2007 AMC 12B Problem 23

Let the legs be $a\le b.$ The condition is $\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right),$ so $$ab-6a-6b=6\sqrt{a^2+b^2}.$$ Squaring and simplifying gives $ab(ab-12a-12b+72)=0,$ hence $(a-12)(b-12)=72.$ The positive integer solutions are $(a,b)=(3,4),$ $(13,84),$ $(14,48),$ $(15,36),$ $(16,30),$ $(18,24),$ $(20,21).$ The pair $(3,4)$ is extraneous: its area is $6,$ while its perimeter is $12$ and three times that is $36.$ So exactly $6$ triangles work. Thus, the correct answer is A.