Let $I_a$ the $A$-excenter of triangle $ABC$, $S$ be the second intersection of the line $PX$ and $\omega$, and $T$ be a point on the line $PX$ such that $\measuredangle TAI_a = \measuredangle I_aAS$. We show that $\triangle ASI_a \stackrel{+}{\sim} \triangle AIT$. Let $K$ be the second intersection of the line $I_aX$ and $\omega$. By Euler's formula we have $$ I_aX \times I_aK = 2Rr_a, $$where $R$ and $r_a$ are the radius of $\omega$ and $\Omega_A$, respectively. Therefore, $I_aK = 2R = KS$. Let $M$ be the midpoint of $I_aS$, then $\measuredangle KMS = 90^{\circ} = \measuredangle KXS$, so $M$ lies on $\omega$. Let $A'$ be the symmetric point of $A$ w.r.t $M$, then $$ \measuredangle I_aA'A = \measuredangle SAM = \measuredangle MAX = \measuredangle I_aSX. $$Hence there exists a point $T'$ on the line $SX$ such that $\triangle I_aAA' \stackrel{+}{\sim} I_aT'S$. Then we have $$ \frac{SA}{SI_a} = \frac{I_aA'}{I_aS} = \frac{I_aA}{I_aT'}, $$and $$ \measuredangle ASI_a = \measuredangle A'I_aS = \measuredangle A'I_aA - \measuredangle SI_aA = \measuredangle SI_aT' - \measuredangle SI_aA = \measuredangle AI_aT'. $$This implies that $\triangle ASI_a \stackrel{+}{\sim} \triangle AI_aT'$. Hence $\measuredangle SAI_a = \measuredangle I_aAT'$ and $T = T'$. Let $\ell$ be a tangent line of $\Omega_A$ different from $BC$ but parallel to $BC$, and $B', C'$ be the intersection of $\ell$ and $AB, AC$, respectively. Consider the transformation $\varphi$ which is the composition of the inversion $\mathcal{I}(A, AI_a^2)$ and the symmetry with respect to $AI_a$. It is known that $\varphi(B) = C'$ and $\varphi(C) = B'$, so $\varphi(ABC) = \ell$. Besides, by the conclusion above we have $\varphi(S) = T$, so $T$ lies on $\ell$. Finally, let $R'$ be the projection of $A$ onto $\ell$, and $O$ be the circumcenter of $ABC$, then by angle chasing we have $$ \measuredangle R'PT = \measuredangle R'AT = \measuredangle SAO = \measuredangle KXA = \measuredangle PAX, $$which implies that $R'P$ is tangent to $\odot(APX)$. Similarly $R'Q$ is tangent to $\odot(AQY)$. Hence $R' = R$ and the statement is proved.

Fun problem Here is a solution quite different from the official one, using Poncelet’s porism and Simson lines. Step 1: Relating $X,Y$ to the common tangents of $\omega,\Omega_A$. First recall Poncelet’s Porism, which says for any $P\in \omega$, if we take $P_1\neq P_2\in \omega$ with $PP_1,PP_2$ tangent to $\Omega_A$, then $P_1P_2$ is tangent to $\Omega_A$ as well. By taking the limit as $P\to X$, we find that the tangent to $\Omega_A$ at $X$ meets $\omega$ at some point $X_1$ lying on a common tangent $\ell_X$ between $\omega,\Omega_A$. Define $Y_1, \ell_Y$ similarly. For reference, in my diagram $X,\ell_Y$ are closer to $B$ and $Y, \ell_X$ are closer to $C$. Step 2: Getting rid of $(APX), (AQY)$. Let $P_1\in \ell_X$ with $PP_1$ tangent to $(APX)$. Then $\angle PP_1X_1 = 180^{\circ} -\angle PX_1P_1 - \angle P_1PX_1 = \angle XAX_1 - \angle XAP = \angle PAX_1$, so $P_1\in (APX_1)$ and therefore $AP_1 \perp \ell_X$. Similarly we can define $Q_1$ as the projection of $A$ onto $\ell_Y$ and we only need to show $R=PP_1\cap QQ_1$ satisfies $AR\perp BC$. Step 3: Reduction to Simson lines. Define $P_2\in \ell_X$ and $P_3\in XX_1$ with $P_2P_3||BC$ and $AX_1P_2P_3$ cyclic. Similarly define $Q_2,Q_3$. Then the Simson line of $A$ with respect to $X_1P_2P_3$ passes through $P_1,P$, and the projection of $A$ onto $P_2P_3$ and similarly for $Y_1Q_2Q_3$. Therefore, if we can show the lines $P_2P_3,Q_2Q_3$ are the same line $\ell || BC$, it will follow that $P_1P\cap Q_1Q$ is the projection of $A$ onto $\ell$, which will finish the problem. Step 4: Identifying $P_3,Q_3$. By Reim’s theorem on $(AX_1P_2P_3), \omega$ with lines $AP_3$ and $X_1P_2$, we know that line $AP_3$ meets $\omega$ at some $X_1’$ with $X_1X_1’||BC$. It follows that if we define $P_3’ = AP_3\cap BC$, $P_3’$ is the inverse of $X_1$ in $\sqrt{bc}$-inversion. Therefore, if $I,I_A$ are the incenter and $A$-excenter of $\triangle ABC$, we know $AP_3’I\sim AI_AX_1$ so $\angle AP_3’I = \angle AI_AX_1$. Since $AI_A$ bisects the angle between $AB,AC$ and $I_AX_1$ bisects the angle between $X_1X, \ell_X$, we have \[\angle (I_AA, I_AX_1) = \frac{1}{2} \angle (AB, XX_1) + \frac{1}{2} \angle (AC, \ell_X)= \frac{1}{4} (\stackrel{\frown}{AX_1} - \stackrel{\frown}{BX} ) +\]\[\frac{1}{4} (\stackrel{\frown}{AX_1} - \stackrel{\frown}{CX_1})= \frac{1}{4} \stackrel{\frown}{AX_1} + \frac{1}{4} ( \stackrel{\frown}{AX_1} - \stackrel{\frown}{XX_1’}) = \frac{1}{2}\angle (AP_3, BC) + \frac{1}{2} \angle (AP_3, P_3X).\]It follows that reflecting $BC$ over $P_3’I$ gives a line parallel to $P_3X$; since this line is tangent to the incircle $(I)$ of $\triangle ABC$, it follows that the homothety sending $(I)$ to $\Omega_A$ also sends $P_3’$ to $P_3$, hence $P_3$ lies on the other tangent to $\Omega_A$ parallel to $BC$. Similarly $Q_3$ lies on this line as well, so $P_2,P_3,Q_2,Q_3$ lie on a common line as desired, and we are done.

