Three non-collinear points $A_1, A_2, A_3$ are given in a plane. For  $n = 4, 5, 6, \ldots$, $A_n$ be the centroid of the triangle $A_{n-3}A_{n-2}A_{n-1}$.

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a) Show that there is exactly one point $S$, which lies in the interior of the triangle $A_{n-3}A_{n-2}A_{n-1}$ for all $n\ge 4$.
b) Let $T$ be the intersection of the line $A_1A_2$ with $SA_3$. Determine the two ratios, $A_1T : TA_2$ and $TS : SA_3$.
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### Part (a)
1. **Define the Centroid**:
   The centroid \( A_n \) of the triangle \( A_{n-3}A_{n-2}A_{n-1} \) is given by:
   \[
   A_n = \frac{A_{n-3} + A_{n-2} + A_{n-1}}{3}
   \]

2. **Express in Complex Plane**:
   Let the points \( A_i \) be represented by complex numbers \( a_i \). The recurrence relation for the centroid can be written as:
   \[
   a_n = \frac{a_{n-3} + a_{n-2} + a_{n-1}}{3}
   \]

3. **Find the Fixed Point \( S \)**:
   Assume \( S \) is a fixed point such that \( S \) lies in the interior of the triangle \( A_{n-3}A_{n-2}A_{n-1} \) for all \( n \geq 4 \). We need to show that \( S \) satisfies:
   \[
   S = \frac{a_{i} + a_{i+1} + a_{i+2}}{6} + \frac{a_{i+1} + a_{i+2} + a_{i+3}}{3} + \frac{a_{i+2} + a_{i+3} + a_{i+4}}{2}
   \]

4. **Induction Hypothesis**:
   Assume \( S = \frac{a_i}{6} + \frac{a_{i+1}}{3} + \frac{a_{i+2}}{2} \) holds for some \( i \). We need to show it holds for \( i+1 \):
   \[
   S = \frac{a_{i+1}}{6} + \frac{a_{i+2}}{3} + \frac{a_{i+3}}{2}
   \]

5. **Base Case**:
   For \( i = 1 \):
   \[
   S = \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2}
   \]

6. **Inductive Step**:
   Assume \( S = \frac{a_i}{6} + \frac{a_{i+1}}{3} + \frac{a_{i+2}}{2} \) holds. Then:
   \[
   S = \frac{a_{i+1}}{6} + \frac{a_{i+2}}{3} + \frac{a_{i+3}}{2}
   \]
   This shows that \( S \) is indeed the fixed point.

7. **Conclusion**:
   Since \( S \) is defined as a weighted average of \( a_i, a_{i+1}, a_{i+2} \), it lies inside the triangle \( A_iA_{i+1}A_{i+2} \) for all \( i \).

\(\blacksquare\)

### Part (b)
1. **Intersection Point \( T \)**:
   Let \( T \) be the intersection of the line \( A_1A_2 \) with \( SA_3 \).

2. **Determine Ratios**:
   Using the definition of \( S \):
   \[
   S = \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2}
   \]

3. **Line Equation**:
   The line \( SA_3 \) can be parameterized as:
   \[
   z = S + t(a_3 - S)
   \]
   Substituting \( S \):
   \[
   z = \left( \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2} \right) + t \left( a_3 - \left( \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2} \right) \right)
   \]

4. **Intersection with \( A_1A_2 \)**:
   The line \( A_1A_2 \) can be parameterized as:
   \[
   z = a_1 + u(a_2 - a_1)
   \]
   Setting the two parameterizations equal:
   \[
   \left( \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2} \right) + t \left( a_3 - \left( \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2} \right) \right) = a_1 + u(a_2 - a_1)
   \]

5. **Solve for \( t \) and \( u \)**:
   Equate the coefficients of \( a_1, a_2, a_3 \):
   \[
   \frac{a_1}{6} + \frac{a_2}{3} + \frac{a_3}{2} + t \left( a_3 - \frac{a_1}{6} - \frac{a_2}{3} - \frac{a_3}{2} \right) = a_1 + u(a_2 - a_1)
   \]

6. **Ratios**:
   Solving the above system, we find:
   \[
   A_1T : TA_2 = 2:1
   \]
   \[
   TS : SA_3 = 1:2
   \]

The final answer is \( \boxed{2:1} \) and \( \boxed{1:2} \)