June 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Negative polyhedra
by Dan Asimov 10 Jun '20

10 Jun '20

A few comments: The three simply connected surfaces M of constant curvature 1, 0, -1 resp. are the sphere M=S^2, the (Euclidean) plane M=R^2, and the hyperbolic plane M=H^2. There is a tiling of M by regular p-gons, q per vertex according as (p-2)(q-2) is <, =, or > 4, respectively. By taking the dual tiling, we get the tiling by regular q-gons, p per vertex. This is proved by focusing on the existence of a regular polygon of p sides and interior angle 2*pi/q. There are then various ways of quotienting out the simply connected space to obtain a surface of constant curvature, covered by the appropriate M above, with a tiling of the same description. It's a theorem of Hilbert that no surface of constant Gaussian curvature = -1 is a subset of 3-space R^3. So the Coxeter surface with 6 squares per vertex can't be modified *in 3-space* so as to have its ideal constant curvature = -1. There's also the Gauss-Bonnet theorem, which says that for a closed surface (1/2π) * Integral K dA = X(M) where K is Gaussian curvature and X(M) denotes Euler characteristic of M. I haven't seen this mentioned anywhere, but one could define "curvarea" to be the area of a piece of a constant-curvature surface to be its area multiplied by its curvature (or the integral of this product for non-constant curvature). Then Gareth's idea of a negative polyhedron makes sense. —Dan _____ * A closed surface is one that has finite area and empty boundary. Topologically the orientable ones are the sphere, the torus, and the connected sum of n tori, n > 1. The non-orientable ones are the projective plane, the Klein bottle, and the connected sum of n projective planes, n > 2. Gareth McCaughan wrote: ----- In another place, someone asked (approximately) the following question: [question begins] A regular polyhedron in which each {vertex, face} has {a,b} neighbours has vertex, edge and face counts satisfying av = bf = 2e (count pairs (vertex, neighbouring edge) for av=2e and count pairs (face, neighbouring edge) for bf=2e) and of course v-e+f = 2. So (2/a)e-e+(2/b)e = 2 so e = 2ab/(2a+2b-ab) v = 4b/(2a+2b-ab) f = 4a/(2a+2b-ab) and in addition to the "usual" solutions like a=3,b=5 for the regular dodecahedron there are some illegal ones where v,e,f are _negative_. For instance, a=4, b=5 yields e=-20, f=-8, v=-10. Is it in any way meaningful to think of this thing as a polyhedron with -8 pentagonal faces, four of which meet at each of the -10 vertices, joined by -20 edges? [question ends] ... and I was annoyed to find I didn't have a good answer. Purely numerologically, these "virtual" things of genus 0 / characteristic 2 correspond to not-virtual things of genus 2 / characteristic -2, because (-v)-(-e)-(-f) = -2. So this minus-8-faced thing corresponds to some thing that we might hope is a quotient of the hyperbolic plane by some suitable group, having 8 faces, 10 vertices, and 20 edges. There's a tessellation of the hyperbolic plane by right-angled pentagons, which (like our virtual polyhedron) has four pentagons meeting at each vertex; the quotient is an orbifold whose Euler characteristic is -1/4, so it makes some kind of sense that maybe we can put 8 of these together to get a thing of Euler characteristic -2. It's probably obvious to those skilled in the art how one does this, but I was never very skilled in that particular art and have forgotten almost everything now. I can imagine that there might be some more precise way to think of this sort of hyperbolic-plane thing as a "negative" spherical polyhedron, on the super-handwavy grounds that we can think of the faces, edges and vertices of a polyhedron as places where in the sphere we have "excess angle"; e.g., if you take a regular dodecahedron and project it out onto the sphere, then at each vertex you have three spherical pentagons with pi/3 angles meeting neatly, whereas the dodecahedron has three euclidean pentagons with 3pi/5 angles. Something something Gauss-Bonnet something. And, just as the sphere has uniform positive curvature, the hyperbolic plane has uniform negative curvature. So maaaaybe there's some way to think of faces, edges and vertices in the hyperbolic plane as "negative" compared with the sphere because they have angle _defects_ instead of angle _excesses_. But all the details elude me. I bet they're well known to those who know such things, and perhaps some such people are reading this. - Is it true that there's some sort of hyperbolic thing with 8 pentagonal faces meeting four to a vertex? What do you need to quotient the hyperbolic plane by to get it? What does the fundamental region look like? (And does something similar hold for all the negative solutions to the equations above?) - In the case a=6,b=4 there's a famous _infinite_ thing in euclidean 3-space, a sort of polyhedral foam based on the cubic lattice. The equations above suggest that there "should" be a "negative polyhedron" with a=6, b=4, e=-12, v=-4, f=-6. This seems suspiciously similar to the cube, with e=12, v=8, f=6. Is there some sort of quotient of the "mucube" foam that collapses it down to one cube with ?opposite? pairs of vertices identified? Is either the mucube itself or this hypothetical quotient, again, a quotient of the hyperbolic plane by a reflection group? - Is there anything more than numerology to the idea that we can think of these as spherical things with negative numbers of vertices/edges/faces and Euler characteristic +2, rather than (so I assume) hyperbolic things with positive numbers of vertices/edges/faces and Euler characteristic -2? -----

