Let the other excenters be $I_B$ and $I_C$, and consider a negative inversion with radius $\sqrt{IA \cdot II_A}$ and denote the images with $\bullet'$. Then the problem becomes Inverted Version wrote: $\triangle I_AI_BI_C$ has orthocenter $I$ and orthic triangle $\triangle ABC$. $E'$ is the second intersection of $\overline{IC}$ with $(II_BI_C)$. $F'$ is defined similarly. $X'$, $Y'$ and $Z'$ are circumcenters of $\triangle IE'F'$, $\triangle IF'A$ and $\triangle IE'A$. Show that $(X'Y'Z')$ is tangent to $(I_AI_BI_C)$. By the orthocenter lemma, $E'$ and $F'$ are reflections of $I_B$ and $I_C$ over $B$ and $C$. Also notice that $X'Z' \perp BE'$, so $X'Z' \parallel I_AI_C$. Similarly, $X'Y' \parallel I_AI_B$ and $Y'Z' \parallel I_BI_C$ so $\triangle X'Y'Z'$ and $\triangle I_AI_BI_C$ are homothetic. Therefore, it suffices to show that their homothetic center lies on one of the circles. Let $P$ be the point on $\overline{I_BI_C}$ such that $PE' \perp IE'$. and let $T$ be the antipode of $I_A$ in $(I_AI_BI_C)$. Obviously, $Z'$ is the midpoint of $PI$. Moreover, since $B$ is the midpoint of $E'I_B$, we see that $I_C$ is the midpoint of $PI_B$. Hence $Z'I_C \parallel II_B \Rightarrow \angle Z'I_CI_A = 90^\circ$, so $\overline{Z'I_C}$ passes through $T$. Similarly, $\overline{Y'I_B}$ also passes through $T$ so we are done.

What a beautiful problem and configuration. Crazyguy and rafaello helped me to solve this problem. Does proposer have aops account (if yes what is his nickname)? $\textbf{Claim:}$ $PIYC$ (similarly $PIZB$) is a cyclic quadrilateral. Proof: $EYCI_A$ is a cyclic quadrilateral because $EYI_A=EII_A=ICI_A$. If $T$ is on $AC$ such that $AIT=90$ then $PITC$ is a cyclic quadrilateral because $IPC=90-A/2$. $EIT=ECI=C/2 \implies EI$ is tangent to $PITC$. $IYC=EYC-EYI=180-EI_AC-EIY=180-EIC=180-IPC$. $\square$ Let $BK$ cut $(ABC)$ at $U$. We have $YPC=YIC=KIC=KBC \implies P-Y-U$, similarly $C-L-V, P-Z-V$. $ZXY=LFI+IEK=180-FIL-EIK=180-ZPY \implies P \in (XYZ)$. Because $BLKC$ is cyclic we have $ZY\parallel LK \parallel UV \implies (PYXZ)$ and $(PABC)$ are tangent. $Note:$ If you know more about this configuration (maybe contest problem or your finding), please let me know.

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SerdarBozdag wrote: Does proposer have aops account (if yes what is his nickname)? Nope. TWN2 of this year is the proposer.

SerdarBozdag wrote: What a beautiful problem and configuration. Crazyguy and rafaello helped me to solve this problem. Does proposer have aops account (if yes what is his nickname)? $\textbf{Claim:}$ $PIYC$ (similarly $PIZB$) is a cyclic quadrilateral. Proof: $EYCI_A$ is a cyclic quadrilateral because $EYI_A=EII_A=ICI_A$. If $T$ is on $AC$ such that $AIT=90$ then $PITC$ is a cyclic quadrilateral because $IPC=90-A/2$. $EIT=ECI=C/2 \implies EI$ is tangent to $PITC$. $IYC=EYC-EYI=180-EI_AC-EIY=180-EIC=180-IPC$. $\square$ Let $BK$ cut $(ABC)$ at $U$. We have $YPC=YIC=KIC=KBC \implies P-Y-U$, similarly $C-L-V, P-Z-V$. $ZXY=LFI+IEK=180-FIL-EIK=180-ZPY \implies P \in (XYZ)$. Because $BLKC$ is cyclic we have $ZY\parallel LK \parallel UV \implies (PYXZ)$ and $(PABC)$ are tangent. $Note:$ If you know more about this configuration (maybe contest problem or your finding), please let me know. GOOD JOB！ But could you please tell me why $PXYZ$ is a cyclic quadrilateral？

$ZXY=LFI+IEK=180-FIL-EIK=180-ZPY \implies P \in (XYZ)$.