Let $O$ and $I_A$ be the centers of $\omega$ and $\Omega_A$, $P',Q',R'$ be the reflections of $A$ across $P,Q,R$, $X',Y'$ be on $\omega$ with $XX',YY'$ tangent to $\Omega_A$, $Z=XP'\cap YQ'$, and $O'$ be the circumcenter of $P'Q'Z$. Then, we have that $P',Q'$ are the reflections of $A$ across $XX,YY'$ and $P'R'\perp P'X,Q'R'\perp Q'Y$, so $R'$ is the reflection of $Z$ over $O'$. Fix the circumcircle and excircle of $ABC$, and vary triangle $ABC$ so that they have the same circumcircle and excircle, which is possible by Poncelet's Porism. Varying $A$ at a fixed angular velocity, we note that $P'$ and $Q'$ vary at the opposite angular velocity on fixed circles as they are reflections of $A$ across fixed lines. Since $\angle P'XY$ and $\angle Q'YX$ vary at the same velocity, $Z$ lies on a fixed circle through $X$ and $Y$ and varies at the same angular velocity as $P'$ and $Q'$. Because $\angle P'ZQ'$ is fixed, $P'O'Q'$ is similar to a fixed triangle, and hence $O'$ also lies on a fixed circle and moves at the same angular velocity. Finally, as $R'$ is the reflection of $Z$ over $O'$, the same is true for $R'$: it lies on a fixed circle and moves at an angular velocity opposite to that of $A$. We now apply complex numbers. Let $\omega$ be the unit circle, $a$ be $A$, $j$ be the center of $\Gamma_A$, $r$ be $R'$. By the previous paragraph, we have that there exist constants $u$ and $v$ with $r=u+\frac va$. The condition that $AR=AR'\perp BC$ is equivalent to $AI_A$ bisecting $\angle R'AO$, or $\frac{P(a)}{Q(a)}=\frac{a(a-r)}{(a-j)^2}\in\mathbb R\iff P(a)\overline Q(a)=\overline P(a)Q(a)$, where $P$ and $Q$ are fixed quadratics. To verify that this is true for all $a$ on the unit circle, it suffices to check five cases. When $A$ lies on the symmetry axis of $\omega$ and $\Omega_A$, the result follows by symmetry. When $A=X$, we have that $B=C=X'$ so $P'=X$. Hence, $XP'$ is the reflection of the tangent at $X$ to $\omega$ over $XX'$. Then, it is clear that $AR'=XP'\parallel OX'\perp BC$. When $A=X'$, we have that $BC=XX'$ so $P'=X'$. Then $AR'=X'R'\perp P'X=BC$.

But you've only checked 3 cases, not 5 Actually if $|a|=1$ and $P(1/a)Q(a)=P(a)Q(1/a)$ where $P,Q$ are quadratics, the two polynomials have the same "leading term" so $a^2(P(1/a)Q(a)-P(a)Q(1/a))$ should be a poly of degree 3? So does that mean you check 4 cases?

Looks like I accidentally left out the last sentence while copy pasting: The cases $A=Y$ and $A=Y'$ follow analogously, so we are done. I’m not sure how many cases are actually needed beyond five being sufficient. There is also actually a sixth trivial case which comes from the other point on the symmetry axis.

The problem is equivalent to sin(AXY-2AIaY)/sin(AYX-2AIaX) = (AY/AX)*( cos(AYIa)/cos(AXIa) )^2 ---- (*) Let A' be the antipodal of A, U,V be the second intersection of IaX,IaY and (ABC) Then IaU=IaV=2R Therefore UV, tangent at A' wrt (ABC)，perpendicular bisector of AIa are concurrent Then use Menelaus's theorem we can get（*)

