It's true in general that if you take any polyhedron, in n dimensions if you wish, and for each face associate a normal vector whose length is the (n-1)-dimensional area of that face, the sum is 0. Your statement about the "hypoteface" translates to the usual Euclidean formula for length of a vector in R^3. Even more generally: (a) for any closed hypersurface in n dimensions, the integral of the unit normal vector field wrt n-1-dimensional area is 0, and (b) for any k-dimensional polygon in n-space, its k-dimensional volume is the square root of the sum of squares of the projections to all k-dimensional coordinate planes. (key words exterior algebra, or wedge products). I forget whatever I might once have known about trying to generalize Heron's formula, but note that the volume is 0 whenever two sides flatten out to a plane, as when 3 sides flatten out to the fourth. Unfortunately you can't detect this kind of flattening from the areas of the faces: they could all be equal, just as for the regular tetrahedron, so you need something else. Bill On Apr 26, 2006, at 4:23 PM, Schroeppel, Richard wrote:
Somewhat more seriously: If we have a right tetrahedron, where one of the corners matches a cube's corner, there are three leg-faces and one hypote-face. It turns out that the square of the hypoteface area is equal to the sum of the squares of the three legface areas. (I don't know why, or if it works in more dimensions.) Any triangle can be a hypoteface, and the lengths of the X,Y,Z legs are easy to calculate: If the hypoteface edges are A,B,C, then X2+Y2 = A2, etc. and X2 = (A2+B2-C2)/2. The legface areas are XY/2 etc., the area squares are X2Y2/4, and it's not hard to check that the sum of squares of the three legface areas matches Hero^2.
Rich
-----Original Message----- From: Schroeppel, Richard Sent: Wednesday, April 26, 2006 1:26 PM To: 'math-fun' Cc: Schroeppel, Richard; 'rcs@cs.arizona.edu' Subject: RE: [math-fun] Hero[n]'s formula
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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