Let $P,Q$ be the tangency points of $(O_1)$ and $(O_2)$ with $(O).$ It's well-known that $AP$ and $AQ$ go through the exsimilicenter $U$ and insimilicenter $V$ of $(O)$ and the incircle $(I)$ (for this use Monge & d'Alembert theorem for $(I),(O),(O_1)$ and similarly for $(I),(O),(O_2)$). Since $(O,I,U,V)=-1,$ then the pencil $A(A',M,P,Q)$ is harmonic $\Longrightarrow$ quadrilateral $PMQA'$ is harmonic $\Longrightarrow$ $X \in MA'.$

My solution: Let $ I, I_a $ be the Incenter, A-excenter of $ \triangle ABC $, respectively . Let $ T=AI \cap BC $ and $ H $ be the projection of $ A $ on $ BC $ . Let $ Y, Z $ be the tangent point of $ \odot (I), \odot (I_a) $ with $ BC $, respectively . Let $ Y^*, Z^* $ be the tangent point of $ \odot (O_2), \odot (O_1) $ with $ \odot (O) $, respectively . Since $ \{ AY, AY^* \}, \{ AZ, AZ^* \} $ are isogonal conjugate of $ \angle A $ (well-known) , so from $ (A, T;I, I_a)=-1 \Longrightarrow A(H, T; Y, Z)=-1 \Longrightarrow A(A', M; Y^*, Z^*)=-1 $ , hence we get $ A'Y^*MZ^* $ is a harmonic quadrilateral $ \Longrightarrow X, M $ and $ A' $ are collinear . Q.E.D

$X$ is the radical center of three circles,that is,to prove $A'M$ is radical axis of $O_1,O_2$,and it is not hard by using mannheim theorem and chicken claw theorem.

Let $I,J$ be incenter and A-excenter of $\triangle ABC$. Let $U,V$ be the points of tangency of $O_1,O_2$ and $\overrightarrow{AB}$. Let $f$ be the inversion followed by reflection about $AM$ such that $f(B)=B,f(C)=C$. Then $f$ brings $O_1,O_2$ into the A-excircle and incircle of $\triangle ABC$ respectively, and also $f(I)=J$, so we have that $IU,JV \perp AM$. Since $M$ is the midpoint of $IJ$, it follows that if $W$ is the midpoint of $UV$, $MW \perp AM$. $W$ and its corresponding point on $\overrightarrow{AC}$ are on the radical axis of $O_1,O_2$. Since $A'M \perp AM$, it follows that $M,A',X$ are all on this radical axis.

Let $O_1,O_2$ touch $\odot ABC$ in $P,Q$.$O_1,O_2$ are the mixtlinear in,ex circle and hence the touch point lie on isogonals of Nagell and Gergonne cevian.Take the inversion $\Psi$ with radius $\sqrt{bc}$ composed with reflection $A$ angle bisector. Let $\Psi(P),\Psi(Q),\Psi(A_1),\Psi(M)$ be the touch point of incircle,excircle,the foot of A-altitude,the foot of $A$-angle bicestor,respectively.Show that $(\Psi(P),\Psi(Q);\Psi(M),\Psi(A'))=-1$ Proof. Let $W$ be the midpoint of $A\Psi(A')$ than it's well-known that $W,I,\Psi(Q)$ are collinear.Projecting $I(A,\Psi(A');W,\infty)=-1$ on $BC$ we're finished.

Telvcohl, can you give me a simply proof for AZ, AZ* is isogonal conjugate of angle A.

nguyenvanthien63 wrote: Telvcohl, can you give me a simply proof for AZ, AZ* is isogonal conjugate of angle A. Check out EGMO 2013

YaWNeeT wrote: mannheim theorem and chicken claw theorem. Which theorems are these? Can you share any link or something?

Wizard_32 wrote: YaWNeeT wrote: mannheim theorem and chicken claw theorem. Which theorems are these? Can you share any link or something? Mannheim wrote: Let $ABC$ be a triangle, and let $L,M,N$ be points on $BC,CA,AB$ respectively. Let $A', B', C'$ be points on $(AMN), (BNL), (CLM)$, and denote $K \equiv AA' \cap BB'$. Then if $K \in CC'$, $A',B',C',K$ are concyclic. Chicken claw theorem is probably Fact 5, a.k.a. the Trident theorem.

