The Panama palindrome By its true and real author, Dan Hoey Copyright 2001 by Dan Hoey. ______________________________________________________________ Jim Saxe was perhaps the first person to put a cat in the canal. A man, a plan, a cat, a canal; Panama! -- Jim Saxe, plan file @ CMU, 9 October 1983 Guy Jacobson added several items later that year. A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal--Panama! -- Guy Jacobson, plan file @ CMU late 1983 Guy's palindrome appears on page 127 of Common Lisp, the Language (page 170 of the 2nd edition). The 2nd edition of Common Lisp, the Language also contains the remarkable: A man, a plan, a canoe, pasta, heros, rajahs, a coloratura, maps, snipe, percale, macaroni, a gag, a banana bag, a tan, a tag, a banana bag again (or a camel), a crepe, pins, Spam, a rut, a Rolo, cash, a jar, sore hats, a peon, a canal--Panama! which is I'd guess is the work of Guy Steele, the book's author. But I haven't asked either of those Guys to make sure. The appearance of the first two variants intrigued me, and I wondered just how much could be put into the canal. I wrote some code in 1984 to work with it using the Unix "spell" dictionary. By the time I had it working, I was mostly tired of the problem, so I settled for the fairly modest 543-word version that you see. The problem of making the longest possible palindrome of this type is NP-complete, but there are approaches that I expect would generate palindromes many thousands of words long. A trickier problem is getting a good vocabulary, especially one with parts of speech. It took a lot of work to prune out the palindromes that were full of things like "a how, a do, a for, an ochre, a was, an of, ...." I was never quite sure whether I should have put in many of the things that I did. Just what would a myriad look like in a canal? Or a tort? There's a short article on my approach in Expert C Programming: Deep C Secrets by Peter van der Linden. Unfortunately, when the book got reviewed by Stan Kelly-Bootle in Unix Review he included the palindrome but not only misspelled my name, but credited the algorithm to Jim Saxe. As far as publication history goes, I put the palindrome in my plan file on host CMU-CS-A.ARPA on 26 July 1984. My recollection is that my program had produced it earlier that same day. In those days the plan files were collected and redistributed by email to a number of CMU and ex-CMU people, as well as being available by finger. On 29 Feb 1988 I gave permission for it to be posted to the Stanford bulletin board. I don't know of any Usenet appearance before I posted it in May, 1990. It was in January, 1992 that I first saw it posted without attribution, in a collection on rec.humor. I have sent a few dozen notices to people asking them to add attributions, but it's not terribly pleasant work--who likes to hear complaints?--so I'm not very reliable about letting people know. But that's led to even more unattributed reposting. --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 6 Jan 2001 22:36:13 -0500 Subject: Pi is a constant Heidi Burgiel writes: > It is my understanding that the value of Pi depends on the geometry > of the space you're computing it in. Once and for all (I wish) that is not the case. Pi has one and only one value in all of mathematics. Pi is the imaginary part of log(-1). Pi is the square root of 6(1/1 + 1/4 + 1/9 + 1/16 + ...). Pi is not dependent on geometry. There may be geometries with things analogous to circles, whose circumference-to-diameter ratio is not Pi. In those geometries, the value of Pi is the same. The circumference-to-diameter ratio is something else. > Changing the number of > dimensions of the space doesn't affect the value of Pi. True. > If you assume we're living in a space that is curved by gravity, as > described by the theory of relativity, the value of Pi is different > from the one for Euclidean space that you learned in college. No again. The value of Pi is the same unless you assume we live in a space with different arithmetic. Dan haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 12 Jan 2001 11:52:22 -0500 Subject: Circumference to diameter ratios in Lp metrics Recall that for 1 <= p <= infinity there is a metric on R^2 called Lp, for which the distance is given as d(p,q) = (|p_x-q_x|^p + |p_y-q_y|^p) ^ (1/p). In the L2 (Euclidean) metric, the ratio of a disc's circumference to its diameter is Pi. But in the L1 (taxicab) metric the ratio is 4. The Linfinity metric is proportional to L1 rotated 90 degrees, so of course its ratio is 4 also. In the Lp metric the ratio is Pi/4 p(p-1) p(p-1) (cos t) + (sin t) (1/p) R(p) = 4 Integral ( ----------------------------- ) dt p p (p+1) 0 ( (cos t) + (sin t) ) according to Mathematica. The only p for which I've been able to integrate R(p) to anything recognizable are 1, 2, 1/2, 1/3, -1, and -1/2, though I haven't gone seriously into the inverse symbolic calculator. Not that L_p is a metric for p<1 , but it's still neat that R(1/3) = 4 + 8 Gamma[4/3] Gamma[5/3] , R(1/2) = 4 + Pi , R(-1) = Pi , R(-1/2) = 8 Gamma[4/3] Gamma[5/3]. at least numerically. Can any other integraters do anything with this, either in general or for other special cases? The other interesting thing I've found is that, numerically at least, R(p) = R(p/(p-1)) for p>1, R(p) = R(p/(p-1)) - 4 for 1>p>0, and R(p) = R(p/(p-1)) + 4 for p<0. At least the first one is suggested by the L1 disc rotating to the rescaled Linfinity disc, and the L2 disc being fixed under rotation, but in general the Lp norm is not just the L_(p/(p-1)) norm scaled and rotated. So what gives? Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 7 Feb 2001 14:20:57 -0500 Subject: Re: geometry John Conway wrote: > On 7 Feb 2001, ADAM wrote: > > Give an example of an arrangement of 10 points in the plane, > > not all on one line, such that htere are exactly 10 connecting > > lines possible. > * > * * * * > * * > * > * * > Sorry I haven't manged to draw it perfectly! John Conway I count fifteen lines connecting those points--five four-point lines, five extremal lines, and five lines through the center. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 24 Apr 2001 00:27:45 -0400 Subject: Re: Trapezoid with 3 equal sides Jim Totten wrote: > In a trapezoid with three equal sides, the square of the lesser > base plus the product of the bases equals the square of a > diagonal. Guy Brandenburg replied: > Appears to be true. But do you know why? One plus two is three. That also appears to be true, but I'm not sure I know "why". Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 24 Apr 2001 11:21:58 -0400 Subject: Re: Trapezoid with 3 equal sides Jim Totten wrote: > In a trapezoid with three equal sides, the square of the lesser > base plus the product of the bases equals the square of a > diagonal. Guy Brandenburg replied: > Appears to be true. But do you know why? Last night I wrote something snide because I didn't read the question carefully. I apologize, regret, and am sorry. One correct explanation is that a trapezoid with 3 equal sides has vertices that are equally spaced on a circle. So the bases are (r/2) sin x and (r/2) sin 3x = (r/2) (4 cos^2 x - 1) sin x and the diagonal is (r/2) sin 2x = (r/2) (2 sin x cos x). The result follows algebraically. There must be a more enlightening solution, though. Dan Hoey haoyuep@aol.com --- Newsgroups: sci.math From: haoyuep@aol.com (Dan Hoey) Date: 25 Apr 2001 07:33:50 -0400 Subject: Re: Matrix with integer entries Gil Liber asks: > A is an n by n matrix such that every entry of A is an integer m > in the range 1 <= m <= n and for every integer value m in this > interval there are exactly n entries of A equal to m. > How it can be proved that there is a row or a column of A with no > more than 2*sqrt(n) identical entries ? It's pretty easy to make such a matrix whose every row and column has at least ceiling(n/2) identical entries, by filling the matrix with a diagonal of n "L" shapes, e.g. for n=15: 15 15 15 15 15 15 15 8 9 10 11 12 13 14 15 14 14 14 14 14 14 7 8 9 10 11 12 13 14 14 13 13 13 13 13 6 7 8 9 10 11 12 13 13 13 12 12 12 12 5 6 7 8 9 10 11 12 12 12 12 11 11 11 4 5 6 7 8 9 10 11 11 11 11 11 10 10 3 4 5 6 7 8 9 10 10 10 10 10 10 9 2 3 4 5 6 7 8 9 9 9 9 9 9 9 1 2 3 4 5 6 7 8 8 8 8 8 8 8 8 1 2 3 4 5 6 7 7 7 7 7 7 7 7 15 1 2 3 4 5 6 6 6 6 6 6 6 6 14 15 1 2 3 4 5 5 5 5 5 5 5 5 13 14 15 1 2 3 4 4 4 4 4 4 4 4 12 13 14 15 1 2 3 3 3 3 3 3 3 3 11 12 13 14 15 1 2 2 2 2 2 2 2 2 10 11 12 13 14 15 1 1 1 1 1 1 1 1 9 10 11 12 13 14 15 Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 26 Apr 2001 12:42:55 -0400 Subject: Trapezoid with 3 equal sides The problem is: > > In a trapezoid with three equal sides, the square of the lesser > > base plus the product of the bases equals the square of a > > diagonal. Thanks, Jim, for providing the enlightening solution I was hoping for. Somehow the use of similar triangles always seems to make the argument neater. And the subtended/central angle step is very nice, too. But at the risk of quibbling, I'd like to correct the statement of the problem. Where you say "the square of the lesser base" you should refer to the base that is different from the sides-- perhaps the "odd", "unmatched", or "unique" base, or maybe there's a better adjective. In a trapezoid of sides 1,3,3,3, the square of the diagonal is 12, not 4. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 29 Apr 2001 13:43:30 -0400 Subject: Re: blocking sets in PG(3,4) Michael Toalster writes: > For some time I have been meaning to have another look at the > subject of my maths diploma, blocking sets in PG(d,q) for > finite d and q, and, especially, the question whether such > a set exists in PG(3,4). That sounds interesting, and I hope to understand it, but I'm somewhat hampered in a similar way. I was able to find out from the Internet that a blocking set is a set that meets every member of a family of sets, without containing all of any member. Unfortunately, I haven't been able to find an explicit definition of PG(3,4) online. I see it's a 3 dimensional projective space, which I imagine is a family of subsets of some set S, such that any three points of S lie in exactly one member of the family, and such that the intersection of any three members is a single point. At least I guess that's what it is. But beyond that, I'm not sure what the parameter (4) indicates. I'd appreciate a definition if anyone has it handy. I hope someone answers some of your questions about how amateur mathematicians may best search through the literature, since I am also in somewhat of that position. I suspect there may be some possibilities in the math archive (or arxiv?) that gets mentioned now and again on sci.math.research. Best of luck in your search, Dan --- Newsgroups: alt.radio.networks.npr From: haoyuep@aol.com (Dan Hoey) Date: 02 Jun 2001 15:37:46 GMT Subject: What has happened to npr.org? [Also e-mailed to yourt...@npr.org and ombuds...@npr.org] I just now went looking to replay an article I heard on Weekend Edition, because I thought it was so good I wanted to make sure I hadn't missed any details. To my amazement, the npr.org web page has been removed. Instead there are a bunch of selections about what some web designer thinks I might be interested in, but lacks any kind of index to help me find what I'm looking for. The only thing that looked like it would give me a choice about what I wanted to hear was a link to http://audible.com/npr/. So I clicked on it and it started up a new browser process (who said I wanted a new browser process?) and said my browser was incapable of Javascript (a lie, my browser will do Javascript just fine if I don't disable it by choice) and suggested that if I have an account I should log in (I'm never going to get an account; I refuse to have my web browsing monitored.) So far, I haven't gotten very close to finding the archives of Weekend Edition, and my opinion of your new Web design is wonderment at why you broke it. Did audible.com give you a lot of money for spying rights on your customer base? You have turned a valuable web resource into a corporate scam trap. I hope there are enough of us who can tell the difference that you are forced to put the old one back. A public apology for the inconvenience wouldn't be out of place, either, provided it occurs after you remove the inconvenience. Dan Hoey Washington DC The E is silent; my name rhymes with Boy. --- Newsgroups: alt.radio.networks.npr From: haoyuep@aol.com (Dan Hoey) Date: 03 Jun 2001 13:31:37 GMT Subject: Re: What has happened to npr.org? Paul L. Madarasz writes: >Hmm... No problem connecting with npr.org at 9.45 am here in >Arizona... Might have been some transitory glitch somewhere. I can connect to www.npr.org, too. But is there a link to Weekend Edition? Is there a link to the list of programs? Is there a link to the list of people inside NPR? To the external web pages of public radio shows that appear on NPR? The only link on www.npr.org that suggested being able to look up archives was a link to audible.com. Have you tried navigating audible.com? Have you found Weekend Edition archives there? Did you have to register (how much information do they want?). Did you have Javascript enabled? Dan Hoey haoyuep@aol.com --- Newsgroups: alt.radio.networks.npr From: haoyuep@aol.com (Dan Hoey) Date: 03 Jun 2001 13:41:04 GMT Subject: Re: (yourturn@npr.org automated reply) What has happened to npr.org? When I sent my message to "yourt...