Re: [math-fun] polyhedra with all faces congruent
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
There's a possible lacuna here: The number of vertices of a polygon can be ambiguous, for the purposes of the Euler formula, if one side can be split so as to match two other sides. There are a couple of strange plane-tiling pentagons where three copies of the tile meet, but the "vertex" where they meet lies along an edge of one copy. Perhaps some version of this phenomenon could spoil the obviousness of the triangle or pentagon cases. Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, November 18, 2008 12:40 PM To: math-fun Subject: Re: [math-fun] polyhedra with all faces congruent 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If you're not concerned with convexity, can't you, say, glue platonic solids together at their faces? For example, join two dodecahedra at a face to get a 22-face pentagonahedron(?). This can be done ad infinitum to make tree-shaped hedra.
DW> If you're not concerned with convexity, can't you, say, glue platonic solids
together at their faces? For example, join two dodecahedra at a face to get a 22-face pentagonahedron(?). This can be done ad infinitum to make tree-shaped hedra.
Right, nonconvex is easy. E.g., that 3D "Snowflake". Or worse: flat triangular grids on the faces of a tetrahedron. But hey, isn't the following a fairly ball-shaped 360hedron?: "Start" with a pentakis dodecahedron = 60 equilateral triangles (which are not quite equilateral when projected onto the circumsphere). Then "triakis" that with very shallow triangular pyramids. Finally perpendicularly bisect the long sides and lift the midpoints even more slightly. Since not all vertices are on the sphere, it's hard to claim moral superiority over the dipyramid, except perhaps on volume. Rejecting that, there's "DisdyakisTriacontahedron", which looks to me like it 120-sects its circumsphere. (It's hard to tell without rebooting this stupid laptop, which stubbornly refuses to reset to 1440x900.) rwg>If you unite each hexagon with three nonconsecutive
neighbor triangles, you get equilateral triangles enclosing the pentagons in an appealing camera-shutter arrangement that maybe we could sell to a soccer ball manufacturer.
I think you can open and close the twelve shutters and continuously morph between an icosahedron and a dodecahedron. --rwg ANODISED ADENOIDS
rwg>there's "DisdyakisTriacontahedron", which looks to me like
it 120-sects its circumsphere.
Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm --rwg PHENOCRYSTIC PYROTECHNICS
Apologies -- I lost track of this conversation. What's our current understanding of the possible numbers of congruent faces on a convex polyhedron? Certainly all the "fair dice" qualify -- those are the polyhedra whose symmetry groups act transitively on their faces, so of course the faces are all congruent. That gives us the "2n for all n" families (which are actually a 2-parameter family whose generic members have faces with no symmetry, but which also includes the duals of the prisms and antiprisms). It also covers the 120-hedraon, which is a "DisdyakisTriacontahedron" if you want to cut rhombuses into quarters, but you can also visualize it as the barycentric subdivision of the icosahedron or dodecahedron, which is probably more accessible ("poke up" the center of each face & edge, so that each n-gon becomes 2n scalene triangles). But if you want all faces to be congruent *by symmetry*, then those are the best you can do: we know all the finite subgroups of SO(3), and those are the biggest ones available. So what was the 180-hedron that sparked this conversation? If it really exists, then it's going to have to have faces which are congruent to each other despite being in different symmetry classes, and I'd very much like to understand what that looks like -- but right now I don't. --Michael Kleber On Wed, Nov 26, 2008 at 7:01 AM, <rwg@sdf.lonestar.org> wrote:
rwg>there's "DisdyakisTriacontahedron", which looks to me like
it 120-sects its circumsphere.
Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm --rwg PHENOCRYSTIC PYROTECHNICS
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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On Wed, Nov 26, 2008 at 4:01 AM, <rwg@sdf.lonestar.org> wrote:
rwg>there's "DisdyakisTriacontahedron", which looks to me like
it 120-sects its circumsphere.
Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm
:) Can you make it so you have to twist them as they come apart? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
On 11/26/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm --rwg
You dismantle the thing by putting a balloon inside and then inflating it. Uh-huh. So how do you put it together in the first place? WFL
On 11/26/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Application: A puzzle box requiring an explosive decompression chamber. See http://gosper.org/slab.htm --rwg
You dismantle the thing by putting a balloon inside and then inflating it.
Uh-huh.
So how do you put it together in the first place? WFL
Make a mold for ice "cubes" exactly the shape and size that the pieces enclose as they are about to fall apart. Moisten the pieces and stick them to the "cube". Bind it up with rubber bands and let the ice melt. Actually, with low friction these things might be too easy to disassemble--just pull on opposite sides. Even with Mike Stay's evil twist. But I suspect a subtle tongue and groove detent mechanism could fix that. --rwg TURTLE ISLAND TRUNDLE-TAILS
Downrate my excitement over the Abel-Plana formula, inf inf inf ==== / / \ [ h(%i y) - h(- %i y) [ h(0) > h(n) = %i I ------------------- dy + I h(x) dx + ----, / ] 2 %pi y ] 2 ==== / %e - 1 / n = 0 0 0 which is like Euler-Maclaurin but gives exact identities instead of divergent approximations. Reason: it can be wrong! E.g., for h(n):=q^n^2*e^(a*n), inf inf inf ==== 2 / / 2 \ a n n [ sin(a y) [ x a x 1
%e q = - 2 I ------------------- dy + I q %e dx + - / ] 2 ] 2 ==== / y 2 %pi y / n = 0 0 q (%e - 1) 0
is only meaningful for q on the unit circle, (e^(i b)). (Is Limbaugh still on?). But even though everything then converges, it's wrong! According to http://dlmf.nist.gov/2/10/ Sufficient conditions for the validity of this [...] result are: 1.(a) On the strip 0<=Re(z)<oo, h(z) is analytic in its interior, <the derivatives of h are> continuous on its closure, and h(z)=o(exp(2 pi Im(z)) as Im(z)->±oo, uniformly with respect to 0<=Re(z)<oo. (b) h(z) is real when 0<=z<oo. Geez, Edith, I think those shysters voided our warranty. --rwg PS, rtw>Some detail regarding <paleoballistics>:
The telegram with Gosper's P30 gun coordinates was dated 4 Nov70.
