Nice DDIT exercise!

Here is an relatively easy to find hybrid solution, which makes the problem a bit unsuitable for this position (say, in comparison with ISL 2021 G8, this is much easier).

Very nice problem! Here's my solution: Let $\Gamma$ be the common circumcircle of $ABC,A'B'C'$, let incenters of $ABC,A'B'C'$ be $I,I'$ and let incircles of $ABC,A'B'C'$ be $\omega,\omega'$ Let $T$ be the ex-similicenter of $\omega,\omega'$. By Monge on $\omega,\omega'$ and the circle tangent to $BY,YB',B'X$ and lying on the same side as $B,X$ w.r.t. $B'Y$, it suffices to show that $T,B,B'$ are collinear. Let the direct common tangents of $\omega,\omega'$ be $t_1,t_2$. Let $t_1$ meet $\Gamma$ at $P_1,P_2$ and let $t_2$ meet $\Gamma$ at $Q_1,Q_2$. By Poncelet porism, $\exists P_3,P_4 \in \Gamma$ s.t. $\omega$ is the incircle of $P_1P_2P_3$ and $\omega'$ is the incircle of $P_1P_2P_4$. $\therefore P_1,P_2, I,I'$ are concyclic on a circle with the center as the midpoint of arc $P_1P_2$ on $\Gamma$. Similarly, $Q_1,Q_2,I,I'$ are concyclic. $\therefore$ By radical axis, $T,I,I'$ are collinear and $TI\cdot TI' = \mathrm{Pow}_\Gamma (T)$ $\;\;\dots(1)$ Now, let $BI,B'I'$ intersect $\Gamma$ again at $L,L'$ respectively. By observing that $L,L'$ are midpoints of arcs $AC,A'C'$ in $\Gamma$, and that $II'$ is the angle bisector of $AC,A'C'$; angle chasing gives $LL'\parallel II'$. $\therefore$ By Reim's $B,I,I',B'$ are concyclic on a circle say $\Omega$ By $(1)$; $T$ has same power w.r.t. $\Omega,\Gamma$ and hence lies on their radical axis which is the line $BB'$ and we are done!

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=mediumblue; pen sec=purple; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=paleblue; pair O, A1, C1, A, C, P, I, I1, I2, I3, N, N1, M, M1, B, B1, F, F1, Q, Y, L, X; O = (0, 0); C = dir(30); A1 = dir(150); A = dir(200); C1 = dir(310); P = intersectionpoint(A--C,A1--C1); N = (A+C)/2; N1 = (A1+C1)/2; I2 = incenter(A1, P, C); I3 = incenter(A, P, C1); M = intersectionpoint(N--(N*-100), circumcircle(C, A, C1)); M1 = intersectionpoint(N1--(N1*-100), circumcircle(C, A, C1)); I = intersectionpoint(P--(P+(I2-P)*100),circle(M, distance(M, A))); I1 = intersectionpoint(P--(P+(I3-P)*100),circle(M1, distance(M1, A1))); B = intersectionpoint((M+(I-M)*0.1)--(M+(I-M)*100), circumcircle(C, A, C1)); B1 = intersectionpoint((M1+(I1-M1)*0.1)--(M1+(I1-M1)*100), circumcircle(C, A, C1)); F = foot(I, A, C); F1 = foot(I1, A, C); Q = intersectionpoint(B--M, B1--M1); Y = intersectionpoint(B--(B+(C-B)*100), B1--(B1+(C1-B1)*100)); L = P + 3(I - P); X = intersectionpoint(A--B, A1--B1); draw(circumcircle(C,A,C1), pri); draw(A--C,pri); draw(A1--C1,pri); draw(circle(M, distance(M,A)), dashed+sec); draw(circle(M1, distance(M1, A1)), dashed+sec); draw(A--B, pri); draw(C--B, pri); draw(A1--B1, pri); draw(C1--B1, pri); draw(circle(I, distance(I, F)), tri); draw(circle(I1, distance(I1, F1)), tri); draw((P+(I-P)*100)--(P+(I1-P)*100),sec); draw(B--Y, pri); draw(B1--Y, pri); draw(B--M,dotted+tri); draw(B1--M1,dotted+tri); label("$\ell$",L,dir(120)); label("$A'$",A1,dir(150)); label("$C'$",C1,dir(300)); label("$A$",A,dir(190)); label("$C$",C,dir(60)); label("$P$",P,dir(180)); label("$I$",I,dir(120)); label("$I'$",I1,dir(30)); label("$M'$",M1,dir(60)); label("$M$",M,dir(300)); label("$B'$",B1,dir(240)); label("$B$",B,dir(120)); label("$Q$",Q,dir(0)); label("$X$",X,dir(180)); path clip[]; clip=circle((0,0),(distance(O,A))*1.5); clip(currentpicture,clip); [/asy][/asy] (very similar to above solution) Let $I, I'$ be the incenters of $\triangle{ABC}, \triangle{A'B'C'}$. It suffices to show that $BI \cap B'I' = Q \in$ bisector of $\angle{CYC'}$, since this would imply that $Q$ is equidistant from all four sides of $XBYB'$, and thus that $XBYB'$ is tangential. Let $M, M'$ be the midpoints of $\widehat{AC}$ including $B', C'$ and $\widehat{A'C'}$ including $B$, respectively, and let $(M), (M')$ be the circles centered at $M, M'$ passing through $A, C$ and $A', C'$ respectively. Then, by Fact 5, $I \in (M), I' \in (M')$. Let $\ell$ be the line through $AC \cap A'C' = P$ bisecting $\angle{APC'}$; then $I, I' = MB, M'B' \cap \ell$. Now note that $Q \in$ bisector of $\angle{CYC'}$ if and only if $BQ \sin \angle{CBQ} = B'Q \sin \angle{C'B'Q}$, and since $\frac{\sin \angle{CBQ}}{\sin \angle{C'B'Q}} = \frac{CM}{C'M'} = \frac{MI}{M'I'}$, it suffices to show that $BQ \cdot MI = B'Q \cdot M'I'$. By Power of a Point at $Q$, $BQ \cdot MQ = B'Q \cdot M'Q$, so it suffices to show that $\frac{MI}{MQ} = \frac{M'I}{M'Q}$, or $MM' \parallel II'$, or that $MM'$ is parallel to the bisector of the angle between $AC, A'C'$. However, this is true by angle chasing since the measure of the acute angle between $MM', AC$ is equal to $\frac{\widehat{MM'}}{2}$, which is also equal to the acute angle between $MM', A'C'$. $\square$

