45th Mongolian Mathematical Olympiad

Answer: $ABCD$ be a convex and circumscribed quadrilateral.

First case, $ABCD$ be a not circumscribed quadrilateral. Hence, $I$ doesn't lie on the internal bisector of $A$, thus which bisector must be intersects with $BI$ and $CI$. Assume that, bisector of $A$ and line $DI$ meet at $I^{\prime }$. (Figure).



Suppose that $M\equiv B$ and $N\equiv I^{\prime }$ hence satisfying following conditions $M\in [BI)$, $N\in [DI)$ and $\angle MAN=\angle BA{I}^{\prime }=\frac{1}{2}\angle A$. But not satisfy $\angle MCN=\frac{1}{2}\angle C$. Because assume contrary, $\angle MCN=\frac{1}{2}\angle C=\angle BCI$, so $C{I}^{\prime }$ is bisector of angle $C$. Hence $ABCD$ be a circumscribed quadrilateral. This leads a contradiction.

Second case, $ABCD$ be a circumscribed and convex quadrilateral. Thus the bisector of angles $A$, $B$, $C$ and $D$ meets at $I$. (Figure)



In the figure we signed that angles $\alpha ,\beta ,\gamma ,\phi ,x,y,z$. By the Sinus's theorem, we get $$\begin{array}{rl}\frac{BM}{MI}\cdot \frac{DN}{NI}& =\frac{BM}{MA}\cdot \frac{MA}{MI}\cdot \frac{DN}{NA}\cdot \frac{NA}{NI}=\\ & =\frac{\sin x}{\sin \frac{\beta }{2}}\cdot \frac{\sin \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)}{\sin \left(\frac{\alpha }{2}-x\right)}\cdot \frac{\sin \left(\frac{\alpha }{2}-x\right)}{\sin \frac{\phi }{2}}\cdot \frac{\sin \left(\frac{\alpha +\phi }{2}\right)}{\sin x}=\\ & =\frac{\sin \frac{\alpha +\beta }{2}\cdot \sin \frac{\alpha +\beta }{2}}{\sin \frac{\beta }{2}\cdot \sin \frac{\phi }{2}}.\end{array}\text{\hspace{1em}(1)}$$

Analogously, $$\frac{BM}{MI}\cdot \frac{DN}{NI}=\frac{BM}{MC}\cdot \frac{MC}{MI}\cdot \frac{DN}{NC}\cdot \frac{NC}{NI}=$$ $$=\frac{\sin (\frac{\gamma }{2}-y)}{\sin \frac{\beta }{2}}\cdot \frac{\sin (\frac{\beta +\gamma }{2})}{\sin y}\cdot \frac{\sin (\frac{\gamma }{2}-z)}{\sin \phi }\cdot \frac{\sin (\frac{\alpha +\beta }{2})}{\sin z}.(2)$$ If we observe that $\frac{\alpha +\beta }{2}=\frac{\pi }{2}-\frac{\gamma +\phi }{2}$; $\frac{\alpha +\phi }{2}=\frac{\pi }{2}-\frac{\beta +\gamma }{2}$ hence, from (1) and (2) we get $$\frac{\sin (\frac{\gamma }{2}-y)}{\sin y}=\frac{\sin z}{\sin (\frac{\gamma }{2}-z)}=\frac{\sin (\frac{\gamma }{2}-(\frac{\gamma }{2}-z))}{\sin (\frac{\gamma }{2}-z)}$$ If and only if $\mathrm{ctg}y=\mathrm{ctg}(\frac{\gamma }{2}-z)$, because of the function $f(x)=\mathrm{ctg}x$ has $\pi k$ periodic, thus $y=\frac{\gamma }{2}-z$. In other words $\angle MCN=y+z=\frac{\gamma }{2}=\frac{1}{2}\cdot \angle C$.