This is more or less the first part of LMAO 2021 P5

Let $MN$ meet the sides $AB$ and $AC$ at points $D$ and $E$, respectively. We'll show incircle of $\triangle ABC$, is the $A$-excircle of $\triangle ADE$. By shooting lemma, we have $AM^2=AI^2=AD\cdot AB \implies AI$ is tangent to $(BDI)$ and similarly, $AI$ is tangent to $(CEI)$. Therefore, $\angle DIE = \angle DIA + \angle EIA= \frac{\angle B + \angle C}{2}=90- \frac{\angle A}{2}.$ So $I$ is the $A$-excenter of $\triangle ADE.$ and we are done.

We will use inversion and angle chasing: Denote $\mathcal I(A, AI)$ the inversion with center $A$ and the radius $AI$. Note that $\mathcal I(M)=M$ and $\mathcal I(N)=N$. Now $BMNC$ is cyclic and hence $\mathcal I(B)$ and $\mathcal I(C)$ must lie on $MN$ but note that the inversed point is colinear with the original point and the center of invertion and hence: $\mathcal I(B)=MN \cap AB=B'$ and $\mathcal I(C)=MN \cap AC=C'$ and then $B'C'CB$ is cyclic. Now the problem is equivalent to prove that the angle bisectors on $B'C'CB$ are concurrent at $I$. Since $B'C'CB$ is cyclic $\angle ABC=\angle AC'B'=180º-\angle B'C'C$ and $\angle ACB=\angle AB'C'=180º-\angle BB'C'$. By another propety of inversion $AI^2=AC' \cdot AC \implies \triangle AC'I \sim \triangle AIC \implies \angle AC'I=\angle AIC$ and $AI^2=AB' \cdot AB \implies \triangle AB'I \sim \triangle AIB \implies \angle AB'I=\angle AIB$. Now by angle chasing: $\angle AIC=\angle AC'B'+\angle B'C'I=\angle ABC+\angle B'C'I=90º+\frac{\angle ABC}{2} \implies \angle B'C'I=\angle IC'C=90º-\frac{\angle ABC}{2}$ This means $\frac{\angle B'C'C}{2}=90º-\frac{\angle ABC}{2}=\angle B'C'I=\angle IC'C$ and that means $CI$ bisects $\angle B'C'C$. By the same way you get $BI$ bisects $\angle BB'C'$ and then $B'C'CB$ has an incircle. In fact the incircle of $\triangle ABC$ and the incircle of $B'C'CB$ are the same and that means $B'C'$ tangent to the incircle and that means $MN$ tangent to the incircle of $\triangle ABC$. Thus we are done

Also same as Peru TST 2015.

My solution in the contest: Let $U$ be the midpoint of the smaller arc $BC$ in the circle $(ABC)$. Let $I'$ be the incenter in the triangle $UMN$, $r_A$ be the radius of the incircle in the triangle $ABC$ and $r_U$ be the radius of the incircle of the triangle $UMN$, $R$ be the radius of $(ABC)$. From the Incenter-Excenter Lemma it follows that $I=I'$. From Euler's relation in the triangles $ABC$ and $UMN$ it follows that $R^2-2Rr_A=OI^2=R^2-2Rr_U$, which implies $r_U=r_A$. From this and $I=I'$ it follows that the triangles $ABC$ and $UMN$ have the same incircle, and the problem is solved.

Let $H= AI\cup(ABC)$. To prove $MN$ is tangent to the incircle $\iff \triangle HMN$ shares the same incircle with $ \triangle ABC $. So since we know that $A$ is the $H$ south pole of $ \triangle HMN$, if we prove $ \angle MIN= \frac{\pi+ \angle MHN}{2} $ we are done because $I$ is the only point inside $ \triangle HMN$ that holds that property. Proof: $ \angle MIN=a, \angle MAN=2(\pi-a)= \pi- \angle MHN \iff \pi+ \angle MHN= 2a \iff \angle MIN= \frac{\pi+ \angle MHN}{2}$ So $I$ is the incenter of $ \triangle HMN$ and so $MN$ is tangent to the incircle.

Here's a solution I wrote back in 2015.