January 2003 - mathfun@mailman.xmission.com

[math-fun] <Blush!>, solid angle derivation sketch
by R. William Gosper 04 Jan '03

04 Jan '03

I'll respond to George Hart's highly embarrassing hexagon question as soon as I unwedge my head on the subject. (In case anyone besides Shel is interested.) Given a tetrahedral apex with face angles (trihedral components) a, b, c, the solid angle is just the area on the unit sphere of the projection of the tetrahedron's base (opposite face) when the apex is at the center of the sphere. By the Spherical Triangle entry in Eric's Encyclopedia, this area is just the spherical excess, Omega = C + B + A - pi, where A, B, C are the vertex angles of the spherical triangle, i.e., the dihedral angles between the other three faces. Erics' Spherical Trigonometry entry derives the dihedrals from the face angles: cos(a) A = acos(------------- - cot(b) cot(c)), sin(b) sin(c) symmetrically for B and C. So Omega = acos + acos + acos - pi, which is maybe what most people use. To get 2 (cos(c) + cos(b) + cos(a) + 1) Omega = acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) I TRIGEXPANDed cos(Omega), then tediously massaged the resulting mess with TRIGSIMP, FACTOR, and my own trivial DESINIFY and DECOSIFY, and eventually got lucky and tried subtracting and adding (1+cos+cos+cos)^2 from/to the numerator. --rwg

1 1

[math-fun] Helaman Ferguson vacates sculpture barn
by David H Bailey 03 Jan '03

03 Jan '03

Sculptor Solves Massive Problem Tons of Equipment, Stone Moved From Md. Studio on Time By Eugene L. Meyer Washington Post Staff Writer Friday, January 3, 2003; Page B03 Packing up posed some sizable problems for mathematician-sculptor Helaman Ferguson, who was facing eviction from his studio barn at the Maryland Science and Technology Center in Bowie. Ferguson had a Jan. 1 deadline to move out, and he needed a way to transport two granite blocks, 11 and 13 tons; 24 pallets with additional pieces of stone, each weighing at least a ton; a security fence and yard lights; a computer-controlled milling machine; and about 8,000 pounds of tools, computers and other equipment. Ferguson, who also works for the Institute for Defense Analyses at the Bowie office park, had been given use of the barn in return for creating a fountain sculpture, which is now located there in the middle of a reservoir. The sculpture's form is derived from an 800-year-old mathematical formula, and it was Ferguson's dream to complete a "mathematical theme park" with 14 more pieces around the lake. Ferguson recently finished a piece for Merck, the pharmaceutical company, and is overdue on another commission, for Macalester College in St. Paul, Minn. But the landlord, MIE Properties, wants to develop the site. The company agreed to one extension of Ferguson's deadline in the fall but insisted that he be out by Jan. 1. After word of Ferguson's situation spread, many hands turned out to help with the New Year's Eve move. "What do you do when 50 friends and neighbors and all kinds of people show up and want to help?" Ferguson said. "It worked. It's happened." Ferguson has a good working relationship with the Springfield Rental Crane Corp., whose president, Randy Thompson, provided five flatbed trucks and free storage at the company's Springfield facility until the sculptor can relocate. "We just got through unloading the last trailer," Thompson said yesterday afternoon, adding that Ferguson's stuff fills about 40 by 75 feet of storage space. "I've done a lot of work for him. He's a pretty good guy. He was in a pinch." Ferguson has been offered a two-acre site elsewhere on the Bowie property, but he would have to build a studio from scratch. For the moment, he seems elated just to have the monstrous move behind him. "The barn is totally empty now," he said in a telephone message to The Washington Post Tuesday night. I've mothballed my studio. . . . I have to tell you I feel like a free man." Â© 2003 The Washington Post Company