@npr.org" I got an automated reply. Here is my response: Hello, "NPR Web team", I ususally do not answer automated replies, but your reply was simply amazing. In it, you mentioned the URL http://www.npr.org/search/ which does not work, presumably because when you sold out your web site to audible.com they didn't want anyone to be able to get to the useful archives. I hesitate to admit that the other URL you mention, http://www.npr.org/members/ actually does work in the intended way, allowing access to the list of stations. I hope that my mentioning that this link works does not lead you to wreck it as well. Of course, you also list your now-useless home page, http://www.npr.org/. Now in the automated message you claim that > Due to the overwhelming amount of mail we receive, we cannot > personally respond to each email, but we DO read each message. Given the obsolete nature of much of the rest of the message, please favor me with a quick personal reply, only to assure me that the "NRL Web team" that signed the message actually does exist, and that there is still someone left to actually carry out your claim that you "DO read each message." I have no trust in your assurance that my email is being read, when the assurance comes from an automated process that could be left running long after the people have been sacked. If you prefer, you may post your assurance to the Usenet newsgroup "alt.radio.networks.npr", where the message will be open to the public. There are other messages in the thread "What has happened to npr.org?" that you may find interesting. Dan Hoey haoyuep@aol.com The "e" is silent; "Hoey" rhymes with "boy". --- Newsgroups: alt.radio.networks.npr From: haoyuep@aol.com (Dan Hoey) Date: 04 Jun 2001 15:59:18 GMT Subject: Re: What has happened to npr.org? Lee Hollaar writes: >Go to the box at the top that says "NPR Programs A-Z". >When you click on it, you'll get a menu.... Thanks. That's the way it works today. Yesterday the "NPR Programs A-Z" box was missing, which is what I was complaining about. I've gotten a nice letter from NPR Vice President MJ Bear, who explained that some things got misplaced in the redesign, and they've put the NPR Programs index back. Thanks again for your assistance, Lee. Dan Hoey posted and e-mailed haoyuep@aol.com P.S. Oh well, I see this thread has been invaded by the local axe-grinder. Pity. The price of free speach is interminable gibbering. Never mind. --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 18 Jun 2001 13:14:20 -0400 Subject: Re: dissection > On Mon, 18 Jun 2001, Darvasi Gyula wrote: > > Dear Sir, > > do you know a general dissection proof the mean property of the > > altitude to the hypotenuse of the right triangle: h^2=p*q with > > usual notations. > > Thank you verymuch. > > Gyula Darvasi Perhaps you'll find the following pair of dissections of the right triangle satisfactory. ._______ ...|\ h | .......| \ | _ - |.......|h \ | __-_____|.......|___\___| p-h q .____. | | | | | | | |p _ - |\ | __-_____| \ | .......p-h..|h \ | ................|___\| q In each case the dotted part of the triangle is removed and rotated to abut the smaller base. Dan haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 6 Jul 2001 14:36:26 -0400 Subject: Re: exact values of sines Andrew Weimholt writes: > ...I believe that some rational values of a/b in > sin((a/b)*pi) still yield transcendental numbers. > Someone correct me if I'm wrong on this. I will; you are. In fact, 2 sin((a/b)*pi) is an algebraic integer. First, note that z=exp(2*pi*i*(a/b)) satisfies z^b=1, so z is an algebraic integer. Algebraic integers are closed under conjugation, addition, and multiplication, so 2 sin((a/b)*pi) = (-i * (z + zbar)) is an algebraic integer. I don't recall off-hand how to find polynomials for sums and products, given polynomials for their operands, but I seem to recall it's mechanical. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 16 Jul 2001 13:50:54 -0400 Subject: Re: exact value of sin(pi/7) Andrew Weimholt writes: > You probably weren't dreaming. Eric Weistein's "CRC Concise > Encyclopedia of Mathematics" gives several exact values for > sin(pi*a/b) including sin(pi/7). Unfortunately the text is > no longer available online, and I don't have my copy of the > book with me. I recall it was a fairly complicated expression > cointaining nested roots. I believe the text also showed the > derivation. Well, there's ((-1)^(5/14) - (-1)^(9/14))/2, but that seems like cheating to me. I wonder if there's a solution in real roots. If you find your treasure trove, please let us know what it said. Dan haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 18 Jul 2001 01:54:12 -0400 Subject: Re: exact value of sin(pi/7) Randall L. Rathbun wrote: > sin(Pi/n) = ( (-1)^((n-2)/2n) - (-1)^((n+2)/2n) )/2 > so we're definitely on to something here. Where does this > relationship show up? It definitely works. Sorry, I didn't mean to be mysterious. First, note that sin(pi - x) = sin(x) = cos(pi/2 - x) = -cos(pi/2 + x) for all x, which can be proved in any number of ways. So 2 sin(x) = cos(pi/2 - x) - cos(pi/2 + x) = (cos(pi/2 - x) + i sin(pi/2 - x)) - (cos(pi/2 + x) + i sin(pi/2 + x)) = (-1)^(1/2 - x/pi) - (-1)^(1/2 + x/pi) from Euler's formula. The reason it's cheating is that calculating (-1)^(5/14) and (-1)^(9/14) is essentially the same as calculating the sine and cosine of pi/7. On the other hand, the formula from Kevin Brown's web site involves only taking complex cube roots (and square roots) so it's like trisecting and bisecting angles. It's surprising it can be done so easily. But see the Conway and Guy's _Book of Numbers_ for constructions of the regular heptagon and 13-gon with trisector, compass, and straightedge. