Re: [math-fun] Random slice of a cube
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
I'm pretty sure it's not possible to get a polygon with fewer than four sides from a plane's intersection with an octahedron, or with fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
Since nobody else seems to have done so, I wrote a quick program to generate (pseudo-)random planes and count sides (for the cube only, not for other regular polygons).
I meant regular polyhedrons, not polygons.
* If I set Max_D (the maximum distance from the center of the cube to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
Unfortunately, my program cannot easily be adapted to other regular polygons.
Again, I meant polyhedrons, not polygons.
I could easily collect statistics on the side lengths and angles of the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4. This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square. I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4. I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried. Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.) I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question. Paging George Hart (one of the original designers of that exhibit): is this something you already knew? Jim On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
I'm pretty sure it's not possible to get a polygon with fewer than four sides from a plane's intersection with an octahedron, or with fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
Since nobody else seems to have done so, I wrote a quick program to generate (pseudo-)random planes and count sides (for the cube only, not for other regular polygons).
I meant regular polyhedrons, not polygons.
* If I set Max_D (the maximum distance from the center of the cube to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
Unfortunately, my program cannot easily be adapted to other regular polygons.
Again, I meant polyhedrons, not polygons.
I could easily collect statistics on the side lengths and angles of the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Hi Jim, I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube. It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n. (So it holds for any parallelepiped, not just cubes.) Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.) I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron... George http://georgehart.com/ On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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For other regular solids except the tetrahedron: I'm not sure if this helps, but (of course), for any particular vertex, the vector sum of the edges for any path from the vertex to its antipode is the same. Rich ----- Quoting George Hart <george@georgehart.com>:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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My original question involved the same measure that George is talking about, but averaged over all orientations. I greatly prefer the stronger result in which the orientation is fixed, and one merely translates the plane. (At some point soon I'll post Warren's cute proof, which he sent to me by private email.) Jim Propp On Wednesday, October 18, 2017, George Hart <george@georgehart.com> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto: kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Also, the measure of the range of offsets that intersect one EDGE of the cube (namely one parallel to the vector a) is n.a/n.(a+b+c), and for the other three parallel edges it is the same. Adding up all twelve edges of the cube we get n.(4a+4b+4c)/n.(a+b+c) = 4, which tells us that the cross-section has exactly 4 vertices on average. (Of course we knew that, since for any polygon the number of vertices equals the number of sides!) I believe that for all k and n, George's argument can be applied to compute the expected number of k-dimensional facets in the intersection of a random n-dimensional hypercube (or more generally parallelotope) with a codimension-1 plane of fixed orientation, chosen uniformly from all translates that intersect the n-cube nontrivially. That's right, isn't it? It might be possible to bring a third parameter into the story, by using a random translate of a codimension-m plane with m > 1, but here my geometric intuition fails me. I think the smallest interesting case for m=2 would be n=4. A randomly translated 2-plane that meets a 4-cube nontrivially would have some sort of polygon as the intersection. Warren's approach (if I'm not mistaken) gives the answer 4 semi-rigorously; can George's vector-algebra approach be applied? Jim On Wednesday, October 18, 2017, George Hart <george@georgehart.com> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto: kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Here is Warren's lovely approach: "OK, so here is a vague half-proof that for your problem, the exact expectation equals 4. Do not use just 1 cube. Instead cover all of 3-space with a grid of equal cubes, usual fashion. Now cut that with a random plane [or a random plane with prescribed orientation]. "Now, consider all the polygons your plane is thus divided into (restricting attention to a huge ball on your plane to keep things finite, in limit as radius large). "We want to prove the average number of edges is 4. Well, the line-segments bounding your polygonal cells form a planar graph each of whose valencies is (generically) 4. The dual planar graph also (by Euler's formula, edge-count to vertex-count ratio, mumble) therefore must have expected valency 4." As Warren pointed out in a later email to me: "My proof works (appropriately altered) for any polyhedron which tiles 3-space in a face-to-face manner with a constant number K of cells sharing each edge. For the cube, K=4, for the rhombic dodec, K=3, and for infinitely long eq-triangular prisms, K=6. "So a random plane cut thru a rhombic dodec should be an N-gon with mean(N)=6 exactly." Hopefully some of the emails I've sent to math-fun make more sense now! (Sorry for the odd font; I'd fix it if I knew how to do it on my iPhone.) Jim
