July 2019 - 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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[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] How does a dull spoon manage to cut a stringy piece of meat
by Bill Gosper 14 Jul '20

14 Jul '20

while sliding back an forth against the inside of a slippery bowl? --rwg

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[math-fun] Another pic of Hilbert's spacefill
by Bill Gosper 19 Apr '20

19 Apr '20

to accompany should've been "sampHilbert" <http://www.tweedledum.com/rwg/sampeano.htm> Hilbert's [spacefilling] function is dense with quadruple points, i.e. points in [0,*i*] with four distinct preimage points in [0,1]: gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, 41/48, 43/48}. This picture connects, in the order they were swept, all the points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and 43/12288. The white segments are retraced boundary, retroflexed in the middle, (making quadruple points). (Spacefilling functions map closed intervals to closed sets.) The Heighway Dragon is dense with mere triple points. Here's one: Hi res Heighway <http://gosper.org/dragtrip!.png> . —rwg

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[math-fun] Heighway triple points redux
by Bill Gosper 21 Mar '20

21 Mar '20

I thought OEIS A260482, A260747..A260750 completely described the (initial) Dragon triple points, all of which have preimages a(n)/15/2^k. But I just noticed an editor's remark that there is another infinite family of the form a(n)/7/2^k. These are much less "photogenic" than the old set (e.g. gosper.org/dragtrip!.png) <http://gosper.org/dragtrip!.png> because two preimages always(?) differ by 2/7/2^k, and ~75% of them have the 3rd preimage within 4/7/2^k. But they can get "arbitrarily" lopsided: In[508]:= undrag[(69 + 27 I)/160] Out[508]= {1539/3584, 2551/3584, 2553/3584} (Differences presumably limited to 40%.) Also, Julian has privately remarked that there are no irrational triple points. --rwg

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[math-fun] Hot tea sucks.
by Bill Gosper 25 Nov '19

25 Nov '19

I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? —Bill

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[math-fun] Dean Hickerson
by James Propp 22 Oct '19

22 Oct '19

Anyone have a current email address for Dean Hickerson? "dean at math daht ucdavis daht edu" and "dean at ucdmath daht ucdavis daht edu" aren't working anymore. Thanks, Jim Propp

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[math-fun] ries near miss
by Bill Gosper 04 Oct '19

04 Oct '19

Out[10]= {-(Log[7]/(2 ProductLog[-1, -(((5/(1 + E^3 Sqrt[\[Pi]]))^(1/4) Log[7])/( 2 \[Pi]))])), (8 Sqrt[3/23] \[Pi] Sinh[(Sqrt[23] \[Pi])/6])/(-1 + 2 Cosh[(Sqrt[23] \[Pi])/3])} In[11]:= N@% Out[11]= {0.368412535931434, 0.368412535931434} In[12]:= Equal @@ %% Out[12]= False The 2nd expression is, amazingly, Integrate[QPochhammer[q,q],{q,0,1}]. https://en.wikipedia.org/wiki/Euler_function I can't quite coax it out of ries. Does anyone have a closed form for Integrate[DedekindEta[Log[q]/2/I/π],{q,0,1}] ~ .3416968346182687559823? It will probably involve Gamma, which is not yet in ries. —rwg

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Re: [math-fun] Draft of short essay on Scott Kim's motley dissections
by Bill Gosper 04 Sep '19