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 18 Jul 2001 11:49:20 -0400 Subject: Re: exact value of sin(pi/7) Randall L. Rathbun wrote: > Dan Hoey wrote: > > Well, there's ((-1)^(5/14) - (-1)^(9/14))/2, but that > > seems like cheating to me. [...] > sin(Pi/n) = ( (-1)^((n-2)/2n) - (-1)^((n+2)/2n) )/2 > so we're definitely on to something here. Where does this > relationship show up? It definitely works. [ Sorry if this shows up twice. forum.swarthmore seems to have swallowed the copy I posted last night. ] Sorry, I didn't mean to be mysterious. To see why this works first note that sin(pi-x) = sin(x) = cos(pi/2 - x) = -cos(pi/2 + x) for all x. So 2 sin(x) = cos(pi/2 - x) - cos(pi/2 + x) = (cos(pi/2 - x) + i sin(pi/2 - x)) - (cos(pi/2 + x) + i sin(pi/2 + x)) = (-1)^(1/2 + x/pi) - (-1)^(1/2 - x/pi) by Euler's formula (since exp(i pi)=-1). This is why I meant it was cheating--taking fourteenth roots of -1 is essentially the same as computing the sine and cosine of pi/7. On the other hand, Kevin Brown's page (referenced elsewhere in what should have been this same thread) shows an interesting way of calculating sin(pi/7) using complex cube roots and square roots, which is tantamount to using only trisection and bisection to construct an angle of pi/7. I still think it's surprising that it can be done, though my surprise is probably due to the superficiality of my understanding. See also Conway and Guy's _The Book of Numbers_, which gives constructions of the heptagon and 13-gon using trisector, compass, and straightedge. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 2 Aug 2001 17:07:21 -0400 Subject: Re: systems of pivoting lines John Steinberger writes: > Let's define a 'system of n pivoting lines' like so: > ...(Got me? the axis lines serve as tracks > for the intersections of the pivot lines). Well, I'll explain it the way I got it, with some alterations that make it a little easier to work with (provided I got it right in the first place.) My alterations mostly consist of adding an axis line in the beginning, and ignoring the pivot lines except as a constraint on the linkage points. In the plane we are given a set of n axis lines {a_1, ..., a_n} and a set of n pivot points {p_1, ..., p_n}. With respect to this system, we define a linkage set to be a set of points {q_1, ..., q_n} such that Point q_i lies on line a_i, and Points q_i, p_i, q_(i+1) are collinear. We want to express q_n as a function of q_1. When point q is on line a and point p is not, I define the parameter of q with respect to a, p as Q(q,a,p) = tan(angle "n p q"), where n is the point on a closest to p. I choose a sign for the angle so that Q is a linear map from the line a onto the reals--this is equivalent to choosing a direction for line a. Let x = Q(q_1,a_1,p_1) be the parameter for the first point of a linkage set; we want to compute f(x) = Q(q_n,a_n,p_n). Let's look at the first step. We can choose a direction for line a_2 so that Q(q_2,a_2,p_1) = tan(c_1 + Atan(x)), where c_1 is the angle "n p_1 m", with n and m the closest points to p_1 on lines a_1 and a_2, respectively. Then because two linear functions on a line are linearly related, we have Q(q_2,a_2,p_2) = d_1 Q(q_2,a_2,p_1) + e_1 for some real constants d_1, e_1. So Q(q_2,a_2,p_2) = d_1 tan(c_1 + Atan(x)) + e_1 x (d_1 cos(c_1) - e_1 sin(c_1)) + d_1 sin(c_1) + e_1 cos(c_1) = --------------------------------------------------------------- -x sin(c_1) + cos(c_1) which is a linear fractional transform (and in the case where lines a_1 and a_2 are equidistant from p_1, the determinant is 1. It is possible that I am abusing the term LFT when the determinant is not 1, but if so I don't know the right word for a general function of the form f(x) = (A x + b) / (C x + d) ). We can continue choosing directions for each a_i in turn to produce an LFT for each Q(q_i,a_i,p_i) with respect to Q(q_(i-1),a_(i-1),p_(i-1)). The LFT's for each step can be composed by matrix multiplication to produce a LFT for the whole system, so that Q(q_n,a_n,p_n) = (A x + B)/(C x + D) for some real constants A, B, C, D. I hope this answers your question in enough detail to be useful. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 3 Aug 2001 14:18:00 -0400 Subject: Re: more questions First, the silliness of my problem with the determinant came to me last night, so let me fix that. Unless a function f(x) = (ax + c) / (bx + d) is constant, it will have a nonzero determinant ad - bc . If the determinant is not equal to 1 , we can take it to be 1 by dividing through by sqrt(ad - bc) . I'm not sure what people usually do if this ends up with imaginary numbers--if you don't like them you can leave the determinant -1 . But in our case we can choose the direction of the axis lines a_i so that our determinants are all positive. Then the transfer function from x = Q(q,a,p) to f(x) = Q(q',a',p') is (r cos(t) - s/r sin(t)) x + (r sin(t) + s/r cos(t)) f(x) = --------------------------------------------------- (-sin(t)/r) x + (cos(t)/r) where t is the angle "n p n'" between the closest points to p on lines a and a' (or maybe supplement of that angle?), r^2 is the ratio d(p,m) / d(p',n') between the distances of p and p' to their closest points on a' , and s is something like the tangent of angle "p n' m". Second, you'll note I've reordered the coefficients above. That's so that so that we can multiply matrices left-to-right for function composition. To be explicit, if f(x) = (ax+c)/(bx+d) and g(x) = (px+r)/(qx+s) , then g(f(x)) = ((ap+br)x + (cp+dr)) / ((aq+bs)x + (cq+ds)) . This corresponds to the matrix multiplication [ a b ] [ p q ] [ ap+br aq+bs ] [ ] [ ] = [ ] . [ c d ] [ r s ] [ cp+dr cq+ds ] The other parameterization goes with right-to-left composition. John Steinberger asks: > (I in fact suffered from a severe lack of trigonometric > functions manipulations in high school and would be grateful > if you showed me how you took tan(c + atan(b)) ). Sure, I have sympathy because I had so much non-trigonometry in college that I forgot all my high school trig except tan(x) = sin(x)/cos(x) , sin(x + y) = sin(x)cos(y) + cos(x)sin(y) , cos(x + y) = cos(x)cos(y) - sin(x)sin(y) , and of course sin(x)^2 + cos(x)^2 = 1 . But you only need the first three of these to do tan(c + atan(b)) = (sin(c + atan(b))) / (cos(c + atan(b))) sin(c)cos(atan(b)) + cos(c)sin(atan(b)) = --------------------------------------- cos(c)cos(atan(b)) - sin(c)sin(atan(b)) sin(c) + cos(c)tan(atan(b)) = --------------------------- cos(c) - sin(c)tan(atan(b)) = (sin(c) + b cos(c)) / (cos(c) - b sin(c)) . > One