Sorry, but in this graph the circle tangent to the edges of $BXB'Y$ is outside of the quadrilateral. By Poncelet, we construct tangents from $A,C$ to the circle $I'$, they intersect at $D$, a point on the common circumcircle. Define $D'$ similarly. Let $BB'$ intersect $II'$ at $K$, it suffices to prove that $K$ is the homothetic center of $\odot I,\odot I'$ since using reverse to Monge will prove the desired statement. Now we prove $\frac{IK}{I'K}=\frac{r_1}{r_2}$. Let $M,N$ be the midpoints of arc $AC,A'C'$. We have $B,I,M;B',I',N;D,I',M;D',I,N$ collinear. By Pascal, $D,K,D'$ collinear. Notice that it is well known that $A'C'\parallel BD,AC\parallel B'D'$ by $AII'C$ concyclic and some angle chasing. Now we start to compute, denoting $r_1,r_2$ as radius of $\odot I,\odot I'$: $\frac{IK}{I'K}=\frac{BI \sin IBK}{B'I'\sin I'B'K}=\frac{r_1\sin A'B'N \sin DNM}{r_2\sin ADM \sin BMN}=\frac{r_1}{r_2}\frac{A'N \sin DNM}{AM \sin BMN}=\frac{r_1}{r_2}\frac{NI \sin DNM}{MI \sin BNM}=\frac{r_1}{r_2}\frac{II'/2}{II'/2}=\frac{r_1}{r_2}$.

Call $\Omega =(AA'BCC'B')$, $\omega$ and $\omega'$ (with centers $I$ and $I'$) the incircles of $\triangle ABC$ and $\triangle A'B'C'$ respectively. It's a well-known fact that if we have two circles, $\omega$ inside $\Omega$, which are in a Poncelet configuration, the map $\Omega\rightarrow \omega$ given by $P\rightarrow P^*$ where $\triangle PQR$ is the Poncelet triangle of $P$ and $QR$ is tangent to $\omega$ at $P^*$, is projective. Consider the map $f: \Omega \rightarrow \omega\rightarrow \omega'\rightarrow \Omega$ given by $P\rightarrow P^* \rightarrow Q^* \rightarrow Q$, where the map from $\omega$ to $\omega'$ is the negative homothety composed with symmetry across $II'$. Thus, $f:\Omega\rightarrow\Omega$ is both projective and $f\circ f =\text{id}$ so $f$ is an involution, meaning $PQ$ passes through a fixed point. Firstly, notice that $f(B)=B'$. It's enough to prove that the fixed point is the eximilicenter of $\omega$ and $\omega'$. Indeed, taking $P^*Q^*$ to be a common external tangent of $\omega$ and $\omega'$, since there are two choices, it's enough to prove that for the corresponding $P$ and $Q$, $PQ$ passes through the eximilicenter. The problem now reads: Well-known wrote: Let $ABCD$ be a cyclic quadrilateral with $I$ and $J$ the incenters of $\triangle ACD$ and $\triangle BCD$ respectively. Prove that $AB$, $CD$ and $IJ$ are concurrent. The proof is straight-forward by radical center, since both $AIJB$ and $DIJC$ are cyclic.