Gosper's puffer train and P46 gun discovery reported Aug/Sep 71.
Cordership reported Dec 71. [Actually, c/12 puffer]
Schick's puffers reported early (Feb?) 72. R. Wainwright
Also, correction: guns of all true periods >=60 are known, *except* 61. INTERSPECIES REST IN PIECES
I don't recall seeing special values of thetas at pi/3. After a long sequence of "miraculous" simplifications, pi - pi/3 pi - pi/3 theta (--, %e ) = theta (--, %e ) 3 3 4 6 1/8 1 3 sqrt(sqrt(3) - 1) Gamma(-) 4 = -------------------------------. 1/4 3/4 2 2 pi rwg> [...] According to http://dlmf.nist.gov/2/10/
Sufficient conditions for the validity of this [Abel-Plana] result are:
1.(a) On the strip 0<=Re(z)<oo, h(z) is analytic in its interior, <the derivatives of h are> continuous on its closure, and h(z)=o(exp(2 pi Im(z)) as Im(z)->±oo, uniformly with respect to 0<=Re(z)<oo.
(b) h(z) is real when 0<=z<oo.
I should have punted (b). I think it's only there because DLMF for some reason wrote - 2 Im(h(i y)) instead of i (h(i y) - h(-i y)). --rwg POLAR ANGLE ANAL PROLEG PROLEGOMENA OREGON MAPLE
rwg>I don't recall seeing special values of thetas at pi/3. After a long
sequence of "miraculous" simplifications,
pi - pi/3 pi - pi/3 theta (--, e ) = theta (--, %e ) 3 3 4 6 1/8 1 3 sqrt(sqrt(3) - 1) Gamma(-) 4 = ------------------------------- 5/4 3/4 2 pi Foo, pi pi 6 theta (--, q) = theta (--, q) = sqrt(3) eta(q ) 2 6 1 3 and we can do eta(e^-(pi sqrt r)) for probably any rational r. E.g., pi pi 5/8 3/2 1 - ------- - ------- 3 Gamma (-) pi sqrt(3) pi sqrt(3) 3 theta (--, e ) = theta (--, e ) = ---------------- 1 3 2 6 4/3 2 pi --rwg AUTOGENOTYPES GET ONE PAST YOU
I added a codicil to http://www.tweedledum.com/rwg/pizza.htm partially rescuing the Kapteyn (Bessel) series approach. Unrelated: A kid showed me a math homework with a dire warning to show all work or else. Two of the problems were to distribute a monomial across a binomial. I couldn't see any work to show, except maybe "2 - 1 = 1" type exponent simplifications. So I suggested he go ahead and do it in his head, and attach a functional PET scan of his brain. --rwg CHEMOSORPTIONS PROCESSION MOTH ROCHON PRISM PROCHRONISM
And the codicil provides a wonderful segue into series acceleration, and your mysterious irregular convergence of the pi = 4 arctan 1 series. Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] On Behalf Of rwg@sdf.lonestar.org [rwg@sdf.lonestar.org] Sent: Tuesday, December 09, 2008 11:38 PM To: math-fun Subject: [math-fun] Luchshe pizza v'rukye ... I added a codicil to http://www.tweedledum.com/rwg/pizza.htm partially rescuing the Kapteyn (Bessel) series approach. Unrelated: A kid showed me a math homework with a dire warning to show all work or else. Two of the problems were to distribute a monomial across a binomial. I couldn't see any work to show, except maybe "2 - 1 = 1" type exponent simplifications. So I suggested he go ahead and do it in his head, and attach a functional PET scan of his brain. --rwg CHEMOSORPTIONS PROCESSION MOTH ROCHON PRISM PROCHRONISM _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
rcs(privately)>Gosper: one more possible way to solve the pizza/3 problem:
the function z e^(-z) has a nice power series inversion,
I.e., Lambert W
and converges out to the expected radius 1/e or whatever. Maybe turn this into a Kepler reversion?
If you expand the inner (Kapteyn) series in Lagrange's solution to Kepler's equation and then sumswap, you get a sum of derivatives of Lambert Ws. The derivation assumes epsilon (eccentricity) < 2/e, but it seems to work fine for epsilon = 1. So http://www.tweedledum.com/rwg/pizza.htm is tweaked yet again. Re Segway, a 17 yr old mechanical genius at the local TechShop built himself a wild one that has slightly injured even experienced riders. I'll try to learn what he spent, and add an estimate for speed limitation! (Keep this quiet or we'll have yet another helmet law.) --rwg UNBRACED BEANCURD TRACKSHOE SHORTCAKE COLINEAR POINTS IN LOCO PARENTIS
participants (7)
-
Dan Asimov -
David Wilson -
Fred lunnon -
Michael Kleber -
Mike Stay -
rwg＠sdf.lonestar.org -
Schroeppel, Richard