# 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.

• 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. – poetasis Aug 5 at 17:36

There are naturals $$m$$ and $$n$$ such that $$m>n$$, $$a=m^2-n^2$$, $$b=2mn$$ and $$c=m^2+n^2$$.
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.