thing that's interesting is to know when the LFT will > boil down to a linear function, i.e. when c_1 == 0, i.e. two > axes are [parallel]. Are there cases when not all the axis > lines are [parallel] and the LFT boils down to just a linear > function? That's something I want to ponder a bit more. I > guess it's equivalent to asking: when multiplying a series of > 2-2 matrices, under what conditions do we obtain a zero in > the lower left position? ... Well, given any nonpathological linkage, you get an LFT with nonzero determinant, so you can calculate the inverse function by inverting the matrix. You can then figure out what LFT you need to append to get any LFT you want out of the system. I think you can construct a linkage for any nonpathological f(x)=(ax+c)/(bx+d) of the right sign by taking r = 1/sqrt(b^2 + d^2) , s = r sqrt(a^2 + c^2) - r^2 , and t = atan(-b/d) . I think this method will get most of your answers, but for > Another question is about the angles: what are the values for > A, B, C, D such that atan((A x + B)/(C x + D)) is a linear > function of atan(x)? where I really have no idea. I somewhat doubt if there is any such LFT other than the identity. You may have to trade in your levers for some gears. Dan --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 4 Aug 2001 20:19:34 -0400 Subject: Re: ellipse or parabola? Hop David wrote: > Imagine a boy throwing a rock. What looks like a > parabolic trajectory is actually part of a very > eccentric ellipse with the earth's center of gravity > at the far focus. Yes, as long as we assume an earth of uniform (or sufficiently symmetric) density, so that we can model the gravity field as attraction to a point having all of the earth's mass. This model is only accurate outside the earth's surface; beneath the surface, the point must be modeled as shedding all the mass at a greater elevation than the ball. (I have been told that the density irregularities of the actual Earth swamp the difference between ellipse and parabola, but I haven't verified that.) > If the boy throws straight up, the "ellipse" has > eccentricity 1. It's a line segment whose endpoints > are the apex of the throw and the earth's center of > gravity. See model failure above, but let's imagine a point mass at the earth's center that swings the ball around in an orbit whose cometariness approaches ever-so-much. > I'd be inclined to call this a parabola, but it has 2 > foci - the endpoints. It's major axis is also of > finite length which goes against my notion of a parabola. > OTOH, take any point p and d(p,f1) + d(p,f2) = d(f1,f2) > which seems to fit the definition of ellipse. It's an ellipse--the only mystery I see your inclination. A parabola is modeled by a uniform gravity field, and so a degenerate parabola is a ray--as if the ball just keeps falling to infinity. A line segment is a degenerate ellipse (and a line minus a segment is a degenerate hyperbola). > side note: taking the limit of these trajectories it > seems weird to me that the end point would be the earth's > center of gravity, it _seems_ like the rock would go > zooming _past_ the center of gravity (assuming an > ideal point source of gravity and a rock that passes > through the earth's matter). With those assumptions, the end point would approach the center of gravity as eccentricity increased. What actually happens when a theoretical point mass falls theoretically straight into another theoretical point mass depends on the axioms of your theory. > I guess the rock would approach infinite acceleration as > it neared the gravity point source. Yes (if you can axiomatize the behavior in some consistent way), there would be infinite acceleration away from the boy, then twice as much infinite acceleration back towards the boy, then the original amount of infinite deceleration as the rock starts back up, so that at all noninfinitesimal distances from the center the rock has finite speed. There's a slight problem of where the infinite _angular_ acceleration happens if the rock never goes sideways, but that's one of the problems with these proposed but never axiomatized schemes that never seem to deal with degenerate conditions in any completely consistent way. > It's an image that works better in a Newtonian universe. I think you need more than Newton to deal with this. It's one of those things that seems reasonable over a few beers, but doesn't work too well on paper. Malt does than Newton can, with apologies to Housman (and Milton). Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.research From: haoyuep@aol.com (Dan Hoey) Date: 9 Aug 2001 09:20:53 -0400 Subject: Re: Stick knot Steve Gray writes: > Question: How to characterize algebraically > an ordered set of 6 points (XiYiZi), 0<=i<=5, in > 3-D such that when connected with straight segments > the resulting polygon is knotted (that is, cannot ^^^^^^^ > be continuously deformed into a convex plane > polygon. (Only "overhand" knots are possible with > six segments.) I wouldn't call a non-planar path a polygon. Let's at least use scare-quotes. I am not sure of the answer, but I have a plausible possibility that may turn out to be true. If you calculate the volume of a tetrahedron given the coordinates of its four vertices, you get the absolute value | S(A,B,C,D) | of a cubic polynomial; I call S(A,B,C,D) itself the "signed volume" of the tetrahedron. Signed volume is negated by exchanging two vertices or any other opposite isometry, and preserved by direct isometries. I'm virtually certain that the knot type of a "polgon" defined by a sequence of n points is determined by the signs of the (n choose 4) tetrahedra formed from its vertices. In the case of a six-point "hexagon" ABCDEF, I suspect you get an overhand knot just when S(A,B,C,D), S(E,B,C,D), S(E,F,C,D), S(E,F,A,D), S(E,F,A,B), and S(C,F,A,B) are either all positive or all negative. I expect this (or the true answer) is a classical result that any real knot theorist could tell you immediately. If you want to look up an answer you can trust, I'd look for a text that deals with knots on simplicial complexes. Dan Hoey haoyuep@aol.com --- From Hoey@aic.nrl.navy.mil Wed Aug 15 14:20:50 2001 Date: Wed, 15 Aug 2001 14:22:08 -0400 (EDT) From: Dan Hoey To: Keith Lynch Cc: Bob MacIntosh <[redacted]> Subject: My E-mail address Keith, I think this is yours. Please remove my e-mail address from the CapClave web page http:/www.wsfa.org/capc01/index.htm Thanks. Dan --- From Hoey@aic.nrl.navy.mil Thu Aug 16 08:18:18 2001 Date: Thu, 16 Aug 2001 08:19:48 -0400 (EDT) From: Dan Hoey To: "Keith F. Lynch" Cc: [redacted] Subject: Re: My E-mail address "Keith F. Lynch" wrote: > > Please remove my e-mail address from the CapClave web page > > http:/www.wsfa.org/capc01/index.htm > Done. > What shall I put in its place? The lack of an e-mail address, just like you have it. Thanks for the quick change. > See you Friday. See you tomorrow night. Dan P.S. I was prompted to check for my address on the page by receiving the enclosed message. I'm forwarding it to you (in spite of its amusing prohibition) in case either of you didn't get your own copy and want to respond. Cheers! ================================================================ From: Trisha.Tunis@hartfordlife.com Subject: Shipping Freebie Flyers? To: Don't nobody give their right name Date: Wed, 15 Aug 2001 12:23:30 -0400 Greetings, I was just wondering if you were planning on having a freebie table at CAPCLAVE? And if so, where and to who's attention I could send a batch of flyers to display there? Thanks! Trisha Tunis Minister of Propaganda RSE International, Inc. http://www.rsempire.org ************************************************************************* PRIVILEGED AND CONFIDENTIAL: This communication, including attachments, is for the exclusive use of addressee and may contain proprietary, confidential and/or privileged information. If you are not the intended recipient, any use, copying, disclosure, dissemination or distribution is strictly prohibited. If you are not the intended recipient, please notify the sender immediately by return e-mail, delete this communication and destroy all copies. ************************************************************************* ================================================================ --- Newsgroups: rec.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 18 Oct 2001 00:56:18 GMT Subject: Re: The man in the park Mark Brader writes: > Quoted text is spoiler space. Jonathan Rivet writes: >> You are walking through the park one day when a man you >> have never met before comes up to you and shows you a >> folded up piece of paper. >> "I have written on this piece of paper a number, an integer, from >> 0 to 119," he says. "For $10, I will allow you to choose one of the >> groups I am about to list; if my number is part of the group you >> choose, I will give you $30 back. If not, I will keep your money. >> Each group contains 60 of the 120 numbers from 0 to 119." >> You figure a 50 percent chance at earning $20 is worth possibly >> parting with the 10-dollar bill burning a hole in your pocket, so >> you accept. He lists the groups as follows: >> 1. The numbers from 0-59 >> 2. The numbers from 60-119 >> 3. Even numbers >> 4. Odd numbers >> 5. Numbers which have a remainder of 0 or 1 when divided by 4 >> 6. Numbers which have a remainder of 2 or 3 when divided by 4 >> 7. Numbers which have a remainder of 0, 1, or 2 when >> divided by 6 >> 8. Numbers which have a remainder of 3, 4, or 5 when >> divided by 6 >> 9. Numbers which have a remainder of 0, 1, 2, or 3 when >> divided by 8 >> 10. Numbers which have a remainder of 4, 5, 6, or 7 when >> divided by 8 >> 11. Numbers which have a remainder of 0, 1, 2, 3, or 4 when >> divided by 10 >> 12. Numbers which have a remainder of 5, 6, 7, 8, or 9 when >> divided by 10 >> So. Which group do you choose? > If I'm going to play this game, I first have to assume that if > I do win he's actually going to pay off honestly, which already > seems an unlikely bet to me, but I'll go with it for purposes > of the puzzle. Beyond that, I have to that it's a scam based > on the man's having written a number in the permitted range > which he present to me as a different number in the range by > inverting the paper. That's a really neat idea, and I'll take it as the starting point for an analysis of the payoff. The situation has the stranger making a choice of what to write on the paper, and you picking one of the twelve subsets, and then you reveal your choices to each other determining a zero-sum payoff. This is a classical game theory problem, so we should be able to find optimal mixed strategies for the two players. As the stranger told you, you have twelve choices of subset to pick. The stranger wants you to believe he has 120 choices, but he actually has 113 choices--106 unambiguous numbers, and 7 pairs that he can manipulate, possibly to avoid the subset you pick. I first looked at the simpler game where he always picks one of the seven ambiguous pairs; by eliminating dominated strategies I got him choosing between 6/9, 16/91, and 19/61 while you choose subset 7, 9, or 12. His fourth possibility of choosing something in none of subsets 7, 9, nor 12 gives him the edge. Here are the mixed strategies I came up with. The Stranger's Strategy The stranger can guarantee that he wins 3/4 of the time by adopting the strategy of writing: 6 (or 9) 1/4 of the time, 16 (or 91) 1/4 of the time, 19 (or 61) 1/4 of the time, and 100 1/4 of the time. If you choose subset 1 2 3 4 5 6, then he wins unless he wrote 6/9 100 100 19/61 100 19/61; If you choose subset 7 8 9 10 11 12, then he wins unless he wrote 19/61 100 16/91 100 100 6/9. Your Strategy You can ensure that you win at least 1/4 of the time by choosing subset: 7 1/4 of the time, 8 1/12 of the time, 9 1/4 of the time, 10 1/12 of the time, 11 1/12 of the time, and 12 1/4 of the time. If he wrote 6/9 16/91 18/81 19/61 You win if you chose subset 12 9 9 or 12 7; If he wrote 66/99 68/89 86/98 You win if you chose subset 9 9 7 or 11; If he chose a single number in the intersection of sets 8, 10, and 11, you win 1/4 of the time, and if he chose any other single number you win more often. Summary Given the two strategies, the game-theoretic probability of you winning the game is 1/4. In 1/4 of the games you win $20, and in 3/4 of the games you lose $10, so on average you lose $2.50 a game. That seems a fairly cheap lesson in game theory. Dan Hoey posted and e-mailed haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: Wed, 24 Oct 2001 04:11:36 +0000 (UTC) Subject: Re: Carts of Coal Barry R. Clarke wrote: > In Victorian London, there was a coal yard which customers would > visit with their horse and cart. The yard sold two different size > bags of coal, one being a two-digit number ab (i.e. 10a+b) times > the weight of the other. One day, a trader