This does look pretty convincing, though I shan't really believe it until I can manage to implement it. The apparently unavoidable involvement of the complex plane in the argument raises a further question: is an analogous theorem true for Heronian tetrahedra? Or Heronian simplices in n-space --- always assuming such things can actually be constructed ... WFL On 11/18/11, Michael Reid <reid@gauss.math.ucf.edu> wrote:
Theorem. Any Heronian triangle can be embedded in the plane so its vertices have integer coordinates. Proof: First position the triangle so that A is at (0, 0) , B is at (a, 0) . It is easy to show that vertex C has rational coordinates. Consider the points as complex numbers; the vertices are all in Q(i) , the fraction field of Z[i] . It is well-known that Z[i] , which is the ring of Gaussian integers, is a PID, and even a Euclidean domain.
Write the coordinate(s) of C as alpha / beta , in lowest terms. We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') (where ' denotes complex conjugation, and N(z) is the norm of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, we have beta divides alpha alpha' . As it is relatively prime to alpha , we have beta divides alpha' , and thus beta' divides alpha , i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta is relatively prime to alpha , it follows that beta is also relatively prime to beta' .
Also, |BC|^2 = N(a - alpha / beta) = a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') is an integer, so it follows that a (alpha / beta + alpha' / beta') = a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that beta is relatively prime to alpha beta' and therefore to alpha beta' + alpha' beta . Take complex conjugates to see that beta' is also relatively prime to alpha beta' + alpha' beta , and therefore beta beta' is relatively prime to alpha beta' + alpha' beta . However, beta beta' divides a (alpha beta' + alpha' beta) . This now implies that beta beta' divides a , so we may write a = b beta beta' , where b is a rational integer.
Finally, rotate the triangle by multiplying by beta / beta' , which has magnitude 1 . The vertices of the triangle rotate to 0 , a beta / beta' = b beta^2 and alpha / beta' = gamma , which are Gaussian integers. Therefore the vertices of the rotated triangle have integer coordinates. QED
Note that this gives a constructive way to find such an embedding.
Michael Reid
Yet another elementary program glitch caused misprinted vertex coordinates --- for [a, b, c] = [5, 29, 30] the lattice vertices should have read [0, 0], [3, 4], [21, -20] .
The data list sent out subsequently was similarly garbled, and a corrected version will follow [embarrassed groan]. WFL
On 11/17/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
As far as I can see, splitting a Heronian triangle (by its altitude) into two Pythagorean (right-angled, with rational edge-lengths, hence areas) constructs merely a pose with rational coordinates.
For example, suppose one edge joins [0, 0] and [a, 0]; then the third vertex lies at [(a^2+b^2-c^2) / 2 a, 2 d / a], where d denotes the area. This will not be an integer unless some edge-length a happens to divide 2d .
For triangles such as [a, b, c] = [5, 29, 30], where s, d = 32, 72, there is no such edge. However, the rotated pose with vertices [0, 0], [-4, 3], [-20, 21] still lies on the lattice. A similar apparently happy accident overtakes every case with s <= 200.
Fred Lunnon
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