I think $M$ should be the midpoint of arc $BAC$

ayeen_izady wrote: I think $M$ should be the midpoint of arc $BAC$ Oops

I think the remark should be $AP\cap MO$ and not $AP\cap MP$

In this solution I will use a lot of properties of mixtilinear incircles that you can read about here: https://blog.evanchen.cc/2015/08/11/the-mixtilinear-incircle/. Let $N$ be the $A$-superpoint, and let $T=MI\cap \Omega$. It is well known that $T$ is the $A$-mixtilinear tangency point. Let $\omega$ be the $A$-mixtilinear incircle. Let $\omega$ be tangent to $AB,AC$ at $G,H$ respectively and let the incircle be tangent to $AB, AC$ at $K,L$. Redefine $P$ as the intersection of $EF$ and $AT$. We will show $DP\parallel AI$ Claim 1: lines $EF, GH, BC, NT$ concur at a point $X$. Proof: It is well known that $G,I,H$ collinear and $BC, GH, NT$ concur. Note that $AN\perp GH$, so line $GH$ is the common tangent of the $A$-supercircle and $(IFE)$. Thus lines $EF,GH,BC$ concur at the radical center of $\Omega, (IEF), (BIC)$, which proves the claim. Claim 2: $P,T$ are inverses WRT $(IFE)$, and $EF$ is tangent to the incircle at $P$ Proof. Consider inversion around $(IFE)$. First, $EF$ and $\Omega$ are swapped. as $\angle AIG=90^{\circ}=\angle AKI$ we have that $K,G$ and $L,H$ are 2 pairs of inverses. This is enough to force that $\omega$ and the incircle are swapped. By the definition of $P$ we have that $P,T$ are inverses, and as $\omega$ and the incircle are swapped and $EF$ and $\Omega$ are swapped, $EF$ has to be tangent to the incircle at $P$ as inversion preserves intersections. claim 3: $PIDTX$ is cyclic. Proof: $\measuredangle ITX=\measuredangle MTN=90^\circ=\measuredangle IPX=\measuredangle IDX$ by claim 2. To finish, note that by claim 2 we have $AT\times AP = AI^2$ so $\measuredangle TPD =\measuredangle TID=\measuredangle TAI$ , so $DP\parallel AI$

Let the midpoint of $\widehat {BC}$ (not containing $A$) be $N$ and let the tangent lines of incircle from $N$ intersect $\Omega$ at $E’,F’$ and intersect $BC$ at $ Q_{1},Q_{2}$. By Poncelet Porism, $E’F’$ is tangent to the incircle. Let the tangency point be $P’$. Thus $I$ is the incenter of $\triangle NE’F’$. Also, $A,I,N$ are collinear, so $A$ is the midpoint of arc $E’F’$ (not containing $N$), and we have $AI=AE’=AF’$, hence $E’=E, F’=F$. Let $M_{1},M_{2}$ be the intersection of $AN$ and $BC,EF$ ,respectively. So we have $\measuredangle M_{1}M_{2}P’ =\measuredangle AM_{2}P’ =\widehat {AN}=\measuredangle DM_{1}N=\measuredangle DM_{1}M_{2}$. (P.S. $\widehat {AN}$ can be referred to the one containing $M$ or the one not containing $M$, decided on how you use angle-chasing.) Let $Z$ be the intersection of $BC$ and $EF$, we have $\measuredangle ZM_{1}M_{2}=\measuredangle M_{1}M_{2}Z$ and $ZM_{1}=ZM_{2}$. Also $ZD=ZP’$, so $M_{1}M_{2} \parallel DP’$, $P’=P$. By 3-point DDIT, we have that $(NB, NC),(NQ_1, NQ_2),(ND, NA)$ are involution pairs and project it onto $BC$ we also have $(B,C),(Q_1,Q_2),(D, M_1)$ are involution pairs on $BC$. Thus, $(NBC), (N Q_{1}Q_{2}), (NDM_{1}) $ are coaxal. Let the radical axis be $NT (T \neq N) $. Well-known by 2014 Taiwan TST third round M3, $T$ is the tangency of $(ABC)$ and $A$-mixtilinear incircle. Let $Z$ be the intersection of $NT$ and $BC$, we have $Z$ lies on the line passes through $I$ and perpendicular to $AI$. (proof: Let $G,H$ be the intersection of the line which passes through $I$ and perpendicular to $AI$ and $AB,AC$ respectively, $GH$ cut $BC$ at $Z’$. By homothety $TG, TH$ passes through the midpoints of $\widehat {AB}, \widehat {AC}$ so $TG, TH$ bisects $\angle BTA, \angle CTA$. By Menelaus, $1=\frac{HA}{HC} \times \frac{GB}{GA} \times \frac{Z’C}{Z’B}=\frac{TA}{TC} \times \frac{TB}{TA} \times \frac{Z’C}{Z’B}= \frac{TB}{TC} \times \frac{Z’C}{Z’B}$, so $TW$ is the external bisector of $\angle BTC$ so $Z’,T,N$ are collinear, $Z’=Z$.) Similarly, $Z$ is also the concurrency of $EF$ and the line which passes through $I$ and perpendicular to $AI$. Let the tangent lines of incircle from $A$ intersect $EF$ at $R_{1}, R_{2}$, we have that $(E,F),(R_{1},R_{2}),(P, M_{2})$ are involution pairs on $EF$. Project this involution onto $\Omega$, we have that $(E,F),(B,C),(AP \cap \Omega, N)$ are involution pairs and $EF,BC,(AP \cap \Omega)N$ are concurrent at point $Z$, thus $AP \cap \Omega =T=MI \cap \Omega$, we are done.