George, Your suspicion about the rhombic dodecahedron was 100% right, as Warren's argument shows. Is there a slick way to prove it with vectors, as in your proof for the cube? Jim On Wednesday, October 18, 2017, George Hart <george@georgehart.com> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto: kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Jim, Yes, I believe my method works for any convex polyhedron bounded by parallelograms. These are the "generic zonohedra", i.e., the polyhedra that can be derived as the Minkowski sum of k line segments in general position. For any slicing plane direction, the average number of sides in the slice (weighing uniformly on the range of offsets that intersect the polyhedron) is 2(k-1). This includes the cube and all parallelepipeds (k=3), the rhombic dodecahedron (k=4), and much more. For example, the rhombic enneacontahedron is the Minkowski sum of the ten long diagonals of a regular dodecahedron, so k=10, it has 90 faces, and the average number of sides when it is sliced is 18. As you suggest, this includes all 3D shadows of k-dimensional hypercubes (and affine transformations of them). (Most of these are not space fillers, so this method of proof at first seems more general than Warren's, but perhaps his method can be generalized via a packing in k-space.) My method of proof works once you note (from the combinatorics of zonotopes) that there are k(k-1) faces, one opposing pair for each of the k-choose-2 ways of choosing two of the generating segments, and sum their measures. Equivalently, you can count the vertices as you suggest, after noting that for each of the k generating segments, the Minkowski sum has 2(k-1) parallel edges, e.g., the four groups of six parallel edges in a rhombic dodecahedron, and sum their measures. If three (or more) of the line segments are co-planar, their Minkowski sum has some faces which are hexagons (or octagons...) and the formula for average number of faces in the slice needs to be reduced with an adjustment which depends on the geometry of the polyhedron and the slice direction. For example if k=4 but three of the directions are coplanar, the Minkowski sum is a hexagonal prism. Let a, b, and c be the coplanar directions that span the hexagon and d the remaining direction, so the prism has 6 edges in the d direction but only 4 in each of the a, b, c, directions. Let n be the slice normal. Then the average number of sides on a slice of the hexagonal prism works out to 6 - 2 n.(a+b+c)/n.(a+b+c+d) [The directions of segments a, b, c, d should be taken to have positive projection in the n direction.] George http://georgehart.com/ On 10/19/2017 9:01 AM, James Propp wrote:
George,
Your suspicion about the rhombic dodecahedron was 100% right, as Warren's argument shows.
Is there a slick way to prove it with vectors, as in your proof for the cube?
Jim
On Wednesday, October 18, 2017, George Hart <george@georgehart.com <mailto:george@georgehart.com>> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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I'm still getting the hang of thinking this way. Part of what I'm gleaning is that there's a sense in which the cube and the rhombic dodecahedron are "combinatorially constant-width zonohedra": that is, if you take two generic parallel planes of support, touching the zonohedron at points P and P', then the path from P to P' on the 1-skeleton of the zonohedron is independent of which planes of support you chose, modulo (1) replacing an edge by a parallel edge, (2) reversing the direction of an edge, and (3) permuting the edges. This means that the distance between the planes can be written in the form |a.n| + |b.n| + |c.n| + ..., where a, b, c, ... are vectors associated with line-segments whose Minkowski sum is the zonohedron. Have I got that right, George? Let's think about those absolute-value signs for a minute, because they're missing from George's formulas. I think what he's doing is choosing an orientation of those vectors that's consistent with the choice of the supporting planes. (That is, if you think of the planes as level-sets of some linear function f on R^3, then the edges should be oriented in the direction of increasing f.) Then he can write |a.n| + |b.n| + |c.n| + ... as a.n + b.n + c.n + ..., which becomes (a+b+c+...).n. Moreover, if (as I prefer) we count intersections of intervening planes with the 1-skeleton (that is, if we count the vertices, rather than the sides, of each polygonal cross-section), then we get a.n, b.n, c.n, ... to measure how "many" of the parallel planes intersect each edge of type a, each edge of type b, each edge of type c, etc. So, no need for absolute value signs. BUT: this hinges over making a choice of orientations of the edges that are compatible with the choice of supporting planes. And I don't see how to see which zonohedra are "combinatorially of constant width" and which aren't. For instance, I would've initially guessed that ALL zonohedra (including the rhombic dodecahedron) have this property, but George's example of the hexagonal prism shows me that my intuition is faulty, and that makes me less secure in thinking that the rhombic dodecahedron has this property. Can someone help me think about this more clearly? Jim On Thu, Oct 19, 2017 at 11:19 AM, George Hart <george@georgehart.com> wrote:
Jim,
Yes, I believe my method works for any convex polyhedron bounded by parallelograms. These are the "generic zonohedra", i.e., the polyhedra that can be derived as the Minkowski sum of k line segments in general position. For any slicing plane direction, the average number of sides in the slice (weighing uniformly on the range of offsets that intersect the polyhedron) is 2(k-1).
