49th Mathematical Olympiad in Ukraine

Let us prove that $AK$ is the bisector of $\angle BAM$. Point $P$ is equidistant from the lines $BC$ and $AC$, and also from the lines $BM$ and $BC$, since $\angle ABM=\angle MBC=\angle CBP=60^{\circ }$ (fig. 13). Thus $P$ is equidistant from the lines $BM$ and $MC$, which means that $P$ belongs to the bisector of $\angle BMC$. Therefore $K$ is equidistant from the lines $AC$ and $BM$, and also from $BM$ and $AP$, which proves that $AK$ is the bisector of $\angle BAC$. Denote $\angle BAK=\angle KAC=\alpha$. Then step by step we can calculate the following:

$$\begin{array}{rl}\angle KMC& =\frac{1}{2}\angle BMC=\frac{1}{2}(60^{\circ }+2\alpha )=30^{\circ }+\alpha .\\ \angle BCA& =60^{\circ }-2\alpha .\\ \angle BCP& =\frac{1}{2}(180^{\circ }-60^{\circ }+2\alpha )=60^{\circ }+\alpha ,\\ \angle ACP& =60^{\circ }-2\alpha +60^{\circ }+\alpha =120^{\circ }-\alpha .\\ \angle MPC& =180^{\circ }-(120^{\circ }-\alpha +30^{\circ }+\alpha )=30^{\circ };\\ \angle AKM& =180^{\circ }-\angle AMK-\alpha =\\ & =\angle KMC-\alpha =30^{\circ }+\alpha -\alpha =30^{\circ }=\angle MPC,\text{ what was to be proved.}\end{array}$$

Fig. 13