Date: Mon, 15 Dec 1997 17:20:58 -0500 (EST) From: John Conway Subject: Heron To: Peter Doyle I thought about proofs of Heron's formula and came up with a few, but couldn't find your email address so didn't send them to you. To my mind the "real" proof says (somehow) that the squared area is a quadratic in a^2, b^2, c^2 that vanishes when +-a+-b+-c = 0, and so ... . I found several ways of proving the first half of the sentence, but non of them were really pleasing, and you will doubtless prefer your own. The other way I like is to prove the half-angle formulae, (they are very easy from the geometry), and then use area = bc sin(A/2)cos(A/2). I also looked in my "Workman" and found, as I had vaguely remembered, two distinct analogues of Heron for spherical triangles. The neater one was l'Huiller's, but the other was also interesting. I'll try to bring Workman into Fine Hall, so that next time I can quote it exactly. [I like his occasional remarks to the student about what should be learned. I'll quote some of those too!] JHC From doyle Wed Dec 17 19:47:46 1997 To: conway@math.princeton.edu, res@math.umd.edu Subject: Heron's formula Content-Length: 1471 John, How about this: If a quadratic form takes values A, B, C for the three vectors of a superbase, then the determinant of the form relative to the superbase is 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) (Symmetric of degree 2; vanishes when A=0, B=C.) Any triangle congruent to the image of the standard equilateral triangle with vertices {1, omega, omega^2} under a linear mapping. Its squared area is proportional to the determinant of the quadratic form Q you get by pulling back the standard norm in the image under this linear mapping. This pulled-back form Q takes values A=a^2, B=b^2, C=c^2 for the three vectors of the superbase {I, I omega, I omega^2}. Thus the determinant of Q is 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) = 1/4 (b+c-a) (a+b-c) (a+c-b) (a+b+c) and the squared area of the triangle is 1/16 (b+c-a) (a+b-c) (a+c-b) (a+b+c) This can be seen directly, without computation, by noting that the determinant of a quadratic form is certainly some quadratic function of the norms of any three independent vectors, and hence the area of a triangle is quadratic in a^2, b^2, c^2, and vanishes when +-a+-b+-c=0. Brahmagupta's formula says that the squared area of a quadrilateral inscribed in a circle is 1/16 (a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a) This reduces to Heron's formula when d=0. To verify this formula, we need only show that the squared area is a quartic function of the side lengths. Why is this true, I wonder? Peter Date: Tue, 23 Dec 1997 15:02:44 -0500 (EST) From: John Conway Subject: Re: Heron's formula To: Peter Doyle Cc: res@math.umd.edu On Wed, 17 Dec 1997, Peter Doyle wrote: > John, > > How about this: > > If a quadratic form takes values A, B, C for the three vectors of a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > Thus the determinant of Q is > > 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) > = > 1/4 (b+c-a) (a+b-c) (a+c-b) (a+b+c) > > and the squared area of the triangle is > > 1/16 (b+c-a) (a+b-c) (a+c-b) (a+b+c) > > This can be seen directly, without computation, by noting that the > determinant of a quadratic form is certainly some quadratic function > of the norms of any three independent vectors, and hence the area of a > triangle is quadratic in a^2, b^2, c^2, and vanishes when +-a+-b+-c=0. Well, ... yes, I suppose so. I found a few other reasons to justify the quadraticity-in-squares, at least one of which seemed easier. > Brahmagupta's formula says that the squared area of a quadrilateral inscribed > in a circle is > > 1/16 (a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a) > > This reduces to Heron's formula when d=0. You may not know the generalization to arbitrary quadrilaterals, namely area^2 = (s-a)(s-b)(s-c)(s-d) - ?abcd.cos^2(u) where u is the semi-sum of two opposite angles. (This shows, inter alia, that the area is maximized in the cyclic case.) > To verify this formula, we > need only show that the squared area is a quartic function of the > side lengths. Why is this true, I wonder? > > Peter Dunno at the moment, but surely this can't be hard. I have a nice little book that gives some other nice stuff about cyclic quadrilaterals. Namely, by putting a,b,c,d