I saw this problem about a year ago, then it suddenly got deleted after I solved it, quite surprise to see it again Let $L$ be the reflection of $I$ with $O$ Let $D$ be the midpoint of arc $BC$ of $(O)$ not contain $A$ and $(J)$ tangent to $BC$ at $E$ From Euler's formula $JO^2=R^2+2RR_a$, with $R, R_a$ circumradius of $(ABC)$ and $(J)$ Or $JO^2-R^2=R.JE$, or $JD.JA=R.JT$ note that $JL = 2R$ and $JI = 2JD$ $\Rightarrow$ $JI.JA=JL.JT$ so $A, I, L, T$ are concyclic By Miquel theorem $I, X, E, L$ are concyclic so $IX \perp IO$ From the definition of $Y$ we have $AY \perp IO$ too Let $DY$ meet $BC$ at $K$ $\angle DKB = \angle YBD = \angle YAD = \angle IYA = \angle YIX$ so $K, I, X, Y$ are concyclic We have $DI^2 = DY. DK$ $\Rightarrow$ $DI$ is tangent to $(IXY)$

Attachments:

Nice! Here's my solution: Let $M$ be the midpoint of $IJ$, and let $I'$ be the reflection of $I$ in line $BC$. Also suppose lines $AI,MY$ meet line $BC$ at $L,K$. Invert about $M$ with radius $MI$. Using Shooting Lemma, $Y \mapsto K$ and $A \mapsto L$. Then, using inversive distance formula ($r=MI$) and the fact that $IA=IY$, we have $$\frac{IK}{IL}=\frac{r^2 \cdot IY}{MI \cdot MY} \div \frac{r^2 \cdot IA}{MI \cdot MA}=\frac{MA}{MY}=\frac{MK}{ML}$$Since $L \in MI$, so this means that $KL$ bisects $\angle IKM$. In particular, point $I'$ is the reflection of $I$ in the angle bisector of $\angle IKM$, and so $I' \in MK$. Suppose $\odot (IKY)$ meets line $BC$ again at $X'$. We claim that $X' \equiv X$. Note that this finishes the problem, since we have $MK \cdot MY=MI^2$, i.e. $\odot (IKY)$ is tangent to $MI$. Thus it suffices to prove that $X' \in \odot (AIP)$. Let $H$ be the orthocenter of $\triangle JBC$, and let $H'$ be its reflection in line $BC$. Note that $HH'II'$ is a rectangle, and $M$ lies on the common perpendicular bisector of $HI',H'I$. In particular, this gives us $$\measuredangle YIX'=\measuredangle YKX'=\angle (I'M,BC)=\angle (BC,MH)$$Also, $\sqrt{JB \cdot JC}$ inversion about $J$ followed by antiparallelism in $\angle BJC$ gives that $$JM \cdot JT=JB \cdot JC=JH \cdot JA \Rightarrow \frac{JM}{JA}=\frac{JH}{JT}$$where we use the fact that $\{H,A\}$ and $\{M,T\}$ are swapped in this process. This gives that $AT \parallel MH$, and so $$\measuredangle APX'=\angle (BC,MH)=\measuredangle YIX'$$Finally, since $KX'$ bisects $\angle IKY$, so by Fact 5, $X'I=X'Y$. Thus, we have $$\measuredangle AIX'=\measuredangle X'YI=\measuredangle YIX'=\measuredangle APX' \Rightarrow X' \in \odot (AIP) \quad \blacksquare$$ NOTE: The first paragraph is basically this KoMaL problem.

