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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