September 2011 - mathfun@mailman.xmission.com

[math-fun] NASA Discovers 'Star Wars'-Like Planet With Two Suns
by Henry Baker 16 Sep '11

16 Sep '11

FYI -- I wonder how stable this system is? Could this planet orbit for billions of years? http://www.huffingtonpost.com/2011/09/15/kepler-16b-planet-two-suns_n_96479… Kepler 16b: NASA Discovers 'Star Wars'-Like Planet With Two Suns (PHOTOS, VIDEO) The Huffington Post Timothy Stenovec First Posted: 9/15/11 07:23 PM ET Updated: 9/15/11 07:23 PM ET NASA reported on Thursday the discovery of Kepler 16b, a planet with two suns that evokes images of Luke Skywalker's home planet, Tatooine, from the "Star Wars" franchise. The planet, which is 200 light years away and roughly the size of Saturn, is the first circumbinary planet -- meaning that it orbits two suns -- ever discovered, according to NASA. ILLUSTRATION AND ANIMATIONS BELOW "This discovery confirms a new class of planetary systems that could harbor life," William Borucki, the principal investigator of the Kepler telescope said in a statement. "Given that most stars in our galaxy are part of a binary system, this means the opportunities for life are much broader than if planets form only around single stars. This milestone discovery confirms a theory that scientists have had for decades but could not prove until now." Kepler 16b was discovered by the Kepler space vehicle, a satellite telescope that searches for habitable planets. Scientists, however, do not believe that life exists on the planet. The temperatures are about minus 150 degrees Fahrenheit, according to The Wall Street Journal. Researchers say that Kepler 16b is composed of roughly half rock and half gas. The suns that the planet orbits are smaller -- and cooler -- than Earth's sun; one of the stars is about 69% the mass of the sun and the other is about 20%. According to The New York Times, one of the stars is orange, and the other is red, and the planet orbits both stars every 229 days. The research will be published on Friday in the journal Science. In a HuffPost blog post published on Thursday, Dr. Laurance Doyle, an astrophysicist at the SETI Institute and the lead author of the paper, referenced Star Wars: "But today science fiction has become science fact, and that galaxy far, far away has become our own galaxy. A whole new kind of planetary system has been shown to exist and -- like Luke in the story -- the adventure is just getting started." LOOK: An artist's illustration of Kepler 16b: WATCH: Three Eclipsing Bodies: Animation of Kepler 16b orbiting its two suns: http://www.nasa.gov/multimedia/videogallery/index.html?media_id=111428821 WATCH: A Dance of Two Suns and One Planet: Animation of the orbital pattern of the stars: http://www.nasa.gov/multimedia/videogallery/index.html?media_id=111428361

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[math-fun] Incentres, etc
by Richard Guy 14 Sep '11

14 Sep '11

Dear all, I've received 5 messages from Lunnon, one from Rich, one from myself, one from Salamin, one from Asimov, but attempts to answer them, or forward relevant messages from Peter Moses, seem to fail. I'm copying to rcs, in case this also fails. R.

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Re: [math-fun] Incircles problem (fwd)
by Richard Guy 14 Sep '11

14 Sep '11

Here's the second message. R. ---------- Forwarded message ---------- Date: Fri, 9 Sep 2011 15:23:28 +0100 From: "Moses, Peter J. C." <mows(a)mopar.freeserve.co.uk> To: Richard Guy <rkg(a)cpsc.ucalgary.ca> Subject: Re: Incircles problem Hi Richard, > 6. If we apply Conway's extraversion to the problem, are there any selections of six excentres which lie on a conic? Well,in the case of the Fermat point, yes. (See attached) Have not looked at any others. Do you have The Geometer's Sketchpad ? Best regards, Peter. ----- Original Message ----- From: "Richard Guy" <rkg(a)cpsc.ucalgary.ca> To: "math-fun" <math-fun(a)mailman.xmission.com> Cc: "Julian Salazar" <julian.salazar(a)shaw.ca> Sent: Thursday, September 08, 2011 9:43 PM > Dear all, > Panic over! Stanley Rabinowitz has > used Geometer's sketch pad (see attached) to > show that the statement is not true! Sorry to > have bothered you with it. R.