Proof: Define $M$ as the midpoint of $\widehat{BC}$. We break the problem into several claims. Claim 1: If $H= \overline{IO} \cap \overline{JT}$ then $\overline{IO}=\overline{OH}$. Proof: Notice that since $\overline{JT} \perp \overline{BC}$ and also $\overline{OM} \perp \overline{BC}$ $\implies$ $\overline{OM} \parallel \overline{JH}$. Now by fact 5 we have that $M$ is the midpoint of $\overline{IJ}$ and so by thales theorem we are done $\square$. Claim 2: $(AIHT)$ is cyclic. Proof: Denote by $R$ the circumradius. For this firstly notice that $$\begin{cases} JI=4R \sin \left(\frac{A}{2} \right) \\ JA=4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right) \\ JH=2R \\ JT=8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right) \end{cases}$$Now we are all set to bash. Notice that we just need to show $$JI \cdot JA=JH \cdot JT$$$$\iff 4R \sin \left(\frac{A}{2} \right) \times \left(4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right)\right)=2R \times 8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)$$$$\iff\sin \left(\frac{A}{2} \right)=\cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right) $$$$\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=4 \cos \left(\frac{A}{2} \right) \cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-4\cos \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$$$$\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=\sin A+\sin B+\sin C- \sin B-\sin C+\sin A$$which is trivial. $\square$. Now back to the main problem. Observe that $$\angle XIJ=\angle APX=90^\circ-\angle PTJ=90^\circ-\angle HIJ=\angle AYI=\angle GYI$$where $G=\overline{AY} \cap \odot(IYX)$. Now this implies that $\overline{GX} \parallel \overline{AI}$ $\implies$ $\angle YGX=\angle YAI=\angle IYA=\angle IYG$ $\implies$ $\overline{YG} \parallel \overline{IX}$. This implies the required tangency $\blacksquare$.

Edit

@above Why does this imply tangency?

Nice! Here is a synthetic solution Firstly, we are going to prove the following $\textbf{lemma.}$ Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. Let $M=AI \cap (ABC)$ and $L$ be the reflection of $I$ with respect to $O$. If $K$ was the $A$-excircle touch point with $BC$, then $ALKM$ is cyclic. $\textcolor{black}{\textit{\textbf{proof.}}}$ Let $R=AI \cap BC$, $N=MO \cap (ABC) $ and $J$ the $A$-excenter. The perpendicular to $AI$ at $I$ meet $BC$ at $Q$, finally $S=QM \cap (ABC)$. If $D$ was the projection of $I$ on $BC$ then clearly $K$ is the reflection of $D$ with respect to $MN$, thus $LK \perp BC$ which implies that $L,K,J$ are collinear. Some angle chaising gives $$ \angle{QIB} = 90 - \angle{QIM} = 90 - \frac{\angle{A}}{2} - \frac{\angle{B}}{2} = \frac{\angle{C}}{2} = \angle{ICB}$$Hence $QI^2 = QB \cdot QC = QS \cdot QM$, therefore $$\angle{MSI} = 180 - \angle{QSI} = 180 - \angle{QIM} = 90 = \angle{MSN} $$giving that $N,I,S$ are collinear. But notice that $O$ is the midpoint of $IL,MN$ thus $MINL$ is a parallelogram, so we get $ML \perp MQ$ and $$ \Longrightarrow \angle{QML} = 90 = \angle{QKL} $$yields that $QMKL$ is cyclic. It is clear that $\angle{QIJ} = 90 = \angle{QKJ}$, so $QIKJ$ is cyclic giving that $$ QR \cdot RK = \underbrace{IR \cdot RJ = BR \cdot RC}_{IBJC \cdots cyclic} = AR \cdot RM $$Thus $QMKA$ is cyclic, as we proved earlier $QMKL$ is cyclic, so both $A,L \in (PMK)$ and therefore $ALKM$ is cyclic, as required. $\blacksquare$ Now let $T$ be the reflection of $J$ with respect to $BC$, clearly $K$ is the midpoint of $JT$, but $M$ is the midpoint of $JI$, thus $IT \parallel MK$. Some angle chaising gives $$ \angle{JTI} = \underbrace{\angle{JKM} = \angle{JAL}}_{AMKL \cdots cyclic} $$Eventually, $AILT$ is cyclic. 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Since $IA=IY$, $OA=OY$, then $OI$ is the perpendicular bisector of $AY$. Some angle chaising gives $$ \angle{TPC} = 90 - \angle{ATL} = 90 - (180 - \angle{AIL}) = \angle{AIL} - 90 = \frac{360-\angle{AIY}}{2} - 90 = \frac{180-\angle{AIY}}{2} = \angle{IYA} =\angle{IAY} $$Now let the circle that passes through $Y,I$ and tangent to $AI$ meet $BC$ at $X'$, and define $Z = MX' \cap (ABC)$. It is easy to get that $$\angle{MZB} = \frac{\angle{A}}{2} = \angle{MBX'} $$Thus $MI^2 = MB^2 = MX' \cdot MZ$. Hence $MI$ is tangent to $(X'IZ) $, but $MI$ by defenition is tangent to $(X'IY)$, therefore $X'YZI$ is cyclic. By angle chaising $$\angle{X'IY} = \angle{X'ZY} = \angle{MZY} = \angle{MAY} = \angle{IAY}$$Thus $$ \angle{APX'} + \angle{AIX'} = \angle{TPC} + \angle{YIA} + \angle{X'IY} = \angle{IAY}+ (180 - 2\angle{IAY}) + \angle{IAY} = 180 $$So $X' \equiv (AIP) \cap BC \equiv X$, as desired. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -394.07964514897344, xmax = 3123.5977929092946, ymin = -1732.9666354926225, ymax = 663.187963228231; /* image dimensions */ pen wwqqcc = rgb(0.4,0,0.8); pen ffdxqq = rgb(1,0.8431372549019608,0); pen qqccqq = rgb(0,0.8,0); pen ffqqtt = rgb(1,0,0.2); /* draw figures */ draw((-190.97556956311763,-53.03741002300586)--(-90.56751623206671,175.74320132216192), linewidth(0.8)); 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I use "$J'$" instead of "$T$" Let $M=AI \cap \odot(ABC) \ne A$ , $I'$ be the reflection of $I$ across $BC$ , $A'$ be the antipode of $A$ and incircle touch $BC$ at $D$. Since, $\frac{AI}{AA'}=\frac{ID}{IM}$ We get $\triangle AIA' \sim \triangle IDM$ then $\triangle AIO \sim \triangle II'M$. By angle-chasing, $I',M,Y$ are collinear. Let $S=AI \cap BC$. By power of points, $A,I',M'J'$ are concyclic. $X'=MI' \cap BC$. $\angle SAP = \angle SI'X'=\angle SIX'$. Therefore, $AP \parallel IX'$. By power points, $\frac{SI}{SX}=\frac{SP}{SA}=\frac{SX'}{SI},SI^2=SX \cdot SX'$. So, $\omega_1=\odot(IXX')$ tangents to $AI$ and $\omega_2=\odot(IYX)$ tangents to $AI$ ($MI^2=MX' \cdot MY$). So, $\omega_1=\omega_2$. Therefore, $\odot(IXY)$ tangents to $AI$.