04 Sep '19

On Thu, Jul 11, 2019 at 3:14 AM rwg <rwg(a)ma.sdf.org> wrote: > > > -------- Original Message -------- > Subject: Re: [math-fun] Draft of short essay on Scott Kim's motley > dissections > Date: 2019-07-10 12:59 > From: James Propp <jamespropp(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > Beautiful! > > Thanks, Fred! > > Jim > I'll second that! How hard would it be to twist the whole scene so that the front face doesn't rotate? We can change the problem so that the answer is an ancient Macsyma animation most of you should recall. Problem: Reshape the six "squares" so they can simply rotate, while four "triangles" alternately swell, then shrink away, while the other four complementarily follow suit in the opposite phase. The edges need no longer lie on great circles. Spoiler Lissajous arcs <http://gosper.org/lisspump.gif> On Wed, Jul 10, 2019 at 3:41 PM Fred Lunnon <fred.lunnon(a)gmail.com> > wrote: > > > I have posted an animation at > > https://www.dropbox.com/s/5yrmef1djjuatfc/pseudo_movie.gif > > of a spherical pseudo-cuboctahedron (SPCO) morphing continuously via > > octahedron -- big triangles -- cuboctahedron -- big squares -- cube, > > then back again via the mirror-images; Maple program is available > > on request. > > > > This should run continuously when the link is opened in a browser: > > please advise me of any problems! > > > > Time is proportional to `twist' angle t between (plane containing) > > SPCO edge arc and associated edge of scaffolding cube, modulo pi . > > SPCO vertices are given by 24 cube symmetries (with even resigned > > permutations of components) of point with Cartesian coordinate > > P(t) = (a^2 + a*b, a^2 - a*b, a^2 + b^2)/d , > > where > > b = cos t , a = (1/sqrt2) sin t , d^2 = 1 - a^4 . > > > > For octahedron t = 0 , for cube t = pi/2 ; for cuboctahedron > > t = arccos(-1/3)/2 . For Jim Propp's case, where edge plane meets > > 4 vertices of the scaffold cube, rather unexpectedly t = > > arccos(+1/3)/2 > > --- just pi/2 minus the cuboctahedron angle! > > > > While I was busily upstaging his previous version, Christian has > > quietly replaced that static CGI view with a 3D-printed solid model: > > see > > https://www.thingiverse.com/thing:3726912 > > > > Hmmm ... a mechanically functional solid morphing model, anybody? > The only hope would be van Deventer. —rwg > > > > Finally, note that a similar method should cope straightforwardly > > with morphing pseudo-octahedron and pseudo-icosidodecahedron. > > > > Fred Lunnon > > > > > > On 7/6/19, James Propp <jamespropp(a)gmail.com> wrote: > > > I've figured out one way to make a 3D realization of a > pseudo-octahedron, > > > by decorating the faces of a tetrahedron. (Maybe this was obvious to > you > > > all, but I didn't see it.) See > > > > > > http://faculty.uml.edu/jpropp/mathenchant/semi-pseudo-octahedron.JPG > > > > > > which shows how two of the four faces should be decorated, with three > > > vertices per face. If you breathe into the tetrahedron to inflate it to > > > become a sphere, the pattern of decorations on the tetrahedron should > > > become a motley dissection of the sphere with 12 vertices, 6 > > pseudo-edges, > > > and 8 three-sided pseudo-faces. > > > > > > Can Christian (or someone else) create an image of this for me? I'd > love > > to > > > use it in my next mini-essay (~ 1000 words) for the Big Internet > > Math-Off, > > > as well as in the longer Mathematical Enchantments piece on the subject > > of > > > motley dissections that I plan to write later this summer or in the > Fall. > > > > > > Thanks, > > > > > > Jim > > > > > > > > > > > > On Thu, Jul 4, 2019 at 10:39 AM ed pegg <ed(a)mathpuzzle.com> wrote: > > > > > >> Related: > > >> Moritz W. Schmitt, On Space Groups and Dirichlet-Voronoi Stereohedra > > >> https://refubium.fu-berlin.de/handle/fub188/10176 > > >> > > >> I want to see pictures of all of these space filling plesiohedra. > I've > > >> been wanting to see Engel's 38-sided spacefiller for years. > > >> --Ed Pegg Jr On Thursday, July 4, 2019, 07:58:05 AM CDT, Christian > > >> Lawson-Perfect <christianperfect(a)gmail.com> wrote: > > >> > > >> Scott, I'm interested in giving it a go. I've got a couple of student > > >> working with me over the summer on 3d printing mathematical objects, > so > > >> this'd be great. > > >> > > >> On Thu, 4 Jul 2019 at 06:46, Scott Kim <scott(a)scottkim.com> wrote: > > >> > > >> > The pseudo octahedron as drawn has bar 3 symmetry, meaning 4-fold > > >> > rotational symmetry plus top and lower hemispheres congruent but > > mirror > > >> > images (I think...not sure). The pseudo-cube needs to be redrawn to > > work > > >> > nicely on a sphere, as does the pseudo-dodecahedron. The 3d models I > > >> really > > >> > want to have 3d printed are