RU-ALLRUSSIAN 2021 Problem 96

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 $TM/MK$ .