Attachments:

GeoMetrix wrote:

Proof: Define $M$ as the midpoint of $\widehat{BC}$. We break the problem into several claims. Claim 1: If $H= \overline{IO} \cap \overline{JT}$ then $\overline{IO}=\overline{OH}$. Proof: Notice that since $\overline{JT} \perp \overline{BC}$ and also $\overline{OM} \perp \overline{BC}$ $\implies$ $\overline{OM} \parallel \overline{JH}$. Now by fact 5 we have that $M$ is the midpoint of $\overline{IJ}$ and so by thales theorem we are done $\square$. Claim 2: $(AIHT)$ is cyclic. Proof: Denote by $R$ the circumradius. For this firstly notice that $$\begin{cases} JI=4R \sin \left(\frac{A}{2} \right) \\ JA=4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right) \\ JH=2R \\ JT=8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right) \end{cases}$$Now we are all set to bash. Notice that we just need to show $$JI \cdot JA=JH \cdot JT$$$$\iff 4R \sin \left(\frac{A}{2} \right) \times \left(4R\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)+4R \sin \left(\frac{A}{2} \right)\right)=2R \times 8R \sin \left(\frac{A}{2} \right) \cos \left (\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)$$$$\iff\sin \left(\frac{A}{2} \right)=\cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-\sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right) $$$$\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=4 \cos \left(\frac{A}{2} \right) \cos \left(\frac{B}{2} \right) \cos \left(\frac{C}{2} \right)-4\cos \left(\frac{A}{2} \right) \sin \left(\frac{B}{2} \right) \sin \left(\frac{C}{2} \right)$$$$\iff 4\sin \left(\frac{A}{2} \right) \cos \left(\frac{A}{2} \right)=\sin A+\sin B+\sin C- \sin B-\sin C+\sin A$$which is trivial. $\square$. Now back to the main problem. Observe that $$\angle XIJ=\angle APX=90^\circ-\angle PTJ=90^\circ-\angle HIJ=\angle AYI=\angle GYI$$where $G=\overline{AY} \cap \odot(IYX)$. Now this implies that $\overline{GX} \parallel \overline{AI}$ $\implies$ $\angle YGX=\angle YAI=\angle IYA=\angle IYG$ $\implies$ $\overline{YG} \parallel \overline{IX}$. This implies the required tangency $\blacksquare$.