turned up with his horse > Neddy and asked for a number c of small bags and a number d of > large bags to make the number e of cwt (hundred-weight) that his > horse could pull. Shortly afterwards, another trader appeared with > his horse Dobbin and asked for a number f of small bags and a > number g of large bags to total the number h of cwt that Dobbin > could pull. Each of a,b,c,d,e,f,g,h was a single digit, no two > digits being equal and none of them being zero or one. What were > the weights of the large and small bags? If s is the weight of a small bag in cwt, we are given s(c + d(10a+b)) = e and s(f + g(10a+b)) = h. Solving both equations for s, we get e/(c + d(10a+b)) = h/(f + g(10a+b)) which can be rearranged to form ef-hc = (hd-eg)(10a+b) . At this point, we could write a computer program to try all 40320 assignments of {2,3,...,9} to {a,b,...,h}, but such a method would be an anachronism in Victorian London. So let us examine the possibilities by hand. Table 1 lists the products achievable by multiplying two different numbers from {2,3,...,9} in order. Table 1 : products of two different numbers from {2,3,...,9} 6 = 2x3 15 = 3x5 24 = 3x8 = 4x6 35 = 5x7 48 = 6x8 8 = 2x4 16 = 2x8 27 = 3x9 36 = 4x9 54 = 6x9 10 = 2x5 18 = 2x9 = 3x6 28 = 4x7 40 = 5x8 56 = 7x8 12 = 2x6 = 3x4 20 = 4x5 30 = 5x6 42 = 6x7 63 = 7x9 14 = 2x7 21 = 3x7 32 = 4x8 45 = 5x9 72 = 8x9 If the two-term expressions ef-hc and hd-eg were both zero, then the products ef=hc and hd=eg would have to come from the three duplicate factorizations 2x6=3x4, 2x9=3x6, and 3x8=4x6. Then one of the terms would be 3x6 or 4x6, and its two factors would appear in separate terms of the other two-term expression, which is not possible. So none of the terms in ef-hc = (hd-eg)(10a+b) is zero. 10a+b is at least 10x2+3 = 23 and |ef-hc| is at most 8x9-2x3 = 66, so |hd-eg| is either 1 or 2. From examining nearby entries in Table 1, we see this can arise in 14 ways, as listed in the first column of Table 2. The choices of {h,d,e,g} restrict our choices for {a,b}, so there is a new lower bound for 10a+b in the second column. The third column has upper bounds for |ef-hc| (remembering that ef-hc and hd-eg have the same sign). The eight ways eliminated by these bounds are marked in the fourth column. Table 2 : small values for |hd-eg| |hd-eg| min(10a+b) max(|ef-hc|) 3x4-2x5 = 2 67 9x5-6x3 = 27 no 2x7-3x4 = 2 56 9x4-5x2 = 26 no 3x5-2x7 = 1 46 9x7-4x3 = 51 2x8-3x5 = 1 46 9x5-4x2 = 37 no 3x6-2x8 = 2 45 9x8-4x3 = 60 no 4x5-2x9 = 2 36 8x9-3x4 = 60 no 4x5-3x6 = 2 27 9x6-2x4 = 46 no 3x7-4x5 = 2 26 9x5-2x3 = 39 no 4x7-3x9 = 1 25 8x9-2x4 = 64 5x6-4x7 = 2 23 9x7-2x5 = 53 4x8-5x6 = 2 23 9x6-2x4 = 46 4x9-5x7 = 1 23 8x7-2x4 = 48 6x7-5x8 = 2 23 9x8-2x6 = 60 7x8-6x9 = 2 23 5x9-2x7 = 31 no For the remaining six possibilities, we try all possibilities for |ef-hc| down to min(10a+b) |hd-eg|, and list {a,b} as the remaining two available digits: |hd-eg| min(|ef-hc|) |ef-hc| {a,b} 3x5-2x7 = 1 46 9x7-4x3 = 51 {6,8} 4x7-3x9 = 1 25 8x9-2x4 = 64 {5,6} 8x9-2x7 = 58 {5,6} 8x9-5x4 = 52 {2,6} 8x9-5x7 = 37 {2,6} 8x9-6x4 = 48 {2,5} 8x9-6x7 = 30 {2,5} 6x9-2x4 = 46 {5,8} 6x9-2x7 = 40 {5,8} 6x9-5x4 = 34 {2,8} 5x9-2x4 = 37 {6,8} 5x9-2x7 = 31 {6,8} 5x6-4x7 = 2 46 9x7-2x5 = 53 {3,8} 9x7-2x6 = 51 {3,8} 9x7-3x5 = 48 {2,8} 8x7-2x5 = 46 {2,9} 4x8-5x6 = 2 46 9x6-2x4 = 46 {3,7} 4x9-5x7 = 1 23 8x7-2x4 = 48 {3,6} 8x7-2x9 = 38 {3,6} 8x5-2x4 = 32 {3,6} 8x7-3x4 = 44 {2,6} 8x7-3x9 = 29 {2,6} 8x5-3x4 = 28 {2,6} 8x7-6x4 = 32 {2,3} * 6x7-2x4 = 34 {3,8} 6x7-2x9 = 24 {3,8} 6x7-3x4 = 30 {2,8} 6x7-5x8 = 2 46 9x8-2x6 = 60 {3,4} 9x8-2x7 = 58 {3,4} 9x8-3x6 = 54 {2,4} 9x8-3x7 = 51 {2,4} 9x8-4x6 = 48 {2,3} The only solution is that 10a+b=32. A small bag is 1/42 cwt and a large bag is 16/21 cwt. One horse pulled 7 cwt in 6 small bags and 9 large bags. The other horse pulled 4 cwt in 8 small bags and 5 large bags. Dan Hoey haoyuep@aol.com --- Newsgroups: rec.puzzles From: haoyuep@aol.com (Dan Hoey) Date: 16 Nov 2001 13:40:17 GMT Subject: Re: Painting a cube Nick Wedd n...@maproom.co.uk writes: [a very nice case analysis of painting a cube with three colors]. But earlier, you wrote: > >What about if you have n colors, instead of three? > 27*Cn,2 + 30*Cn,3 + 80*Cn,4 + 60*Cn,5 + 30*Cn,6 > or, if mirror images are considered the same, > 27*Cn,2 + 29*Cn,3 + 52*Cn,4 + 30*Cn,5 + 15*Cn,6 That doesn't even agree with the three-color case, though maybe you meant 9Cn,2. But that's still wrong, and there are 3 other errors. I'll give you my analysis (with no gratutitous whitespace*). The answer as a polynomial is almost immediate from the Cauchy-Frobenius aka Polya-(not)Burnside theorem. For each of the 24 rotations of the cube (or the 48 rotations and reflections) each cycle of faces must be monochrome, so it contributes n^c/|G| where n is the number of colors, c is the number of cycles, and |G| is the size of the group (24 or 48). So we add up representative number of contribution contribution rotation/ equivalents counting including reflection (conjugates) rotations reflections (B)(F)(T)(D)(L)(R) 1 n^6/24 n^6/48 (B,F)(T,D)(L)(R) 3 3 n^4/24 3 n^4/48 (B,T,L)(F,D,R) 8 8 n^2/24 8 n^2/48 (B,F)(T,L)(D,R) 6 6 n^3/24 6 n^3/48 (B,T,F,D)(L)(R) 6 6 n^3/24 6 n^3/48 (B,F)(T,D)(L,R) 1 n^3/48 (B,F)(T)(D)(L)(R) 3 3 n^5/48 (B,T,L,F,D,R) 8 8 n /48 (B,T)(F,D)(L)(R) 6 6 n^4/48 (B,T,F,D)(L)(R) 6 6 n^6/48 so there are (n^6 + 3n^4 + 12n^3 + 8n^2)/24 paintings up to rotation, or (n^6 + 3n^5 + 9n^4 + 13n^3 + 14n^2 + 8n)/48 up to rotations and reflections. Converting to binomial coefficents may be done by dividing out by their polynomial representations Cn,1 = n Cn,2 = (n^2 - n)/2 Cn,3 = (n^3 - 3n^2 + 2n)/6 Cn,4 = (n^4 - 6n^3 + 11n^2 - 6n)/24 Cn,5 = (n^5 - 10n^4 + 35n^3 - 50n^2 + 24n)/120 Cn,6 = (n^6 - 15n^5 + 85n^4 - 225n^3 + 274n^2 - 120n)/720 to get 30Cn,6 + 75Cn,5 + 68Cn,4 + 30Cn,3 + 8Cn,2 + Cn,1 counting only rotations, or 15Cn,6 + 45Cn,5 + 52Cn,4 + 29Cn,3 + 8Cn,2 + Cn,1 including reflection. Alternatively, we can generalize your case analysis: Color counts subcase arrangements (rot or rot+ref) 6 Cn,1 5 1 2Cn,2 4 2 ad 2Cn,2 4 2 op 2Cn,2 3 3 clump Cn,2 3 3 strip Cn,2 4 1 1 ad 3Cn,3 4 1 1 op 3Cn,3 3 2 1 clump 6Cn,3 3 2 1 1 midstrip 6Cn,3 3 2 1 1 end strip 6Cn,3 2 2 2 op op op Cn,3 2 2 2 op ad ad 3Cn,3 2 2 2 ad ad ad 2Cn,3 or Cn,3 3 1 1 1 clump 8Cn,4 or 4Cn,4 3 1 1 1 strip 12Cn,4 2 2 1 1 op op op 6Cn,4 2 2 1 1 op ad ad 12Cn,4 2 2 1 1 ad ad op 6Cn,4 2 2 1 1 ad ad ad 24Cn,4 or 12Cn,4 2 1 1 1 1 op 15Cn,5 2 1 