Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse. 
I came up with 2 equations which are:
$\dfrac{ab}2 = 2(a+b+c)\;\;$  and  $\;\;a^2+b^2=c^2$
but I'm not sure how to proceed and solve for $a, b, c$.

With Euclid's formula $F(m,k)$, we can solve $R=$area/perimeter for $k$ and test a defined range of $m$-values to see which yield integers.
$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$
$$R=\frac{area}{perimeter}=\frac{AB}{2P} =\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}$$
\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 $m> k$.
$$R=2\implies  \lceil\sqrt{8(2)}\rceil=4\le m \le (2(2)+1)=5 \\ 
m\in\{ 4\}\implies k\in\{ 2\}
 \qquad\land\qquad
m\in\{ 5\}\implies k\in\{ 4,1\}
$$
$$F(4,2)=(12,16,20)\qquad\land\qquad\frac{96}{48}=2\\
   F(5,4)=(\space 9,40,41)\qquad\land\qquad\frac{180}{90}=2\\
   F(5,1)=(24,10,26)\qquad\land\qquad\frac{120}{60}=2\\
$$
We can see that only one of these is primitive but this formula finds non-primitives as well.