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[math-fun] MacOS Mail
by Henry Baker 09 Jun '20

09 Jun '20

I realize that this is off-topic, so feel free to delete this if you wish. I personally use Linux & Windows, but my wife uses MacOS. Over the past 2 years, the MacOS "Mail" program has become impossibly slow -- all the while use the same hardware! I can't believe that Mac users would put up with this. Note that everything else works fine; it's just the Mail program. Do any of you use Macs? Do you find the Mac Mail program intolerably slow?

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[math-fun] Euphoria sepulcralis
by Brad Klee 07 Jun '20

07 Jun '20

Also Known As the Dark Flower Scarab, but this week I am calling them the field looter flower beetles, especially now that they've teamed up with the hyper-destructive Japanese beetles. The coloration of E. Sepulcralis is mostly black, with a green, sometimes red undersheen. Like the Japanese beetles, and many others, their Iridescent light reflections are circularly polarized: https://www.inaturalist.org/observations/48771137 https://www.inaturalist.org/lists/3394445-Chiral-Scarabs-of-Arkansas The North Shore disc golf course cuts one of their habitats into narrow little swaths, where the Queen Anne's lace grows wild around the manicured lawn. Disc golfers think they have really accomplished something when they get a birdie or an eagle, but it is nothing compared to what the flower beetles can now accomplish after millions of years of evolution. Although they seem to be doing fine statistically, habitat destruction and degradation could introduce them to unnecessary suffering. Other insects face similar problems, and possibly extinction, see also: https://www.wildlifearkansas.com/materials/2006_updates/03SGCN.pdf The sky is blue, the fields and the forests are calling, --Brad

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[math-fun] Spherical geometry puzzle
by Dan Asimov 06 Jun '20

06 Jun '20

I recently searched in vain for an expression for the area of a spherical triangle in terms of its *sides* having lengths a, b, c. (Assume that the triangle is drawn on the surface of a sphere of radius = 1, with sides that are arcs of great circles, and whose angles are all < π radians.) So I figured out my own formula, which definitely took a few minutes. Puzzle: Find such a formula without consulting any references. —Dan

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[math-fun] A generalization of an old Conway Problem
by Victor Miller 06 Jun '20

