It is easy to see that any Heronian triangle (with integer side-length and area) can be posed in 2-space so that its vertices have rational Cartesian coordinates. A natural question is whether it can be posed with its vertices on the integer lattice. I earlier proposed a counterexample having no lattice pose; however this claim later turned out to result from an elementary programming error. An amended run found lattice poses for every Heronian triangle with semi-perimeter s <= 200. It now seems reasonable to conjecture that every Heronian triangle may be posed on the lattice. If this is true, it must surely be obvious --- but to me, not at the moment! Did Minkowski have something to say about this, I wonder? Fred Lunnon On 10/29/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
I've been taking a desultory look at "rational" test triangles lately, which turn out to have some interest, as well as connections with a well-known notion --- see (tho' overlapping & slightly inaccurate)
http://mathworld.wolfram.com/HeronianTriangle.html http://en.wikipedia.org/wiki/Integer_triangle http://en.wikipedia.org/wiki/Heronian_triangle
A triangle T is "Heronian" means it has integer sides a,b,c and rational semi-perimeter s, area d. Sample results:
A triangle has rational Cartesian coordinates and sides iff it is similar to a Heronian triangle. Not every Heronian triangle can be posed to have integer coordinates: s,d,a,b,c = 75, 420, 26, 51, 73 is the smallest counterexample.
For T Heronian: an odd number of a,b,c are even; s is an integer, d is an even integer; when T is proper (ie. d <> 0) s is composite; semi-angle tangents are rational: tan C/2 = d/(s-a)(s-b), etc.
Bucholtz' parameterisation should read a = (n^2 + k^2)m, b = (m^2 + k^2)n, c = (m + n)(m n - k^2), s = (m + n)m n, d = k m n(m + n)(m n - k^2); constraints GCD(m,n,k) = 1; m > n > 0 & 0 < k <= sqrt( m n^2/(m + 2 n) ) enforce one (nonprimitive) triangle in each Heronian similarity class.
The inverse mapping (modulo similarity) is given by k = d, n = s(s-b), m = s(s-a). Bucholtz may sometimes yield a primitive triangle: s,d,a,b,c = 70, 630, 25, 52, 63 is given directly by m,n,k = 5, 2, 1 .
"Swinging" either side around the vertex C (say) so as to meet the line of c again yields a new (nonprimitive, unordered) triangle with sides [a, b, c] -> [ac, bc, b^2 - a^2] . For instance, [3,4,5] -> [7,15,20] -> [35,44,75] -> ... Unfortunately the sides grow exponentially; I conjecture that a proper Heronian triangle is the smallest swinger in town (as it were) just when no two sides have a common factor, or when isosceles.
A couple of questions for you: Is this stuff worth writing up [I'm undecided]? What about writing up the conics through six centres conjectures, which seem to be genuinely difficult?
Fred Lunnon [29/10/11]
On 10/29/11, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Fred wins his bet! I've just submitted a paper, ``Five-point circles, the 76-point sphere, and the Pavillet tetrahedron'' to the Monthly. The theme is that triangle geometry is not dead. I would attach a copy if attachments were allowed; I will entertain individual requests. Of course, I'm hoping that the paper will be accepted, and I also now hope to include a stop press paragraph with the following theorem:
The external angle-bisectors of a triangle meet the opposite edges in collinear points. Call it the zero-sum line because: The sum of the directed distances to the edges of a triangle from points on its zero-sum line is zero.
So go ahead and spoil my fun and tell me that that's well known to those who well know it. Certainly the first part is well known since it follows immediately from (the converse of) Menelaus's theorem, so the zero-sum line may already have some other name. The Gergonne line would be a good guess, but that's not quite right.
Note that the zero-sum line (and the Gergonne line) are the line at infinity if the triangle is equilateral. Best, R.