Perimeter=Area

Let each cathetus be $a$ and $b$ and the hypotenuse be $c$ . Then:

$\large \displaystyle \frac{ab}{2} = a + b + \sqrt{a^2+b^2}$

$\large \displaystyle \frac{ab}{2} - a - b = \sqrt{a^2+b^2}$

$\large \displaystyle \frac{(ab)^2}{4} + a^2 +b^2 - a^2b - ab^2 + 2ab = a^2+b^2$

$\large \displaystyle \frac{(ab)^2}{4} - a^2b - ab^2 + 2ab = 0$

$\large \displaystyle \frac{ab}{4} - a - b + 2 = 0$

$\large \displaystyle b = \frac{4a-8}{a-4}$

$\large \displaystyle b = 4 + \frac{8}{a-4}$

So, for $b$ to be an integer, $a-4$ should be $\pm 1, \pm2, \pm 4, \pm 8$ . For $a-4$ equal to $-8$ or $-4$ , $a \leq 0$ . For $a-4$ equal to $-2$ or $-1$ , $b \leq 0$ .

For $a-4 = 1$ , $a = 5, b=12$ . For $a-4 = 8$ , $a = 12, b=5$ . This is the same right triangle, with $\color{#20A900} \boxed{c=13}$ .

For $a-4 = 2$ , $a = 6, b=8$ . For $a-4 = 4$ , $a = 8, b=6$ . This is the same right triangle, with $\color{#20A900} \boxed{c=10}$ .

So, the sum of all posible hypotenuses is $\color{#3D99F6} 13 + 10 = \boxed{23}$

By Euclid's formula, the pythagorean triples are generated by $k(m^2-n^2), k(2mn), k(m^2+n^2)$ where $m>n, k$ are positive integers.

Using these parameters, the perimeter $P$ and the area $A$ are given by $P = k(m^2-n^2) + k(2mn) + k(m^2+n^2) = 2mk(m+n) \\ A = \frac{1}{2}\cdot k(m^2-n^2) \cdot k(2mn) = mnk^2(m+n)(m-n)$ so by setting them equal and canceling common [necessarily nonzero] factors: $2mk(m+n) = P = A = mnk^2(m+n)(m-n) \\ \implies \quad 2 = nk(m-n)$ so one of $n, k, m-n$ equals $2$ and the other two equal $1$ .

• If $n=2, k=1, m-n=1,$ then $m=1+2 = 3.$ This gives the pythagorean triple $\Big\{k(m^2-n^2), k(2mn), k(m^2+n^2)\Big\} = \{5, 12, 13\}$
• If $n=1, k=2, m-n=1,$ then $m=1+1 = 2.$ This gives the pythagorean triple $\Big\{k(m^2-n^2), k(2mn), k(m^2+n^2)\Big\} = \{6, 8, 10\}$
• If $n=1, k=1, m-n=2,$ then $m=2+1 = 3.$ This gives the pythagorean triple $\Big\{k(m^2-n^2), k(2mn), k(m^2+n^2)\Big\} = \{8, 6, 10\}$ (which is a repeat)

Therefore the sum of the hypotenuses of the distinct triangles is $13 + 10 = \boxed{23}$

$10+13==23$

Hmmmm, I think I posted this one here

But it's okay, I don't mind.