December 2007 - mathfun@mailman.xmission.com

Re: [math-fun] A New Year's Puzzle
by Michael Reid 15 Dec '07

15 Dec '07

> Thanks, but this is only putting geometrically > what I put algebraically, and doesn't really show > what's going on, because your eventual equation > is one between four-dimensional volumes. There > ought to be a solution which stays in at most two > dimensions. R. there ought?? that the final relation > > [BPQ]^2 = [ ... ] > > = ([ABQ] + [BCP] + [PDQ])^2 - 4 [ABQ] [BCP] . is quadratic in areas suggests that there won't be such a solution. my solution had the mild benefit that being quadratic in areas arose from ratios of edge lengths. but yes, after that it's all algebraic. by the way, we went down a similar path some time ago when someone asked a related question about hero(n)'s formula: is there a reasonable 4-dimensional interpretation for s(s - a)(s - b)(s - c) ? i don't think anyone posted one. mike > > B----F---------C > > /|\_ / / > > / / /_ / > > / | / \_ / > > / / / \_ / > > / | / \/ > > E----X------_--P > > / |/ __-- / > > / /_-- / > > A----Q---------D

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[math-fun] A New Year's Puzzle
by Schroeppel, Richard 14 Dec '07

14 Dec '07

A New Year's Puzzle that's been circulating here ... B--------------C /|\_ / / / \_ 29 / / | \_ / / / \_ / / | ? \/ / / __--P / 6 | __-- / / /_-- 17 / A----Q---------D ABCD is a parallelogram. P is on CD, Q is on AD. The areas of the triangles ABQ, BCP, and PDQ, are 6, 29, and 17. What's the area of triangle BPQ? Rich

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Re: [math-fun] A New Year's Puzzle
by Michael Reid 14 Dec '07

14 Dec '07

> but how does one SEE what is going on ??? hopefully this "explains" where it comes from: draws lines through P and through Q parallel to AD and AB respectively; suppose they intersect at point X and meet the opposite sides of the parallelogram at E and F respectively. my attempt to mark up the original ascii art: B----F---------C /|\_ / / / / /_ / / | / \_ / / / / \_ / / | / \/ E----X------_--P / |/ __-- / / /_-- / A----Q---------D then we have [ABCD] - 2 [BPQ] = [EBFX] , where i use [ABC...Z] to denote the area of the polygon ABC...Z . we also have [EBFX] / [ABFQ] = EB / AB = [EBCP] / [ABCD] , so that [EBFX] [ABCD] = [ABFQ] [EBCP] = 4 [ABQ] [BCP] . multiply the first equation by [ABCD] and substitute this last one to get [ABCD]^2 - 2 [ABCD] [BPQ] = 4 [ABQ] [BCP] , which is ([ABCD] - [BPQ])^2 - [BPQ]^2 = 4 [ABQ] [BCP] , so that [BPQ]^2 = ([ABCD] - [BPQ])^2 - 4 [ABQ] [BCP] = ([ABQ] + [BCP] + [PDQ])^2 - 4 [ABQ] [BCP] . for the numbers in the original problem, we have [BPQ]^2 = (6 + 29 + 17)^2 - 4 * 6 * 29 = 2008 . mike

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Re: [math-fun] A New Year's Puzzle
by Dan Asimov 13 Dec '07

13 Dec '07

Richard Guy asks (of the sqrt(2008) New Year's puzzle): << Where is The Book solution? Which brings up an interesting question (that has undoubtedly been asked before): Can the concept of a Book Proof be formalized? --Dan

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[math-fun] That last post was meant only for Jim Propp
by Dan Asimov 12 Dec '07

12 Dec '07

Sorry about that. --Dan

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[math-fun] mathematics question re tiling & much else
by Dan Asimov 12 Dec '07

12 Dec '07

Dear Jim, I've been thinking about a rather interesting math picture that starts with a strange tiling-ish thing on the plane -- and quickly encompasses a bunch of other stuff, including cyclotomic rings and hyperbolic geometry. I'm far from understanding the situation fully but would like to discuss it with you if you'd care to listen. (Especially when I realized that tilings is one of your specialties, from your home page.) And incidentally, I realized that there's an old computer hack I wrote years ago which suggests the existence of an interesting stochastic process -- and since I now know that such things are also among your interests, I'd like to discuss that, too. Please let me know if you have any time to discuss one or both of these, and if so, I'd like to chat on the phone if your schedule permits. Regards, Dan P.S. Thanks for the acknowledgment in your recent joint paper on entropy of monomial maps (though I can hardly imagine what I contributed).

