Points $A, B$ are given. Find the locus of points $C$ such that $C$, the midpoints of $AC, BC$ and the centroid of triangle $ABC$ are concyclic.
1. **Define the points and the centroid:**
   - Let \( A \) and \( B \) be given points.
   - Let \( C \) be a point such that the midpoints of \( AC \) and \( BC \) and the centroid of triangle \( ABC \) are concyclic.
   - Denote by \( G \) the centroid of triangle \( ABC \).
   - Denote by \( M \) the midpoint of segment \( AB \).

2. **Identify the homothety:**
   - Consider the homothety \(\mathcal{H}(A, 2)\) centered at \( A \) with a ratio of 2. This homothety maps the centroid \( G \) to a point such that \( GC = GX \), where \( X \) is the second intersection of \( CM \) with the circumcircle \(\omega\) of triangle \( ABC \).

3. **Calculate the power of point \( M \) with respect to the circumcircle \(\omega\):**
   - The power of point \( M \) with respect to the circumcircle \(\omega\) is given by:
     \[
     p(M, \omega) = \frac{1}{4} AB^2
     \]
   - This is because \( M \) is the midpoint of \( AB \), and the power of a point with respect to a circle is the square of the distance from the point to the center minus the square of the radius.

4. **Relate the power of point \( M \) to the median \( CM \):**
   - The power of point \( M \) can also be expressed in terms of the segments created by the median \( CM \):
     \[
     p(M, \omega) = \frac{1}{3} CM \cdot \frac{2}{3} CM
     \]
   - This follows from the fact that the centroid \( G \) divides the median \( CM \) in the ratio 2:1.

5. **Equate the expressions for the power of point \( M \):**
   - Equating the two expressions for the power of point \( M \), we get:
     \[
     \frac{1}{4} AB^2 = \frac{1}{3} CM \cdot \frac{2}{3} CM
     \]
   - Simplifying this, we find:
     \[
     \frac{1}{4} AB^2 = \frac{2}{9} CM^2
     \]
     \[
     CM^2 = \frac{9}{8} AB^2
     \]
     \[
     CM = \frac{3}{2\sqrt{2}} AB
     \]

6. **Conclusion:**
   - The median \( CM \) has a fixed length, which describes the desired locus of points \( C \). This fixed length ensures that the points \( C \), the midpoints of \( AC \) and \( BC \), and the centroid \( G \) are concyclic.

The final answer is the locus of points \( \boxed{ C } \) such that the median \( CM \) has a fixed length \(\frac{3}{2\sqrt{2}} AB\).