Very beautiful and very challenging geometry! Here is my solution. Let the tangents to $\Omega_A$ at $X$ and $Y$ meet each other at $T$ and meet $\omega$ again at $X_1,Y_1$. Let $AT$ meet $\omega$ again at $K$. The main lemma is the following. Quote: Lines $KY_1$, $XX_1$, $BC$ are concurrent at $U$. Similarly, lines $KX_1$, $YY_1$, $BC$ are concurrent at $V$. To prove this lemma, we define more points. Let $N$ be the midpoint of arc $BAC$. Let $\Omega_A$ touch $BC,CA,AB$ at $D,E,F$. Claim: $X_1Y_1, AN, BC$ are concurrent. Proof. By Poncelet's Porism on degenerate $\triangle XX_1X_1$, we get that one common external tangent of $\omega, \Omega_A$ passes through $X_1$. Similarly, the other common external tangent passes through $Y_1$. Therefore, by a well-known lemma, we get that $X_1Y_1$ passes through $BI\cap AC$ and $CI\cap AB$. The conclusion then follows through harmonic. $\blacksquare$ Claim: $K,D,N$ are colinear. Proof. First, we prove that $KK, XY, BC$ are concurrent. Applying DDIT on the degenerate quadrilateral $XXYY$ and point $A$, we get an involution swapping $(AX, AY)$, $(AT, AT)$, and $(AB, AC)$. Projecting this involution onto $\omega$ gives the conclusion. Now, let the concurrency point be $U$. Notice that $UB\cdot UC = UK^2 = UD^2$. Thus, $D$ lies on $K$-Appollonius circle of $\triangle BKC$, meaning that $KD$ bisects $\angle BKC$, implying the conclusion. $\blacksquare$ Claim: Let $U = XX_1\cap BC$. Then, $X_1D$ and $AU$ meet at $L\in\omega$. Proof. Use DDIT again but on point $X_1$ and quadrilateral $ABDC$. We get an involution swapping $(X_1B, X_1C)$, $(X_1A, X_1D)$, and $(X_1X_1, X_1X)$. Projecting this involution on to $\omega$, we get an involution swapping $(B,c)$ ,$(A,X_1D\cap\omega)$, and $(X_1,X)$. This gives the conclusion. $\blacksquare$ Finally, the requested lemma follows from Pascal on $ALX_1Y_1KN$. Reduction to the lemma: Let $PQ$ meet $\odot(APR)$ and $\odot(AQR)$ again at $P_1$ and $Q_1$. Now, let $AT$ meet $\odot(KUV)$ again at $K_1$ and let $K_1'$ be the isogonal conjugate of $K_1$ w.r.t. $\triangle TUV$. A straightforward angle chasing gives that $$\measuredangle K_1'VT = \measuredangle UVK_1 = \measuredangle UKK_1 = \measuredangle Y_1KA = \measuredangle QYA = \measuredangle RQA = \angle RQ_1A,$$and $\angle K_1'UT = \angle AP_1R$. Moreover, since $$\measuredangle K_1'TV = \measuredangle UTK_1 = \measuredangle PTA = \measuredangle PQA = \measuredangle Q_1QA = \measuredangle Q_1RA,$$so $\triangle K_1'TV \stackrel{-}{\sim} \triangle ARQ_1$. Similarly, $\triangle K_1'TU\stackrel{-}{\sim} \triangle ARP_1$. Therefore, quadrilaterals $K_1'TUV$ and $ARP_1Q_1$ are similar. Hence, $$\measuredangle RAP = \measuredangle RP_1P = \measuredangle RP_1Q_1 = -\measuredangle TUV$$As $AP\perp UT$, the conclusion follows.

Denote by $S$ intersection of perpendicular to $BC$ through $A$ with tangent $\ell$ to $\Omega_A,$ parallel to $BC.$ Let $XP$ meet $\omega$ again at $X_1$ - by Poncelet theorem we easily conclude that different from $XX_1$ tangent to $\Omega_A$ through $X_1$ also tangent to $\omega.$ Let this common tangent intersect $RX$ at $P_1,$ so we get $$\measuredangle APP_1=\measuredangle AXX_1=\measuredangle AX_1P_1\implies A\in \odot (PP_1X_1)\implies AP_1\perp X_1P_1.$$Let $X_1P_1\cap \ell =X_2,$ $XX_1\cap \ell =X_3,$ $\tau$ denotes reflection over bisector of angle $BAC.$ Claim. $\tau (AX)=AX_2.$ Proof. Let $AX_2$ meet $\omega$ again at $Z,$ and $\phi$ denotes involution which swaps $(B;C),(XX_1\cap BC;AX_2\cap BC).$ DIT on $AXZX_1$ gives $\phi (AX_1\cap BC) =XZ\cap BC.$ By DDIT on $BC,XX_1,X_1X_2,\ell$ there exist involution which swaps $$(AB;AC),(\overline{A(XX_1\cap BC)},AX_2),(A\infty_{BC};AX_1)$$so projecting on $BC$ we get $\phi (AX_1\cap BC)= \infty_{BC}$ and in fact $XZ\parallel BC,$ thus we are done $\Box$ By DDIT on $XX_1X_2X_3$ we get $\tau (AX_1)=AX_3\implies \measuredangle X_2AX_3=\measuredangle X_1AX=\measuredangle X_2X_1X_3\implies A\in \odot (X_1X_2X_3)$. By Simson theorem $S\in \overline{RPP_1}.$ Analogously $S\in RQ,$ therefore $S=R,$ and we are done.

This problem was proposed by Burii.

