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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.

Michael Rozenberg
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Aleksandr
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  • 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 05 '19 at 17:36

2 Answers2

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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.

Michael Rozenberg
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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 $56$

poetasis
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