Let $N$ be the midpoint of arc $BAC$ in $O$, and let $T_1$ and $T_2$ be the tangency points between $O, O_1$ and $O, O_2$, respectively. It's well-known($\sqrt{bc}$) that $N, I, T_1$ and $N, I_A, T_2$ are collinear, where $I$ and $I_A$ are the incenter and $A$-excenter of $\triangle ABC$, respectively. Then $N(T_1, T_2; M, A') = (I, I_A; M, P_\infty) = -1$, where $P_\infty$ is the point at infinity along $AM$, so $(T_1MT_2A')$ is harmonic and we are done.

Let $T_1$, $T_2$ be the points of tangency of $O_1$, $O_2$ with $O$, $N_1$, $N_2$ the points of tangency of the excirlce and incircle $(I)$ with $BC$. It's well known that the point symmetric to $N_2$ wrt $I$ ($U$) lies on $AN_1$ (homothety) and $AT_1,AN_1$ and $AT_2,AN_2$ are isogonals wrt angle $BAC$ ($\sqrt{bc}$ inversion). It's enough to prove $(A'M;T_1T_2)=-1 \iff (AO,AM;AT_1,AT_2)=-1$ reflecting in the $AM$ we're left to prove that $(AH,AM;AN_1,AN_2)=1$, where $AH \perp BC$. Projecting this to $IN_2$ we get $(P_{\infty}I;UN_2)=-1$, so we're done.

Nice problem. Here's my solution: Invert the figure about $A$ with radius $\sqrt{AB \cdot AC}$, followed by reflection in the angle bisector of $\angle BAC$. Then we get the following equivalent problem:- Inverted problem wrote: Let $P$ and $Q$ be the $A$-intouch and $A$-extouch point in $\triangle ABC$, $X$ be the foot of internal angle bisector of $\angle BAC$, and $D$ be the foot of the $A$-altitude. Let $H_A$ be the $A$-Humpty point of $\triangle APQ$. Show that $ADXH_A$ is cyclic. Let $H$ be the orthocenter of $\triangle APQ$. We wish to show that $\measuredangle XH_AA=\measuredangle XDA=90^{\circ}$, or equivalently that $H,X,H_A$ are collinear. From here, this is equivalent to proving that $(D,X;P,Q)=-1$. Now, Let $I$ and $I_A$ be the incenter and $A$-excenter of $\triangle ABC$. Also, Let $P_{\infty}$ be the point at infinity on line $AD$. Then, as $BI$ and $BI_A$ are internal and external angle bisectors of $\angle ABX$, we have that $-1=(A,X;I,I_A) \overset{P_{\infty}}{=} (D,X;P,Q)$. Hence, done. $\blacksquare$

Restated problem wrote: Let $O$ be the circumcircle of the triangle $ABC$. Two circles $O_1,O_2$ are tangent to each of the circle $O$ and the rays $\overrightarrow{AB},\overrightarrow{AC}$, with $O_1$ interior to $O$, $O_2$ exterior to $O$. The common tangent of $O,O_1$ and the common tangent of $O,O_2$ intersect at the point $X$. Let $M$ be the midpoint of the arc $BC$ (not containing the point $A$) on the circle $O$, and the segment $\overline{AA'}$ be the diameter of $O$. Denote \(T_1, T_2\) as the touchpoints of \(O_1, O_2\) resp. with \((O)\) \(; I, I_A\) resp. as the incenter and \(A-\) excenter and \(N\) the midpoint of \(\overarc{BAC}.\) Then \(T_1 = IM \cap \odot O, T_2 = I_AM \cap \odot O\). Prove that \(\overline{A',M,X}\) So, we have to prove that \(A'X\) is the symmedian of \(\Delta A'T_1T_2 \iff \Delta A'T_2T_1 \sim \Delta MII_A \iff \angle A'T_2T_1 = \angle MII_A \iff 180^{\circ}- \angle A'T_2T_1= \angle A'AT_1 = 180^{\circ}-\angle MII_A\) which is obvious. Darn

let $X_1,X_2$ the intouch,extouch point of inMixtilinear and exMixtilinear with $(ABC)$ first solution by me let $I,I_a$ be the incenter A-excenter and let $K,K_a,K' $ be on $AB$ such that $\angle{IKA}=90$ , $\angle{K_aI_aA}=90$ , $\angle{MK'A}=90$ and so define $L,L_a,L'$ since $M$ is the midpoint of $I,I_a$ we have $K',l'$ are the midpoints of $KK_a,LL_a$ but it's well-known that $L,K$ are on the A-mixtilinear circle which is $O_1$ and similary $L_a,K_a$ also on the exterior A-mixtilinear which is $O_2$ so $M'K'$ is the radical axis of $O_1,O_2$ but $A',M,L',K'$ are collinear so $X,M,A',L',K'$ are colinear and we win $\blacksquare$ a one-line solution via Fouad Al-mouine let $N$ the midpoint of arc $BAC$ it's well-konwn that $N,I,X_1$ and $N,I_a,X_2$ are collinear $-1=(M,P_{\infty};I,I_a) \stackrel{N}{=} (X_1,X_2;M;A')$ done