LoloChen wrote:

How do you think of this approach?

Does anyone have a moving points approach. Based on post #8, it seems like such a solution might exist. In any case, I will edit in a solution on this post when I find one.

Let $\omega$ and $\omega'$ be the incircles of $\triangle ABC, \triangle A'B'C'$ respectfully. Suppose the exsimillicenter of $\omega$ and $\omega'$ is on $\overline{BB'}$. Then consider the circle $\Omega$ obtained when dilating $\omega$ at $B$ until it is tangent to $\overline{B'C'}$. The exsimillicenter of $\Omega$ and $\omega'$ is on $\overline{B'C'}$. By Monge on $\Omega$ $\omega$, and $\omega'$, $B'$ would be the exsimillicenter of $\Omega$ and $\omega'$, so $XBYB'$ has $\omega$ as an incircle. Thus we will prove the exsimillicenter of $\omega$ and $\omega'$ is on $\overline{BB'}$. Let $I$ be the center of $\omega$, let $M=\overline{BI}\cap (ABC)$, $L=\overline{MO}\cap (ABC)$, $T=\overline{LI}\cap (ABC)$, and $G=\overline{AT}\cap\overline{IO}$. It is known that $G$ is the exsimillicenter of $\omega$ and $(ABC)$, for example because $T$ is the $B$-mixtilinear touchpoint of $\triangle ABC$ and monge on $\omega$, $(ABC)$, and the $B$-mixtilinear incircle of $\triangle ABC$. Define $I',L',T'$, and $G'$ similarly. Define $M'$ similarly as well. Define $O$ as the center of $(ABC)$. First, note that $\overline{II'}\parallel \overline{MM'}$ because $\overline{II'}$ is an angle bisector of $\overline{AC},\overline{A'C'}$ and $\triangle OMM'$ being isosceles with $\overline{OM}\perp\overline{AC},\overline{OM'}\perp\overline{A'C'}$ means that $\overline{MM'}$ is also parallel to an angle biector $\overline{AC},\overline{A'C'}$. Recall that I don't care about configuration issues. Now we organize the rest into a list because yes. $\overline{MM'}\parallel \overline{LL'}$ because $LML'M'$ is a rectangle, so $\overline{II'},\overline{MM'},\overline{LL'}$ are concurrent. Thus Desargues on $\triangle IML, \triangle I'M'L'$ means that $O$, $\overline{BI}\cap\overline{B'I'}, \overline{LI}\cap {L'I'}$ are collinear. Pascal's theorem on $BTLB'T'L'$ means that $\overline{BG}\cap \overline{B'G'},\overline{LI}\cap\overline{L'I'},\overline{B'L}\cap\overline{BL'}$ are collinear. Pascal's theorem on $BMLB'M'L'$ means that $\overline{BI}\cap\overline{B'I'},O,\overline{BL'}\cap\overline{B'L}$ are collinear. The above gives us that $O,\overline{BG}\cap\overline{B'G'},\overline{BI}\cap\overline{B'I'}$ are collinear. Desargues on $\triangle BIG, \triangle B'I'G'$ means that $\overline{BB'},\overline{II'},\overline{GG'}$ concur. By Monge on $\omega, \omega',(ABC)$, $\overline{GG'}\cap\overline{II'}$ is the exsimillicenter of $\omega, \omega'$, as desired.