This includes the cube and all parallelepipeds (k=3), the rhombic dodecahedron (k=4), and much more. For example, the rhombic enneacontahedron is the Minkowski sum of the ten long diagonals of a regular dodecahedron, so k=10, it has 90 faces, and the average number of sides when it is sliced is 18.
As you suggest, this includes all 3D shadows of k-dimensional hypercubes (and affine transformations of them). (Most of these are not space fillers, so this method of proof at first seems more general than Warren's, but perhaps his method can be generalized via a packing in k-space.)
My method of proof works once you note (from the combinatorics of zonotopes) that there are k(k-1) faces, one opposing pair for each of the k-choose-2 ways of choosing two of the generating segments, and sum their measures.
Equivalently, you can count the vertices as you suggest, after noting that for each of the k generating segments, the Minkowski sum has 2(k-1) parallel edges, e.g., the four groups of six parallel edges in a rhombic dodecahedron, and sum their measures.
If three (or more) of the line segments are co-planar, their Minkowski sum has some faces which are hexagons (or octagons...) and the formula for average number of faces in the slice needs to be reduced with an adjustment which depends on the geometry of the polyhedron and the slice direction. For example if k=4 but three of the directions are coplanar, the Minkowski sum is a hexagonal prism. Let a, b, and c be the coplanar directions that span the hexagon and d the remaining direction, so the prism has 6 edges in the d direction but only 4 in each of the a, b, c, directions. Let n be the slice normal. Then the average number of sides on a slice of the hexagonal prism works out to 6 - 2 n.(a+b+c)/n.(a+b+c+d)
[The directions of segments a, b, c, d should be taken to have positive projection in the n direction.]
George http://georgehart.com/
On 10/19/2017 9:01 AM, James Propp wrote:
George,
Your suspicion about the rhombic dodecahedron was 100% right, as Warren's argument shows.
Is there a slick way to prove it with vectors, as in your proof for the cube?
Jim
On Wednesday, October 18, 2017, George Hart <george@georgehart.com <mailto:george@georgehart.com>> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Jim, I'm not so keen on the term "combinatorially constant-width zonohedra", but your formulas are correct. All paths from P to P' which don't "back up" visit one of each type of edge. They are all the same length. (You cross each zone exactly once, hence the "zone" of "zonohedra".) My footnote was intended to indicate the sign convention: "The directions of segments a, b, c, d should be taken to have positive projection in the n direction." In your example, use the P'-P direction. The non-generic zonohedra (e.g., the hexagonal prism example) also have this property. It's just that they have fewer paths (unless you draw dotted lines on the hexagons to dissect them into three coplanar parallelograms and restore the missing paths). George http://georgehart.com/ On 10/20/2017 10:17 AM, James Propp wrote:
I'm still getting the hang of thinking this way.
Part of what I'm gleaning is that there's a sense in which the cube and the rhombic dodecahedron are "combinatorially constant-width zonohedra": that is, if you take two generic parallel planes of support, touching the zonohedron at points P and P', then the path from P to P' on the 1-skeleton of the zonohedron is independent of which planes of support you chose, modulo (1) replacing an edge by a parallel edge, (2) reversing the direction of an edge, and (3) permuting the edges. This means that the distance between the planes can be written in the form |a.n| + |b.n| + |c.n| + ..., where a, b, c, ... are vectors associated with line-segments whose Minkowski sum is the zonohedron.
Have I got that right, George?
Let's think about those absolute-value signs for a minute, because they're missing from George's formulas. I think what he's doing is choosing an orientation of those vectors that's consistent with the choice of the supporting planes. (That is, if you think of the planes as level-sets of some linear function f on R^3, then the edges should be oriented in the direction of increasing f.) Then he can write |a.n| + |b.n| + |c.n| + ... as a.n + b.n + c.n + ..., which becomes (a+b+c+...).n. Moreover, if (as I prefer) we count intersections of intervening planes with the 1-skeleton (that is, if we count the vertices, rather than the sides, of each polygonal cross-section), then we get a.n, b.n, c.n, ... to measure how "many" of the parallel planes intersect each edge of type a, each edge of type b, each edge of type c, etc. So, no need for absolute value signs.
