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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