Let $\omega ,\omega '$ be incircles of $ABC,A'B'C'$ with centers $I,I'$ and radii $r,r'$ respectively. Let $\Omega$ denotes the circumcircle of hexagon and $D=AB\cap A'C',$ $D'=A'B'\cap AC.$ Claim 1. $BB',CC',DD'$ concur. Proof. By Pascal on $ABCA'B'C'$ point $Z=AC'\cap A'C$ lies on $XY.$ Let $CD,C'D'$ meet $\Omega$ again at $E,E',$ so $$C(C'A'BD)=(C'A'BE)_{\Omega}\stackrel{D}{=}(A'C'AC)_{\Omega} =(CAC'A')_{\Omega}\stackrel{D'}{=}(ACE'B')_{\Omega} =(CAB'E')_{\Omega}=C'(CAB'D"),$$implying that $CD,C'D'$ meet on $ZY.$ Desargues theorem on $BCD,B'C'D'$ finishes $\Box$ Claim 2. $T=BB'\cap II'$ is the exsimilicenter of $\omega ,\omega'.$ Proof. Let $\tau_1, \tau_2, \tau_1 ',\tau_2 '$ be tangents from $T$ to both circles respectively. Note that $S=AC\cap A'C'$ is the insimilicenter of these circles. Let $W=TD'\cap A'C'.$ Due to the first claim converse of Pascal on $TWC'CAB$ gives that this hexagon has a circumscribed conic; by DIT on $ABCT$ there exist involution by $A'C'$ with reciprocal pairs $$(W;C'),(D;TC\cap A'C'),(S;TB\cap A'C').$$By DDIT on both $CBDS,$ $C'B'D'S$ projection of this involution from $T$ gives an involution by pencil with reciprocal pairs $$(\overline{TII'};\overline{TBB'}),(TC;TD),(TC';TD'),(\tau_1;\tau_2), (\tau_1 '; \tau_2 ')\implies$$$$\implies (TI,TB,\tau_1,\tau_1 ')=(TB,TI,\tau_2,\tau_2 ')=(TI,TB,\tau_2 ',\tau_2).$$Since $\overline{TII'}$ bisects angles between $\tau_1,\tau_2$ and between $\tau_1 ',\tau_2 '$ the last equality gives $$\left \{ \tau_1, \tau_2\right \} = \left \{ \tau_1 ',\tau_2 '\right \} \text{ } \Box$$ Finish. If $XBYC$ is convex, then lines $BI,B'I'$ meet at none-infinite point $O.$ By Menelaus on $OBB'$ $$\frac{\text{d}(O,BC)\cdot r'}{r\cdot \text{d}(O,B'C')}=\frac{|OB|\cdot |I'B'|}{|IB|\cdot |OB'|}=\frac{|TI'|}{|TI|}=\frac{r'}{r}\implies \text{d}(O,BC)=\text{d} (O,B'C')$$(note that this occurs even when $T$ lies at infinity). Therefore $O$ is the incenter of $BXCY,$ we are done.

Sourorange wrote: By reflection $\angle (OD, II')=\angle (OD', II')$, so $II' // DD'$. So $\frac{d_1}{d_2}=\frac{BI_0sin\angle ABD}{B'I_0sin\angle A'B'D'}=\frac{D'I_0*AD}{DI_0*A'D'}=\frac{D'I_0*ID}{DI_0*I'D'}=1$. Nice solution! It amazes me how you came up with the same exact wording and diagram as post #9 as well!

By Monge on the two incircles and the circle tangent to $XB'$, $B'Y$, and $BY$ it suffices to show that the exsimilicenter of the two incircles lies on $BB'$. Let $I$, $I'$ be the incenters and $F$, $G$ be the exsimilicenter and insimilicenter. Let $BI$ and $B'I'$ intersect $(ABC)$ at $D$ and $D'$ respectively. Let $A'C'$ and $AC$ intersect $DD'$ at $H$ and $J$. We have \[\angle I'GH=\frac12 \angle AGC'=\frac12 \left(\angle GHJ+\angle GJH\right)\]but from angle bisector stuff we have $\angle GHJ=\angle GJH$. Thus, $II'\parallel DD'$. By Reim's, $II'B'B$ is cyclic. Therefore, it suffices to show that $FI\cdot FI'$ is equal to the power of $F$ with respect to $(ABC)$. Let $r$ and $r'$ be the inradii of $ABC$ and $A'B'C'$. Let $O$ and $R$ be the circumcenter and circumradius of $ABC$. From the Law of Cosines, we have \[\frac{IF^2+OF^2-OI^2}{2IF\cdot OF}=\frac{I'F^2+OF^2-OI'^2}{2I'F\cdot OF}=\cos(\angle IFO)\]We can cross out $2OF$ from both denominators. We have \begin{align*} \left(OF^2+\frac{d^2r^2}{(r'-r)^2}-R^2+2Rr\right)(r')&=\left(OF^2+\frac{d^2r'^2}{(r'-r)^2}-R^2+2Rr'\right)(r)\\ OF^2r+\frac{d^2r'^2r}{(r'-r)^2}-R^2r'+2Rrr'&=OF^2r'+\frac{d^2r^2r'}{(r'-r)^2}-R^2r+2Rrr'\\ OF^2(r'-r)&=\left(\frac{d^2rr'}{(r'-r)^2}+R^2\right)(r'-r) OF^2&= FI\cdot FJ+R^2 \end{align*}and we are done.