BUT: this hinges over making a choice of orientations of the edges that are compatible with the choice of supporting planes. And I don't see how to see which zonohedra are "combinatorially of constant width" and which aren't. For instance, I would've initially guessed that ALL zonohedra (including the rhombic dodecahedron) have this property, but George's example of the hexagonal prism shows me that my intuition is faulty, and that makes me less secure in thinking that the rhombic dodecahedron has this property.
Can someone help me think about this more clearly?
Jim
On Thu, Oct 19, 2017 at 11:19 AM, George Hart <george@georgehart.com> wrote:
Jim,
Yes, I believe my method works for any convex polyhedron bounded by parallelograms. These are the "generic zonohedra", i.e., the polyhedra that can be derived as the Minkowski sum of k line segments in general position. For any slicing plane direction, the average number of sides in the slice (weighing uniformly on the range of offsets that intersect the polyhedron) is 2(k-1).
This includes the cube and all parallelepipeds (k=3), the rhombic dodecahedron (k=4), and much more. For example, the rhombic enneacontahedron is the Minkowski sum of the ten long diagonals of a regular dodecahedron, so k=10, it has 90 faces, and the average number of sides when it is sliced is 18.
As you suggest, this includes all 3D shadows of k-dimensional hypercubes (and affine transformations of them). (Most of these are not space fillers, so this method of proof at first seems more general than Warren's, but perhaps his method can be generalized via a packing in k-space.)
My method of proof works once you note (from the combinatorics of zonotopes) that there are k(k-1) faces, one opposing pair for each of the k-choose-2 ways of choosing two of the generating segments, and sum their measures.
Equivalently, you can count the vertices as you suggest, after noting that for each of the k generating segments, the Minkowski sum has 2(k-1) parallel edges, e.g., the four groups of six parallel edges in a rhombic dodecahedron, and sum their measures.
If three (or more) of the line segments are co-planar, their Minkowski sum has some faces which are hexagons (or octagons...) and the formula for average number of faces in the slice needs to be reduced with an adjustment which depends on the geometry of the polyhedron and the slice direction. For example if k=4 but three of the directions are coplanar, the Minkowski sum is a hexagonal prism. Let a, b, and c be the coplanar directions that span the hexagon and d the remaining direction, so the prism has 6 edges in the d direction but only 4 in each of the a, b, c, directions. Let n be the slice normal. Then the average number of sides on a slice of the hexagonal prism works out to 6 - 2 n.(a+b+c)/n.(a+b+c+d)
[The directions of segments a, b, c, d should be taken to have positive projection in the n direction.]
George http://georgehart.com/
On 10/19/2017 9:01 AM, James Propp wrote:
George,
Your suspicion about the rhombic dodecahedron was 100% right, as Warren's argument shows.
Is there a slick way to prove it with vectors, as in your proof for the cube?
Jim
On Wednesday, October 18, 2017, George Hart <george@georgehart.com <mailto:george@georgehart.com>> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch <kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Okay, I'll write "Hart property" instead of "combinatorially constant width property" until someone suggests a better term. :-) It seems to me that (if I understood what George wrote in his earlier message) the hexagonal prism, although it is a zonohedron, does NOT have the Hart property, since 6 - 2 n.(a+b+c)/n.(a+b+c+d) depends on n. What am I missing? Jim On Fri, Oct 20, 2017 at 11:16 AM, George Hart <george@georgehart.com> wrote:
Jim,
I'm not so keen on the term "combinatorially constant-width zonohedra", but your formulas are correct. All paths from P to P' which don't "back up" visit one of each type of edge. They are all the same length. (You cross each zone exactly once, hence the "zone" of "zonohedra".) My footnote was intended to indicate the sign convention: "The directions of segments a, b, c, d should be taken to have positive projection in the n direction." In your example, use the P'-P direction.
The non-generic zonohedra (e.g., the hexagonal prism example) also have this property. It's just that they have fewer paths (unless you draw dotted lines on the hexagons to dissect them into three coplanar parallelograms and restore the missing paths).
George http://georgehart.com/
On 10/20/2017 10:17 AM, James Propp wrote:
I'm still getting the hang of thinking this way.
