Area........Perimeter

Let $P$ be the perimeter and $A$ the area. So, $P=a+b+c$ and $A=\dfrac{ab}{2}$ . We need that $A=2P$ , so:

$\dfrac{ab}{2}=2(a+b+c)$

Also, we know that $c=\sqrt{a^2+b^2}$ , so:

$ab=4(a+b+\sqrt{a^2+b^2})$

Rearrange to delete that square root, and express $a^2+b^2$ as $(a+b)^2-2ab$ :

$ab-4(a+b)=4\sqrt{(a+b)^2-2ab}$

Square both sides:

$(ab)^2-8ab(a+b)+16(a+b)^2=16(a+b)^2-32ab$

Rearrange again, simplify and solve for $a$ :

$(ab)^2+32ab=8ab(a+b)$

$ab+32=8(a+b)$

$ab-8a=8b-32$

$a=\dfrac{8b-32}{b-8}$

Now, let's assume that $a>b$ . The only three pairs of positive integers $a$ and $b$ that satisfy the equation and the condition are: $(a,b)=(16,12),(24,10),(40,9)$

From these pairs, $(a,b)=(40,9)$ maximizes the area, so the maximum area is:

$A=\dfrac{ab}{2}=\dfrac{40 \times 9}{2}=\boxed{180}$

Using the formula for Pythagorean triples, let

$a={ (m }^{ 2 }-{ n }^{ 2 })k$
$b=2mnk$
$c={ (m }^{ 2 }+{ n }^{ 2 })k$
Then the condition $\dfrac { 1 }{ 2 } ab=2(a+b+c)$

becomes, after re-arranging and factoring

$2km(m+n)(k{ n }^{ 2 }-kmn+4)=0$

so solving for $m$ we have $m=\dfrac { k{ n }^{ 2 }+4 }{ kn }$

which means that $kn$ must divide into $4$ .

Now, the area becomes $8k{ n }^{ 2 }+\dfrac { 64 }{ k{ n }^{ 2 } } +48$

which has the maximum value of $180$

when $k=1$ and $n=4$

Let the sides of right triangle ABC be expressed by the integral values:

$a = p^2 - q^2$ , $b = 2pq$ , $c = p^2 + q^2$

where $p,q \in \mathbb{N}$ and $p > q$ . Knowing the area is twice the perimeter, we arrive at:

$\frac{ab}{2} = 2(a + b + c) \Rightarrow \frac{(p^2 - q^2)(2pq)}{2} = 2[(p^2 - q^2) + 2pq + (p^2 + q^2)];$

or $(p+q)(p-q)pq = 2(2p^2 + 2pq);$

or $(p+q)(p-q)pq = 4p(p+q);$

or $(p-q)q = 4;$

or $p = \frac{4}{q} + q$ .

If p is to be a positive integer, then we require $q = 1, 2, 4 \Rightarrow (p,q) = (5,1); (4,2); (5,4)$ . The corresponding triplets $(a,b,c)$ for right triangle ABC are computed to be $(a,b,c) = (24,10,26); (12,16,20); (9,40,41)$ . Of these three choices, the $(9,40,41)$ primitive triplet yields the maximum area of $\boxed{180}.$

$Working ~out~ the~ list~ of ~Pythagorean~ Triples.\\ With~ all~ letters~ representing ~integers,~~\\ Primitive~ triple~ (a,b,c),~ P=perimeter, ~T=area,~ and~ k=\dfrac T P, ~a ~constant~ for~ a~ given~ (a,b,c). \\ Value ~of~ k~ increases~ as~ triangles ~become~ bigger.\\ First \ 7\ primitive\ triples\ are, \\ (3,4,5),\ \ \ (5,12,13),\ \ \ (8,15,17),\ \ \ (7,24,25),\ \ \ (9,40,41),\ \ \ (11,60,61),\ \ \ (12,35,37),\\ \color{#EC7300}{\text{We are interested in k=2, and maximum area. So, we start from the bigger triples.}\\ (12,35,37)\ gives\ \ \ \dfrac {T_1}{P_1}=\dfrac {210}{84}=2.5=k>2. \text{ So we should go for smaller triangle.}\\ (11,60,61)\ gives\ \ \ \dfrac {T_1}{P_1}=\dfrac {330}{132}=2.5=k>2. \text{ So we should still go for smaller triangle.}\\ (9,40,41)\ gives\ \ \ \dfrac {T_1}{P_1}=\dfrac {180}{90}=2. \ \ OK.\\ So\ the\ area\ required \ is\ \ \Large\ \ \ \color{#D61F06}{180}.}\\ \ \ \\ \\ For~triple~(a,b,c),~~ T=\frac 1 2 *a*b,~~P=a+b+c,\\ \therefore~k=\dfrac{ \frac 1 2 *a*b}{a+b+c}.\\ for~triple~(na,nb,nc),~~ T_n=\frac 1 2 *na*nb,~~P_n=n(a+b+c),\\ \therefore~\dfrac {T_n}{ P_n}=n*k.\\ Just\ to\ further~investigate\ non-ptimitive~SMALLER~triples. \\ (7,24,25),\ gives\ \ \ k=\dfrac T P=\dfrac {84}{56}=1.5<2.0\ \\ For~n=2,\ \ \dfrac {T_2}{P_2}=nk=3>2.\ \ \text{So no solution with this triple.}\\ (8,15,17),\ gives\ \ \ k=\dfrac {T_1}{P_1}=\dfrac {60}{40}=1.5<2,\ \ so \ just\ as\ above.\ No\ solution\ here\ also.\\ (5,12,13), \ gives\ \ \ k=\dfrac {T_1}{P_1}=\dfrac {30}{30}=1<2.\\ So\ if~n=2,~~\ \ nk=2,....\ and\ (10,24,26), ......then~T_2=\color{#3D99F6}{120<180\ we\ got~above.}\\ (3,4,5),\ gives\ \ k= \dfrac {T_1}{P_1}=\dfrac {6}{12}=0.5.\\ So\ if~ n=4,\ \ nk=4*.5=2 ,\ and\ (12,16,20),\ \ T_4=\color{#3D99F6}{96<180 \ ~we \ got~above.} \\ So~maximum~area~180~only~with~primitive~triples.$

I used the all the triplets and eventually 9-40-41 triplet matched