Given is a non-isosceles triangle $ABC$ with $\angle ABC=60^{\circ}$, and in its interior, a point $T$ is selected such that $\angle ATC= \angle BTC=\angle BTA=120^{\circ}$. Let $M$  the intersection point of the medians in $ABC$. Let $TM$ intersect $(ATC)$ at $K$. Find $TM/MK$.
1. **Define Point \( K \)**:
   Define \( K \) as the point on ray \( TM \) such that \(\frac{TM}{MK} = \frac{1}{2}\). We aim to prove that \( K \) belongs to the circumcircle of \(\triangle ATC\) by showing that \(\triangle AKC\) and \(\triangle BTA\) are similar.

2. **Similarity Condition**:
   To show similarity, we need to prove:
   \[
   \frac{AK}{BT} = \frac{KC}{TA} = \frac{AC}{BA}
   \]

3. **Define Point \( D \)**:
   Let \( D \) be the intersection of \( KN \) and \( BT \), where \( N \) is the midpoint of \( AC \). Given \(\frac{MT}{MK} = \frac{MN}{MB} = \frac{1}{2}\), it follows that \( TN \parallel BK \). Also, \(\frac{TN}{BK} = \frac{TM}{MK} = \frac{1}{2}\).

4. **Midpoint Relationships**:
   Since \( TN \parallel BK \), we have:
   \[
   \frac{DT}{DB} = \frac{DN}{DK} = \frac{TN}{BK} = \frac{1}{2}
   \]
   This implies \( N \) is the midpoint of \( DK \) and \( T \) is the midpoint of \( BD \).

5. **Similarity of Triangles**:
   We will prove that \(\triangle ATD\) and \(\triangle DTC\) are similar to \(\triangle ABC\).

6. **Angle Calculation**:
   \[
   \angle BAT = \angle BAC - \angle TAC = \angle BAC - \angle TSC
   \]
   Since \(\angle TSC = 180^\circ - \angle SBC - \angle ACB - \angle ACS\), and given \(\angle ABC = 60^\circ\), we have:
   \[
   \angle BAT = \angle BAC - (180^\circ - \angle SBC - \angle ACB - 60^\circ) = \angle BAC - (\angle BAC - \angle SBC) = \angle SBC
   \]
   Thus, \(\angle BAT = \angle SBC\).

7. **Similarity Conclusion**:
   Given \(\angle ATB = \angle BTC = 120^\circ\), \(\triangle ATB\) and \(\triangle BTC\) are similar. Hence:
   \[
   \frac{AT}{BT} = \frac{AB}{BC}
   \]
   \[
   \frac{TB}{TC} = \frac{AB}{BC}
   \]

8. **Midpoint Proportions**:
   Since \( T \) is the midpoint of \( BD \):
   \[
   \frac{AT}{TD} = \frac{AB}{BC}
   \]
   \[
   \frac{TD}{TC} = \frac{AB}{BC}
   \]

9. **Similarity of Triangles**:
   These relationships, along with \(\angle ATD = \angle DTC = 60^\circ = \angle ABC\), imply \(\triangle ATD\) and \(\triangle DTC\) are similar to \(\triangle ABC\).

10. **Parallelogram \( AKCD \)**:
    Since \( AKCD \) is a parallelogram:
    \[
    \frac{AK}{BT} = \frac{DC}{DT} = \frac{AC}{BA}
    \]
    because \(\triangle DTC\) is similar to \(\triangle ABC\).

11. **Final Similarity**:
    Similarly:
    \[
    \frac{KC}{TA} = \frac{AD}{AT} = \frac{AC}{BA}
    \]
    because \(\triangle ATD\) is similar to \(\triangle ABC\).

12. **Conclusion**:
    Thus, we have:
    \[
    \frac{AK}{BT} = \frac{KC}{TA} = \frac{AC}{BA}
    \]
    which means \(\triangle AKC\) and \(\triangle BTA\) are similar. Therefore, \(\angle AKC = \angle ATB = \angle ATC = 120^\circ\), proving that \( K \) belongs to the circumcircle of \(\triangle ATC\).

\(\blacksquare\)

The final answer is \(\boxed{\frac{TM}{MK} = 2}\).