Let $P_0$ be a point in the plane of triangle $A_0A_1A_2$. Define $P_i\ (i=1,\ldots ,6)$ inductively as the point symmetric to $P_{i-1}$ with respect to $A_k$, where $k$ is the remainder when $i$ is divided by $3$.

a) Prove that $P_6\equiv P_1$.
b) Find the locus of points $P_0$ for which $P_iP_{i+1}$ does not meet the interior of $\triangle A_0A_1A_2$ for $0\le i\le 5$.
### Part (a): Prove that \( P_6 \equiv P_0 \)

1. **Define the Symmetry Transformation:**
   Let \( P_0 \) be a point in the plane of triangle \( A_0A_1A_2 \). Define \( P_i \) (for \( i = 1, \ldots, 6 \)) inductively as the point symmetric to \( P_{i-1} \) with respect to \( A_k \), where \( k \) is the remainder when \( i \) is divided by 3.

2. **Calculate \( P_1 \):**
   Since \( P_1 \) is symmetric to \( P_0 \) with respect to \( A_1 \), we have:
   \[
   \frac{P_0 + P_1}{2} = A_1 \implies P_1 = 2A_1 - P_0
   \]

3. **Calculate \( P_2 \):**
   Since \( P_2 \) is symmetric to \( P_1 \) with respect to \( A_2 \), we have:
   \[
   \frac{P_1 + P_2}{2} = A_2 \implies P_2 = 2A_2 - P_1 = 2A_2 - (2A_1 - P_0) = 2A_2 - 2A_1 + P_0
   \]

4. **Calculate \( P_3 \):**
   Since \( P_3 \) is symmetric to \( P_2 \) with respect to \( A_0 \), we have:
   \[
   \frac{P_2 + P_3}{2} = A_0 \implies P_3 = 2A_0 - P_2 = 2A_0 - (2A_2 - 2A_1 + P_0) = 2A_0 - 2A_2 + 2A_1 - P_0
   \]

5. **Calculate \( P_4 \):**
   Since \( P_4 \) is symmetric to \( P_3 \) with respect to \( A_1 \), we have:
   \[
   \frac{P_3 + P_4}{2} = A_1 \implies P_4 = 2A_1 - P_3 = 2A_1 - (2A_0 - 2A_2 + 2A_1 - P_0) = 2A_1 - 2A_0 + 2A_2 - 2A_1 + P_0 = 2A_2 - 2A_0 + P_0
   \]

6. **Calculate \( P_5 \):**
   Since \( P_5 \) is symmetric to \( P_4 \) with respect to \( A_2 \), we have:
   \[
   \frac{P_4 + P_5}{2} = A_2 \implies P_5 = 2A_2 - P_4 = 2A_2 - (2A_2 - 2A_0 + P_0) = 2A_2 - 2A_2 + 2A_0 - P_0 = 2A_0 - P_0
   \]

7. **Calculate \( P_6 \):**
   Since \( P_6 \) is symmetric to \( P_5 \) with respect to \( A_0 \), we have:
   \[
   \frac{P_5 + P_6}{2} = A_0 \implies P_6 = 2A_0 - P_5 = 2A_0 - (2A_0 - P_0) = P_0
   \]

Thus, we have shown that \( P_6 = P_0 \).

\(\blacksquare\)

### Part (b): Find the locus of points \( P_0 \) for which \( P_iP_{i+1} \) does not meet the interior of \( \triangle A_0A_1A_2 \) for \( 0 \le i \le 5 \)

1. **Condition for \( P_iP_{i+1} \) not meeting the interior:**
   For \( P_iP_{i+1} \) to not meet the interior of \( \triangle A_0A_1A_2 \), \( P_0 \) must lie on the boundary of the triangle. This is because if \( P_0 \) is inside the triangle, the symmetry operations will cause the points \( P_i \) to traverse through the interior of the triangle.

2. **Locus of \( P_0 \):**
   The locus of points \( P_0 \) for which \( P_iP_{i+1} \) does not meet the interior of \( \triangle A_0A_1A_2 \) is the boundary of the triangle \( \triangle A_0A_1A_2 \).

\(\boxed{\text{The boundary of } \triangle A_0A_1A_2}\)