1 1 1 ad 60Cn,5 or 30Cn,5 1 1 1 1 1 1 30Cn,6 or 15Cn,6 which yields the same result. Dan Hoey Posted and -emailed haoyuep@aol.com * Why? Because I think it's silly. --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: Sat, 17 Nov 2001 17:23:51 +0000 (UTC) Subject: Re: equilateral triangle Rouben Rostamian wrote: > The story so far: > > manuel wrote: > > I need an hint for this problem: > > Considerer an equilateral triangle ABC and a point P on the plane. > > Prove that |AP|+|BP|>=|CP|. > > Can we have the equality? [...] > Lemma: Consider an equilateral triangle ABC and let P be any point > on the arc AB of its circumcircle. Then PA + PB = PC. [...] > See if you can find a better proof. I hope this counts. Remember that 2 r sin (x/2) is the length of a chord subtending an angle x in a circle of radius r. Let the circumcircle have radius 1/2 and let 2x be the angle POA, so angles POB = 2pi/3 - 2x and POC = 2pi/3 + 2x. Then AP = sin x = 2 cos (pi/3) sin x because cos (pi/3) = 1/2. Then BP = sin (pi/3 - x) = sin (pi/3) cos x - cos (pi/3) sin x, CP = sin (pi/3 + x) = sin (pi/3) cos x + cos (pi/3) sin x = BP + AP, proving the lemma. When I wrote that, I hadn't got a proof that equality can't hold elsewhere, but I noticed that the triangle inequality implies that equality can only hold in the trianguloid max(AP,BP) <= AB <= CP . > Remark: None of the observations above solve Manuel's original > question on whether PA + PB >= PC in general. I believe that this > assertion is correct but I haven't found a proof. Eureka! (I've always wanted to say that.) Actually, Manuel's original question asks for a hint. Here's a hint: What is odd about the exponents of the polynomial (r+s+t)(-r+s+t)(r-s+t)(r+s-t) ? Application of this hint has enabled me to prove the inequality algebraically, but I'd certainly like a geometric proof. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: Tue, 20 Nov 2001 14:58:06 +0000 (UTC) Subject: Re: equilateral triangle > > manuel wrote: > > I need an hint for this problem: > > Considerer an equilateral triangle ABC and a point P on the plane. > > Prove that |AP|+|BP|>=|CP|. > > Can we have the equality? I hinted: > What is odd about the exponents of the polynomial > (r+s+t)(-r+s+t)(r-s+t)(r+s-t) ? > Application of this hint has enabled me to prove the inequality > algebraically, but I'd certainly like a geometric proof. Well, darn it, I knew where that polynomial came from, but I didn't see until now how it will yield a geometric proof. Try this: Erect equilateral triangle AA'P (resp BB'P) in the half-plane of AP (BP) that includes the ray PC. (If P is outside the circumcircle, the erected triangles will overlap.) Prove that CA'PB' is a parallelogram. I've never been able to prove these Napoleonic things geometrically, but I know they can be done in principle. I hope it's easier than phonying up axes to simulate the (easy) algebraic proof. Does anyone know the provenance or any history of the problem? Dan Hoey haoyuep@aol.com --- Newsgroups: sci.math From: haoyuep@aol.com (Dan Hoey) Date: Fri, 23 Nov 2001 01:58:26 +0000 (UTC) Subject: Re: Equilateral triangle problem Thanks to all on who discussed the problem on geometry.puzzles, especially to Dan Ross who called my attention to the discussion here on sci.math. I'm very glad to be reminded of the big hammer that can be used to derive the inequality from the connectedness of the complement of {P : CP = AP + BP} . (As David Eppstein implies, path-connectedness is not needed.) In case anyone finds a direct geometric proof of the inequality enlightening, here's one I came up with (though I'm sure it's been done before). Let ABC be an equilateral triangle and P a point in the plane. Prove: AP + BP >= CP. Proof: Erect equilateral triangle APD, with the angles = CD + DP = BP + AP, QED. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: Fri, 23 Nov 2001 02:14:35 +0000 (UTC) Subject: Re: equilateral triangle (Spoiler) Thanks to all on geometry.puzzles, and to Andrei Ismail on sci.math who identified this as the "Van Schooten theorem," though I'm not quite sure whether that name refers to the inequality as well as the equality. Here's my simple geometric proof of the inequality (though I'm sure it's been done before). Let ABC be an equilateral triangle and P a point in the plane. Prove: AP + BP >= CP. Proof: Erect equilateral triangle APD, with the angles = CD + DP = BP + AP, QED. Dan Hoey haoyuep@aol.com --- Newsgroups: geometry.puzzles From: haoyuep@aol.com (Dan Hoey) Date: Wed, 5 Dec 2001 18:12:01 +0000 (UTC) Subject: Re: Reply to tangent Rouben Rostamian wrote: > "Javier Orman" wrote: > > Does anyone know the geometric (not analytical) definition > > of tangent to a curve? [...] > As I wrote somewhere else in this thread, a non-analytic > definition will have a very hard time coping with a complicated > curve [....] I think you have to essentially duplicate the analytic version in geometry. For instance, let C be a curve in the plane containing point P. For any positive number N, let us define an N-rectangle of C at P to be a rectangle QRST with center P, for which C intersects triangles PQR and PST only at P, and for which segment QR is N times as long as segment RS. Then a line L is said to be tangent to C at P when for every N there is an N-rectangle QRST of C at P for which L intersects segments RS and QT. It is amusing that the proposition that P is on L is now a theorem rather than part of the definition. Dan Hoey haoyuep@aol.com --- Newsgroups: sci.math.research From: Dan Hoey Date: 11 Dec 01 13:08:06 -0500 (EST) Subject: Re: Constant-area tetrahedron Robert Israel writes: [...] > The intersection with z=0 is > |y| + |b x - a y| + |b x + (1-a) y - b| = g. > If g > b this will be the equation of a hexagon: on each ray > extending a side of the triangle in one direction there will be a > vertex. and the sides of the hexagon will be parallel to the sides of the triangle. Geometrically, this is easy. The area of a tetrahedron that has been degenerated into the plane is twice the area of the convex hull of its vertices. So we are keeping the convex hull of a finite set of points constant, varying one of them. The variable point P moves keeping the area of a triangle PAB constant, where A and B are the two extreme points of the hull of the other points as seen from P. So P moves parallel to AB as long as PA and PB don't go through the interior of the hull. Dan Hoey haoyuep@aol.com ---