I have now verified this conjecture for semiperimeter s <= 303; in all 208 essentially distinct cases where the obvious method placing one side along an axis is inapplicable, there is a rotated pose with lattice vertices. My (simpleminded) program is bogging down by this point; I don't know if it's worth polishing up a tad. Anybody wishing to inspect the data is welcome to get in touch. Fred Lunnon 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 a pose with merely 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
On 11/16/11, Allan Wechsler <acwacw@gmail.com> wrote:
If I am understanding Fred's original query, he wants a congruent pose, not just a similar one; if that's the case, then scaling is not permitted, and that would be what Gene is missing. But it looks like the theorem Richard attributes to Brahmagupta affirms Fred's conjecture.
On Wed, Nov 16, 2011 at 5:52 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
Brahmagupta showed that any Heron triangle can be thought of as the adjunction of two Pythagorean ones. R.
On Wed, 16 Nov 2011, Eugene Salamin wrote:
Fred, I'm confused about this. If a triangle is posed with rational
coordinates, then scaling by the lcm of all six denominators will make the coordinated integers, while the triangle continues to have integer sides and area. Am I missing something?
-- Gene
______________________________**__
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com**> Sent: Wednesday, November 16, 2011 2:31 PM Subject: Re: [math-fun] Elementary triangle puzzle
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
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