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Re: [math-fun] Least prime divisor probabilities (or, The Lurch of the SubGenius)
by Eugene Salamin 12 Dec '07

12 Dec '07

----- Original Message ---- From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Tuesday, December 11, 2007 11:52:05 PM Subject: [math-fun] Least prime divisor probabilities (or, The Lurch of the SubGenius) ... So: suppose each prime 2,3,5,7,...,p_k < sqrt(n) does not divide n while there remain primes p_r in the range p_k < p_r <= sqrt(n). The greater such k, the greater the chance that such p_r | n. But also the greater such k, the greater is the chance that n is prime. How do these two possibilities relate to each other? (I.e., what should I expect, or bet on, as k increases -- that n is prime, or that some larger test prime <= sqrt(n) divides n ???) (For starters, let p_r denote the largest prime <= sqrt(n).) ... I would expect divisibility by different primes to be independent, i.e. knowing that N is divisible by p, or that N is not divisible by p, should not change the probability (1/q) that N is divisible by a different prime q. Each time a new prime is found to not divide N, the probability that N is prime increases. Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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[math-fun] Least prime divisor probabilities (or, The Lurch of the SubGenius)
by Dan Asimov 12 Dec '07

12 Dec '07

[*begin discursiveness warning*] In trying to mentally factor a 4-digit number (7989, if you must know). After dividing by the obvious 3, the next step was to find the least prime factor of 7989/3 = 2663. I tried the only factoring algorithm I know: naively try each prime in order no greater than the square root of the number (in this case < 52). I soon realized (what everyone else probably knows well): that if 2663 were not prime, then the more primes that were eliminated, the more likely it became that the next test prime <= sqrt(2663) would divide it. But of course there was always the chance the 2663 is in fact prime. So: suppose each prime 2,3,5,7,...,p_k < sqrt(n) does not divide n while there remain primes p_r in the range p_k < p_r <= sqrt(n). The greater such k, the greater the chance that such p_r | n. But also the greater such k, the greater is the chance that n is prime. How do these two possibilities relate to each other? (I.e., what should I expect, or bet on, as k increases -- that n is prime, or that some larger test prime <= sqrt(n) divides n ???) (For starters, let p_r denote the largest prime <= sqrt(n).) [*end discursiveness warning*] --J.R. "Dan" Dobbs

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Re: [math-fun] church of the sub-genus?
by Dan Asimov 11 Dec '07

11 Dec '07

Marc wrote: << This is doubtless crankery, but... In computing the Euler characteristic I notice that handles get twice the weight of cross-caps. >> Lessee. Suppose we have a surface N and "add a handle" -- remove (the interior of) a disk from each of N and a torus T, then identify the resulting two circular edges -- to get a new surface that is the "connected sum" of N & T, denoted N#T. If X (chi) stands for Euler characteristic, then we have X(N#T) = (X(N) - 1) + (X(T) - 1) = X(N) + X(T) - 2 = X(N) - 2, because circles and tori have X = 0, but a disk D has X(D) = 1. By contrast, "adding a crosscap" to N means likewise creating the connected sum of N and a projective plane P. The calculation is the same except that while X(T) = 0, we have X(P) = 1 (since P is double-covered by the sphere S, well-known to have X(S) = 2). So there is an increase of 1: X(N#T) = X(N)-2 but X(N#P) = X(N)-1. But this because of this calculation, not because of a crosscap getting "twice the weight" of a handle. (In fact, a handle may be thought of as a torus minus a disk, and a crosscap as a projective plane minus a disk (aka a Moebius band), with X(T-D) = -1, while X(P-D) = 0.) << Plugging p=1/2 in for the "genus" in the chromatic formula floor((7+sqrt(1+48*p))/2) gives 6, the number for the Mobius strip. Does that figure, or is it just an accident? Is there some general chromatic expression that covers non-orientable surfaces? If we can have "fractional dimension", can we somehow have "fractional genii"? Happy Holey-days! >> There are uses for fractional Euler characteristics in describing orbifolds, but as far as I know ordinary surfaces have integer genera. The formula for the Euler char of an orientable surface of genus g (connected sum of a sphere and g tori for g >= 0) is X = 2-2g. The Euler char of a nonorientable surface of nonorientable genus q (connected sum of a sphere and q projective planes (q >= 1) is X = 2-q. --Dan

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[math-fun] church of the sub-genus?
by Marc LeBrun 11 Dec '07

11 Dec '07

This is doubtless crankery, but... In computing the Euler characteristic I notice that handles get twice the weight of cross-caps. Plugging p=1/2 in for the "genus" in the chromatic formula floor((7+sqrt(1+48*p))/2) gives 6, the number for the Mobius strip. Does that figure, or is it just an accident? Is there some general chromatic expression that covers non-orientable surfaces? If we can have "fractional dimension", can we somehow have "fractional genii"? Happy Holey-days!

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