Wow. Did this with geogebra. Let $\measuredangle$ denote directed angles $\!\!\!\mod 180^\circ$. Let $\ell$ be the line parallel to $\overline{BC}$ tangent to $\Omega_A$. Redefine $R$ as the foot of the altitude from $A$ to $\ell$. It's sufficient to show that $\overline{PR}$ is tangent to the circumcircle of $APX$. Step 1: Removing $P$ and $R$. Since $\angle APZ=\angle ARZ=90^\circ$, $APRZ$ is cyclic. We claim that proving that the circumcircle of $AXZ$ is tangent to $\overline{RZ}$ solves the problem. Indeed, if that was true, then we have $\measuredangle APR=\measuredangle AZR=\measuredangle AXZ=\measuredangle AXP$, as desired. Thus, we can ignore $P$ and $R$ in the diagram. Step 2: Reduction to $\overline{BC} \parallel \ell$. Let $\overline{XZ}$ intersect $\omega$ at $U \ne X$. Claim: $U$ is on a common tangent of $\omega$ and $\Omega_A$. Proof: By Poncelet's porism on $ABC$ and degenerate triangle $UUX$, the tangent at $U$ to $\omega$ is tangent to $\Omega_A$. Consider the following problem: Converse of ISL 2021 G8 wrote: Let $UZV$ be a triangle with incircle $\Omega_A$ and $V$-intouch point $X$. Let the circle through $U$ and $X$ tangent to $\overline{UV}$ intersect the circumcircle of $UZV$ at $A$. Let the tangents from $A$ to $\Omega_A$ intersect the circumcircle of $AUX$ at $B \ne A$ and $C \ne A$. Prove that $\overline{BC} \parallel \overline{ZV}$. We claim that this problem implies the original problem. Ignore the condition $\ell \parallel \overline{BC}$, and consider all positions of $\ell$ that make $\Omega_A$ the incircle of $UZV$. Since $\angle AVZ$ is monotonic and ranges from $0^\circ$ to $180^\circ$, there is exactly one position of $\ell$ that makes $AUVZ$ cyclic. Choose this position of $\ell$. By Miquel's theorem on $UZV$, the circumcircle of $AXZ$ is tangent to $\overline{ZV}$, so we want to prove that $\ell \parallel \overline{BC}$. Step 3: Poncelet's porism and DDIT finish. We will now solve the modified problem in step 2. Let $\overline{AB}$ and $\overline{AC}$ intersect the circumcircle of $AUVZ$ at $D \ne A$ and $E \ne A$. By Poncelet's porism on $UZV$ and $ADE$, $\overline{DE}$ is tangent to $\Omega_A$. Let $I_A$ be the center of $\Omega_A$, and let $UV$ intersect $\Omega_A$ at $X'$. Since $\measuredangle BUC=\measuredangle BAC=\measuredangle DAE=\measuredangle DUE$, the angle bisectors of $\angle BUE$ and $\angle CUD$ coincide. Consider a line $\ell_1$ perpendicular to the angle bisector of $\angle BUE$, and let $UB$, $UC$, $UD$, $UE$, $UX$, and $UX'$ intersect $\ell_1$ at $B^*$, $C^*$, $D^*$, $E^*$, $X^*$, and $X'^*$, respectively. By symmetry, $B^*C^*=D^*E^*$ in directed lengths, so the midpoints of $B^*E^*$ and $C^*D^*$ coincide at a point we call $M$. By DDIT, there exists a pencil involution at $U$ sending $\overline{UB}$ to $\overline{UE}$, $\overline{UC}$ to $\overline{UD}$, and $\overline{UX}$ to $\overline{UX'}$. Therefore there exists an involution sending $B^*$ to $E^*$, $C^*$ to $D^*$, and $X^*$ to $X'^*$. The involution must be a reflection over $M$, so the angle bisector of $\angle BUE$ also coincides with the angle bisector of $\angle XUX'$. Thus, Let $\overline{UB}$ intersect the circumcircle of $AUVZ$ at $B' \ne U$. Then, $\measuredangle B'UZ=\measuredangle VUE$, so $\overline{B'E} \parallel \ell$. By Reim's theorem, we also have $\overline{B'E} \parallel \overline{BC}$, so $\ell \parallel \overline{BC}$, as desired. Remark: The claim in step 2 is a lemma in https://artofproblemsolving.com/community/c6h2313676. The construction of $V$ is motivated by reverse engineering Miquel's theorem and trying to make $\Omega_A$ the incircle of something. The construction of $D$ and $E$ is motivated by a solution to Taiwan TST 2014/3/3.

Here is more synthetic solution I wrote during TST. Let us redefine point R as intersection of tangent at $D'$ and $A$-altitude and prove $PR$ is tangent to $\omega_{XPA}$. Let us denote some points: $D'D'\cap AB = B'$,$D'D'\cap AC = C'$,$D'D'\cap XX = B'$,$XX\cap AB = X'$ and $CX\cap AP' = M$. Now let's write Brianchon's theorem on hexagon $DD'XXB"C"$: Number 1 $DD\cap D'D' = \infty$ Number 2 $D'D'\cap XX = P'$ Number 3 $XX\cap XX = X$ Number 4 $XX\cap B"B" = X'$ Number 5 $B"B"\cap C"C" = A$ Number 6 $C"C"\cap DD = C'$ and lines $14,25,36$ must be concurrent. Since $AP'\cap CX = M$ we can imply that $MX'||BC$ By angle chase $\angle{MX'A} = \angle{P'B'A}=180 - \angle{AB'C'}=180- \angle{ABC}=180- \angle{AXC} = \angle{MXA}$ so $MXX'A$ is concyclic. Again by angle chase $\angle{MAX}=\angle{MX'P'}=\angle{X'P'B'}=\angle{PAR}$ (since $P'APR$ is concyclic) thus $\angle{XAP}=\angle{P'AR} = \angle{P'PR}$ and we are done.

i3435 wrote: ('cause every night I lie in bed...) spamming ddit fills my head... (geometry is keeping me awake...) [asy][asy] //21SLG8 //setup size(13cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn pair A,B1,C1,Ia,D,D1; A=(14.92,11.07); B1=(0,0); C1=(14,0); Ia=incenter(A,B1,C1); D=foot(Ia,B1,C1); D1=2*Ia-D; pair B,C; B=extension(A,B1,D1,D1+(1,0)); C=extension(A,C1,D1,D1+(1,0)); path W,wa; W=circumcircle(A,B,C); wa=incircle(A,B1,C1); pair X,R,V,P; X=intersectionpoints(W,wa)[1]; R=foot(A,B1,C1); V= extension(X,X+rotate(90)*(X-Ia),B1,C1); P=foot(A,X,V); pair U,X1,V1; U=extension(X,V,B,C); X1=extension(A,X,B,C); V1=extension(A,V,B,C); //draw filldraw(A--B1--C1--cycle,blu1,blu); draw(C1--V^^B--V1,blu); draw(wa,blu); draw(A--R,dashed+blu); draw(W,purple); draw(circumcircle(A,P,X),magenta+dashdotted); draw(circumcircle(A,X,V),red+dotted); draw(X--A--V,red); draw(R--P--V,magenta); clip((-2,-2)--(-2,15)--(16,15)--(22,-2)--cycle); //label label("$A$",A,dir(90)); label("$B$",B,dir(90)); label("$C$",C,-dir(40)); label("$B'$",B1,-dir(90)); label("$C'$",C1,-dir(90)); label("$X$",X,dir(-70)); label("$D$",D,dir(-90)); label("$R$",R,dir(-90)); label("$V$",V,dir(-90)); label("$P$",P,dir(50)); label("$U$",U,dir(60)); label("$V'$",V1,dir(10));label("$X'$",X1,dir(110)); label("$\omega$",(10.5,11.2),purple); label("$\omega_a$",(8.31,2.73),blu); [/asy][/asy] Let the antipode of the $A$-extouch point be $D$; let the tangent to $\omega_a$ at $D$ intersect $\overline{AB},\overline{AC}$ at $B',C'$ respectively. Let line $x$ be tangent to $\omega_a$ at $X$, $U=x\cap\overline{BC}$, and $V=x\cap\overline{B'C'}$. Finally, let $X'=\overline{AX}\cap\overline{BC}$, $V'=\overline{AV}\cap\overline{BC}$. Claim 1: $AXUV'$ cyclic. Proof. Apply DDIT to $A$, $UXV\infty_{BC}$ with inconic $\omega_a$, and project onto $\overline{BC}$, to obtain the involution $(BC;UV';\infty_{BC}X')$-- or equivalently, $X'B\cdot X'C=X'U\cdot X'V'$. By power of a point, $X'B\cdot X'C=X'A\cdot X'X$, so the claim follows from power of a point converse on $X'U\cdot X'V=X'A\cdot X'X$. $\qquad\qquad\square$ Claim 2: $\overline{DV}$ is tangent to $(AXV)$. Proof.Angle chase using previous claim, and the fact that $\overline{BC}\parallel \overline{B'C'}$: \[\measuredangle XAV \overset{\text{claim }1}= \measuredangle XUV'=\measuredangle XVD.\qquad\qquad\square\]Redefine $R$ as the foot from $A$ to $\overline{B'C'}$. It remains to show, Claim 3: $\overline{PR}$ touches $(APX')$. Proof.Since $\angle VPA=\angle VRA=90^\circ$, $APRV$ cyclic, so we may anglechase as follows: \[\measuredangle APR=\measuredangle AVR \overset{\text{claim }2} =\measuredangle AXV=\measuredangle AXP.\qquad\qquad\square\]

