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 $TM/MK$ & Write down The value of $TM+MK$ in the answer box.