May 2005 - mathfun@mailman.xmission.com

RE: [math-fun] fun with A := solid_angle(a,b,c)
by Schroeppel, Richard 26 May '05

26 May '05

There's a plane geometry theorem that the sum of the vertex angles of a polygon is pi*(#vertices-2). Is there a 3D analog? Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of R. William Gosper Sent: Wednesday, May 25, 2005 11:17 PM To: math-fun(a)mailman.xmission.com Subject: [math-fun] fun with A := solid_angle(a,b,c) Recall cos(c) + cos(b) + cos(a) + 1 A = 2 acos ----------------------------, a b c 4 cos(-) cos(-) cos(-) 2 2 2 where a,b,c are the (trihedral) angles at the vertex. Problem 1: What is t, the tetrahedral bond angle? Dissect tbe tetrahedron into four congruent squat ones whose apex angles are t. Then each has (c136) solid_angle(t,t,t) = 4*%pi/4) 3 cos(t) + 1 (d136) 2 acos(------------) = %pi 3 t 4 cos (-) 2 taking cos of half of both sides, (c137) cos(%/2) 3 cos(t) + 1 (d137) ------------ = 0 3 t 4 cos (-) 2 I.e., t = asec(-3). Problem 2: What is the solid angle of the apex of a regular polygonal pyramid with all n vertex angles = v? Dissect the pyramid into n congruent tetrahedra, then multiply n by the apical solid angle of one of them: (c138) (asin(sin(v/2)/sin(%pi/n)),n*solid_angle(v,%%,%%))$ (Cleaning up cos^2 of a half angle) (c139) subst(v/2,v,halfangles(subst(2*v,v,%))) 2 v sin (-) 2 cos(v) + 2 sqrt(1 - ---------) + 1 2 %pi sin (---) n (d139) 2 n acos(----------------------------------) 2 v sin (-) v 2 2 cos(-) (sqrt(1 - ---------) + 1) 2 2 %pi sin (---) n Problem 3: What is the circumradius of a regular dodecahedron with unit face circumradius? Use the preceding formula for one of the twelve component pyramids, with n=5, r unknown: (c140) subst([n=5,v=2*asin(print(side_regular_polygon(5,1))/2/r)],%) =4*%pi/12 %pi 2 sin(---) 5 %pi sin(---) 1 5 2 sqrt(1 - --) + cos(2 asin(--------)) + 1 2 r r %pi (d140) 10 acos(------------------------------------------) = --- 2 %pi 3 sin (---) 1 5 2 (sqrt(1 - --) + 1) sqrt(1 - ---------) 2 2 r r Clear the acos and square: (c141) halfangles(cos(trigexpand(%/10)))^2 2 %pi 2 sin (---) 5 1 2 (----------- - 2 sqrt(1 - --) - 2) %pi 2 2 cos(---) + 1 r r 15 (d141) ------------------------------------- = ------------ 2 %pi 2 sin (---) 1 2 5 4 (sqrt(1 - --) + 1) (1 - ---------) 2 2 r r Express (sin pi/5)^2 ito cos pi/15: (c142) (sin(3*x)^2,subst(%pi/15,x,%%=desinify(factor(trigexpand(%%))))) 2 %pi 2 %pi 2 2 %pi (d142) sin (---) = (1 - 4 cos (---)) (1 - cos (---)) 5 15 15 Find an polynomial equation for cos pi/15: (c143) cos(%pi/15) = c %pi (d143) cos(---) = c 15 (c144) block([%piargs:all],factor(multiply_conjugates(?meqhk(''%)))) 4 3 2 16 c + 8 c - 16 c - 8 c + 1 (d144) ------------------------------ 16 (The function ?meqhk is just lhs-rhs.) Find a polynomial equation for r in terms of c: (c145) factor(multiply_conjugates(ratnumer(subst([d142,d143],?meqhk(d141)))),%) 3 2 3 (d145) - (24 c - 8 c - 16 c + 1) (2 r - 8 c + 10 c + 1) 3 2 3 2 (2 r + 8 c - 10 c - 1) (4 r + 24 c + 8 c - 16 c - 7)/64 Which factor makes numeric sense? (c146) sfloat(subst(reverse(d143),args(part(%,1,1)))) 2 (d146) [0.15621, 2.0 r + 3.29456, 2.0 r - 3.29456, 4.0 r + 7.46457] Only the third, so (c147) combine(expand(trigreduce(subst(reverse(d143),solve(part(d145,1,1,3),r)) ))) %pi 1 - sqrt(5) (d147) [r = 2 cos(---) + -----------] 15 4 Check: (c148) subst(%,d140) %pi %pi (d148) --- = --- 3 3 (How did Macsyma do that?-) --rwg PS, DPThurston: (Recall that if you look at a cube from a corner you see a hexagon.) If you tile the plane with those hexagons, their cubes mechanically lock together in an airtight sheet. Similarly for octagons resting on a face, and dodecahedra stood on a vertex (www.ippi.com/rwg/dodex.gif) Tetrahedra stood on edge lock together in a square grid. Icosahedra on a face lock but leak. _______________________________________________ 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] Counting function
by David Wilson 26 May '05