Let $\ell \ne BC$ be the line parallel to $BC$ and tangent to $\Omega_A$. The tangent line to $\Omega_A$ at $X$ meets $\ell$ at $S$ and meets $\omega$ for a second time at $K$. By Poncelet's porism, there is a common tangent line to $\omega, \Omega_A$ through $K$. Let this common tangent meet $\ell$ at $T$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.806963058530559, xmax = 27.409045556859237, ymin = -2.1522151990399934, ymax = 18.2287489970944; /* image dimensions */ pen evevev = rgb(0.8980392156862745,0.8980392156862745,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); filldraw((11.070163074164075,13.579393494621243)--(9.85222486650305,9.002512170668734)--(15.403417278955729,9.002512170668734)--cycle, evevev, linewidth(1.2)); /* draw figures */ draw((9.85222486650305,9.002512170668734)--(15.403417278955729,9.002512170668734), linewidth(1.2)); draw(circle((13.411126821667693,4.372710352455849), 4.629801818212886), linewidth(1.2)); draw(circle((12.62782107272939,10.714399117792942), 3.261056764264046), linewidth(1.2)); draw((15.403417278955729,9.002512170668734)--(15.159988564460608,8.659497900319954), linewidth(1.2)); draw((11.070163074164075,13.579393494621243)--(11.070163074164139,-0.25709146575662056), linewidth(1.2) + dotted); draw((11.070163074164075,13.579393494621243)--(13.541896000323954,10.211634652335071), linewidth(1.2) + dotted); draw((11.070163074164075,13.579393494621243)--(14.633491267851694,14.834954496727654), linewidth(1.2) + dotted); draw((11.070163074164139,-0.25709146575662056)--(14.633491267851694,14.834954496727654), linewidth(1.2) + qqzzqq); draw((7.1695628133431075,9.002512170668735)--(9.85222486650305,9.002512170668734), linewidth(1.2)); draw((15.403417278955729,9.002512170668734)--(16.68858884337307,9.002512170668735), linewidth(1.2)); draw((11.070163074164075,13.579393494621243)--(-0.7218396074105807,-0.25709146575704056), linewidth(1.2)); draw((14.633491267851694,14.834954496727654)--(19.951268167004887,-0.2570914657570312), linewidth(1.2)); draw((-0.7218396074105807,-0.25709146575704056)--(19.951268167004887,-0.2570914657570312), linewidth(1.2)); draw((11.070163074164075,13.579393494621243)--(8.93702727489717,5.563297653545161), linewidth(1.2) + blue); draw((11.070163074164075,13.579393494621243)--(16.773148093569667,7.555771433114837), linewidth(1.2) + blue); draw((-0.7218396074105807,-0.25709146575704056)--(15.703530280369414,11.798144516763893), linewidth(1.2) + red); /* dots and labels */ dot((11.070163074164075,13.579393494621243),linewidth(4.pt) + dotstyle); label("$A$", (10.829860095768108,13.798104606630401), N * labelscalefactor); dot((9.85222486650305,9.002512170668734),linewidth(4.pt) + dotstyle); label("$B$", (9.314865562254608,8.395576930516235), N * labelscalefactor); dot((15.403417278955729,9.002512170668734),linewidth(4.pt) + dotstyle); label("$C$", (15.10333488119387,9.367460216166403), NE * labelscalefactor); label("$\Omega_A$", (8.544480158746174,2.10972890532501), N * labelscalefactor); label("$\omega$", (9.029392007598816,12.140654148690581), NE * labelscalefactor); dot((10.671757998305733,8.105125565896055),linewidth(4.pt) + dotstyle); label("$X$", (10.429672860500391,7.423693644866068), NE * labelscalefactor); dot((13.541896000323954,10.211634652335071),linewidth(4.pt) + dotstyle); label("$P$", (13.859849162795108,9.681893043876752), N * labelscalefactor); dot((11.070163074164139,-0.25709146575662056),linewidth(4.pt) + dotstyle); label("$R$", (10.829860095768108,-0.8944838881986001), NE * labelscalefactor); dot((-0.7218396074105807,-0.25709146575704056),linewidth(4.pt) + dotstyle); label("$S$", (-1.004248147148664,-0.54881459493244466), SE * labelscalefactor); dot((15.703530280369414,11.798144516763893),linewidth(4.pt) + dotstyle); label("$K$", (15.717861326538078,12.083016455483047), E * labelscalefactor); dot((19.951268167004887,-0.2570914657570312),linewidth(4.pt) + dotstyle); label("$T$", (20.119920914482964,-1.0088230982750905), NE * labelscalefactor); dot((7.1695628133431075,9.002512170668735),linewidth(4.pt) + dotstyle); label("$S'$", (6.799402940571816,9.138781796013422), NE * labelscalefactor); dot((16.68858884337307,9.002512170668735),linewidth(4.pt) + dotstyle); label("$T'$", (16.804083822264737,9.167366598532544), E * labelscalefactor); dot((14.633491267851694,14.834954496727654),linewidth(4.pt) + dotstyle); label("$H$", (14.745978040887907,14.99866631243355), NE * labelscalefactor); label("$\ell$", (9.572128784926711,-0.8658990856794776), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 1: Quadrilateral $AKST$ is cyclic. Proof. Let lines $AS, KT$ meet line $BC$ at $S', T'$. Since $\ell \parallel BC$, it suffices to show that quadrilateral $AKS'T'$ is cyclic. By the dual of Desargues' involution theorem for quadrilateral $SKT'\infty_{BC}$ with inconic $\Omega_A$, there is an involution swapping $(AB, AC), (AS, AT'), (AK, A\infty_{BC})$. Projecting onto line $BC$, we obtain an involution swapping $(B, C), (S', T'), (Z, \infty_{BC})$, where $Z := AK \cap BC$. Since $Z$ is swapped with the point at infinity, this involution is an inversion with pole $Z$. We conclude using power of a point: \[ ZS' \cdot ZT' = ZB \cdot ZC = ZA \cdot ZK. \quad \blacksquare \] Claim 2: The Simson line of $A$ w.r.t. $\triangle KST$ is tangent to $\odot(APX)$ at $P$. Proof. Let $H$ be the projection of $A$ onto line $KT$ (so the Simson line is $PH$). Since $APKH$ is cyclic (with diameter $\overline{AK}$) and line $KH$ is tangent to $\omega$, we obtain \[ \measuredangle HPA = \measuredangle HKA = \measuredangle KXA = \measuredangle PXA. \quad \blacksquare \] We conclude by observing that the projection $R'$ of $A$ onto $\ell$ lies on the Simson line of $A$ w.r.t. $\triangle KST$. By claim 2, line $R'P$ is tangent to $\odot(APX)$. Similarly, line $R'Q$ is tangent to $\odot(AQY)$. Therefore, $R' = R$, and $AR \perp BC$.

