November 2008 - mathfun@mailman.xmission.com

[math-fun] polyhedra with all faces congruent
by Michael Reid 21 Nov '08

21 Nov '08

> > A 180-hedron! Is there any limit to the number of congruent faces > > on a polyhedron? > > No. Take a skew prism -- lots of congruent triangular faces -- and > put a cap on each end made out of triangles of the same dimensions. just a minor nit. this cannot always be done; it depends upon the antiprism. er, ... did i interpret this correctly "skew prism" = "(what i've always known as an) antiprism"? anyway, it can be done if the isosceles triangles are sufficiently tall, and that suffices for the "no" answer above. this construction gives a polyhedron with 4n faces. however, the "belt" around the antiprism is not needed! simply take a 2 point suspension of a regular n-gon, which gives a polyhedron with 2n triangular faces. (alternatively, start with a "regular" antiprism, and extend the triangular faces. the planes of the upward-pointing triangles all meet at one point, and the planes of the downward-pointing triangles all meet at another point. this gives a polyhedron with 2n quadrilateral faces. surely there's a picture online ... ) [edit: pictures of stellated "dipyramids" at: http://www.georgehart.com/virtual-polyhedra/pyramids-info.html ] are there examples of polyhedra with all faces congruent, that have an odd number of faces? (i did not see any parity restriction, although the faces must have an even number of sides.) mike

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Re: [math-fun] tetraroller volume
by rwg＠sdf.lonestar.org 21 Nov '08

21 Nov '08

> DanA wrote: > > << > was able to > figure out the face structure of the four equal cylinders > (which could be said to intersect as do the great diagonals > of a cube). Same method should apply to the higherhedra. >>> > > Here's the method (but I've patented it, and you may need to > pay me royalties): Yeah but you didn't say <below> is the projection of a cuboctahedron onto a concentric sphere, so I don't need no steenking license agreement! > > Suppose n solid infinite cylinders of unit radius are > arranged in 3-space so their axes each pass through the > origin. > > Then the solid K where they all intersect has boundary surface > N, where N has "faces" that are maximal portions of 2D unit > cylinders. >DanA> Imagine the intersection of N with the unit >> sphere S^2 about the origin. That will be a number of great >> circles of S^2. >This isn't right; rather it's just each 2D cylinder >that intersects the unit sphere in a great circle. >Now look at how each of these great circles on S^2 >is cut by the others; each resulting arc corresponds >to one "face" of the boundary of the intersection solid. >I hope that was clearer and truer. >--Dan > These circles cut each other up into > (uncut) great circle arcs. Each of these occurs in just > one face of N. So # faces of N = # largest uncut great > circle arcs on S^2. > > The faces will resemble Voronoi regions of the great > circle arcs. In a symmetrical situation, e.g., where > the axes are the diagonals of a higherhedron, this will > be exact (if the faces of N are radially projected onto S^2). > > E.g., for 4 main diagonals of a cube, we can pretend the cube > is a unit sphere, and the great circles will be equators w.r.t. > opposite pairs of cube vertices -- giving us 4 regular hexagons > inscribed on the cube. Each hexagon is cut in sixths by the > other hexagons, giving us 24 "great circle" arcs, hence 24 > faces of the intersection solid K. The Voronoi regions on > the sphere are each a quadrilateral with one axis of symmetry. > > --Dan I believe that's a deltoidal icositetrahedron with faces bulged out to cylindrical. It would need a pretty shallow grade and some friction not to start rolling like a sphere. --rwg > _____________________________________________________________________ > "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] Inside a spherical mirror
by Dan Asimov 20 Nov '08

20 Nov '08

What would it look like if one were inside a spherical mirror silvered on the inside? Suppose a you are suspended in mid-air inside the mirror, a bit off-center but with your eye in the same plane, holding a candle for light and looking in the direction of the center. What would you see? I don't know the answer. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] polyhedra with all faces congruent
by Michael Reid 19 Nov '08

19 Nov '08

thanks dan! that settles the question for convex polyhedra. the case of non-convex polyhedra and positive genus is still in doubt, but perhaps grunbaum's method sheds some light on it. mike > Here's the Math Reviews review of a paper > by Branko Grunbaum that answers the > "Odd?" question in the negative. > > > << > Branko Grunbaum > On polyhedra in E^3 having all faces congruent. > Bull. Res. Council Israel Sect. F 8F 1960 215–218 (1960) > > The polyhedra having all faces congruent include the reciprocals (duals) of all the “uniform” polyhedra, > namely, the five Platonic solids, the dipyramids, the reciprocals of the thirteen Archimedian > solids, and the reciprocals of the antiprisms. These last have, for any n, 2n kite-shaped faces. (It is > unfortunate that the author uses, instead of “kite”, the name “deltoid”, which belongs more properly > to a curve, the three-cusped hypocycloid.) The author proves that, if all the faces of a convex > polyhedron are congruent, their number is even. Since the faces cannot all have more than five > sides, and the result is obvious when the faces are triangular or pentagonal, the problem is quickly > reduced to the case of quadrangular faces, and then to kites or parallelograms. The kites are dealt > with by an ingenious combination of metrical and topological arguments. Finally, if the faces are > parallelograms, their number is equal to the product of two consecutive integers [Coxeter, Regular > polytopes, Methuen, London, 1948, p. 27; MR0027148 (10,261e)]. > > > > _____________________________________________________________________ > "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele >

