A triangle with sides $a, b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s=(a+b+c) / 2$. Construct a triangle with sides $s-a, s-b$, and $s-c$. This process is repeated until a triangle can no longer be constructed with the sidelengths given.
For which original triangles can this process be repeated indefinitely?
Answer: Only equilateral triangles.
The perimeter of each new triangle constructed by the process is $(s-a)+(s-b)+(s-c) = 3s - (a+b+c) = 3s - 2s = s$, that is, it is halved. Consider a new equivalent process in which a similar triangle with side lengths $2(s-a), 2(s-b), 2(s-c)$ is constructed, so the perimeter is kept invariant.
Suppose without loss of generality that $a \leq b \leq c$. Then $2(s-c) \leq 2(s-b) \leq 2(s-a)$, and the difference between the largest side and the smallest side changes from $c-a$ to $2(s-a) - 2(s-c) = 2(c-a)$, that is, it doubles. Therefore, if $c-a > 0$ then eventually this difference becomes larger than $a+b+c$, and it's immediate that a triangle cannot be constructed with the side lengths. Hence the only possibility is $c-a = 0 \Longrightarrow a = b = c$, and it is clear that equilateral triangles can yield an infinite process, because all generated triangles are equilateral.