A triangle with sides $a,b,c$ and perimeter $2p$ is given. Is possible, a new triangle with sides $p-a$, $p-b$, $p-c$ is formed. The process is then repeated with the new triangle. For which original triangles can this process be repeated indefinitely?
To determine for which original triangles the process can be repeated indefinitely, we need to ensure that the new sides formed after each transformation still satisfy the triangle inequality. Let's break down the problem step-by-step.

1. **Initial Triangle Inequality Conditions:**
   For a triangle with sides \(a\), \(b\), and \(c\), the triangle inequality must hold:
   \[
   a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b.
   \]

2. **Transformation to New Triangle:**
   Given the perimeter \(2p = a + b + c\), the new sides after the first transformation are:
   \[
   p - a, \quad p - b, \quad \text{and} \quad p - c.
   \]
   We need to check if these new sides form a valid triangle by satisfying the triangle inequality.

3. **Triangle Inequality for New Sides:**
   The new sides must satisfy:
   \[
   (p - a) + (p - b) > (p - c), \quad (p - b) + (p - c) > (p - a), \quad \text{and} \quad (p - c) + (p - a) > (p - b).
   \]
   Simplifying these inequalities:
   \[
   2p - a - b > p - c \implies p > a + b - c,
   \]
   \[
   2p - b - c > p - a \implies p > b + c - a,
   \]
   \[
   2p - c - a > p - b \implies p > c + a - b.
   \]
   Since \(p = \frac{a + b + c}{2}\), we substitute \(p\) into the inequalities:
   \[
   \frac{a + b + c}{2} > a + b - c,
   \]
   \[
   \frac{a + b + c}{2} > b + c - a,
   \]
   \[
   \frac{a + b + c}{2} > c + a - b.
   \]

4. **Simplifying the Inequalities:**
   Simplify each inequality:
   \[
   \frac{a + b + c}{2} > a + b - c \implies a + b + c > 2a + 2b - 2c \implies c > a + b - c,
   \]
   \[
   \frac{a + b + c}{2} > b + c - a \implies a + b + c > 2b + 2c - 2a \implies a > b + c - a,
   \]
   \[
   \frac{a + b + c}{2} > c + a - b \implies a + b + c > 2c + 2a - 2b \implies b > c + a - b.
   \]

5. **Final Conditions:**
   The conditions simplify to:
   \[
   c > a + b - c, \quad a > b + c - a, \quad \text{and} \quad b > c + a - b.
   \]
   These are equivalent to:
   \[
   2c > a + b, \quad 2a > b + c, \quad \text{and} \quad 2b > c + a.
   \]

6. **Conclusion:**
   The process can be repeated indefinitely if the original triangle satisfies:
   \[
   2c > a + b, \quad 2a > b + c, \quad \text{and} \quad 2b > c + a.
   \]

The final answer is \( \boxed{ 2c > a + b, \quad 2a > b + c, \quad \text{and} \quad 2b > c + a } \).