into the three essentially different orders, we obtain 3 such, whose diagonals have only 3 lengths, which my book calls e,f,g (in the wrong order, but I'll use the right one). Then e^2 = (bd+ca)(cd+ab)/(ad+bc) etc., and I think that area is something like efg/4R, from which (when corrected) you can deduce a formula for R. [This is only a sample.] Ptolemy's theorem shows that AD.BC, BD.CA, CD.AB are the sides of a triangle, which degenerates to a straight line precisely when the quadrilateral ABCD is cyclic. I would love to know just how this triangle is related to ABCD - for instance do its angles have any geometrical interpretation. Which quadrilaterals correspond to the same triangle? I think I've seen a TINY bit of this stuff, but would like to know it ALL. JHC Date: Mon, 10 May 1999 09:42:17 -0400 (EDT) From: John Conway To: Peter Doyle Subject: triangle trigonometry Peter - you once asked did I know a nice proof of Heron's formula, to which the answer was that I didn't. Well, I still don't, but (in connection with my proposed "Triangle Book") am feeling a bit happier, because I've worked out a really nice ab initio development of trigonometry. By "the Somecenter picture" I mean the picture obtained by joining the Somecenter to the vertices, and dropping perps from it to the sides. Then in the incenter picture the sides get cut into pieces of lengths sa,sb,sc say (sx is really s_x), giving the equations sb + sc = a sc + sa = b sa + sb = c which we semi-add to get sa + sb + sc = (a+b+c)/2 = s, say, whence sa = s-a, sb = s-b, sc = s-c. In the circumcenter picture, the sides get bisected into pieces of lengths RsinA, RsinB, RsinC, so we get the sine rule a/sinA = b/sinB = c/sinC = 2R. In the orthocenter picture, the b-edge is cut into pieces of lengths a.cosC and c.cosA, so we get ( 2Rsin(C+A) = ) 2RsinB = 2R( sinA.cosC + sinC.cosA ) whence the addition-formula for sines, and we can similarly get the addition-formula for cosines by looking at the way the altitude splits up. But also in that picture draw squares on the sides, and continue the altitudes through them, so cutting the b-square into pieces of areas bc.cosA and ab.cosC = SA and SB, say. (SX is really S_X). Then we get the equations SB + SC = aa SC + SA = bb SA + SB = cc, which semi-add to give SA + SB + SC + (aa+bb+cc)/2 = S, say, whence SA = S - aa = (bb + cc - aa)/2, the cosine formula. Now we go into algebraic mode. We find ssa = (b+c)^2/4 - aa/4 = (bc + bc.cosA)/2 = bc.cos^2(A/2) sbsc = aa/4 - (b-c)^2/4 = (bc - bc.cosA)/2 = bc.sin^2(A/2) whence the half-angle formulae sin(A/2) = root(sbsc/bc) cos(A/2) = root(ssa/bc) tan(A/2) = root(sbsc/ssa) and also, taking the geometric mean, Heron's formula: root(ssasbsc) = bc.sin(A/2).cos(A/2) = bc.sinA/2 = DELTA A pretty neat treatment, eh? But I'd still like to reduce the algebraic part, and would absolutely LOVE to get a "symmetrical" proof of Heron's formula. JHC Date: Fri, 3 Mar 2000 12:03:49 -0500 (EST) From: John Conway To: "Peter G. Doyle" Subject: Heron You were once keen to know the best proof of Heron. The best I've found is this: DELTA^2 = ro.so.ra.sa = so.sa.sb.sc since from the diagram below we have ra/sb = sc/ro, the triangles Ia X C and C Y Io being similar in view of the angles Io C V = C/2 and Ia C X the complement of this. Ia | | |ra Io | |ro ----X----C-------Y------A sb sc I write so,sa,sb,sc for the usual s,s-a,s-b,s-c, and remark that the formulae DELTA = rx.sx (x=0,a,b,c) have well-known very simple proofs. It would be nice to find a simple symmetric 4-dimensional proof, but I've not yet managed to do so. Regards, JHC From doyle Wed Jun 28 10:17:04 2000 To: conway Subject: Cantor's worm-eaten paradise John, I liked your Heron proof very much. Can you modify it to get Brahmagupta's formula? I've just got a new digital tape recorder, and I've been copying the Conway tapes. Looking them over, I can't believe I didn't copy them earlier. I mean, what if my house had burned down? Of course I want to make the tapes available on the web. But the first concern is to assure that the priceless wisdom on these tapes is preserved. I came across the following snippet, delivered after the memorable `ordinal walk' field trip in your `Romance of Numbers' class. PGD: Tell me about the benefit of being brought up in a religious household. JHC: Well, in a religious household of this particular kind