60th Belarusian Mathematical Olympiad

Let $P$ be the point of intersection of the medians $AM$ and $BK$. Draw the line $l$ through $A$ parallel to $BC$. Let $L$ be the point of intersection of $l$ and the line $BK$. The triangles $AKL$ and $BKC$ are equal ($AK=KC$, $\angle AKL=\angle BKC$, $\angle KAL=\angle BCK$), so $AL=BC=2AB$. Since $\angle BAL=180^{\circ }-\angle ABC=180^{\circ }-120^{\circ }=60^{\circ }$, we see that $ABL$ is a right-angled triangle and $\angle ABP=\angle ABL=90^{\circ }$.

Since $BM=MC=0.5BC=AB$, it follows that $ABM$ is an isosceles triangle and $$\angle BAM=\angle BMA=0.5(180^{\circ }-\angle ABC)=0.5(180^{\circ }-120^{\circ })=30^{\circ }.$$ $$\text{Therefore, }\angle BPA=180^{\circ }-\angle PAB-\angle ABP=180^{\circ }-30^{\circ }-90^{\circ }=60^{\circ }.$$