26 May '05

For n >= 0, we have |{floor(n/k): k in Z+}| = floor(sqrt(4n+1)). Can we find a similar formula for a(n) = |{floor(n/(k^2)): k in Z+}|? For which n is a(n) > a(n-1)? - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] list of "challenge sequences"
by N. J. A. Sloane 26 May '05

26 May '05

I have collected most of the sequences that have been proposed as good challenge problems in a new entry in the Index to the OEIS - see under sequences that need extending: challenge problems the url is http://www.research.att.com/~njas/sequences/Sindx_Se.html#extend suggestions for further entries will be welcomed NJAS

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[math-fun] fun with A := solid_angle(a,b,c)
by R. William Gosper 26 May '05

26 May '05

Recall cos(c) + cos(b) + cos(a) + 1 A = 2 acos ----------------------------, a b c 4 cos(-) cos(-) cos(-) 2 2 2 where a,b,c are the (trihedral) angles at the vertex. Problem 1: What is t, the tetrahedral bond angle? Dissect tbe tetrahedron into four congruent squat ones whose apex angles are t. Then each has (c136) solid_angle(t,t,t) = 4*%pi/4) 3 cos(t) + 1 (d136) 2 acos(------------) = %pi 3 t 4 cos (-) 2 taking cos of half of both sides, (c137) cos(%/2) 3 cos(t) + 1 (d137) ------------ = 0 3 t 4 cos (-) 2 I.e., t = asec(-3). Problem 2: What is the solid angle of the apex of a regular polygonal pyramid with all n vertex angles = v? Dissect the pyramid into n congruent tetrahedra, then multiply n by the apical solid angle of one of them: (c138) (asin(sin(v/2)/sin(%pi/n)),n*solid_angle(v,%%,%%))$ (Cleaning up cos^2 of a half angle) (c139) subst(v/2,v,halfangles(subst(2*v,v,%))) 2 v sin (-) 2 cos(v) + 2 sqrt(1 - ---------) + 1 2 %pi sin (---) n (d139) 2 n acos(----------------------------------) 2 v sin (-) v 2 2 cos(-) (sqrt(1 - ---------) + 1) 2 2 %pi sin (---) n Problem 3: What is the circumradius of a regular dodecahedron with unit face circumradius? Use the preceding formula for one of the twelve component pyramids, with n=5, r unknown: (c140) subst([n=5,v=2*asin(print(side_regular_polygon(5,1))/2/r)],%) =4*%pi/12 %pi 2 sin(---) 5 %pi sin(---) 1 5 2 sqrt(1 - --) + cos(2 asin(--------)) + 1 2 r r %pi (d140) 10 acos(------------------------------------------) = --- 2 %pi 3 sin (---) 1 5 2 (sqrt(1 - --) + 1) sqrt(1 - ---------) 2 2 r r Clear the acos and square: (c141) halfangles(cos(trigexpand(%/10)))^2 2 %pi 2 sin (---) 5 1 2 (----------- - 2 sqrt(1 - --) - 2) %pi 2 2 cos(---) + 1 r r 15 (d141) ------------------------------------- = ------------ 2 %pi 2 sin (---) 1 2 5 4 (sqrt(1 - --) + 1) (1 - ---------) 2 2 r r Express (sin pi/5)^2 ito cos pi/15: (c142) (sin(3*x)^2,subst(%pi/15,x,%%=desinify(factor(trigexpand(%%))))) 2 %pi 2 %pi 2 2 %pi (d142) sin (---) = (1 - 4 cos (---)) (1 - cos (---)) 5 15 15 Find an polynomial equation for cos pi/15: (c143) cos(%pi/15) = c %pi (d143) cos(---) = c 15 (c144) block([%piargs:all],factor(multiply_conjugates(?meqhk(''%)))) 4 3 2 16 c + 8 c - 16 c - 8 c + 1 (d144) ------------------------------ 16 (The function ?meqhk is just lhs-rhs.) Find a polynomial equation for r in terms of c: (c145) factor(multiply_conjugates(ratnumer(subst([d142,d143],?meqhk(d141)))),%) 3 2 3 (d145) - (24 c - 8 c - 16 c + 1) (2 r - 8 c + 10 c + 1) 3 2 3 2 (2 r + 8 c - 10 c - 1) (4 r + 24 c + 8 c - 16 c - 7)/64 Which factor makes numeric sense? (c146) sfloat(subst(reverse(d143),args(part(%,1,1)))) 2 (d146) [0.15621, 2.0 r + 3.29456, 2.0 r - 3.29456, 4.0 r + 7.46457] Only the third, so (c147) combine(expand(trigreduce(subst(reverse(d143),solve(part(d145,1,1,3),r))))) %pi 1 - sqrt(5) (d147) [r = 2 cos(---) + -----------] 15 4 Check: (c148) subst(%,d140) %pi %pi (d148) --- = --- 3 3 (How did Macsyma do that?