the pseudo 5-cell (easy) and pseudo > > 16-cell > > >> > (hard), which have exteriors that are concave curivlinear > tetrahedra. > > I > > >> can > > >> > provide drawings if anyone's interested in giving them a try. — > Scott > > >> > > > >> > On Wed, Jul 3, 2019 at 9:07 PM James Propp <jamespropp(a)gmail.com> > > wrote: > > >> > > > >> > > And it would be good to have a picture of the pseudo-cube as an > > actual > > >> > > motley dissection of the sphere, but I have no idea how to realize > > it. > > >> > I've > > >> > > put Scott's sketch of the combinatorial structure of the > pseudo-cube > > >> > > at > > >> > > http://mathenchant.org/pseudo-cube.png; in both this picture and > > the > > >> > > pseudo-octahedron picture, it's important to keep in mind that the > > >> > > noncompact region in the picture is also a pseudo-face. In the > > >> > symmetrical > > >> > > pseudo-cube, the six pseudo-faces would be centered at the > vertices > > of > > >> an > > >> > > octahedron, but I don't see how the pseudo-faces fit together. Can > > any > > >> of > > >> > > you figure this out? > > >> > > > > >> > > Jim > > >> > > > > >> > > On Wed, Jul 3, 2019 at 9:28 PM James Propp <jamespropp(a)gmail.com> > > >> wrote: > > >> > > > > >> > > > It would also be good to have a picture of the > pseudo-octahedron. > > >> I've > > >> > > put > > >> > > > Scott Kim's picture of it at > > >> > > http://mathenchant.org/pseudo-octahedron.png; > > >> > > > there should be a realization of it as a motley dissection of > the > > >> > surface > > >> > > > of the sphere that has all the rotational symmetries of the > > >> octahedron > > >> > > (but > > >> > > > none of the reflection symmetries). Can Christian (or anyone > else) > > >> > > > construct a virtual model? In some ways it might be harder to > > >> visually > > >> > > > parse the pseudo-octahedral motley dissection than the > > >> > > pseudo-cuboctahedral > > >> > > > motley dissection, even though the former is simpler, because > the > > >> > visible > > >> > > > hemisphere offers less information. But an interactive rotatable > > >> > version > > >> > > > (or just a GIF that shows the object rotating) would solve that > > >> > problem. > > >> > > > > > >> > > > Jim > > >> > > > > > >> > > > On Wed, Jul 3, 2019 at 2:49 PM James Propp < > jamespropp(a)gmail.com> > > >> > wrote: > > >> > > > > > >> > > >> Wow; this is great! > > >> > > >> > > >> > > >> Is there a way to rotate the view? > > >> > > >> > > >> > > >> Jim > > >> > > >> > > >> > > >> On Wed, Jul 3, 2019 at 3:38 AM Christian Lawson-Perfect < > > >> > > >> christianperfect(a)gmail.com> wrote: > > >> > > >> > > >> > > >>> I've made a 3d model in OpenSCAD and uploaded it to > thingiverse: > > >> > > >>> https://www.thingiverse.com/thing:3726912. I've put a simple > > >> > rendering > > >> > > >>> of > > >> > > >>> the shape on that page; I'm going to both 3d print it and do a > > >> nicer > > >> > > CGI > > >> > > >>> rendering now. > > >> > > >>> > > >> > > >>> On Tue, 2 Jul 2019 at 19:24, James Propp < > jamespropp(a)gmail.com> > > >> > wrote: > > >> > > >>> > > >> > > >>> > The link to the video apparently didn't survive the > conversion > > >> from > > >> > > >>> HTML to > > >> > > >>> > PDF, so here's the URL for Scott Kim's video: > > >> > > >>> > > > >> > > >>> > https://youtu.be/xK1QA0Oi7iE > > >> > > >>> > > > >> > > >>> > Jim > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > On Tue, Jul 2, 2019 at 12:03 PM James Propp < > > >> jamespropp(a)gmail.com> > > >> > > >>> wrote: > > >> > > >>> > > > >> > > >>> > > At > > http://faculty.uml.edu/jpropp/mathenchant/motley-draft3.pdf > > >> > > >>> you'll > > >> > > >>> > > find a draft of an essay that will be published on July > > 11th. > > >> > > >>> Comments > > >> > > >>> > are > > >> > > >>> > > welcome, but even more welcome would be a good picture of > > the > > >> > > >>> "spherical > > >> > > >>> > > pseudo-cuboctahedron"! In principle I know how to make > such > > a > > >> > > >>> picture, > > >> > > >>> > sort > > >> > > >>> > > of: take the motley dissection of the surface of the cube > > >> (shown > > >> > in > > >> > > >>> the > > >> > > >>> > > essay), push all the vertices out onto a sphere, and draw > > >> > spherical > > >> > > >>> arcs > > >> > > >>> > > joining up those vertices. This seems like a job for > > >> Mathematica, > > >> > > >>> but my > > >> > > >>> > > mastery of 3D graphics isn't up to the job. I'll > gratefully > > >> > > >>> acknowledge > > >> > > >>> > > anyone who can create a picture I can use! > > >> > > >>> > > > > >> > > >>> > > Thanks, > > >> > > >>> > > > > >> > > >>> > > Jim Propp > > >> > > >>> > > > > >> > > >>> > _______________________________________________ >

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