Part of what I'm gleaning is that there's a sense in which the cube and the rhombic dodecahedron are "combinatorially constant-width zonohedra": that is, if you take two generic parallel planes of support, touching the zonohedron at points P and P', then the path from P to P' on the 1-skeleton of the zonohedron is independent of which planes of support you chose, modulo (1) replacing an edge by a parallel edge, (2) reversing the direction of an edge, and (3) permuting the edges. This means that the distance between the planes can be written in the form |a.n| + |b.n| + |c.n| + ..., where a, b, c, ... are vectors associated with line-segments whose Minkowski sum is the zonohedron.
Have I got that right, George?
Let's think about those absolute-value signs for a minute, because they're missing from George's formulas. I think what he's doing is choosing an orientation of those vectors that's consistent with the choice of the supporting planes. (That is, if you think of the planes as level-sets of some linear function f on R^3, then the edges should be oriented in the direction of increasing f.) Then he can write |a.n| + |b.n| + |c.n| + ... as a.n + b.n + c.n + ..., which becomes (a+b+c+...).n. Moreover, if (as I prefer) we count intersections of intervening planes with the 1-skeleton (that is, if we count the vertices, rather than the sides, of each polygonal cross-section), then we get a.n, b.n, c.n, ... to measure how "many" of the parallel planes intersect each edge of type a, each edge of type b, each edge of type c, etc. So, no need for absolute value signs.
BUT: this hinges over making a choice of orientations of the edges that are compatible with the choice of supporting planes. And I don't see how to see which zonohedra are "combinatorially of constant width" and which aren't. For instance, I would've initially guessed that ALL zonohedra (including the rhombic dodecahedron) have this property, but George's example of the hexagonal prism shows me that my intuition is faulty, and that makes me less secure in thinking that the rhombic dodecahedron has this property.
Can someone help me think about this more clearly?
Jim
On Thu, Oct 19, 2017 at 11:19 AM, George Hart <george@georgehart.com> wrote:
Jim,
Yes, I believe my method works for any convex polyhedron bounded by parallelograms. These are the "generic zonohedra", i.e., the polyhedra that can be derived as the Minkowski sum of k line segments in general position. For any slicing plane direction, the average number of sides in the slice (weighing uniformly on the range of offsets that intersect the polyhedron) is 2(k-1).
This includes the cube and all parallelepipeds (k=3), the rhombic dodecahedron (k=4), and much more. For example, the rhombic enneacontahedron is the Minkowski sum of the ten long diagonals of a regular dodecahedron, so k=10, it has 90 faces, and the average number of sides when it is sliced is 18.
As you suggest, this includes all 3D shadows of k-dimensional hypercubes (and affine transformations of them). (Most of these are not space fillers, so this method of proof at first seems more general than Warren's, but perhaps his method can be generalized via a packing in k-space.)
My method of proof works once you note (from the combinatorics of zonotopes) that there are k(k-1) faces, one opposing pair for each of the k-choose-2 ways of choosing two of the generating segments, and sum their measures.
Equivalently, you can count the vertices as you suggest, after noting that for each of the k generating segments, the Minkowski sum has 2(k-1) parallel edges, e.g., the four groups of six parallel edges in a rhombic dodecahedron, and sum their measures.
If three (or more) of the line segments are co-planar, their Minkowski sum has some faces which are hexagons (or octagons...) and the formula for average number of faces in the slice needs to be reduced with an adjustment which depends on the geometry of the polyhedron and the slice direction. For example if k=4 but three of the directions are coplanar, the Minkowski sum is a hexagonal prism. Let a, b, and c be the coplanar directions that span the hexagon and d the remaining direction, so the prism has 6 edges in the d direction but only 4 in each of the a, b, c, directions. Let n be the slice normal. Then the average number of sides on a slice of the hexagonal prism works out to 6 - 2 n.(a+b+c)/n.(a+b+c+d)
[The directions of segments a, b, c, d should be taken to have positive projection in the n direction.]
George http://georgehart.com/
On 10/19/2017 9:01 AM, James Propp wrote:
George,
Your suspicion about the rhombic dodecahedron was 100% right, as Warren's argument shows.
Is there a slick way to prove it with vectors, as in your proof for the cube?
Jim
On Wednesday, October 18, 2017, George Hart <george@georgehart.com <mailto:george@georgehart.com>> wrote:
Hi Jim,
I had not seen this question or result before: For any orientation of the slicing plane, the average number of sides on a cube slice is 4, where the slice is distributed uniformly among all offsets that intersect the cube.
