October 2020 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 16 Nov '20

16 Nov '20

Yes (I presume this means in R^3). To get on with all faces rectangles, drill a hole straight through a rectangular solid and look at its boundary. Now apply a random linear mapping and all the rectangles will probably turn into parallelograms. —Dan Jim Propp wrote: ----- George Hart’s Celebration of Mind talk (just concluded) makes me wonder: is there a non-self-intersecting torus surface made of parallelograms? -----

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[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

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[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

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[math-fun] A pretty surface with surprising symmetry
by Dan Asimov 31 Oct '20

31 Oct '20

I'm always struck by instances in math unexpected symmetry. For instance, consider ways of coloring the tiles of the hexagonal tessellation of the plane with n colors so that no two tiles of the same color share an edge. One way to do this involves 7 colors. It could be like this: 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 (and repeat). So here's a tessellation by 6-sided figures that is colored with 7 colors, each color forming a subset of the hexagons isometric to that of any other color. Or consider the number 73. Not particularly conducive to symmetry, you might think. But recall the regular 4D polytope called the 24-cell: It's 24 octahedra arranged face-to-face to form the 3-dimensional sphere S^3. In fact it has 24 vertices, 96 edges, 96 2-dimensional faces, and 24 3-dimensional octahedra. This makes it self-dual. So if you draw the vertices and edges of this 24-cell on the 3-sphere S^3, you can also draw the *dual* vertices and edges as well — they will form a disjoint 1-dimensional figure in S^3. Now if you removed both of these "1-skeleta" from S^3, the portion of S^3 that remains is topologically equivalent to the product of a surface M_g of genus g with the real line R. And it turns out that the genus of g M_g is 73. Even nicer, it turns out that the surface of genus 73 that is halfway between the two dual 1-skeleta in S^3 is a *minimal surface*, meaning that if you drew a small simple closed curve C about any of its points, the surface minimizes the area among all surfaces in S^3 whose boundary is the curve C. Even nicer yet, this minimal surface M_73 has an isometry with itself for any isometry of the 24-cell. One of the octahedra can be taken to any of the 24 octahedra, and one of its triangles can go to any of the 8 triangular faces in any of 6 dihedral way, making for 24 x 24 x 2 = 1052 isometries. Nicest of all, the two halves of S^3 created when M_73 is removed can be interchanged, which carries M_73 to itself, giving it another 1052 isometries for a grand total of 2104 altogether. Who said 73 wasn't symmetrical. —Dan

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Re: [math-fun] Lampshade weave.
by Steve Witham 30 Oct '20

30 Oct '20

Replying to Allan Wechsler and Dan Asimov. > From: Allan Wechsler <acwacw(a)gmail.com> > Date: 10/29/20, 10:54 PM > > Maybe my visualization skills are tripping me up, but it seems to me that > when you flatten the cone to a plane, the angle between the two edges must > be 90 degrees. Yes, it's 90 degrees. (*Not* 270.) I suppose this pie-slice angle should be theta. This raises another question. What's the angle you get if you slice the cone with a plane through its vertical axis-- call this phi--as a function of theta? Let's see, theta=0 -> phi = 0; theta=2pi -> phi = pi. So phi = theta/2, right? > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 10/30/20, 10:56 AM > > In order to think about this, I find it useful to imagine > first cutting open the cone by a cut from its vertex Â— making > it into a sector of the plane Â— and then sewing together > infinitely many copies of this sector, edge to edge, to that > the whole thing has the topology of an infinite parking ramp > (or a log surface in the complex plane) and laying it flat > in the plane. > > Now any piece of a geodesic on the original cone can be > transferred to this surface and despite its being flat, > you can extend the geodesic indefinitely and it will > always be in the correct corresponding location on one of > the (infinitely many) sectors, each corresponding to the > original cone. (We've put the vertex of the cone at the > origin.) > > Of course the geodesic once it's on the plane will just be > a straight line if thought of in the plane. The generators > of the cone (straight lines through its vertex) correspond > to rays emanating from the origin in the plane. > > >From this it's possible to see that (assuming the geodesic > doesn't cross the cone's vertex) that it will reach a > closest point and then veer away, and that it cannot > circle the cone infinitely many times. (But by choosing > a cone with arbitrarily small cone angle, one can find > geodesic that do circle the cone arbitrarily many times.) > > Â—Dan Yes! That's what I meant by, "Maybe a nice diagram?" I suppose we can call it the fan diagram. There are two kinds of geodesics on a given cone: straight lines through the vertex, and scaled copies "that curve." The follow-up questions are, what are the relative distances- from-the-origin of the nearest point and any self-crossings, and how many self-crossings are there? --Steve

