Let $MN$ intersect $AB,AC$ at points $P,Q$. It suffices to show that $BPQC$ has an incircle. Thus it suffices to show that $IP$ bisects $\angle BPQ$. Now see that from shooting lemma $AM^2=AP\cdot AB = AI^2$. Thus, $AI$ is tangent ti $\odot (BIP)$. Thus angle chase yields $\angle BPI = 90^\circ - \frac{\angle C}{2}$. Also note since $AM=AN$ we have that $PQ$ is antiparallel to $BC$. Thus, $\angle QPA = \angle C$. Thus we have $\angle QPI = 90^\circ - \frac{\angle C}{2}$. So $PI$ bisects $\angle BPQ$ and so $I$ is the incenter of $BPQC$ which means $MN$ is tangent to incircle. $\blacksquare$. Also, I'm interested to know the source of this problem. EDIT: @below, a simpler proof of the incircle mapping to mixtilinear is here: Let $D,E$ be the intouch points on $AB$ and $D',E'$ be the mixtilinear touch point on $AB,AC$. See that $\angle D'IA = 90^\circ = \angle IDA$. So, $AI$ is tangent to the circle $IDD'$. So, $AI^2=AD\cdot AD'$. So, $D$ maps to $D'$ under this inversion. Also, $AB$ and $AC$ map to themselves. Since incircle is tangent to $AB,AC$ we see that it maps to circle tangent to $AB,AC$ at $D',E'$ which is the mixtilinear incircle.

Which year of Peru TST is this?

Dear Mathlinkers http://www.artofproblemsolving.com/community/c6t48f6h1096581_mn_is_tangent_to_the_incircle https://artofproblemsolving.com/community/c6t48f6h1425177_geometry Sincerely Jean-Louis

Here is my solution for this problem Solution Let $K$, $L$ be intersections of $MN$ with $AC$, $AB$ It's easy to prove that $BCKL$ is cyclic quadrilateral But: $\widehat{AML}$ = $\widehat{ABM}$ then $AI^2$ = $AM^2$ = $AL$ .$AB$ So: $\widehat{ALI}$ = $\widehat{AIB}$ = $90^o$ + $\dfrac{\widehat{ACB}}{2}$ or $\widehat{BLI}$ = $90^o$ $-$ $\dfrac{\widehat{ACB}}{2}$ Hence: $LI$ is internal bisector of $\widehat{BLK}$ Similarly: $KI$ is internal bisector of $\widehat{CKL}$ Therefore: $BCKL$ is tangential quadrilateral or $MN$ tangents ($I$)

dchenmathcounts wrote: Which year of Peru TST is this? Is (Peru) TST IMO 2015 (Day 2, Problem 8) Link: https://onemperu.files.wordpress.com/2018/12/Selectivos-IMO.pdf (In Spanish)

janssv.200603 wrote: Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and $N$. Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle. Peru TST for IMO , lets solve this... 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

Let $T$ be the midpoint of arc $BC$ not containing $A$ Let the tangents to $\omega$ at $T$ intersect $(ABC)$ at $M'$ and $N'$ . by poncelet porism $M'N'$ tangents $\omega$ and $I$ is the incentre of $\triangle TM'N'$ so $A$ is the midpoint of arc $M'N'$ so by fact5, $AI=AM'=AN'$ so $M=M'$ and $N=N'$ and done!