smc

Let the triangle have legs of length $a$ and $b$, so by the Pythagorean Theorem, the hypotenuse has length $\sqrt{a^2+b^2}$. Therefore we require $$\begin{array}{rl}& a+b+\sqrt{a^2+b^2}=\frac{1}{2}ab\\ \Rightarrow & 2\sqrt{a^2+b^2}=ab-2a-2b\\ \Rightarrow & 4a^2+4b^2=a^2{b}^2+4a^2+4b^2-4a^{2}b-4a{b}^2+8ab\\ \Rightarrow & 4a^{2}b+4a{b}^2=a^2{b}^2+8ab.\end{array}$$ Now, as $a$ and $b$ are side lengths of a triangle, they must both be non-zero, so we can safely divide by $ab$ to give $4a+4b=ab+8\Rightarrow b(a-4)=4a-8\Rightarrow b=\frac{4a-8}{a-4}$, so for any value of $a$ other than $4$, we can generate a valid corresponding value of $b$. Notice also that each of these values of $a$ will give a unique corresponding value of $b$, since $\frac{4a-8}{a-4}=4+\frac{8}{a-4}$, and by considering the graph of $y=4+\frac{8}{x-4}$, it is clear that any horizontal line will intersect it at most once. Thus there are infinitely many valid solutions (one for every value of $a$ except $4$), so the answer is $\boxed{\text{infinitely many}}$.