My solution: Let $ \ell_P $ be the tangent of $ \odot (I) $ through $ P $ . Let $ \ell $ be a line passing through $ I $ and perpendicular to $ AI $ . Let $ B'=\ell_P \cap AC, C'=\ell_P \cap AB $ and $ T $ be the tangent point of $ A- $ mixtilinear circle with $ \odot (ABC) $ . Since $ \ell_P $ is the reflection of $ BC $ in $ \ell $ , so $ \ell_P $ is anti-parallel to $ BC $ WRT $ \angle BAC \Longrightarrow \triangle AB'C' \sim \triangle ABC $ , hence $ AP $ is the isogonal conjugate of $ A- $ Nagel line WRT $ \angle BAC \Longrightarrow T\in AP $ . Since It's well-known that $ T \in MI $ (see incenter of triangle) , so we get $ T $ is the intersection of $ AP $ and $ IM $ which is lie on $ \odot (ABC) $ . Q.E.D

Dear Mathlinkers, With another regard… 1. N the antipole of M wrt (O) A*, X the second point of intersection of MI, DN wrt (O) U the second point of intersection of the parallel to BC with (O) 2. according to the Reim’s theorem, A*, X, D, I are concyclic on (2) 3. P’ the second point of intersection of A*A with (2) 4. A*M is the A*-inner bissector of A*AU (http://jl.ayme.pagesperso-orange.fr/Docs/Mixtilinear1.pdf, p. 28) In consequence, P’ is on (I) 5. according to the Reim’s theorem, DP’ // AIN and P’=P. Sincerely Jean-Louis

Let $I_a,I_b,I_c$ be the excenters of $\triangle ABC$. Let the incircle touch $AC,AB$ at $E,F$, $H$ be the orthocenter of $\triangle DEF$, let $J$ be the midpoint of $DH$, $G$ the midpoint of $EF$, $(N)$ the nine-point circle of $\triangle DEF$. It is well-known that $IG//JH$, $IG=JH$ hence $IGHJ$ is a parallelogram $\Rightarrow IJ//GH$. Now $M$ is the second intersection of the nine-point circle of $\triangle I_aI_bI_c$ $\odot(ABC)$ with $I_bI_c$, hence $M$ is the midpoint of $I_bI_c$. We now consider the homothety carrying $\triangle DEF$ to $\triangle I_aI_bI_c$, which also carries $H\rightarrow I$, $G\rightarrow M$. Hence under the homothety we have $HG//IM\Rightarrow J,I,M$ collinear. Note that $P,H$ are reflections of each other across $EF\Rightarrow IGPJ$ is a isoceles trapezium and hence cyclic. We now consider the inversion with respect to the incircle, note that the nine point circle $(N)$ of $\triangle DEF$ is sent to $\odot(ABC)$. Thus $G$ is sent to $A$. $IGJP$ cyclic implies $J$, which lies on $(N)$, is sent to a point on $\odot(ABC)$ which lies on line $AP$, as well as $IJ\equiv IM$.

Invert about the incircle, and denote inverses with a $'$. Suppose the intouch triangle is $DFG$; since the tangents from $F$ and $G$ intersect at $A$, $A'$ is the midpoint of $FG$, i.e. $(ABC)$ inverts to the nine-point circle of $(DFG)$. Note that $MI$ hits the circumcircle again at the tangency point of the $A$-mixtilinear incircle (see #3 here). $\angle IAM=\angle IM'A'=90^{\circ}$ and since $T$ lies on $IM$, then $T'$ is the antipode of $A'$ with respect to said nine-point circle, which is the midpoint of $DH$, where $H$ is the orthocenter of the intouch triangle. Thus, we want to prove that $PA'IT'$ is cyclic. This is equivalent to the following problem: Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Suppose the intersection of $AH$ with the circumcircle is $K$, the midpoint of $BC$ is $E$, and the midpoint of $AH$ is $J$. Prove that $JOEK$ is cyclic. Invert about the circumcircle, and use complex numbers. We set $(ABC)$ as the unit circle. Note that $e'=\frac{2bc}{b+c}$, $k=-\frac{bc}{a}$, and $j'=\frac{1}{\frac{1}{a}+\frac{\frac{1}{b}+\frac{1}{c}}{2}}=\frac{2abc}{ab+ac+2bc}$. Then we just have to show that $e', k, j'$ are collinear which is relatively easy.