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Re: [math-fun] polyhedra with all faces congruent
by Michael Reid 18 Nov '08

18 Nov '08

> > are there examples of polyhedra with all faces congruent, that have an > > odd number of faces? (i did not see any parity restriction, although > > the faces must have an even number of sides.) > > If you don't mind nonconvexity, it's pretty easy to make one out of > squares (all meeting at right angles). i don't mind non-convex polyhedra, though a convex one would be better. however, i don't see your "pretty easy" way to do it. unless you intend to count faces in two different ways: first, as (connected) subsets of planes, and second, each individual unit square counts as its own face. for example, a "notched cube" (figure obtained by removing a 1x1x1 corner from a 2x2x2 cube) has 9 planar faces. but then it's unreasonable to say that all faces are congruent, by re-defining the faces by subdividing the previous 9 faces into 24 unit squares, which are indeed congruent. i doubt this is what you had intended anyway ... unless i missed it, your argument that faces must have fewer than 6 sides doesn't require convexity, (as long as we can't have a vertex of degree 2) but probably requires that the genus is 0 . the higher genus case opens another can of worms. for quadrilateral faces, the sides must occur in two congruent pairs (or all four congruent) by the same parity reasoning. this means either parallelograms or kite faces, and i think parallelogram faces will not work (such polyhedra decompose into parallelepipeds). so ... seems unlikely, but can it be proved? mike

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Re: [math-fun] Doin' the Hilbert walk
by rwg＠sdf.lonestar.org 16 Nov '08

16 Nov '08

DanA>What I wrote was misunderstood. Sorry about that, Dan! >(By the way, rwg, your last two sentences are >quite interesting! I wonder how small, in some >appropriate sense, that uncountable set > S = {x in [0,1]^n | #(finv(x)) > 1} >can be. Rather huge. For the 2D Peano-Hilbert squarefiller, S = KxC U CxK, where C := [0,1] and K := the dyadic rationals therein, with the triply visited being the dense set KxK. You can intuit this by noting that images of closed subintervals of C are unions of closed squares. As the spacefill continues, other closed squares will share the boundaries. If instead we're filling a fudgeflake or a france, then the shared boundaries will be fractals, which are in some sense even larger than the square edges. >prp[p_]:=Block[{q=If[p[[1]]^2>p[[2]]^2,{-p[[3]],0,p[[1]]},{0,p[[3]],-p[[2]]}]},q*Sqrt[(p.p)/(q.q)]] >Stick[p_,q_]:=Block[{m=(p+q)/2,d=prp[q-p]/16,e},e=Cross[q-p,d]/Norm[q-p]; > Table[Polygon[{r,m+Cos[2*(k-1)*Pi/3]*d+Sin[2*(k-1)*Pi/3]*e, > m+Cos[2*k*Pi/3]*d+Sin[2*k*Pi/3]*e}],{r,{p,q}},{k,3}]] >Graphics3D[ > Table[Stick[Treano[(k + 1/3)/256], Treano[(k + 4/3)/256]], {k,0,254}]] These revisitations can blow out the Graphics3D above with Sticks of zero length, unless we add the (gaspy?) line Stick[x_, x_]:={} >produce the graphic at http://gosper.org/treano.png . Oops, I used the long awaited mouse tumble feature to find a much nicer pose, but the Export command captured the original one. Here btw is the 2D version of Treano: (* Exactly and continuously map the rational unit interval onto the rational unit square. Peano redefines itself twice, calls itself in four places, and has no discernable termination condition.*) Peano[t_, a1_: 1, a0_: 0] := Peano[t, b1_: 1, b0_: 0] = (Peano[t, s1_: 1, s0_: 0] = ((a0 - s0)/(s1 - a1)); Module[{t4 = 4*t, n}, n = Floor[t4]; t4 -= n; Switch[n, 0, I*(1 - Peano[1 - t4, -I*a1/2, a0 + a1*I/2])/2, 1, (I + Peano[t4, a1/2, a0 + a1*I/2])/2, 2, (1 + I + Peano[t4, a1/2, a0 + a1*(1 + I)/2])/2, 3, 1 + I*Peano[1 - t4, a1*I/2, a0 + a1]/2, 4, 1]]) Nevertheless, e.g., In[2] := Peano[7/22] 7 17 I Out[2]:= -- + ---- 66 22 >So SW's intial assessment was correct: trivial. Do I need a smiley here? --rwg MENSURATIONAL NEONATURALISM

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Re: [math-fun] Doin' the Hilbert walk
by Dan Asimov 15 Nov '08

