62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

a) as from the statement $△XMC\sim △MBC$, we get $$\begin{array}{rl}\angle AMX& =180^{\circ }-\angle XMC-\angle BMC=\\ & =180^{\circ }-\angle XMC-\angle MXC=\angle MCX\end{array}$$

and $\frac{AM}{MX}=\frac{BM}{MX}=\frac{MC}{CX}\Rightarrow △AXM\sim △MXC\Rightarrow \angle XAM=\angle XMC$, so the claim follows from the equality of the inscribed angles.

Fig. 22

b) Let S be the second point of the intersection of AC and $\omega$. We need to show that the point S lies on NX. From the properties of inscribed angles, we get $\angle SMC=\angle SAM=\angle CAM$. It follows that $△CSM\sim △CMA\Rightarrow \frac{CM}{CA}=\frac{SM}{MA}$. Then, as $CM=2MN$ and $MA=\frac{1}{2}AB\Rightarrow \frac{MN}{CA}=\frac{SM}{AB}$. Together with the condition $\angle SMN=\angle SMC=\angle CAM=\angle CAB$ we get that $△SNM\sim △BCA$. Then $\angle MSN=\angle ABC$. Furthermore, $\angle XSM=180^{\circ }-\angle XAM$. From the claim in a), we also get $\angle XAM=\angle XMC=\angle MBC=\angle ABC$, so $\angle XSM=180^{\circ }-\angle ABC=180^{\circ }-\angle MSN$, and $S\in MN$.