.e. a point inside the sphere that circumscribes the cube. But now I'm wondering what *is* the locus of points such that a plane including Q and perpendicular to the line connecting Q to the center would intersect the cube. The locus obviously includes everything in the cube and nothing outside the circumscribing sphere. But it's not coincident with either the cube or the sphere. What is its shape, its volume, and its surface area? I'm assuming the cube has a side length of 2 and is centered on the origin. I'm pretty sure that parts of the shape are concave. A simpler problem is to reduce it by a dimension. What's the shape, area, and perimeter of the locus of points such that a line through the point and perpendicular to the line connecting the point to the center intersects the square with side length 2 and centered on the origin? Again, I'm pretty sure that parts of the shape are concave.
Keith, if you go back to the fourth message on this thread, you'll find me asking the same question. I also give a diffident guess at the answer: the union of eight spheres, which just circumscribe the eight octants of the original cube (so their common radius is half the square root of 3). In your message you say "the cube root of 3" where I'm certain you meant "the square root of 3". I am more certain that the answer in 2 dimensions is the union of the four circles that circumscribe the four quadrants of the square. On Tue, Oct 24, 2017 at 12:24 AM, Keith F. Lynch <kfl@keithlynch.net> wrote:
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.
My method was to pick a point Q at a distance less than the cube root of 3 from the cube's center, i.e. a point inside the sphere that circumscribes the cube.
But now I'm wondering what *is* the locus of points such that a plane including Q and perpendicular to the line connecting Q to the center would intersect the cube. The locus obviously includes everything in the cube and nothing outside the circumscribing sphere. But it's not coincident with either the cube or the sphere. What is its shape, its volume, and its surface area? I'm assuming the cube has a side length of 2 and is centered on the origin. I'm pretty sure that parts of the shape are concave.
A simpler problem is to reduce it by a dimension. What's the shape, area, and perimeter of the locus of points such that a line through the point and perpendicular to the line connecting the point to the center intersects the square with side length 2 and centered on the origin? Again, I'm pretty sure that parts of the shape are concave.
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Keith F. Lynch