shanelin-sigma wrote:

Bro this is so cursed

Nice problem with short sol: Let $EF$ intersect $AB,AC$ at $S,R$ respectively. $Lemma$: $BCRS$ is both inscribed and circumscribed. $Proof$: $BCRS$ is cyclic since $AS.AB=AE^2=AR.AC$ by similarity; Also $AR.AC=AI^2$ $\implies$ $\angle AIR=\frac{C}{2}$ $\implies$ $\angle IRC=90-\frac{B}{2}$ which implies that $BSRC$ is circumscribed. $Lemma$:$P$ is the image of $I$ in $EF$ $Proof$: Let $P',T,K$ be the touch points of the incircle w.r.t $EF,AB,AC$ respectively. We may show that $DP'\perp TK$. Note that $\angle TDP'=\angle SP'E=\frac{C}{2}$ Proving the lemma since $AI\perp TK$. Now we want to show $A-P-Q$ are collinear where $Q$ is the $A$-mixtilinear point. It's enough to show that $A-L-X_{A}$ are collinear, where $L$ is the reflection of $P$ w.r.t $AI$ and $X_{A}$ is the $A$-excircle touch point.(Note that $AQ$ and $AX_{A}$ are isogonal). Which is simple since if the tangent line to $L$ intersects $AC$ at $V$ then $ITSP \cong IKVL$ since They are just reflections w.r.t $AI$, Thus: $\angle LVA=\angle PSA=C$ Thus by reverse of well know lemma $L$ is the intersection of $AX_{A}$ with the incircle. And we are done.

Under the inversion centered at $A$ with radius $AI$, incircle swaps with $A-$mixtilinear incircle. Since $(AEFBC)$ and $A-$mixtilinear are tangent to each other, $EF$ is tangent to the incircle. Let the tangency point be $K$. Radical axises of $(ABCEF),(BIC),(EFI)$ are concurrent thus, $EF\cap BC=R$ satisfies $RI\perp AI$. Combining this with $KD\perp RI$ gives $KD\parallel AI$ which implies $K=P$. If $MI\cap (ABC)=T$ which is $A-$mixtilinear touch point, then $P$ swaps with $T$ under the inversion at $A$ with radius $AI$ thus, $A,P,T$ are collinear as desired.$\blacksquare$