Let $a,b,c$ be three edge lengths of a triangle whose area is $S$.

Then, here is my question.

Question : Supposing that $a,b,c$ are natural numbers, then does there exists $(a,b,c)$ such that $S=6k$ for any $k\in\mathbb N$?

Motivation : I've known the following fact:

Fact : If $a,b,c,S$ are natural numbers, then $S$ is a multiple of $6$.

Proof : By Heron's formula, we get $$\begin{align}16S^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c).\qquad(\star)\end{align}$$ Hence, we know that in mod $2$ $$a+b+c\equiv -a+b+c\equiv a-b+c\equiv a+b-c.$$ If all of these are odd, we reach a contradiction. So we know that all of these are even. Here, letting $$x=\frac{-a+b+c}{2}, y=\frac{a-b+c}{2}, z=\frac{a+b-c}{2},$$ then, getting $\frac{a+b+c}{2}=x+y+z,$ we know $$(\star)\iff S^2=(x+y+z)xyz.$$

By considering in mod $3$, we know that $S^2$ can be divided by $3$, which means that $S$ can be divided by $3$.

By considering in mod $4$, we know that $S^2$ can be divided by $4$, which means that $S$ can be divided by $2$.

Now we know that $S$ can be divided by $6$ as desired. Hence, the proof is now completed.

This got me interested in the above question, but I'm facing difficulty. Can anyone help?

  • $\begingroup$ Do you mean "... for every $k\in\mathbb N$"? $\endgroup$ – Henning Makholm Oct 3 '13 at 12:51
  • $\begingroup$ The areas are tabulated at oeis.org/A188158 but I don't see a formula. It appears there is no such triangle of area 18, nor 78, nor 102, nor 138, 162, 174, .... Related is oeis.org/A083875 $\endgroup$ – Gerry Myerson Oct 3 '13 at 13:20
  • $\begingroup$ en.wikipedia.org/wiki/Heronian_triangle shows some of the primitive cases. No solutions for some multiples though, and no characterisation of which those multiples are. Of course if $A$ is feasible, so is $4A, 9A, etc$. $\endgroup$ – Macavity Oct 3 '13 at 13:22

Theorem: There is no integer-sided triangle whose area is 18 units.

Proof: Assume such a triangle exists. Then there are integers $x, y, z$ such that $18^2 = xyz(x + y + z)$. Assume without loss of generality that $x \ge y \ge z$. Since $y \ge 1$ and $z \ge 1$, we have $x(x + 2) \le 18^2$ which certainly forces $x < 18$. This leaves us with only 17 possibilities for $x$ and hence $17 \times 18 / 2 = 153$ possibilities for the pair $(x, y)$; for each of these there is at most one $z$ which will work, and a moment's computer calculation shows that none of the 153 cases leads to a solution.

  • $\begingroup$ Thank you very much for nice answer! $\endgroup$ – mathlove Oct 3 '13 at 14:35
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    $\begingroup$ Since both $x$ and $y$ divide $18^2$, there are fewer than 153 possibilities to try; no need to try $x=5,7,8,10,11,13,14,15,16,17$, same for $y$. Anyway, can you prove 78 is not an area? 102? 138? $\endgroup$ – Gerry Myerson Oct 4 '13 at 12:26

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