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Re: [math-fun] Incircles problem (fwd)
by Richard Guy 14 Sep '11

14 Sep '11

Here's the first of four messages from Peter Moses. R. ---------- Forwarded message ---------- Date: Fri, 9 Sep 2011 12:34:44 +0100 From: "Moses, Peter J. C." <mows(a)mopar.freeserve.co.uk> To: Richard Guy <rkg(a)cpsc.ucalgary.ca> Subject: Re: Incircles problem Hi Richard, > Panic over! Well, not quite. For the medians, the statement is indeed false, as Stanley's picture shows. However, for cevians of the incenter the statement is true. The center of this ellipse is not in Clark's ETC. Surprisingly, it is true for the Fermat point.(X(13), though not X(14)) and also for both the Soddy centers! Although certainly for Fermat and the outer Soddy center it is a conic rather than specifically an ellipse. Best regards, Peter. ----- Original Message ----- From: "Richard Guy" <rkg(a)cpsc.ucalgary.ca> To: "math-fun" <math-fun(a)mailman.xmission.com> Cc: "Julian Salazar" <julian.salazar(a)shaw.ca> Sent: Thursday, September 08, 2011 9:43 PM > Dear all, > Panic over! Stanley Rabinowitz has > used Geometer's sketch pad (see attached) to > show that the statement is not true! Sorry to > have bothered you with it. R.

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[math-fun] IMO incentre ellipse problem
by Fred lunnon 14 Sep '11

14 Sep '11

I'm guessing that the cryptic reference to a "problem debunked" was intended to relate to Richard Guy's Cevian variation, which I confirm does not stand up. This isn't very surprising --- if the outer triangle's in-centre is perturbed to a more general "Cevian" point, the subsidiary triangles' in-centres would surely also need to be perturbed in some compatible fashion. The original in-centre statement certainly does appear to hold numerically, and I have to confess that I've not the foggiest notion how to prove it. But the following question occurs to me: is it feasible to bound the number of specific instances (independent in some appropriate manner) for which a claim like this must be (exactly!) verified numerically, in order for its truth to implied in general? Fred Lunnon On 9/9/11, Dan Asimov <dasimov(a)earthlink.net> wrote: > << > Twice again my messages have been bounced back, > so math-fun don't know that the problem has been > debunked by Stan Rabinowitz, using Geometer's Sketchpad. R. > > On Thu, 8 Sep 2011, Dan Asimov wrote: > >> Richard Guy, who is having trouble posting to math-fun, sent this message: >> >> -------------------------------------------------------------------------- >> Apologies for butting in, but my messages to math-fun >> are bounced back by a spam filter, so I don't know how >> to send. What I wanted to say was: >> >> The following problem was given to me by Julian >> Salazar, a high school student, who said he got it second-hand >> from members of the US IMO team. There are several subsidiary >> problems. >> >> The medians of a triangle divide it into six smaller >> triangles. Show that the six incentres lie on an ellipse. >> >> 1. Who originated the problem? >> >> 2. Did he/she have a proof? >> >> 3. Experiment suggests that if the three ``cevians'' >> are drawn through any point, not only the centroid, then the >> result is still true, except that if the point is outside >> the triangle, then the conic is a hyperbola, and if the point >> is on an edge of the triangle, the conic is a pair of straight >> lines. >> >> 4. In the original problem, is the centre of the >> ellipse among Clark Kimberling's three-thousand-odd triangle >> centres? >> >> 5. Same question as 4., but taking the defining >> point to be other of the more familiar triangle centres, >> besides the centroid. >> >> 6. If we apply Conway's extraversion to the >> problem, are there any selections of six excentres which >> lie on a conic? >> >> I must be getting old, because any attempt at >> finding a proof bogs down in a morass between Euclidean and >> projective geometry. Or, just possibly, it isn't true?!? >> >> Happy new academic year to all. R. >>> > > Sometimes the brain has a mind of its own. > > _______________________________________________ > 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] Quickie: is the sextic 1 + x + x^2 + x^3 + x^4 + x^5 + x^6
by Bill Gosper 14 Sep '11