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Re: [math-fun] Lampshade weave.
by Dan Asimov 30 Oct '20

30 Oct '20

In order to think about this, I find it useful to imagine first cutting open the cone by a cut from its vertex — making it into a sector of the plane — and then sewing together infinitely many copies of this sector, edge to edge, to that the whole thing has the topology of an infinite parking ramp (or a log surface in the complex plane) and laying it flat in the plane. Now any piece of a geodesic on the original cone can be transferred to this surface and despite its being flat, you can extend the geodesic indefinitely and it will always be in the correct corresponding location on one of the (infinitely many) sectors, each corresponding to the original cone. (We've put the vertex of the cone at the origin.) Of course the geodesic once it's on the plane will just be a straight line if thought of in the plane. The generators of the cone (straight lines through its vertex) correspond to rays emanating from the origin in the plane. >From this it's possible to see that (assuming the geodesic doesn't cross the cone's vertex) that it will reach a closest point and then veer away, and that it cannot circle the cone infinitely many times. (But by choosing a cone with arbitrarily small cone angle, one can find geodesic that do circle the cone arbitrarily many times.) —Dan Steve Witham wrote: ----- Questions: Imagine you took apart the lampshade, cut straight through the vertical joint (so there was no overlap), and laid the cone material out flat. Maybe fill in the center part where the top hole was cut out. The material would form some pie-chart fraction. Measuring from the center/tip, an arbitrary unfolded cone could cover <180, 180 or >180 degrees of the pie. What is that angle for *my* lampshade? Considering whole cones again, how many times does a geodesic of a cone cross itself? Can that number be infinite? Can it be finite? If finite, what does the number of crossings depend on and how? What are the relative distances of the crossings from the tip of the cone? Are the even and odd crossings spaced differently? What about the non- self-crossing point where the curve is closest to the tip? Maybe a nice diagram? Does the 3D curve have a name? If you draw verticals from the curve down to a horizontal plane, what is that curve? If you project horizontally to a vertical plane what kinds of curves do you get? Can the threads be arranged so that they consistently follow the over-under weave pattern all the way around? If so, show how. Hint: imagine making the cone out of graph paper. Can you imagine a machine to weave it, or even a jig to help weave it by hand? If the consistent weave is impossible, prove it. -----

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[math-fun] Lampshade weave.
by Steve Witham 30 Oct '20

30 Oct '20

I looked over at a lamp I've been using for years.Â The lampshade is conical and covered with loose-woven (in the sense that there are gaps for light between the threads) thin twine cloth glued onto a paper backing. And I said, "No, it can't be..." Yup.Â At the vertical seam where the cloth wraps around and meets itself, the threads on the two sides continue in the same directions-- +45 and -45 degrees (not perfectly matched, but it's clear they could have been). on the far side of the shade, the threads are going vertical and horizontal. Following a bottom thread at the seam it arcs up, comes back down and crosses itself at a right angle at the seam. Everybody knows you can make a cone out of paper. That means, despite the curving pattern, the threads of the weave are geodesics... at least locally. And that means that putting any thread under tension wouldn't make it want to slip up or down the cone. Questions: Imagine you took apart the lampshade, cut straight through the vertical joint (so there was no overlap), and laid the cone material out flat.Â Â Maybe fill in the center part where the top hole was cut out.Â The material would form some pie-chart fraction. Measuring from the center/tip, an arbitrary unfolded cone could cover <180, 180 or >180 degrees of the pie. What is that angle for *my* lampshade? Considering whole cones again, how many times does a geodesic of a cone cross itself? Can that number be infinite? Can it be finite? If finite, what does the number of crossings depend on and how? What are the relative distances of the crossings from the tip of the cone? Are the even and odd crossings spaced differently? What about the non- self-crossing point where the curve is closest to the tip? Maybe a nice diagram? Does the 3D curve have a name? If you draw verticals from the curve down to a horizontal plane, what is that curve? If you project horizontally to a vertical plane what kinds of curves do you get? Can the threads be arranged so that they consistently follow the over-under weave pattern all the way around? If so, show how. Hint: imagine making the cone out of graph paper. Can you imagine a machine to weave it, or even a jig to help weave it by hand? If the consistent weave is impossible, prove it. --Steve

