Answer: $A C=13$.

Solution. (Fig. 5). First, we will prove that $M I=M A$ (trident lemma).

Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equal to the sum of angles $I A C$ and $C A M$. But $\angle I A C=\frac{1}{2} \angle A$, and $\angle C A M=\angle C B M=\frac{1}{2} \angle B$ - as inscribed angles.

From this, it follows that $\angle I A M=\frac{1}{2} \angle A+\frac{1}{2} \angle B=\angle A I M$, and therefore, triangle $A M I$ is isosceles, $M I=M A$. By the condition $M O=M I$, so by the trident lemma $A O=M O=M I=M A$. Thus, triangle $A O M$ is equilateral and $\angle A O M=60^{\circ}$. Since the central angle $A O M$ is twice the inscribed angle $A B M$, we have $\frac{1}{2} \angle B=30^{\circ}$, that is, $\angle B=60^{\circ}$.

By the cosine theorem, $A C^{2}=A B^{2}+B C^{2}-2 \cdot A B \cdot B C \cos 60^{\circ}=13^{2}$.

Criteria. Answer without explanation - 0 points. Proved that angle $B$ is $60^{\circ}-5$ points. Points are not deducted for the absence of proof of the trident lemma. Full solution - 7 points.
3. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$.