14 Sep '11

(=0) solvable in radicals? SPOILER: Sure. It's cyclotomic, so x=exp(2 pi k/7) . But wait--what are 7th roots (of unity, or anything) doing in the solution of a sextic? Especially a reciprocal one. The roots should be (bi?)cubic, at worst. Indeed, here's all six. -1/6 + 1/6 E^(I (1/6 + #1) \[Pi]) Sqrt[7] E^(-((2 I \[Pi])/ 3)) + -(((E^((I 5 \[Pi])/6) (7/3)^(1/6) (9 I - 13 Sqrt[3])^( 1/3)) E^(-((I \[Pi])/3)) (E^((I \[Pi])/3))^#1)/( 6 2^(1/3))) + ( 7^(1/3) (E^((I \[Pi])/6) (-I - 3 Sqrt[3])^(1/3)) E^((2 I \[Pi])/ 3) (E^(-((2 I \[Pi])/3)))^#1)/( 6 2^(1/3)) + (((-(7/3))^(1/6) (-9 I - 13 Sqrt[3])^(1/3)) E^(-(( 2 I \[Pi])/3)) (-E^(((2 I \[Pi])/3)))^#1)/(6 2^(1/3)) + 1/6 7^(1/3) (-E^(((I 5 \[Pi])/6)) (1/2 (I - 3 Sqrt[3]))^(1/3)) E^(-(( 2 I \[Pi])/3)) (E^((2 I \[Pi])/3))^#1 & /@ Range[6]; Check: Rationalize[Abs[#]]*E^(\[Pi]*I*Rationalize[Arg[#]/\[Pi]]) & /@ N[%, 33] {E^((2 I \[Pi])/7), E^((6 I \[Pi])/7), E^(( 4 I \[Pi])/7), E^(-((2 I \[Pi])/7)), E^(-((6 I \[Pi])/7)), E^(-((4 I \[Pi])/7))} So we can express E^((2 I \[Pi])/7) in terms of E^((I \[Pi])/3) and cubic surds. http://mathworld.wolfram.com/TrigonometryAngles.html gives orders and, in prime cases, explicit formulas for the minimal polynomials for sin|cos(2 pi/n). Is it obvious why these (degree < n) polynomials always solve in radicals? (I.e., how does Developer`TrigToRadicals work?) Also, is there a solvable sextic trinomial which is neither bicubic nor triquadratic? An all-night (very inefficient) brute force search is so far fruitless. An even stronger conjecture: There are no true 6th roots of surds in the solutions of sextics! I.e., they always "denest" down to cuberoots. E.g., from that Cos[Pi/21] I sent yesterday, DenestRadicals2[(1/2 I (13 I + 3 Sqrt[3]))^(1/6)] -(-1)^(5/6) (1/2 (I - 3 Sqrt[3]))^(1/3) Even in the monster resolvent 2925951033851274156588135512485165232256823853056 - 2697817290737324449800236848640264992467435520 #1 + 9932351343021963689693473396732415411860992 #1^2 - 1881654619801628210127689611068977299937 #1^3 + 11450425009897563891465337536606118710 #1^4 + 3847649781964086608961673413540069 #1^5 + 378818692265664781682717625943 #1^6 of 8903 + 47647 v + 39672 v^2 + 7192 v^3 - 522 v^4 - 174 v^5 + v^7, the Montgomery|Piezas septic: DenestRadicals2[(1/2 (83845278844261475531122300164572695153 - 19718776166730709919674319096710083663 I Sqrt[3]))^(1/6)] ((-(1/3))^(1/6) (-22872248663753176389 I - 2586380087496171401 Sqrt[3])^(1/3))/2^(1/3) --rwg

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Re: [math-fun] IMO incentre ellipse problem
by Dan Asimov 14 Sep '11

14 Sep '11

Oops, just noticed that Fred & Gene have been discussing centroids. Sorry about that. --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] IMO incentre ellipse problem
by Dan Asimov 14 Sep '11

14 Sep '11

I can prove that the *centroids* of the six smaller triangles do lie on some ellipse. (Could that have been the original problem?) --Dan << The medians of a triangle divide it into six smaller triangles. Show that the six incentres lie on an ellipse. >> Sometimes the brain has a mind of its own.

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Re: [math-fun] A big Grabarchuk Puzzle Book
by Scott Fenton 14 Sep '11

14 Sep '11

However, don't forget that Amazon also owns Abebooks. -Scott On Sep 13, 2011 7:38 PM, "Eugene Salamin" <gene_salamin(a)yahoo.com> wrote:

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Re: [math-fun] A big Grabarchuk Puzzle Book
by Phil Carmody 13 Sep '11

13 Sep '11

From: Robert Baillie <rjbaillie(a)frii.com> > I don't buy from Amazon anymore. You can also buy it at Abe Books: > http://www.abebooks.com/servlet/SearchResults?sts=t&tn=%22The+Big%2C+Big%2C… Which indicates the best source for someone in the UK would be "The Book Depository", where AbeBooks tells me at first "The Book Depository is the UK's largest dedicated online bookseller", but the internet at large tells me "Amazon.com this morning announced a deal to buy The Book Depository online bookseller, based in the UK." Hmmm, OK, so was AbeBooks someone who actually *sold books*, or should I just buy from Amazon after all? (Answer to the latter - "no", as I learned to detest them immensely before and during the period where they were known as "spamazon", which most people probably forget.) Does anyone who I wouldn't consider evil or incompetent actually sell the actual book? Phil

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