## Problem 3

Consider a triangle $ABC$, with $AB < AC$, where point $I$ is the intersection of the angle bisectors and point $M$ is the midpoint of side $[BC]$. If $IA = IM$, determine the smallest possible measure of angle $AIM$.

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## Solution.

Answer: $150^{\circ}$.

Let $\{D\}=A I \cap B C$. Since $A B<A C$, then $D \in(B M)$ and $m(A C B)<m(A B C)$.

We have $m(I D B)=m(D A C)+m(A C B)<m(D A B)+m(A B D)=m(A D C)$, so angle IDB

![](https://cdn.mathpix.com/cropped/2024_06_07_6f474b90df0a4ef4e1ecg-2.jpg?height=389&width=745&top_left_y=825&top_left_x=227)
is acute.

Let $F$ and $E$ be the projections of point $I$ onto lines $A B$ and $B C$, respectively.

We deduce that $E \in(B D) \subset B M$.

Since $\triangle I B F \equiv \triangle I B E$ (CU) and $\triangle I F A \equiv \triangle I E M$ (IC), it follows that $B A=B M=\frac{B C}{2}$ and $\triangle I B A \equiv \triangle I B M$.

We have $m(M I D)=m(I D B)-m(I M B)=m(D A C)+m(A C D)-m(I A B)=m(A C D)$.

Therefore, $m(A I M)=180^{\circ}-m(A C B)$

Let $H$ be the projection of point $B$ onto line $A C$. It follows that $B H \leq A B=\frac{B C}{2}$.

We deduce that $m(A C B) \leq 30^{\circ}$. (2)

From (1) and (2), it follows that $m(A I M) \geq 180^{\circ}-30^{\circ}=150^{\circ}$.