THE DANUBE MATHEMATICAL COMPETITION

Let $\{D\}=AI\cap BC$. As $AB<AC$, $D$ lies between $B$ and $M$ and $\angle ACB<\angle ABC$. We have $\angle IDB=\angle DAC+\angle ACB<\angle DAB+\angle ABD=\angle ADC$, therefore angle $IDB$ is acute. Let $F$ and $E$ be the projections of $I$ onto $AB$ and $BC$, respectively. It follows that $E\in (BD)\subset BM$. Triangles $IBF$ and $IBE$ are congruent and so are triangles $IFA$ and $IEM$, therefore $BA=BM=\frac{BC}{2}$ and triangles $IBA$ and $IBM$ are congruent.



We have: $\angle MID=\angle IDB-\angle IMB=\angle DAC+\angle ACD-\angle IAB=\angle ACD$. It follows that $\angle AIM=180^{\circ }-\angle ACB$ (1). Let $H$ be the projection of $B$ onto the line $AC$. It follows that $BH\le AB=\frac{BC}{2}$, which shows that $\angle ACB\le 30^{\circ }$ (2). From (1) and (2) we obtain that $\angle AIM\ge 180^{\circ }-30^{\circ }=150^{\circ }$.