Let $D$ and $F$ be the points where the incircle of $\triangle{ABC}$ is tangent to $BC$ and $BA$. Because $\angle{IBA} = \angle{IBM}$ and $IA = IM$ and $\triangle{ABI}$ and $\triangle{MBI}$ have a common side $BI$ which means that they are congruent, so $BA = BM = \frac{BC}{2}$ so $BC = 2BA$. Observe that $I$ is the intersection of the perpendicular bisector of $AM$ and the angle bisector of $\angle{ACM}$. Hence, $I$ is the midpoint of arc $AM$ of the circumcircle of $\triangle{AMC}$ so $\angle{AIM} = 180^{\circ} - \angle{C}$. But using the law of cosine in $\triangle{ABC}$ gives \[\cos \angle{C} = \frac{CA^2 + CB^2 - AB^2}{2 \cdot CA \cdot CB} = \frac{b^2 + 4c^2 - c^2}{4bc} = \frac{b^2 + 3c^2}{4bc}\] By AM-GM : \[\cos \angle{C} = \frac{b^2 + 3c^2}{4bc} \geq \frac{2bc\sqrt{3}}{4bc} = \frac{\sqrt{3}}{2}\] Thus, $\angle{C} \leq 30^{\circ} \longrightarrow \angle{AIM} \geq 150^{\circ}$. So, the least possible value of $\angle{AIM}$ is $150^{\circ}$. Equality holds when $\triangle{ABC}$ is a 90-60-30 triangle.

MrRTI wrote: Let $D$ and $F$ be the points where the incircle of $\triangle{ABC}$ is tangent to $BC$ and $BA$. Because $\angle{IBA} = \angle{IBM}$ and $IA = IM$ and $\triangle{ABI}$ and $\triangle{MBI}$ have a common side $BI$ which means that they are congruent, so $BA = BM = \frac{BC}{2}$ so $BC = 2BA$. Observe that $I$ is the intersection of the perpendicular bisector of $AM$ and the angle bisector of $\angle{ACM}$. Hence, $I$ is the midpoint of arc $AM$ of the circumcircle of $\triangle{AMC}$ so $\angle{AIM} = 180^{\circ} - \angle{C}$. But using the law of cosine in $\triangle{ABC}$ gives \[\cos \angle{C} = \frac{CA^2 + CB^2 - AB^2}{2 \cdot CA \cdot CB} = \frac{b^2 + 4c^2 - c^2}{4bc} = \frac{b^2 + 3c^2}{4bc}\] By AM-GM : \[\cos \angle{C} = \frac{b^2 + 3c^2}{4bc} \geq \frac{2bc\sqrt{3}}{4bc} = \frac{\sqrt{3}}{2}\] Thus, $\angle{C} \leq 30^{\circ} \longrightarrow \angle{AIM} \geq 150^{\circ}$. So, the least possible value of $\angle{AIM}$ is $150^{\circ}$. Equality holds when $\triangle{ABC}$ is a 90-60-30 triangle. An alternate (and synthetic) way to finish the problem after finding $BA=BM=\frac{BC}{2}$: Observe that since $AIF$ and $MID$ are similar, then $IMB=IAB\Rightarrow IMC=180-IMB=180-IAB=180-IAC\Rightarrow AIMC$ is a cyclic cuadrilateral. Hence $AIM=180-C$ Now we draw in the cartesian plane in the following way: $B=(0,0),C=(2,0),M=(1,0)$ and $A\in \omega$, where $\omega$ is the circle of center $(0,0)$ and radius $1$. We can see de angle $C$ is maximized when the line $AC$ is tangent to $\omega$, which happens iff $BAC=90\Leftrightarrow AM=MB=AC\Leftrightarrow C =30$. So $C\le 30\Rightarrow AIM\ge 150$. And we know its the smallest value achievable because it takes place when $A=90$ Also, in this problem it doesn't make much difference, but we must be aware we chose $BI<IC\Leftrightarrow BA<CA$, the case $CA>BA$ is symetrical, and the case $BA=CA$ trivially makes $AIM=180$. In another problem this edge case may have cost us some points for not considering it. In conclusion $150$ is the smalles achievable value.

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