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Re: [math-fun] Many cheerful facts about the Hurwitz quaternions
by Dan Asimov 28 Oct '20

28 Oct '20

Aha — interesting. (I was recently reminded that the Z^2 lattice has the same kind of similarity property.) If you start with the Hurwitz integral quaternions J, then those ever-finer copies of the D_4 lattice ought to also be subrings of the entire skew-field of quaternions (which I'll call H). So if we took the union of all those ever-finer copies, the whole thing would be what, J ⊗ Z_2^oo ??? (Where Z_2^oo denotes the ring of rational numbers whose denominator is a power of 2.) —Dan Adam Goucher wrote: ----- Something I noticed about the D_4 lattice is that the union of the lattice together with its deep holes is geometrically similar to the original D_4 lattice, scaled by 1/sqrt(2). This means that if you take the 4-dimensional torus R^4 / D_4 and start greedily placing points to be as far away as possible from the nearest point, you'll end up progressively filling ever-finer copies of the D_4 lattice. (This also works on the more conventional torus R^4 / Z^4, because after the first two points are emplaced they'll form D_4 / Z^4.) -----

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[math-fun] Many cheerful facts about the Hurwitz quaternions
by Dan Asimov 28 Oct '20

28 Oct '20

The 24 quaternions G = {±1, ±i, ±j, ±k, (±1 ±i ±j ±k)/2} form a group under multiplication isomorphic to the binary tetrahedral group 2T, which can be defined as the inverse image of the rotation group T of the regular tetrahedron under the double covering map π : S^3 —> SO(3) from the unit sphere S^3 (the group of unit quaternions) to the rotation group SO(3) of 3-space: 2T = π^(-1)(T), where the map π is defined via π(u) = (q |—> u q u^(-1)) for q = xi + yj + zk in the 3-space of pure quaternions. The subring of the quaternions generated by the 24 elements of 2T is the Hurwitz "integral" quaternions J: J = <2T>. As a lattice in R^4, J is just J = Z^4 ∪ (1/2,1/2,1/2,1/2) + Z^4, the union of the integer points and the so-called half-integer points. This lattice is called the D_4 lattice, aka the 4D checkerboard lattice, and is believed to be the unique lattice in R^4 whose points form the centers of the densest possible sphere-packing in R^4. J forms a maximal order of the quaternions (https://en.wikipedia.org/wiki/Order_(ring_theory)) The points of R^4 closer to the origin than any other lattice point form the interior points of the regular polytope called the 24-cell, which is the convex hull of the points of 2T. All of R^4 is tiled by these "Voronoi polytope" 24-cells, one for each lattice point of D^4. It's also true that the Voronoi 24-cell of the origin is surrounded by 24 other 24-cells, and much like a rosette of 7 hexagons in the plane can tessellate the plane, 4-space can be tessellated by these clusters of 25 24-cells, translates of the original cluster by quaternions in the left ideal (2+i)J of the Hurwitz quaternions. This means that if R^4 is quotiented out by the lattice (2+i)J, the result is a certain 4-dimensional torus R^4/(2+i)J that is itself tessellated by 25 24-cells ... each one sharing a 3-dimensional face (an octahedron) with each other one. Which I find to be very cool! —Dan Fred Lunnon wrote: ----- Hurwitz quaternions work in 3-space --- see for example my write-up for when this algorithm last appeared on math-fun, https://arxiv.org/abs/1202.3198 I never needed to consider 4-space, since nobody has managed to discover a Heronian pentatope! On 10/22/20, Henry Baker <hbaker1(a)pipeline.com> wrote: > Hurwitz quaternions might work for 4D. ..... -----

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