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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.

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  • $\begingroup$ I worked out the opposite, i.e. area:perimeter ratio. Michael Rozenberg's answer is correct but, if you would like to see what I worked out, look here. Inserting $R=3/2$, we get for $n+$, $f(4,3)=(7,24,25), P=56$ and for $n-$, $f(4,1)=(15,8,17), P=40$. So $56$ is the correct answer. $\endgroup$ – poetasis Aug 5 at 17:36
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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.

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  • $\begingroup$ This came out of nowhere for me. Can you elaborate? $\endgroup$ – Aleksandr Aug 5 at 5:57
  • $\begingroup$ @Aleksandr Read about Pythagorean triples. I added the link. I am ready to show, how we can prove a first statement if you want. $\endgroup$ – Michael Rozenberg Aug 5 at 6:00

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