45th Mongolian Mathematical Olympiad

Let $O_1$ and $O_2$ be circumcentres of $△ABM$ and $△CDM$ respectively. Then first, we shall prove that $O{O}_{1}M{O}_2$ is parallelogram. Denote by $\angle ABM=\angle DCM=\phi$; $\Rightarrow \angle AM{O}_1=90^{\circ }-\phi$ and $\angle MCD=\phi \Rightarrow O_{1}M\perp CD$; $O{O}_2\perp CD\Rightarrow O_{2}M\parallel O{O}_2\Rightarrow O_{1}M{O}_{2}O$ is parallelogram. So it is sufficient to show that $$\angle O{O}_2{N}_2=\angle O{O}_{2}N.(1)$$ Indeed, from (1) $△O{O}_1{N}_1=△N_2{O}_{2}O\Rightarrow O{N}_1=O{N}_2$ and $O{M}_1=O{M}_2\Rightarrow M_1{N}_1=M_2{N}_2$.



Denote $\alpha ,\beta$ by $\angle BM{N}_1$; $\angle BAM$ respectively. Then $$\angle O_{1}M{O}_2=\angle O_{1}MA+\angle AMD+\angle O_{2}MD=90^{\circ }+\beta +\phi \Rightarrow$$

$$\begin{gathered} \Rightarrow \angle M{O}_{1}O=\angle M{O}_2)=90^{\circ }-\beta -\phi \Rightarrow \\ \to O{O}_1{N}_1=90^{\circ }-\beta -\phi +2\alpha +2\beta =90+\beta +2\alpha -\phi \end{gathered}$$ and after same computing we can see $\angle O{O}_2{N}_2=90^{\circ }+\beta +2\alpha -\phi =\angle O{O}_1{N}_1$.