15 Nov '08

Thanks for the responses. Yes, Baire category and Lebesgue measure are the two main kinds of largeness measure I was thinking of. (Yet another is Hausdorff dimension.) --Dan Andy wrote: << RWG wrote: <<<< I wrote: <<<<<< By the way, rwg, your last two sentences are quite interesting! I wonder how small, in some appropriate sense, that uncountable set S = {x in [0,1]^n | #(finv(x)) > 1} can be. >>>>>> Rather huge. For the 2D Peano-Hilbert squarefiller, S = KxC U CxK, where C := [0,1] and K := the dyadic rationals therein, with the triply visited being the dense set KxK. >>>> Well, that's huge in the sense that it's uncountable and dense, but it's "small" in at least two senses: It has measure 0, and it's a "meagre" set; that is, a countable union of nowhere dense sets. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] optics / materials science question
by Steve Witham 15 Nov '08

15 Nov '08

>From: James Propp <jpropp(a)cs.uml.edu> > >While we're talking about optics, I was musing the other day >about whether there's a way to make a windshield (or a pair >of sunglasses) that will transmit less of the light emitted >by really bright sources without transmitting less of the >light emitted by dimmer sources. i remember a spaceship windshield that had that ability, in some Larry Niven story. Remembering his style, but not the particular story, he probably had three paragraphs of patent-application- quality prose attached. Bug eyes sometimes have an apparent pupil that's really the subset of lenses looking at your eyes. Maybe something like e-ink cells, with floating directional light blockers that could be aimed. Maybe a fancy Fresnel lens could block light in a single direction (or vertical band) but could be mounted for aiming. >(I'm not talking about glare-reduction of the ordinary sort --- >at least I don't think I am! --- since it's my impression that >glare-reduction makes everything look dimmer.) Vertically-polarized glasses selectively filter out specular reflections from horizontal surfaces. But, also more than half of other light, it's true. --Steve

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[math-fun] First fit optimal collections
by Bernie Cosell 15 Nov '08

15 Nov '08

I was recently playing around with a little programming puzzle to find "optimal" coin-amounts to be able to make any amount up to a dollar. The description distinguished between "intuitive" and "nonintuitive" sets, the former being ones in which first-fit gets you the minimum # of coins needs [and so penny-nickel-quarter is intuitive, while penny-dime- quarter isn't, because .30 is better done with three dimes than a quarter and five pennies]. I can write the program fairly easily to try coin- sets and see what the average number of coins it takes is to make up every amount from .01 to 1.00. BUT: the intuitive requirement has me a bit puzzled. Assuming I do some kind of brute-force combinatorial program to try every combination of two-coin-sets, three-coin-sets, etc, I'm not sure what to do next. Doing a quick first-fit on the 100 amounts to get the average # of coins required is simple enough, but is there some way to look at a collection of coin-sizes and tell if they form an "intuitive" set or not? If not, it looks like the search is going to be VERY hard -- basically you'll have to end up doing pretty much a full tree walk of EVERY possible combination (you can do a bunch of tree-pruning to keep the size of the search manageable, but now this is no longer a 15-minute little programming hack but is starting to sound like a real program..:o)) /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] First fit optimal collections
by Eugene Salamin 14 Nov '08

14 Nov '08

________________________________ From: Bernie Cosell <bernie(a)fantasyfarm.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Friday, November 14, 2008 1:21:43 PM Subject: [math-fun] First fit optimal collections I was recently playing around with a little programming puzzle to find "optimal" coin-amounts to be able to make any amount up to a dollar. The description distinguished between "intuitive" and "nonintuitive" sets, the former being ones in which first-fit gets you the minimum # of coins needs [and so penny-nickel-quarter is intuitive, while penny-dime- quarter isn't, because .30 is better done with three dimes than a quarter and five pennies]. I can write the program fairly easily to try coin- sets and see what the average number of coins it takes is to make up every amount from .01 to 1.00. BUT: the intuitive requirement has me a bit puzzled. Assuming I do some kind of brute-force combinatorial program to try every combination of two-coin-sets, three-coin-sets, etc, I'm not sure what to do next. Doing a quick first-fit on the 100 amounts to get the average # of coins required is simple enough, but is there some way to look at a collection of coin-sizes and tell if they form an "intuitive" set or not? If not, it looks like the search is going to be VERY hard -- basically you'll have to end up doing pretty much a full tree walk of EVERY possible combination (you can do a bunch of tree-pruning to keep the size of the search manageable, but now this is no longer a 15-minute little programming hack but is starting to sound like a real program..:o)) /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <-- _______________________________________________ Ignoring the "intuitive" and "unintuitive" distinction, considering instead the unconstrained optimization, I would use dynamic programming. If an optimum N cent solution consists of more than one coin, then for any partition of these coins into N1 cents and N2 cents, N = N1 + N2, the N1 and N2 cent piles are each also optimum. So solve the problem for N =1, 2, 3, 4, ... in succession. (Minimum Number of Coins)(N) = min( (Minimum Number of Coins)(K) + (Minimum Number of Coins)(N-K), K = 1..N-1). This fails only when the optimum is a single coin, but those cases are clear; they are the single coin values. Gene

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