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Why is $90^\circ-\angle HIJ=\angle AYI$ ????

Beautiful problem! Let $\triangle LMN$ be the circumcevian triangle of $I$, and let $D$ and $D'$ be the incircle and $A$-excircle touch points. Let $Q$ be the reflection of $D$ over $I$ and let $R$ be the reflection of $I$ over $Q$. Finally, let $E$ be the reflection of $X$ over $I$. Claim 1: $E$ lies on $\overline{MN}$. Proof. Consider the homothety $\mathcal{H}$ centered at $A$ taking the incircle to $A$-excircle. This takes $\{I,Q,R\}$ to $\{J,D',T\}$, so $A,R,T$ are collinear. It is also easy to see that $E$ lies on the perpendicular bisector of $IR$. Now angle chasing shows that \[\measuredangle IEQ = \measuredangle IXD = \measuredangle IAR\]which means that $E$ is in fact the circumcenter of $\triangle AIR$. Hence it lies on the perpendicular bisector of $AI$ which is $\overline{MN}$ by the incenter lemma. $\square$ Let $O$ be the circumcenter of $\triangle ABC$. Since $IE=IX$, by Butterfly theorem it follows that $OI \perp XE$. Since $Y$ is the reflection of $A$ over $\overline{OI}$, angle chasing finally gives us \[\measuredangle IYX = -\measuredangle IAE = \measuredangle AIE = \measuredangle XIJ\]and we're done. $\blacksquare$ [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(123); pair B = dir(228); pair C = dir(312); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C); pair L = 2*foot(O,A,I)-A; pair M = 2*foot(O,B,I)-B; pair N = 2*foot(O,C,I)-C; pair J = 2*L-I; pair T = 2*foot(J,B,C)-J; pair P = extension(A,T,B,C); pair D = foot(I,B,C); pair D1 = foot(J,B,C); pair Q = 2*I-D; pair R = 2*Q-I; pair E = circumcenter(A,R,I); pair X = 2*I-E; pair Y = 2*foot(A,O,I)-A; draw(A--B--C--cycle, black+1); draw(A--J); draw(D--R); draw(J--T); draw(A--P); draw(E--X); draw(B--M, mydash); draw(C--N, mydash); draw(M--N); draw(C--P); draw(E--Q); draw(O--I); draw(circumcircle(I,X,Y)); draw(circumcircle(A,B,C)); draw(circle(I, abs(I-D))); draw(circumcircle(B,I,C), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(315)); dot("$I$", I, dir(180)); dot("$O$", O, dir(180)); dot("$L$", L, dir(250)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$J$", J, dir(270)); dot("$T$", T, dir(60)); dot("$P$", P, dir(0)); dot("$D$", D, dir(270)); dot("$Q$", Q, dir(135)); dot("$R$", R, dir(90)); dot("$E$", E, dir(90)); dot("$X$", X, dir(270)); dot("$Y$", Y, dir(Y)); dot("$D'$", D1, dir(315)); [/asy][/asy]

Beautiful problem! similar solution to above, posting for storage Let $O$ denote the circumcenter and $\Delta DEF$ be the intouch triangle. Reflect $D$ across $I$ to $S$ and reflect $I$ across $S$ to $K$. Note that the homothety sending the incircle to the $A$-excircle sends $K$ to $T$, so $K$ lies on $AT$ (and we can erase $T,J$ from the diagram). Let $Z$ be the midpoint of $AI$, then by midpoint theorem we have $ZS \parallel AP$. Furthermore let $L$ be the projection of $A$ on $ID$, then it is easy to show that $\Delta LEF$ and $\Delta ABC$ are oppositely similar, $Z$ is the circumcenter of $\Delta LEF$ and $S$ is the incenter of $\Delta LEF$ by Fact 5. So by similarity we have, in formal sum, \begin{align*} ZS + IO &= EF + BC \\ AP + IO &= (AI + 90^\circ) + BC \\ (AI + BC - XI) + IO &= (AI + 90^\circ) + BC \\ IO - XI = 90^\circ \end{align*}so $\angle XIO = 90^\circ$. Finally, let $BI$ and $CI$ meet $\Omega$ again at $S_b, S_c$, then by Butterfly Theorem we have that the reflection $X'$ of $X$ across $I$ lies on $S_bS_c$. So \[ \measuredangle AIX = \measuredangle AIX' = \measuredangle X'AI = \measuredangle IYX \]which finishes the problem.