Let $\Omega_A$ hit $BC$ at $Z$. $ZZ'$ the diameter in $\Omega_A$, the tangent from $Z'$, hit $AB, AC$ at $B_1, C_1$ respectivily, let the tangent from $X$ to $\Omega_A$ hit $BC, \omega, B_1C_1$ at $K, X', G_X$ respectivily, let $AG_X$ hit $BC$ at $G$, redefine $R$ as a point in $B_1C_1$ such that $AR \perp BC$, by poncelet porism $X'$ lies in one of the common tangents of $\omega, \Omega_A$ so let this mentioned line hit $\Omega_A$ at $T_X$ and $B_1C_1$ at $P_X$, let $AB, AC$ hit $\Omega_A$ at $E,F$. Now by DDIT over $G_XXK \infty_{BC}$ we get that $(AG_X, AK), (A \infty_{BC}, AX), (AE, AF)$ are pairs of involution, projecting onto $BC$ we get $(G, K), (AX \cap BC, \infty_{BC}), (B, C)$ are pairs of involution so by PoP over the center of (negative) inversion which turns out to be $AX \cap BC$ we get $AGXK$ cyclic but by reims over $(AGXK), (AXG_X)$ we get that $(AXG_X)$ is tangent to $B_1C_1$ at $G_X$ so now there is a spiral sim centered at $A$ sending $XX' \to G_XP_X$ so $AX'P_XG_X$ is cyclic, now let $(APX') \cap X'P_X=P'$ by angle chase: $$\angle APP'=\angle AX'P'=\angle AG_XP_X=\angle AXK \implies PP' \; \text{tangent to} \; (APX)$$But by simson line over $\triangle X'P_XG_X$ we get that $P',P,R$ are colinear so $PR$ is tangent to $(APX)$, and in a symetric way u can also get $QR$ tangent to $(AQY)$ hence $AR \perp BC$ (in original problem) as desired!.

Fix $\Omega_A , \omega$ and let triang;e $\triangle ABC$ move along $\omega$. By poncelet's porism $\Omega_A$ remains unchanged.It's well-known that the tangent lines to $\Omega_A$ at $X$ intersects $\omega$ at the common tangents of $\omega$,$\Omega_A$ (or we can say this by Poncelet's again).Now , for an arbitrary point $A$ on $\omega$ , let the perpendiculars through $A$ to those common tangents intersect them at $D',E'$ and tangent line at $P$ to the circumcircle of the triangle $APX$ intersects it at $D$ and define $E$ for $Q$ similarly. by simple angle chasing $(APD)$ passes through the intersection of common tangents of $\omega , \Omega_A$ and $\omega$. So by the definition of $P$ , $D'=D$ and $E'=E$ similarly. So $R=PD \cap QE$. Now we'll use degree bounds . $A$ moves along $\omega$ with degree 2 but since $A=D$ for some case $D$ has degree on. similar for $P$. So $PD$ has degree 2 and $R$ has degree 4 and perpendicular from $A$ to $BC$ has degree 2 since both moves with degree at most two 2 becouse the foot of perpendicular coinsides at some cases. So it's enough to check 5 cases for triangle $ABC$ and there's in fact 6 degenerated cases for the triangle and we're done.

