Find the maximum possible perimeter of a right triangle 
The ratio between the perimeter of a right triangle and its area is 2:3. The sides of the triangle are integers. Find the maximum possible perimeter of the triangle. 

If the sides of the triangle are $A$, $B$ and $C$ (the hypotenuse), I have deduced that:
$$A+B+\sqrt{A^2+B^2}=\frac{AB}{3}$$
I am stuck here. Any hints? From the above I know that $A^2+B^2$ should be a square and that $AB$ should be divisible by 3.
 A: There are naturals $m$ and $n$ such that $m>n$, $a=m^2-n^2$, $b=2mn$ and $c=m^2+n^2$.
See here: https://en.wikipedia.org/wiki/Pythagorean_triple
Thus, $$\frac{m^2-n^2+2mn+m^2+n^2}{mn(m^2-n^2)}=\frac{2}{3}.$$
Can you end it now?
I got $56$ as the answer. 
A: I worked out the opposite, i.e. area:perimeter ratio. Michael Rozenberg's answer is correct.
$$ R=\frac{area}{perimeter}=\frac{AB}{2}/P=\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}
\qquad\qquad\qquad\qquad\qquad $$
\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\quad\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1)
\end{equation}
The lower limit insures that $k\in \mathbb{N}$ and the upper limit ensures that $m> k$.
$$R=3/2\implies  \lceil\sqrt{8(1.5)}\rceil=4\le m \le (2(1.5)+1)=4 \quad\land\quad   m\in\{ 4\}\implies k\in\{ 3,1\}$$
$$F(4,3)=(7,24,25)\quad\land\quad \frac{84}{56}=3/2\qquad\qquad f(4,1)=(15,8,17)\quad\land\quad \frac{60}{40}=3/2$$
We can see that the first triple has the maximum perimeter of $56$
