Circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$, respectively, pass through the centers of each other. Let $A$ be one of their intersection points. Two points $M_1$ and $M_2$ begin to move simultaneously starting from $A$. Point $M_1$ moves along $S_1$ and point $M_2$ moves along $S_2$. Both points move in clockwise direction and have the same linear velocity $v$.

(a) Prove that all triangles $AM_1M_2$ are equilateral.
(b) Determine the trajectory of the movement of the center of the triangle $AM_1M_2$ and find its linear velocity.

### Part (a): Prove that all triangles \( AM_1M_2 \) are equilateral.

1. **Given Information and Setup:**
   - Circles \( S_1 \) and \( S_2 \) have centers \( O_1 \) and \( O_2 \) respectively.
   - Both circles pass through the centers of each other, i.e., \( O_1 \) lies on \( S_2 \) and \( O_2 \) lies on \( S_1 \).
   - \( A \) is one of the intersection points of \( S_1 \) and \( S_2 \).
   - Points \( M_1 \) and \( M_2 \) start at \( A \) and move along \( S_1 \) and \( S_2 \) respectively with the same linear velocity \( v \).

2. **Lemma:**
   - For two circles \( w_1 \) and \( w_2 \) with centers \( O_1 \) and \( O_2 \), intersecting at points \( A \) and \( B \), and points \( M \) and \( N \) on \( w_1 \) and \( w_2 \) respectively, \( \triangle O_1AO_2 \sim \triangle MAN \) if and only if \( M, N, B \) are collinear.

3. **Proof of Lemma:**
   - If \( M, N, B \) are collinear, then:
     \[
     \angle AO_1O_2 = \frac{\angle AO_1B}{2} = \angle AKB = \angle AKN
     \]
     and
     \[
     \angle AO_2O_1 = \angle ANM
     \]
     which implies \( \triangle O_1AO_2 \sim \triangle MAN \).

4. **Application to the Problem:**
   - Since \( S_1 \) and \( S_2 \) have the same radius \( r \), having the same linear velocity \( v \) implies the same angular velocity.
   - As \( M_1 \) and \( M_2 \) both start at \( A \) and move in the clockwise direction, we have:
     \[
     \angle ABM_1 = \angle ABM_2
     \]
     which implies that \( B, M_1, M_2 \) are collinear (where \( B \) is the second intersection point of \( S_1 \) and \( S_2 \)).

5. **Conclusion:**
   - By the lemma, \( \triangle AM_1M_2 \sim \triangle AO_1O_2 \).
   - Since \( O_1O_2 = AO_1 = AO_2 = r \), \( \triangle AO_1O_2 \) is equilateral.
   - Therefore, \( \triangle AM_1M_2 \) is also equilateral.

\(\blacksquare\)

### Part (b): Determine the trajectory of the movement of the center of the triangle \( AM_1M_2 \) and find its linear velocity.

1. **Center of the Equilateral Triangle:**
   - The center of an equilateral triangle is also its centroid.
   - The centroid of \( \triangle AM_1M_2 \) divides each median in the ratio 2:1.

2. **Trajectory of the Centroid:**
   - Since \( M_1 \) and \( M_2 \) move along circles \( S_1 \) and \( S_2 \) respectively, the centroid of \( \triangle AM_1M_2 \) will trace a circular path.
   - The radius of this circular path is \(\frac{2}{3}\) of the radius of the original circles \( S_1 \) and \( S_2 \).

3. **Linear Velocity of the Centroid:**
   - The linear velocity of the centroid is also \(\frac{2}{3}\) of the linear velocity \( v \) of \( M_1 \) and \( M_2 \).

\[
\text{Radius of the centroid's path} = \frac{2}{3}r
\]
\[
\text{Linear velocity of the centroid} = \frac{2}{3}v
\]

The final answer is \( \boxed{ \frac{2}{3}v } \)