-) --rwg PS, DPThurston: (Recall that if you look at a cube from a corner you see a hexagon.) If you tile the plane with those hexagons, their cubes mechanically lock together in an airtight sheet. Similarly for octagons resting on a face, and dodecahedra stood on a vertex (www.ippi.com/rwg/dodex.gif) Tetrahedra stood on edge lock together in a square grid. Icosahedra on a face lock but leak.

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RE: [math-fun] more challenge sequences needed
by N. J. A. Sloane 26 May '05

26 May '05

Rich suggested A_n[n]+1. That is A037181, and a lot of terms are already known! Other similar sequences can be found in the Index to the OEIS under "diagonal sequences": diagonal sequences: A031135*, A037181, A031214, A051070, A091967, A102288 Neil

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[math-fun] Re more challenge sequences needed
by N. J. A. Sloane 25 May '05

25 May '05

Hugo P. points out that if people involved in competing in this competition are members of the math-fun or seq-fan mailing lists, then we shouldn't discuss choosing problems for the competition in these mailing lists! The person who wrote to me is "Jean-Charles Meyrignac" <euler(a)free.fr> and it is fine with me if people who have good problems to propose write directly to him! Neil

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RE: [math-fun] more challenge sequences needed
by Schroeppel, Richard 25 May '05

25 May '05

How about A_n[n]+1? :-) --Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of N. J. A. Sloane Sent: Wednesday, May 25, 2005 10:02 AM To: math-fun(a)mailman.xmission.com; seqfan(a)ext.jussieu.fr Cc: njas(a)research.att.com Subject: [math-fun] more challenge sequences needed To Seqfans, Mathfun: the people who are planning the contest (I should probably not have described them as French, although my contact has a French email address - they could be anywhere in the world) replied that none of the sequences that I gave them were suitable. Either they failed to meet some of the requirements for a good challenge, or else they were too close to problems that had been used in previous challenges. So they are still looking for a good sequence to extend. Something like the Hadamard determinant problem or the Golomb ruler problem would be perfect, but these have already been used. Here are the constraints on the problem: 1) It must be NP-complete 2) It must be simple to describe 3) It should not have been widely explored 4) A solution must be easy to check (= it doesn't require a lot of computation to check) 5) It has to be computed from N=8 to at most 64 Further suggestions would be welcomed! Neil _______________________________________________ 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] RE: more challenge sequences needed
by Rob Pratt 25 May '05

