Problem 9.2. Let $I$ be the incenter of $\triangle A B C$ and $M$ be the midpoint of the side $A B$. Find the least possible value of $\Varangle C I M$ if $C I=M I$.

Svetlozar Doychev, Sava Grozdev
9.2. We may assume that $A C<B C$. Since $\Varangle A C I$ and $\Varangle A M I$ are acute, then $\triangle A C I \cong \triangle A M I$.

Hence $A C=A M$ and $\Varangle A I C=\Varangle A I M$, i.e.

$$
\Varangle C I M=360^{\circ}-2 \Varangle A I C=180^{\circ}-\Varangle A B C .
$$

Note that $\Varangle A B C$ is maximal if $B C$ is tangent to the circle with center $A$ and radius $A M$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-094.jpg?height=316&width=462&top_left_y=560&top_left_x=1182)

Then $\Varangle A C B=90^{\circ}, \Varangle A B C=30^{\circ}$, and hence the least possible value of $\Varangle C I M$ is $150^{\circ}$.