One of the hybrid solutions of all time. Let $\ell\neq\overline{BC}$ be the line parallel to $\overline{BC}$ and tangent to $\Omega_{A}$. We'll show that $R$ is the foot of the perpendicular from $A$ onto $\ell$ (this can be noticed from a large, well-drawn diagram). In order to do this, redefine $R$ as said foot and we'll show that $\overline{RP}$ is tangent to the circumcircle of $\triangle APX$ (the other tangency would follow analogously). Let $S = \overline{PX} \cap \overline{AB}$, $T = \overline{PX} \cap \ell$, $V = \overline{AT} \cap \overline{CX}$. Then $\angle ART = 90^{\circ} = \angle APT$, so $TRPA$ is cyclic. Hence $\overline{RP}$ being tangent to the circumcircle of $\triangle APX$ is equivalent to: \[\angle PAX = \angle RPX = \angle RPT = \angle RAT \iff \triangle TAR \sim \triangle XAP\]The last is equivalent to $\angle RTA = \angle PXA$. However, this holds if $\overline{VS} \parallel \ell$. Indeed, if $\overline{VS} \parallel \ell$, then \[\angle SAX = \angle BAX = \angle BCX = \angle XVS\]which implies that $AXSV$ is cyclic and moreover, $\angle PXA = \angle SVA = \angle RTA$ as desired. Furthermore, $\overline{VS} \parallel \ell$ is equivalent to $\frac{AV}{VT} = \frac{AS}{SW}$. However, point $V$ is not particularly nice to work with. Let $\overline{TS} \cap \overline{AC} = L$, $\overline{AS} \cap \ell = W$, and let $\Omega_{A}$ touch $\overline{BC}$, $\ell$, $\overline{AB}$, and $\overline{AC}$ at points $D, E, F, G$ respectively. By Menelaus's theorem applied on $\triangle ATL$ and line $V, X, C$, we get that: \[\frac{AV}{VT} = \frac{XL}{TX} \cdot \frac{CA}{LC}\]Now in order to prove $\frac{AS}{SW} = \frac{XL}{TX} \cdot \frac{CA}{LC}$ and be done, we employ complex numbers with $\Omega_{A}$ as the unit circle. Let $D = 1$, $E = -1$, and $b, c, x$ denote the complex numbers of points $F, G, X$, respectively. Then we get: \[A = \frac{2bc}{b+c}, \quad L = \frac{2cx}{c+x}, \quad C = \frac{2c}{c+1}, \quad S = \frac{2bx}{b+x}, \quad T = \frac{2x}{1-x}, \quad W = \frac{2b}{1-b}\]so all the points we need are easily computed and now we just have: \[|A-S| = \left|\frac{2bc}{b+c} - \frac{2bx}{b+x} \right| = \frac{2|c-x|}{|b+c|\cdot |b+x|}\]\[|S-W| = \left|\frac{2bx}{b+x} - \frac{2b}{1-b}\right| = \frac{2|x+1|}{|b+x| \cdot |1-b|}\]\[|X-L| = \left| x - \frac{2cx}{c+x}\right| = \frac{|x-c|}{|c+x|}\]\[|T-X| = \left|\frac{2x}{1-x} - x\right| = \frac{|x+1|}{|1-x|}\]\[|C-A| = \left|\frac{2bc}{b+c} - \frac{2c}{c+1}\right| = \frac{2|b-1|}{|b+c| \cdot |c+1|}\]\[|L-C| = \left|\frac{2cx}{c+x} - \frac{2c}{c+1}\right| = \frac{2|x-1|}{|c+x|\cdot |c+1|}\]Combining all of the above, we get that: \[\frac{|A-S|}{|S-W|} = \frac{|x-c|\cdot |b-1|}{|b+c| \cdot |x+1|} = \frac{|X-L|}{|T-X|}\cdot \frac{|C-A|}{|L-C|} = \frac{|A-V|}{|V-T|}\]therefore $\overline{VS} \parallel \ell$, as desired.

We begin with some line redefinition. Let $M$ and $N$ be the touchpoint of the external tangents of $\omega$ and $\Omega_A$ closer to $C$ and $B$, respectively. Then redefine the ``tangent lines to $\Omega_A$ at $X$ and $Y$" as the lines $\overline{XM}$ and $\overline{YN}$, respectively. This works by Poncelet's porism. Let $B_0$ be on $\overline{AB}$ and $C_0$ on $\overline{AC}$ such that $B_0C_0$ is parallel to $BC$ and $\Omega_A$ is the incircle of $BCC_0B_0$. Let $S$ be the intersection of lines $\overline{XM}$ and $\overline{BC}$. Claim: It suffices to prove that $(ASX)$ is tangent to $\overline{B_0C_0}$. Proof. Since $(APR'S)$ is cyclic, $\angle ASX= \angle ASR = \angle AR'P$. Realize that $(AXP)$ and $(ASR')$ intersect at $S=\overline{XP} \cap \overline{SR'}$, so $A$ is the center of the spiral similarity mapping $XP$ to $SR'$. Thus, \[ \angle SAX = \angle XSD' = \angle PSR' = \angle R'AP, \]which implies that $(ASX)$ is tangent to $\overline{B_0C_0}$. Since all steps above are reversible, the claim is true. Let $S' = \overline{AS} \cap \overline{BC}$ and $S_0' = \overline{XM} \cap \overline{BC}$. Claim: $A$, $X$, $S'$, and $S_0'$ are concyclic. Proof. Let $X' = \overline{AX} \cap \overline{BC}$. Apply DDIT on $A$ with respect to $SXS_0'\infty_{BC}$, so that $(AB, AC)$, $(AS', AS_0')$, and $(AX', A\infty_{BC})$ are pairs of an involution. By Desargues assistant theorem, this involution must be an inversion (negative or positive) centered at $X'$. Since $B$ and $C$ are swapped by this inversion, it is a negative inversion with radius $-\sqrt{BX' \cdot CX'} = -\sqrt{XX' \cdot AX'}$. Since $S'$ and $S_0'$ are swapped, the desired concyclicity follows by power of a point. Since $\overline{B_0C_0} \parallel \overline{BC}$, it follows by Reim's theorem that $\overline{B_0C_0}$ is tangent to $(ASX)$, as desired.