of religion, anyway. Well, Cantor's construction is so fantastic---so beautiful---so incredible, that we ought to know what it is. If we start disbelieving it before we've even heard it through, you know, we haven't given it a chance, so to speak. I don't believe it, I really don't believe it... But I think it's very sound policy to lie. I mean, let's remember Groucho Marx's wonderful thing, `If you can fake sincerity you've got it made.' Let's fake sincerity for a bit. And then, you know, after a time we can tell them it's all a bit dubious. PGD: It's a Cantorian paradise, right? JHC: Yes: `Out of this paradise that Cantor has created nobody will expell us.' PGD: Hilbert said that? JHC: Yeah. PGD: Oh, those were the days... JHC: Well, nobody actually has thrown us out! All they've just done---I mean, it's like the problem of America, basically. I mean, there are drug addicts and shooting in the streets. It's dangerous, this paradise. It's got contradictions all over the place, OK? It just a slightly sort of worm-eaten paradise. ... Come on, switch that thing off! At the time I was too young to appreciate what you were saying, but now I think I understand. Like horseshoes, Cantor's theory works whether you believe in it or not. And the less you believe in it, the more astonishing it is how well it works. Peter From conway@math.Princeton.EDU Wed Jun 28 14:28:09 2000 Date: Wed, 28 Jun 2000 14:27:09 -0400 (EDT) From: John Conway To: "Peter G. Doyle" Subject: Re: Cantor's worm-eaten paradise On Wed, 28 Jun 2000, Peter G. Doyle wrote: > I liked your Heron proof very much. Can you modify it to get > Brahmagupta's formula? It's not really my proof - I found it in Casey's "Sequel to Euclid" (of around 1880) and he says upfront that he allows himself to quote stuff from the following list of books without attribution, so it's probably in several of them, and can be regarded as traditional. No, I don't see how to get Brahmagupta (but am still hoping). > PGD: Tell me about the benefit of being brought up in a religious household. ... > It just a slightly sort of worm-eaten paradise. ... Come on, switch > that thing off! Why didn't you?! > At the time I was too young to appreciate what you were saying, but > now I think I understand. Like horseshoes, Cantor's theory works whether > you believe in it or not. And the less you believe in it, the more > astonishing it is how well it works. I like the "like horseshoes"! JHC From doyle Wed Jun 28 15:06:51 2000 To: conway Subject: Horseshoes John, I did turn the machine off, just as you instructed, and thereby deprived posterity of any further pearls you had to cast. The phrase `like horseshoes' is a reference to the story about the quantum mechanic who had a horseshoe nailed above his door. Someone said, `Surely, Prof. Bohr (?), you don't believe in horseshoes?' `No,' said Bohr, `but I understand they work whether you believe in them or not.' Peter From conway@math.Princeton.EDU Wed Jun 28 17:38:10 2000 Date: Wed, 28 Jun 2000 17:37:14 -0400 (EDT) From: John Conway To: "Peter G. Doyle" Subject: Re: Horseshoes On Wed, 28 Jun 2000, Peter G. Doyle wrote: > John, > > I did turn the machine off, just as you instructed, and thereby > deprived posterity of any further pearls you had to cast. > > The phrase `like horseshoes' is a reference to the story about the > quantum mechanic who had a horseshoe nailed above his door. Someone > said, `Surely, Prof. Bohr (?), you don't believe in horseshoes?' > `No,' said Bohr, `but I understand they work whether you believe in > them or not.' Yes, I think you've told me this before. Peter - I read "our" paper on division by 3 for the first time when I was at Northwestern earlier this year, and was disturbed by the way that the point got buried in the fluff. I'll expand on this if you like (but must then naturally allow myself to speak as sharply and sarcastically as I should if we were together!), but basically just want to say that we should get together some time (soon, please!) to write a short and snappy version to be published on permanent paper. I hope you like this plan. How on earth can we hope to prove Brahmagupta? One plan is to show somehow that the answer has to be the square root of a polynomial of a certain form, then remark that its difference from s(s-a)(s-b)(s-c) must vanish when [so-and-so], so must be a multiple of [such-and-such], and then the multiplier must be 1. I imagine I could force through such a proof, and maybe it's the best we can expect? JHC PS: I'm writing a book on the geometry of the triangle (which, by the way, will be the second triangular book that I'll own!), and I'm sure you'll like some of the goodies in there. That's another thing to discuss when we're next together. J From doyle Wed Jun 28 20:05:46 2000 To: conway Subject: Fluff John, Yes, I agree that there is a lot of fluff in the `preliminary version' of the division by 3 paper. I've been meaning to take it out for a long time now. The draft version is on my web page under the heading `in remission'. It was originally listed under the heading `in preparation', but at some point I decided that this was an exaggeration. This draft has just recently gotten some press in John Baez's column `This Week's Finds in Mathematical Physics': http://math.ucr.edu/home/baez/week147.html Two things that deserve to be thought about: (1) In the draft I claim that similar methods will show that if 3a=2b then there is some c so that a=2c and b=3c. However, the last time I thought about this I didn't see how to do it, and I'm not quite sure whether we really ever did figure this out. (2) I believe that this method of dividing by three does not require the power-set axiom. I would like to know if this is indeed the case. Are you free to come visit in Hanover, say for a week or so? We could make a whole bunch more TAPES! Or I could come visit in Princeton, and we could make a whole bunch of TAPES. Peter From conway@math.Princeton.EDU Wed Jun 28 22:23:12 2000 Date: Wed, 28 Jun 2000 22:21:52 -0400 (EDT) From: John Conway To: "Peter G. Doyle" Subject: Re: Fluff On Wed, 28 Jun 2000, Peter G. Doyle wrote: > Yes, I agree that there is a lot of fluff ... > ... I've been meaning to take it out for a long time now. Dear Mr Fluff - I regret to inform you that it isn't possible for you to do that alone. > Two things that deserve to be thought about: > > (1) In the draft I claim that similar methods will show that if 3a=2b > then there is some c so that a=2c and b=3c. However, the last time I > thought about this I didn't see how to do it, and I'm not quite sure > whether we really ever did figure this out. I think we did, and if not, that we can and should. > (2) I believe that this method of dividing by three does not require > the power-set axiom. I would like to know if this is indeed the case. Certainly feels that way - but it's the sort of thing that's best checked through in tandem rather than alone. > Are you free to come visit in Hanover, say for a week or so? We could > make a whole bunch more TAPES! Or I could come visit in Princeton, > and we could make a whole bunch of TAPES. Dear Mr Tapeworm, Thank you very much for your kind invitation. I shall ask my social secretary when I'll be free to accept. Your most humble and obedient master. From conway@Math.Princeton.EDU Wed Jan 17 12:38:27 2001 Date: Wed, 17 Jan 2001 12:39:13 -0500 (EST) From: John Conway To: "Peter G. Doyle" Subject: Re: Visit? On Thu, 11 Jan 2001, Peter G. Doyle wrote: > John, > > I understand from Wendy and Bob that you might be induced to visit Dartmouth > for a week or so sometime at the end of January. Dartmouth has money to > pay expenses (train, hotel, meals). Or are you all booked up now? > If not now, when, and how do I contact your social secretary? I'd love to come, but am afraid I can't do so at the end of January. I hope we can fix up something a bit later. You remember asking for the right proof of Heron's formula? Well, here's a pretty nice one: First you prove in the obvious and usual way that ro.so = ra.sa = rb.sb = rc.sc where so = (a+b+c)/2, sa = so - a, etc., and ro, ra, rb, rc are the radii of the in- and ex- circles. Then you prove (eg.) sb.sc = ro.ra by similar triangles Ic B Io --------------C-------X---------A-----Y------------ Namely, in triangle Io X A we have Io X = ro, X A = sa while in the similar triangle A Y Ic we have A Y = sb, Y Ic = rc. Then DELTA^2 = (ro.so)(ra.sa) = (so.sa)(ro.ra) = (so.sa)(sb.sc). I agree this hasn't got the symmetry we'd like to see, and nor does it do something elegant in four dimensions, as we'd hoped, but I think you'll agree it's pretty good. [It seems to have been the standard proof in the long-ago days when Heron's theorem had a proof.] John Conway