Silk Road Mathematics Competition competition problems

Let's define notations of the points and angles as in the figure. Then
$$\angle C_2AB = \angle C_2CB = \angle C_2CA = \angle C_2A_2A = \gamma,$$
$$\angle A_2CB = \angle A_2AB = \angle A_2AC = \angle A_2C_2C = \alpha$$
and
$$\angle C_2AC = \angle A_2CA = \angle ABC = 2\beta.$$

By the law of sines in $\Delta C_2AC_1$ and $\Delta C_2AI$ we easily obtain:
$$\frac{C_2C_1}{C_1I} = \frac{\sin \gamma}{\sin \alpha} \cdot \frac{\sin \angle C_2IA}{\sin \angle AC_2I} = \frac{\sin \gamma}{\sin \alpha} \cdot \frac{\sin (\alpha + \gamma)}{\sin 2\beta}.$$
Similarly, in $\Delta IA_2C$ we get:
$$\frac{IA_1}{A_1A_2} = \frac{\sin \gamma}{\sin \alpha} \cdot \frac{\sin \angle IA_2C}{\sin \angle A_2IC} = \frac{\sin \gamma}{\sin \alpha} \cdot \frac{\sin 2\beta}{\sin (\alpha + \gamma)}.$$
Then
$$(1) \qquad \frac{C_2C_1}{C_1I} \cdot \frac{IA_1}{A_1A_2} = \left( \frac{\sin \gamma}{\sin \alpha} \right)^2$$
By Ceva's theorem in $\Delta C_2A_2I$ we have
$$(2) \qquad \frac{C_2N}{NA_2} = \frac{C_2C_1}{C_1I} \cdot \frac{IA_1}{A_1A_2} = \left( \frac{\sin \gamma}{\sin \alpha} \right)^2$$
In $\Delta C_2IN$ and $\Delta A_2IN$, using (2) we obtain
$$(3) \qquad \frac{C_2IN}{NIA_2} = \frac{C_2N \sin \alpha}{N_2A \sin \gamma} = \left( \frac{\sin \gamma}{\sin \alpha} \right)^2 \cdot \frac{\sin \alpha}{\sin \gamma} = \frac{\sin \gamma}{\sin \alpha}$$
In $\Delta AIM$ and $\Delta CIM$, using (3) we get
$$AM = \frac{MI \sin \alpha}{\sin \angle AIM} = \frac{MI \sin \alpha}{\sin \angle NIA_2} = \frac{MI \sin \gamma}{\sin \angle C_2IN} = \frac{MI \sin \gamma}{\sin \angle MIC} = MC$$