1 0

[math-fun] More Games of No Chance
by Dan Hoey 02 Jan '03

02 Jan '03

I finally got MGONC last month, after the usual delay between announced and actual publication date. I haven't read much of it yet, but I thought it might be a good time to tell funsters (especially those who were at CGT2) about a minor improvement that David Wolfe was kind enough to sneak into the Calistrate, Paulhus, Wolfe paper (also on http://www.gustavus.edu/~wolfe/dayn/mgonc.pdf) If you look at the Hasse diagram of the lattice of games on day 2 (MGONC p. 28), you'll see that the game * now appears between 0 and *2 . This allows the diagram to be overlaid on a table of day 2 games in a grid indexed by left and right option sets. Such tables appear in R K Guy 1996 (GONC, p. 6) and Fraser, Hirschberg, Wolfe 2003 (http://www.gustavus.edu/~wolfe/dayn/structure.pdf, p. 6). I've put a drawing of the resulting overlaid figure at the end of this message. I suspect that a similar table of day 3 games doesn't exist, because the 98 independent sets of day 2 games aren't totally ordered as the day 1 independent sets are. (There are actually two different orders of the day 1 independent sets, depending on whose options they might be). Still, a 98 by 98 grid of the day 2 independent sets might be start toward developing a picture of the day 3 lattice. Dan Hoey(a)AIC.NRL.Navy.Mil ==================================================================== Day 2 games by options and a Hasse diagram of their partial order : Right Options : Left : : Options : -1 : 0,* : 0 : * : 1 : {} : ......:...........:...............:.......:.......:.......:.......: : HOT/FUZZY/NEXT. . . . .POS/RT : : . . . . . : : . . 1|0 <----------< 1* <---< 2 : : . . L V . .L V . V : 1 : . ./ | . / | . | : : . / | . /. | . | : : . L. | . L . | . | : : +-1 <--------< 1|0,* <----+-----< 1|* . | . | : : V . V . | . V . | . | : ...: . . . . . | . . . . . . . | . . . | . . . | . . . | . . . | . : : | . | . | . | . | . | : : | . | . V . | . V . V : : | . | . up* <-----+----< 1/2 <--< 1 : : | . | . L V . | .L . : 0,* : | . | ./ | . | / . : : | . | / | . | /. . : : V . V L. | . V L . . : : 0,*|-1 <----------< *2 <----+-----< up . . : : L V . L V . | . V . . : ...: . . . / . | . . . . . / . | . . . | . . | . . : : / | . / | . | . | . . : : L | . L | . V . | . . : 0 : 0|-1 <----+----< dn* <----+-----< * . | . . : : V | . V | . . | . . : ...: . | . . . | . . . | . . . | . . . . . . . | . . . . . . . . . : : | | . | | . | : : | V . | V . V : * : | *|-1 <----+----< dn <-----------< 0 : : | L . | L . : ...: . | . / . . . . . | . / . . . . . . . . : : | / . | / . : : V L . V L . : -1 : -1* <-----------< -1/2 . : : V . V . : ...: . | . . . . . . . | . . . . . . . . . . : : | . | . : : V . V . : {} : -2 <------------< -1 . : : NEGATIVE/LEFT . . COLD/ZERO/PREVIOUS : ...:...............................................................:

1 0

[math-fun] Quantum fingers
by Dylan Thurston 02 Jan '03

02 Jan '03

I learned about an amusing game recently which I haven't seen described very much, called "Quantum Fingers". This game is much like "Go Fish", with the crucial twist that the hands dealt are determined only by the statements made by players as the game proceeds. The game is played with N players, N >= 3. Each player starts with 4 fingers, held up for all to see. The fingers belong to suits; there are N suits, and each suit has 4 fingers. Each player A, in turn, asks a question of the form: "<Player B>, do you have any <suit S>?". If B answers "yes", then B gives A one finger of the suit S, and A gets another question. If B answers "no", then A's turn ends and the next player gets a turn. In addition, during A's turn, A can put down 4 fingers of a suit. There are two ways the game can end. If one player gets rid of all her cards, she wins. On the other hand, if a player creates a logical contradiction (either in response to a question or when laying cards down), then that player loses. To avoid trivial wins, there is one initial condition: no player starts with all 4 fingers of the same suit. Here's a sample game, between Alice, Bob, and Carol. Alice: Carol, do you have any primes? Carol: Yes, I do. Here, have one. [Carol gives Alice one prime finger. Now Carol has 3 fingers and Alice has 5.] Alice: Bob, do you have any primes? Bob: No, I do not. Alice: Aw, shucks. [Alice's turn ends.] Bob: Alice, do you have any primes? Alice: Yes, I do. Here, have one. [Alice gives Bob one prime finger, perhaps the finger she got from Carol.] Bob: Alice, do you have any primes? Alice: No, I'm all out. [At this point, Alice loses. The four primes originally in the game were distributed between Alice, Bob, and Carol. Bob did not have any, so Alice and Carol had all the primes. Carol could not have had them all (by the initial condition), so Alice must have had at least one initially, and could not answer "no" at this point.] There are a couple of questions I have about this game: 1) If you actually try playing this game, it seems to get very confusing. One might ask: Given the number of players and a sequence of moves, is it NP complete (in the length of the sequence) whether the moves are consistent? 2) What can you say about good strategies? Two colluding players can usually force a win among themselves (the first player to get a move asks the other for cards to complete two sets). But there does seem to be some intuitive notion of "good play"; often this consists of not letting anyone else win immediately and waiting for someone to make a mistake. One hint about teaching others: when two people already know the game, it helps to just start the game and start asking questions, explaining as you go. Enjoy, Dylan Thurston and Chung-chieh Shan

2 3

[math-fun] Can a polyhedron
by R. William Gosper 31 Dec '02

31 Dec '02

with its vertices on a sphere and its edge lengths all equal have more than 92 faces? I'm looking at one (a doenneacontahedron?) with 12 regular pentagons (pentaga?), 60 equilateral triangles, and 20 non-regular (but equilateral) hexagons. There are 90 vertices, all tetravalent, so maybe there is a C_90 fullerane (as opposed to fullerene). I also made a pentacontahedral candidate for C_48 fullerane with 30 squares, 8 equilateral triangles, and twelve rhombi. --rwg

5 4