No Actual Points?

Consider a triangle as described as described by the problem. Label it's vertices $A,B,C$ so that $A=(a,b)$ , $B$ lies on the $x$ -axis, and $C$ lies on the line $y = x$ . Furthermore, let $D=(-a,-b)$ be the reflection of $A$ in the $x$ -axis, and let $E=(b,a)$ be the reflection of $A$ in the line $y=x$ . Thus, $AB=DB$ and $AC=CE$ , so the perimeter of $ABC$ is $DB+BC+CE \geq DE=\sqrt{(a-b)^2+(a+b)^2} = \sqrt{2a^2+2b^2}$ . This lower bound can be achieved: set $B \ (\text{resp.},C)$ to be the intersection between the segment $DE$ and the $x$ -axis $(\text{resp., the line} \ x=y)$ . Therefore, the minimum perimeter is in fact $\boxed{\sqrt{2a^2+2b^2}}$