Re: [math-fun] Heron's formula/Cayley-Menger determinant
I found Fred's cool paper on arxiv. Perhaps some of my ideas can make a 2D version of the embedding theorem a little more elegant. BTW, my determinant III of complex numbers a,b,c: |0 a b c | |a 0 c' b'| |b c' 0 a'| |c b' a' 0 | when c=-a-b can be simplified with row & column operations to |a a' 0 c'| |b b' 0 0 | |0 0 a b | |0 0 a' b'| thus making obvious the factorization (ab'-a'b)^2. (We note again that ab'-a'b is pure imaginary, thus the square is negative real.) At 03:52 AM 6/30/2012, Fred lunnon wrote:
On 6/30/12, Henry Baker <hbaker1@pipeline.com> wrote:
... Is there a name for triangles whose sides are integers, and _which can also be embedded in the plane with integer coordinates for the vertices_ ?? ]
"Heronian".
If the vertices are rational, then the area is rational via the standard determinant giving area in terms of Cartesian coordinates [which ought to have a name, though I know of none]. If the sides are integers and the area rational, then the triangle is Heronian by definition [and furthermore its area 6x integer].
Conversely, if a triangle is Heronian then it may be embedded with integer vertices via Yiu's theorem --- see the remarkable complex GCD proof by Michael Reid which was discussed in math-fun last year.
An analogous argument applies in 3 dimensions.
Fred Lunnon
See also Susan Marshall and Alexander Perlis "Heronian tetrahedra are lattice tetrahedra" American Mathematical Monthly (to appear?) I was rather hoping nobody would think of looking for the arXiv paper --- to my embarrassment, Marshall & Perlis not only beat me to publication (by weeks, extraordinarily), but found a moderately substantial gap in it. While I have now produced a corrected version, which does incidentally go rather further than their result, I've been too lazy as yet to update the arXiv. Fred Lunnon On 6/30/12, Henry Baker <hbaker1@pipeline.com> wrote:
I found Fred's cool paper on arxiv. Perhaps some of my ideas can make a 2D version of the embedding theorem a little more elegant.
BTW,
my determinant III of complex numbers a,b,c:
|0 a b c | |a 0 c' b'| |b c' 0 a'| |c b' a' 0 |
when c=-a-b can be simplified with row & column operations to
|a a' 0 c'| |b b' 0 0 | |0 0 a b | |0 0 a' b'|
thus making obvious the factorization (ab'-a'b)^2.
(We note again that ab'-a'b is pure imaginary, thus the square is negative real.)
At 03:52 AM 6/30/2012, Fred lunnon wrote:
On 6/30/12, Henry Baker <hbaker1@pipeline.com> wrote:
... Is there a name for triangles whose sides are integers, and _which can also be embedded in the plane with integer coordinates for the vertices_ ?? ]
"Heronian".
If the vertices are rational, then the area is rational via the standard determinant giving area in terms of Cartesian coordinates [which ought to have a name, though I know of none]. If the sides are integers and the area rational, then the triangle is Heronian by definition [and furthermore its area 6x integer].
Conversely, if a triangle is Heronian then it may be embedded with integer vertices via Yiu's theorem --- see the remarkable complex GCD proof by Michael Reid which was discussed in math-fun last year.
An analogous argument applies in 3 dimensions.
Fred Lunnon
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