No $\sqrt bc$ inversion yay! Let $X$ be the second intersection point of $MI$ with the circumcircle, $D'$ be the antipodal of $D$ in the incircle and $E,F$ be the touch points of the incircle with $AC,AB$ respectively. Let $AP$ meet $EF$ at $T$ and $AD'$ meet $EF$ at $Y$. Notice that $X$ is the touch point of the $A$ mixitilinear incircle with the circumcircle. It is well known that $AX$ and $AD'$ are isogonal rays. By some angle chase we can see that $\triangle FPE$ is congruent to $\triangle ED'F$. which gives $AP$ and $AD'$ isogonal so $A,P,T,X$(4869 lol) are collinear. Done.

anantmudgal09 wrote: No $\sqrt bc$ inversion yay! Lol, I (and it looks like leeky), both inverted about the incircle. I think $\sqrt bc$ makes this harder, because then you have to deal with a mixtilinear excircle

hint ; let AH is the altitude and z is the intewrestion AH and PD and Q bethe symetric point of d betwwn i prove that the feodbakh point passes from pQ

My solution: Suppose that the $A$-mixtilinear touches $\odot O$ at $Q$. Then $IM$ passes trough $Q$. So we must say that $AP$ passes trough $Q$. Suppose that $X$ is the intersection point of $A$ excircle and $BC$. Then we have $\angle CAQ=\angle BAX$. So we must say that $\angle CAP=\angle BAX$. Suppose that $K$ is the intersection point of $\odot I$ and $DI$. Then $AX$ passes trough $K$. Suppose that $E$ and $F$ are the intersecting points of $\odot I$ with $AC$ and $AB$ respectively. $DP$ is the altitude in triangle $\triangle EDF$, so $\angle FDK=\angle EDP$. Now from $\triangle AEP=\triangle AFK$ we get that $\angle FAK=\angle EAP$. And we are done.

It is well known that $IM$ intersects the circumcircle again at the $A$-mixtilinear touch point; call this $X.$ We wish to show that $A,P,X$ are collinear. Let $D'$ be the antipode of $D$ on $\odot (I).$ Then $PD' \perp AI$ and $IP=ID'$ so $\triangle PAI\cong \triangle D'AI$ and $AP,AD'$ are isogonal. But it is well known that $AD'$ is the $A$-extouch cevian in $\odot (O)$ and that this is isogonal to $AX,$ from which the conclusion follows.

This follows directly from the fact that circumcevian triangle of the orthocenter of the intouch triangle wrt incircle is homothetic to $\Delta ABC$

Let $\alpha=\frac{\angle A}{2}$, and define $\beta, \gamma$ similarly. Note that $\angle PDC = \alpha + 2 \beta$, so $\angle PDI = \gamma - \beta$, so $\angle (PI, BC)=90^\circ - 2(\gamma-\beta) = 90^\circ - 2\gamma + 2\beta $. Also, $\angle (AO,BC)=90^\circ - 2\gamma + 2\beta = \angle (PI, BC)$, so $AO // PI$. It is not hard to see that $O$ and $I$ lie on the same side of line $AP$ (since $\angle CAP < \angle CAI < \angle CAO < 90^\circ$), so segments $AP,OI$ do not intersect. Now let $X$ be the centre of homothety taking $A$ to $P$, and $O$ to $I$. Note that this is a positive homothety. Since $A$ lies on $\odot(O)$ and $P$ lies on $\odot(I)$, $X$ is in fact the exsimilicenter of $\odot(O)$ and $\odot(I)$. It is then well-known that $APX$ passes through the $A$-mixtilinear intouch point $T$. But $IM$ is also known to pass through $T$, hence we are done. $\square$

It is well-known that $IM\cap(ABC)$ is the $A$-mixti touch, now reflect about $AI$, we see that $P$ is sent to $D'$, $D$-antipode in the incircle by rectangle property. We want $AD'$ to be the Nagel Line but this is well-known.