On 11/19/11, Michael Reid <reid@gauss.math.ucf.edu> wrote:
THEOREM: The "Heronian" triangles can also be characterized as the triangles with integer sides and rational area (they then automatically have integer area).
CORRECTED PROOF:
It is much easier than that. Since 16 (area)^2 = 4(a^2 b^2 + b^2 c^2 + c^2 a^2) - (a^2 + b^2 + c^2)^2 , it follows that if a + b + c is odd, then 16 (area)^2 = 3 mod 4 , which is not a square.
In one line, more or less --- my original proof was half a page long!
What is the situation in 3 dimensions? Can a tetrahedron with integer edge lengths have a volume that is one twelfth (one sixth, one fourth, ... ) of an integer, without being an integer itself?
The test data says not --- but I haven't proved this. WFL
I googled Heronian triangles and a lot of what we've been saying was already known!! Duh!!
Paul Yiu: Heronian triangles are lattice triangles, Amer Math Monthly 108,3 (March 2001) 261-263 http://math.fau.edu/Yiu/AMM2001Heron.pdf Proves "Reid's" theorem. Indeed more strongly shows if the sides are sqrt(integer) and the area is rational, then embeddable on integer lattice.
The stronger result needs an additional hypothesis. Otherwise take a Heronian triangle and scale by a factor of sqrt(3) ; the resulting triangle cannot be embedded with integer coordinates. Indeed, he mentions the condition GCD(x, y, z) = 1 , where x , y and z are the squares of the edge lengths. But he curiously omits it from his definition of "geodetic" at the end.
Michael Reid