How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $2$ times their perimeters?

1. Let the legs be $a,b$ and the hypotenuse $c$, and let the area equal $k$ times the perimeter, with $k=2$: $\frac{ab}2=k(a+b+c)$, so $ab-2k(a+b)=2kc$. Squaring and using $c^2=a^2+b^2$ gives $a^2b^2-4kab(a+b)+8k^2ab=0$, so $ab-4k(a+b)+8k^2=0$, i.e. $(a-4k)(b-4k)=8k^2$.
2. With $k=2$: $(a-8)(b-8)=32$. Both factors negative would need two factors in $[-7,-1]$ with product $32$, which is impossible. The positive factor pairs $1\cdot32$, $2\cdot16$, $4\cdot8$ give $(a,b)=(9,40),(10,24),(12,16)$, each with a positive integer $c$ ($41,26,20$): $3$ triangles.
3. The answer is $\boxed3$.