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 halfangle 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
ContentLength: 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 pulledback 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+ca) (a+bc) (a+cb) (a+b+c)
and the squared area of the triangle is
1/16 (b+ca) (a+bc) (a+cb) (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+cd)(a+b+dc)(a+c+db)(b+c+da)
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+ca) (a+bc) (a+cb) (a+b+c)
>
> and the squared area of the triangle is
>
> 1/16 (b+ca) (a+bc) (a+cb) (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 quadraticityinsquares, 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+cd)(a+b+dc)(a+c+db)(b+c+da)
>
> This reduces to Heron's formula when d=0.
You may not know the generalization to arbitrary quadrilaterals,
namely
area^2 = (sa)(sb)(sc)(sd)  ?abcd.cos^2(u)
where u is the semisum 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 semiadd to get
sa + sb + sc = (a+b+c)/2 = s, say,
whence sa = sa, sb = sb, sc = sc.
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 bedge 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 additionformula for sines, and we can similarly get the
additionformula 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 bsquare 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 semiadd 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  (bc)^2/4 = (bc  bc.cosA)/2 = bc.sin^2(A/2)
whence the halfangle 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
XCYA
sb sc
I write so,sa,sb,sc for the usual s,sa,sb,sc, and remark
that the formulae DELTA = rx.sx (x=0,a,b,c) have wellknown very
simple proofs.
It would be nice to find a simple symmetric 4dimensional 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 wormeaten 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 fantasticso beautifulso
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
doneI 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 wormeaten 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 wormeaten 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 wormeaten 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(sa)(sb)(sc) must vanish when [soandso], so must be
a multiple of [suchandsuch], 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 powerset 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 powerset 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
CXAY
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 longago days when Heron's theorem had
a proof.]
John Conway