Very Instructive Problem, so I will post my solution although same as others. Apparently, I am not used to to using Poncelet Porism again and again lol that's why this took lot of time. Let The line parallel to $BC$ and tangent to excircle intersect $AB,AC$ at $B_1,C_1$, Let the tangent at $X$ intersect $(ABC)$ at $V$ and $B_1C_1$ at $S$. Redefine $R$ to be $AR \perp B_1C_1$ s.t. $R \in B_1C_1$. Note that it suffices to prove that $AXS$ is tangent to $B_1C_1$. Now before proceeding further we prove the following lemma: Lemma: $V$ is the point of tangency of tangents between circumcircle and excircle, $V \in (ABC)$.

Now let the other tangent from $V$ to exircle meet $B_1C_1$ at $T$. Notice that we can restate the converse of problem as follows: Converse of ISL 2021 G8 wrote: Let $SVT$ be a triangle with incircle $\Omega_A$ and $T$-intouch point $X$. Let the circle through $V$ and $X$ tangent to $\overline{ST}$ intersect the circumcircle of $SVT$ at $A$. Let the tangents from $A$ to $\Omega_A$ intersect the circumcircle of $AVX$ at $B \ne A$ and $C \ne A$. Prove that $\overline{BC} \parallel \overline{ST}$. It is easy to see by miquel there is a unique position for $B_1C_1$, hence this implies the orignal problem. Now, the idea is to force poncelet type configuration and apply DDIT. Let $AB,AC$ meet $(SVT)$ at $D,E$ then by Poncelet's porism $(ADE)$ has incircle as $\omega_A$ Notice that the angle bisectors of $BVE$ and $CVD$ coincide. Consider a line $\ell_1$ perpendicular to the angle bisector of $\angle BUE$, and let $VB$, $VC$, $VD$, $VE$, $VX$, and $VX'$ intersect $\ell_1$ at $B^*$, $C^*$, $D^*$, $E^*$, $X^*$, and $X'^*$, respectively. By symmetry, $B^*C^*=D^*E^*$ in directed lengths, so the midpoints of $B^*E^*$ and $C^*D^*$ coincide at a point we call $M$. By DDIT,there exists an involution sending $B^*$ to $E^*$, $C^*$ to $D^*$, and $X^*$ to $X'^*$. THis must be reflection over mispoint of $B*E*$, Now let $VB$ intersect $(SVT)$ at $B'$, so $B'E \parallel \ell$. By reims we're done. Remark: This is a very projective type of problem where we need to extend lines to get poncelet type config, this often leads us to getting symmetry which makes DDIT a good option.

The complex bash 3 posts above is neat, but it contains synthetic steps (including guessing the characterization of $R$). Here's a bash with no synthetic steps whatsoever, which requires no insights except for a burning desire for terms to factor or cancel. The finish is a touch too tedious for my liking, though, and I would appreciate help finding a cleaner method!

I found a cleaner way to finish.

Let $B'C'$ be the points so that $B'C' \parallel BC$ and $\Omega_A$ is the incircle of $\triangle AB'C'$. It suffices to show that $PR$ is tangent to $(APX)$. Then let $XP \cap RB = K$. Since $\angle APK = \angle ARK = 90^\circ$ we have $APRK$ cyclic. Then let $X'$, $P'$, $K'$ be points $AX \cap BC$, $XP \cap BC$ and $AK \cap BC$. By DDIT on $P'XK\infty_{BC}$ from point $A$ gives that $(AP', AK)$, $(AX, A\infty_{BC)}$ and since $P'XK\infty_BC$ has incircle $\Omega_A$ we have that $(AB', AC')$ are involutive pairs as well. Projecting onto $BC$ gives us that $(P', K')$, $(X', \infty_BC)$, $(B, C)$ are involutive pairs. Then since the involution is a negative inversion, and since $(X, \infty_BC)$ get swapped we know that the inversion is centered at $X'$ which gives us that $X'P' \cdot X'K' = X'B \cdot X'C$. By PoP we have $X'B \cdot X'C = X'X \cdot X'A$ so $P'XK'A$ is cyclic. Note that $\measuredangle{APR} = \measuredangle{AKR} = \measuredangle{AK'B} = \measuredangle{AXP}$ which proves $RP$ tangent to $(AXP)$.

dDIT RIZZ We reverse reconstruct by starting with $R$ as the foot onto $B'C'$. Define the points obviously when not said. $P$ is the foot onto the tangent from $A$, $S$ the meeting with $B'C'$. It is enough by uniqueness to prove that $RP$ is a tangent. Then $APRS$ cyclic. Draw $T$ as well. We also observe that proving $(AXV)$ tangent to $B'C'$ finishes the problem after a quick angle chase. Now take dDIT on the triangle with sides $ST$, $TC$, and $SD$ so the invoution gives pairs $A(\infty, X), (T, S), (B,C)$, which when projected onto line $BC$ gives that it is an inversion at $X'$, or that $AS'XT$ is cyclic. This implies that $\angle XAS' = \angle XTS' = \angle XSR$ which gives the tangency, and we are done.

ok unc Use the definitions in ohiorizzler's diagram. We claim that if $R$ is redefined as the foot of the perpendicular from $D$ to the altitude at $A$, then $RP$ is tangent to $(APX)$. Indeed, note that $APRS$ is cyclic from right angles. Now, by Alternate segment theorem, clearly we just wish to show that $\angle RAP=\angle SAX$. Let $D'$ be the extouch with $BC$. Let $E,F$ be the extouch with $AC$ and $AB$ respectively. Brianchon on $DD'EFXX$ means that if $PX\cap AB$ is defined as $G$, then the parallel to $BC$ through $G$, $AS$, and $CX$ concur. Let this point be $H$. Note that $AHGX$ is cyclic as \[180^\circ-\angle HGA=\angle B=\angle AXC=180^\circ-\angle HXA\]By Bowtie and corresponding angles. This means that $\angle SAX=180^\circ- \angle XGH$. Yet this is equivalent to showing that the angle formed by $BC$ and $XP$ same as angle formed by $AP$, $AR$. Easy to see by the right angles.