Belarusian Mathematical Olympiad

Answer: $30^{\circ }$.



Since $L$ lies on the bisector of $\angle KCB$, $L$ is equidistant from the lines $CP$ and $CB$. Similarly, since $L$ lies on the bisector of $\angle KMB$, $L$ is equidistant from the rays $MK$ and $MB$. Therefore, $L$ is an equidistant point for the rays $KM$ and $KP$, so $L$ lies on the bisector of $\angle PKM$. Thus, $\angle PKL=\angle BKM=\angle MKC$. Therefore, each of these angles is equal to $60^{\circ }$. Let $\angle ACB=3x$. Since $\angle LKC=120^{\circ }$, we have $\angle KAC=\angle LKC-\angle ACK=120^{\circ }-x$. Therefore $$\angle ABC=180^{\circ }-\angle BAC-\angle BCA=180^{\circ }-(120^{\circ }-x)-3x=60^{\circ }-2x.$$ So, $$\begin{array}{rl}\angle LMB& =0.5\angle KMB=0.5(180^{\circ }-\angle LKM-\angle ABC)=\\ & =0.5(180^{\circ }-60^{\circ }-(60^{\circ }-2x))=30^{\circ }+x.\end{array}$$ Since $\angle LMB=\angle LCM+\angle MLC$, we obtain $\angle MLC=30^{\circ }+x-x=30^{\circ }$.