8
$\begingroup$

Let $a,b,c$ be integers which are the sides of a triangle with integral area, a so called Heronian triangle. This website attributes to Gauss the result that there must then exist integers $m,n,p,q$ such that

$a = mn(p^2+q^2)$

$b = (mp)^2+(nq)^2$

$c = (m+n)(mp^2-nq^2)$

(where I left out a $4pq$ factor designed to make the radius of the circumscribed circle integral as well). It's not hard to see that the triangle defined by these formulas is indeed Heronian, however I could neither prove nor find a reference for the fact that this parametrization is exhaustive.

Can someone do one of these two things?

Thanks!

(Note: I'm communicating this question on behalf of my dad, who is really the person who looked into that but is not easily capable of asking it himself over here. I may be slow to respond on his behalf if questions come up).

$\endgroup$

2 Answers 2

5
$\begingroup$

Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangle is divided into two right triangles one with hypotenuse $a$ and legs $d,h$, and one with hypotenuse $b$ and legs $e,h$. We have $d+e=c\in \mathbb{Q}$. Also notice that $$d-e=\frac{d^2-e^2}{d+e}=\frac{a^2-b^2}{c}\in \mathbb{Q}$$ so we conclude that $d,e$ are rational. From the pythagorean triples we have relations $$a-d=r,\quad a+d=\frac{h^2}{r},\quad b-e=s,\quad b+e=\frac{h^2}{s}$$ and therefore $$a=\frac{1}{2r}(h^2+r^2),\quad b=\frac{1}{2s}(h^2+s^2),\quad c=\frac{r+s}{2rs}(rs-h^2)$$ This is exactly your parametrization up to scaling.

$\endgroup$
1
  • $\begingroup$ Thanks, Gjergji - this looks great. I'll forward to the actual OP... $\endgroup$
    – Alon Amit
    Dec 29, 2009 at 21:10
1
$\begingroup$

This paper by Sascha Kurz credits a parametrization much like yours (for triangles with integer sides and rational area) to the seventh-century Indian mathematician Brahmagupta. The paper also gives an algorithm for generating Heronian triangles. It doesn't provide a proof that the parametrization gives all Heronian triangles, but has a reasonable reference list which might be a good place for further searching.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.