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.35)); dotfactor *= 1.5; pair A = dir(130), B = dir(200), C = dir(340), I = incenter(A,B,C), D = foot(I,B,C), E = extension(A,I,B,C), M = (B+C)/2, F = 2M-D, G = foot(A,B,C), X = foot(E,A,M); draw(A--B--C--A); draw(incircle(A,B,C), gray); draw(circumcircle(A,D,X)^^arc(circumcenter(A,F,X),circumradius(A,F,X),150,315), heavygreen); draw(A--G^^A--E^^A--M^^A--F, blue); draw(circumcircle(A,G,E), purple+dashed); dot("$A$", A, dir(135)); dot("$B$", B, dir(200)); dot("$C$", C, dir(340)); dot("$D$", D, dir(270)); dot("$E$", E, dir(270)); dot("$F$", F, dir(270)); dot("$M$", M, dir(270)); dot("$G$", G, dir(270)); dot("$X$", X, dir(90)); [/asy][/asy] $\sqrt{bc}$ inversion gives the following: Inverted problem wrote: Let $ABC$ be a triangle with altiude $\overline{AG}$, angle bisector $\overline{AE}$, and median $\overline{AM}$. The incircle touches $\overline{BC}$ at $D$ and the $A$-excircle touches $\overline{BC}$ at $F$. The circle through $A$ tangent to $\overline{BC}$ at $D$ and the circle through $A$ tangent to $\overline{BC}$ at $F$ meet again at $X$. Show that $AGEX$ is cyclic. It is well known that $DM = FM$, so $M$ lies on the radical axis $\overline{AX}$. By USMCA 2019 Premier #3 $X$ is the projection of $E$ on $\overline{AM}$, so $AGEX$ is cyclic with diameter $\overline{AE}$ as desired. $\blacksquare$

It is sufficient to prove $M$ lies on the radical axis of $O_1$ and $O_2$. Let $f\colon \mathbb R^2\to \mathbb R$ be a function of point $\bullet$ such that \[ f(\bullet)=P(\bullet, O_1)-P(\bullet, O_2). \]It is known $f$ is linear. Let $P\in AB$, $Q\in AC$ and $R\in AB$, $S\in AC$ be the tangency point of $O_1$, $O_2$ on $AB$, $AC$ respectively. It is known that $I$, $I_A$ are the midpoints of $PQ$, $RS$ respectively. We want $f(M)=0$. Notice \[ 4f(M)=2\left(f(I)+f(I_A)\right)=f(P)+f(Q)+f(R)+f(S)=0. \]Done.

After $\sqrt{bc}$ inversion, the problem becomes the following: let $ABC$ be a triangle and $P,Q$ be the $A$-intouch and $A$-extouch points respectively. Let $D$ be the foot of the $\angle A$-bisector. Let $\omega_1$ and $\omega_2$ be the circles passing through $A$ tangent to $\overline{BC}$ at $P$ and $Q$ respectively, and let them intersect again at $E$. Prove that $\angle AED=90^\circ$. Let $M$ be the midpoint of $\overline{BC}$, which is also the midpoint of $\overline{PQ}$, so $\overline{AE}$ passes through $M$ because of power of a point. Redefine $E$ to be the foot of the altitude from $D$ to $\overline{AM}$; I will show that $(AEP)$ is tangent to $\overline{BC}$: $(AEQ)$ being tangent is similar. It suffices to show that $MP^2=AM\cdot DM$. WLOG let $b>c$. We can calculate $MP=\tfrac{1}{2}|b-c|$ and $$DM=EM\cos \angle AMB=\left(\frac{a}{2}-\frac{ac}{b+c}\right)\frac{AM^2+a^2/4-c^2}{2AM(a/2)} \implies AM\cdot DM=\frac{b-c}{2(b+c)}\cdot \frac{(2b^2+2c^2-a^2)+a^2-4c^2}{4}=\frac{2(b-c)(b^2-c^2)}{8(b+c)}=MP^2,$$as desired. $\blacksquare$