Let's see now. If I wanted the volume of a tetrahedron, I could start by adding up the areas of three faces and subtracting the fourth. That would give me four terms to multiply together, for a total degree of 8. Maybe throw in the sum of all the areas, for degree 10. Since I want a degree 3 answer, I'll raise it to the 3/10 power. What's wrong with this picture? Rich PS: Doyle's web page looks like a lot of fun. I haven't looked beyond the Hero's formula discussion with JHC, but the titles look intriguing. --R -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Richard Guy Sent: Wednesday, April 26, 2006 12:48 PM To: math-fun Subject: Re: [math-fun] Hero[n]'s formula Here is a heuristic, not a proof. But is very suggestive, and may have influenced people two thousand or more years ago. Note that one doesn't need to think about imaginary numbers --- merely that the factors vanish as the triangle degenerates. Leave out the square roots until the dimension has got too big and needs reducing. R. What is a triangle? a + b > c, etc. Might expect area to contain factors sqrt(b+c-a) sqrt(c+a-b) sqrt(b+c-a) To get dimension up to 2 we need another factor, one that is symmetric in a,b,c. sqrt(a+b+c) -- and we're home within a constant, which an easy case --- say 3,4,5 --- shows to be 1/4. Note also: Conway's `extraversion' gives equal weight to the 4 quantities s s-a s-b s-c On Wed, 26 Apr 2006, Eugene Salamin wrote:

--- Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:

...

As all munsters know, area^2 = s(s-a)(s-b)(s-c) = xyz(x+y+z). Is there a really good proof of this? The best one I've been able to cook up[1] says: let t,u,v = tan A/2 etc.; then xt=r (the inradius) etc., so t:u:v = 1/x:1/y:1/z, and (tangent-sum formula, angle-sum of triangle) uv+vt+tu=1, so t = 1/x / sqrt(1/yz+1/zx+1/xy), so r^2 = 1/(1/yz+1/zx+1/xy) = xyz/(x+y+z) and as rs=area we're done. But this feels a bit too complicated. Is there perhaps a neat proof in 4 or 6 dimensions that looks at the cartesian product of the triangle (or some tetrahedron with the triangle as base) with itself?

[1] Presumably it's been known for centuries.

g

Adjoin two congruent triangles along their c edge to obtain a parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). Using the law of cosines, A^2 can be expressed as a polynomial P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So

P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c).

Q is of degree 1, and since P is symmetric, so is Q. Then Q is a multiple of a+b+c. The fixed constant can be found by consdering e.g. an equilateral triangle.

Gene

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