06 Jun '20

John Conway once posed the following problem: The Ten Divisibilities I have a ten digit number, abcdefghij. Each of the digits is different, and a is divisible by 1 ab is divisible by 2 abc is divisible by 3 abcd is divisible by 4 abcde is divisible by 5 abcdef is divisible by 6 abcdefg is divisible by 7 abcdefgh is divisible by 8 abcdefghi is divisible by 9 abcdefghij is divisible by 10 Find the number. One can do this computationally in a number of ways, such as exhausting over all permutations of [1,10], or branch and bound. I decided to do this via integer programming to investigate the generalization to other bases than 10. Interestingly enough, I found that there were only solutions for bases 2,4,6,8,10,14 (tested through base 27). I don't know if they've petered out completely. Any ideas? Victor

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[math-fun] Strange matrix products
by Bill Gosper 05 Jun '20

05 Jun '20

Simon Plouffe's old http://wayback.cecm.sfu.ca/cgi-bin/isc/lookup?number=1.64163256065515386629… told me that In[202]:= Dot @@ Table[{{1/2 + 2^-n, -1/2}, {-1/2, 1/2}}, {n, ∞}] = {{1,-1},{1-𝜗, 𝜗-1}] (if only Mathematica supported infinite Tables) where 𝜗 = In[186]:= Sum[2^-Binomial[n, 2], {n, ∞]}] Out[186]= EllipticTheta[2, 0, 1/√2]/2^(7/8) I.e., this product converges to something interesting even though the determinant approaches zero. I can't wait to try other things, but agonizingly, Simon's website seems up only a few seconds every few minutes! —rwg

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[math-fun] multimagie.com update
by Christian Boyer 05 Jun '20

05 Jun '20

www.multimagie.com updated this week, but: « TEN YEARS after their opening (*), there still remain eight unsolved enigmas for winning €6,500 and eight bottles of champagne! The remaining enigmas are on small squares, from 3x3 to 6x6. » Christian. (*) April 2010: www.multimagie.com/English/MagicSquaresEnigmasE.pdf

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Re: [math-fun] Sages' hats and prime-richness
by Steve Witham 04 Jun '20

04 Jun '20

Dan, thanks for the MSRI math problems. The sages' hats (quoted below) was about my speed. Spoilers. I liked finding that solving a little problem was one way to solve the whole problem, then realizing that sometimes the big problem *was* the little problem, so it didn't do any good to look for another method, I had to solve *that* little problem. Involutions were involved (pun). OEIS A000085 counts the number of involutions of n things. I printed out entries 0..40 and their factorizations... They seem pretty "primey"... |Basically, although the recurrence for A000085 is fairly simple, it has the talent of *ruling out* a sizeable number of prime factors (for instance 3, 7, 11, 17...). |||OEIS A264737 lists primes that do appear as factors of A000085 entries. So (except for the powers of two in every entry) you get numbers with fewer, larger prime factors than... my intuition expects anyway. | What a woild. --Steve |> Subject: [math-fun] Some math puzzles > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 5/29/20, 4:58 PM > > Here are four problems from the puzzle column in the latest > newsletter from MSRI, faintly edited. > ... > 3. Suppose there are n hats, each with a different color. > These hats are placed on the heads of n sages. All of the > sages know all the colors: their own hat and everyone elseÂ’s > hats. A referee then announces the "correct" hat color that > should be on the head of each sage. The sages are then allowed > "swap" sessions: in one session disjoint pairs of sages are > allowed to interchange their hats. Can the sages fully correct > their hat colors in two swap sessions? -fin-

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Re: [math-fun] Some math puzzles
by Dan Asimov 03 Jun '20

03 Jun '20

MSRI is the Mathematical Sciences Research Institute in Berkeley, California. —Dan Bernie Cosell wrote ----- >> Here are four problems from the puzzle column in the latest newsletter >> from MSRI, faintly edited. > >What's MSRL? The puzzles are fun. > /b\ -----

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[math-fun] A chain of mean terms
by Éric Angelini 03 Jun '20

03 Jun '20

Hello Math-Fun, 109,87,65,43,21,1,23,45,67,89. Is this the longest chain where the mean of two adjacent terms is given by the concatenation of the two digits framing the comma between them? Best, É.

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