It is easy to prove. Label the cube's three edge vectors as a, b, and c. Call the normal to the slicing plane n. The measure of range of offsets which intersect one face is n.(a+b)/n.(a+b+c). For the second face it is n.(a+c)/n.(a+b+c), and the third is n.(b+c)/n.(a+b+c). The three opposite parallel faces are the same. Summing over all six faces, this sums to 4, independent of a, b, c, n.
(So it holds for any parallelepiped, not just cubes.)
Note this doesn't hold if you weigh by volume (which I think your original question asked.) In that case the average number of sides depends on the direction n. For example, choosing n parallel to an edge gives 4 but if n is chosen as the direction of the long (body) diagonal, then the weighted average is 5. (Because the two outer pyramids that produce triangles each have volume 1/6, while the central region producing hexagons has volume 2/3.)
I don't expect anything so nice with the other Platonic solids, but perhaps try the rhombic dodecahedron...
George http://georgehart.com/
On 10/18/2017 8:59 AM, James Propp wrote:
I really like Keith's observation about holding the orientation fixed. (Maybe it was in Warren's email too and I just missed the hint.)
I propose that the theorem (once everyone agrees that the proof is solid) be called the ring of fire (or wall the fire) theorem, since that exhibit at the Museum of Mathematics is what inspired me to ask the question.
Paging George Hart (one of the original designers of that exhibit): is this something you already knew?
Jim
On Tuesday, October 17, 2017, Keith F. Lynch < kfl@keithlynch.net <mailto:kfl@keithlynch.net>> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote: > I'm pretty sure it's not possible to get a polygon with fewer than > four sides from a plane's intersection with an octahedron, or with > fewer than five sides from a plane's intersection with an icosahedron.
I should have specified that I was excluding events with measure zero, such as the plane being identical to the plane of one of the faces. Such "infinitely unlikely" events can have no effect on the average number of sides from intersections with random planes.
> Since nobody else seems to have done so, I wrote a quick program to > generate (pseudo-)random planes and count sides (for the cube only, > not for other regular polygons).
I meant regular polyhedrons, not polygons.
> * If I set Max_D (the maximum distance from the center of the cube > to the intersection with the plane) to 0, I only get 4 or 6 sides.
I meant the maximum distance from the center of the cube to the closest point on the plane.
> Unfortunately, my program cannot easily be adapted to other regular > polygons.
Again, I meant polyhedrons, not polygons.
> I could easily collect statistics on the side lengths and angles of > the intersecting polygons, if anyone is interested.
In retrospect, this wouldn't be that easy, except for triangles, as I'd have to keep track of the order of the vertices.
> The average number of sides is 4.0004.
I now suspect it's exactly 4. And not just averaged over all orientations of the planes, but also for every specific orientation of the planes. In other words, choose any orientation and slice the cube into lots of equally-thin parallel slices, like cheese, and in the limit of thinness the average will always be 4.
This is obviously true if the orientation of the planes is parallel to one of the faces. Every slice will be a perfect square.
I've proven it's also true if the orientation of the planes is perpendicular to a body diagonal of the cube. View the body diagonal as an axis, with opposite vertices being the north and south pole. The cube has two polar regions in which the slices will be equilateral triangles, and an equatorial region in which the slices will be hexagons (a regular hexagon at the equator). Using elementary geometry it's easy to show that each of the three regions takes up exactly a third of the axis, hence that the average number of sides is (3+6+3)/3 = 4.
I haven't proven it's true for other orientations (and I don't plan to try -- it's way above my pay grade), but my program gets answers very close to 4 for every random orientation I've tried.
Now I wonder if something similar is true for the other regular polyhedrons. Obviously not for tetrahedrons, since in some orientations you'll always get equilateral triangles and in others you'll always get rectangles.
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Well, coining "Hart property" was lazy of me. I'll try to come up with something different if I need a name for this property, like "the property in question". On Friday, October 20, 2017, James Propp <jamespropp@gmail.com> wrote:
Okay, I'll write "Hart property" instead of "combinatorially constant width property" until someone suggests a better term. :-)
It seems to me that (if I understood what George wrote in his earlier message) the hexagonal prism, although it is a zonohedron, does NOT have the Hart property, since 6 - 2 n.(a+b+c)/n.(a+b+c+d) depends on n. What am I missing?
What I was missing was something clearly stated in George's email, which I didn't process at the time: the two hexagonal faces can be seen as having three internal edges apiece, and if those are included then the prism *does* have the property in question. This is tantamount to viewing some of the cross-sectional polygons as having adjacent sides that are parallel, i.e. as having extra vertices. Jim Propp
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