25 May '05

I have participated in all of the previous contests, and requirement 4 refers only to the feasibility of a solution--not the optimality. It's okay if optimality is hard to prove. But feasibility should be easy to check so that the contest administrator can quickly give scores for solutions submitted by the contestants. There is a leaderboard posted throughout the contest, and it is updated each time a new entry is submitted. Rob -----Original Message----- From: r.rosenthal(a)web.de [mailto:r.rosenthal@web.de] Sent: Wednesday, May 25, 2005 12:56 PM To: math-fun(a)mailman.xmission.com; seqfan(a)ext.jussieu.fr Cc: njas(a)research.att.com Subject: Re: more challenge sequences needed > ... or the Golomb ruler problem would be perfect, but these have > already been used. Hello Neil, this is far too great a chance for extending my favourite A004137 to let it go without fight :-) The rulers considered here are *not* the Golomb rulers, for which there is indeed a vast literature and extensive search by globally connected people. There is just ONE really clever type of rulers, which probably will be optimal for the next may be 10 or 15 arguments: the type invented by Wich- mann in the 1960's. Computational evidence was quite poor before Peter Luschny did focus on this sequence. We weren't able to prove the validity of any method for reducing the search tree. It would be of great value to have solid knowledge about further members of this sequence, so that theoretical considerations could be proved or disproved more easily. > Here are the constraints on the problem: > > 1) It must be NP-complete Nobody has proved the opposite. > 2) It must be simple to describe Find 1 = x_1 < x_2 < ... < x_n = L such that the differences x_i - x_j don't have holes and such that L is maximal. > 3) It should not have been widely explored I think they should try something, which people are interested in. Otherwise their effort will make nobody happy. And *if* they try something interesting, then surely there *must* have been some exploration done before. The term "widely explored" is widely flexible ;-) > 4) A solution must be easy to check (= it doesn't require a lot of > computation to check) This may be the killer argument: a solution is proved to be one, if there is no solution for L+1. And to prove that, you will need very much time :-( (While on the other hand it is easy to show that such a set {x1,...x_m} doesn't have holes in its differences.) > 5) It has to be computed from N=8 to at most 64 Hmmm... no comment. Still hoping for the election, and thank you very much for your first friendly welcome to that suggestion of mine, best regards, Rainer Rosenthal r.rosenthal(a)web.de

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[math-fun] Re: Challenge sequences
by Gareth McCaughan 25 May '05

25 May '05

> Generating a bunch of random boolean 8x8's, I've gotten > every determinant from 0 to 32, and 34, 35, 36, 42. > 73 different values, when you count the negative ones. 33: 00001111 01011100 11000100 01100001 10010101 10111110 01010110 11011011 40: 11101000 00110110 10001111 01011101 10010001 00011010 01100011 00101001 44: 00101010 10000011 11001000 10011110 11110010 01011011 01100101 10111001 45: 11001011 01110011 10010101 10100110 01010110 00101111 11111100 10011010 48: 11100100 01100011 10000111 01001110 10101010 10111101 11010010 00110110 56: 11001100 01101001 10011011 01010101 01100110 00111100 11110000 10100101 These were found by naive hillclimbing starting from random matrices, which makes it interesting to me how much symmetry there seems to be in the last one's rows. But maybe I'm being taken in by the Strong Law of Small Numbers myself now. Anyone got a proof that bigger than 56 is impossible? Or a way to rule out (e.g.) 38 or 39? -- g

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Re: [math-fun] more challenge sequences needed
by Michael Reid 25 May '05

25 May '05

> a(n) = minimum value of k so that k copies each of cubes of sides 1 > through n can be used to exactly fill some rectangular box. is it obvious that such k exists? mike

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