Any solution to this beautiful problem?(I have one but it involves a lot of computation, and I am currently on mobile, so maybe I'll post it later)

Hint. Try to apply Feuerbach theorem. To do so, prove that $AO_a$ is tangent to circumcircle of $ABC$.

Dr.Sop wrote: Hint. Try to apply Feuerbach theorem. To do so, prove that $AO_a$ is tangent to circumcircle of $ABC$. I got that $AO_a$ is tangent to circumcircle of $ABC$, but how does one use Feuerbach theorem after that? Can you post a full solution please.

math_pi_rate wrote: Dr.Sop wrote: Hint. Try to apply Feuerbach theorem. To do so, prove that $AO_a$ is tangent to circumcircle of $ABC$. I got that $AO_a$ is tangent to circumcircle of $ABC$, but how does one use Feuerbach theorem after that? Can you post a full solution please. Ok. Now it is enough to prove that $AO_a$ is parallel to $O_bO_c$ and similar for $b, c$, because then $O_aO_bO_c$ is medial triangle for a triangle formed by $AO_a, BO_b, CO_c$ and circumcircle of $ABC$ is its incircle.

Dr.Sop wrote: Ok. Now it is enough to prove that $AO_a$ is parallel to $O_bO_c$ and similar for $b, c$, because then $O_aO_bO_c$ is medial triangle for a triangle formed by $AO_a, BO_b, CO_c$ and circumcircle of $ABC$ is its incircle. I had thought of this when I was doing the problem, but I couldn't prove it, and that's why I had to follow a computational way.

Let $\Delta XYZ$ be the tangential triangle and let $\Omega=\odot (AA_BA_C)$ and $P$ be the antipode of $A$ in $\odot (ABC)$ and let $T=\odot (ABC)\cap \Omega$ now obviously $AO_a$ is tangent to $\odot (ABC)$ and $AO$ is tangent to $\Omega $ now since $\measuredangle O_aAO=90$ we have $\odot (ABC),\Omega $ are orthogonal so $OT$ is tangent to $\Omega$ so $-1={(A_C,A_B;A,T)}_{\Omega}\stackrel{A}{=}{(B,C;P,T)}_{\odot (ABC)}$ ,$\implies PT$ is the $P-$symmedian of $\Delta PBC$ so $X,P,T$ are collinear now if $PT\cap YZ=L$ then $\measuredangle ATL=90$ so $L$ lies on $\Omega$ and it is well known that $L$ is excircle touch point in $YZ$ so $O_aL=O_aA$ so $O_a$ is the midpoint of $YZ$ $\blacksquare$

[asy][asy] import olympiad; size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120), B = dir(210), C = dir(330),F=(B+C)/2 ; pair O=circumcenter(A, B, C); pair E=extension(O,F,A,C); pair R=extension(O,F,A,B); draw(A--B--F--C--A,purple); draw(circumcircle(R,E,A),dashed); draw(circumcircle(A,B,C),orange); pair OA=circumcenter(R,E,A); pair K=(A+C)/2; pair BC=extension(K,O,B,C); pair BA=extension(K,O,A,B); draw(R--E--O,dotted); draw(A--R,purple); draw(BA--BC--O,dotted); draw(B--BA,purple); draw(circumcircle(B,BC,BA),dashed); pair OB=circumcenter(B,BC,BA); pair Y=(A+B)/2; pair CA=extension(Y,O,A,C); pair CB=extension(Y,O,C,B); pair OC=circumcenter(C,CA,CB); draw(C--CB); draw(circumcircle(C,CA,CB),dashed); draw(O--CA--CB,dotted); draw(circumcircle(OA,OB,OC),red); pair X=extension(OA,A,OB,B); pair Y=extension(OB,B,OC,C); pair Z=extension(OC,C,OA,A); draw(OB--BC--OA,pink); draw(X--Z--Y--X,green); dot("$A$", A, dir(80)); dot("$B$", B, dir(150)); dot("$C$", C, dir(330)); dot("$O$", O, dir(50)); dot("$A_C$",E,dir(50)); dot("$A_B$",R,dir(100)); dot("$O_A$",OA,dir(100)); dot("$B_C$",BC,dir(100)); dot("$B_A$",BA,dir(160)); dot("$O_B$",OB,dir(100)); dot("$C_A$",CA,dir(60)); dot("$C_B$",CB,dir(100)); dot("$O_C$",OC,dir(100)); dot("$X$",X,dir(120)); dot("$Y$",Y,dir(200)); dot("$Z$",Z,dir(100)); [/asy][/asy] Proof: Firstly notice that $$\angle OA_BA=90^\circ-\angle ABC=\angle OAA_C$$$\implies$ $OA$ is tangent to $\odot(A_BAA_C)$ and similiarly the others. So we have that $\triangle{XYZ}$ is the extouch triangle of $\triangle{ABC}$. Now notice that $O$ is the orthocentre of $\triangle{CA_CB_C}$ $\implies$ $$\angle A_CB_CC=90^\circ-\angle OCB_C=\angle CAB$$$\implies$ $(A_CB_CBA)$ is cyclic. Now $$\angle O_BB_CB=\angle O_BBB_C=\angle BAC=180^\circ-\angle BB_CA_C$$$\implies$ $\overline{(O_B,B_C,A_C)}$ is a collinear triple and similiarly obtain that $\overline{(O_B,B_C,A_C,O_A)}$ is a collinear triple. Now we state a crucial claim. Claim: $A_CB_C$ is the midline w.r.t $\triangle{XYZ}$. Proof: For this notice that we colud embed this w.r.t to the extouch triangle and rephrase it as follows. Rephrased lemma wrote: Let $I$ be the incentre of $\triangle{ABC}$ and let $\triangle{DEF}$ be the intouch triangle where $D,E,F$ are in $BC,CA,AB$ respectively. Let $BI \cap DE=X$ and $CI \cap DF=Y$. Then show that $XY$ is the middle line w.r.t $\triangle {ABC}$. Clearly $XY \parallel BC$. Now we'll show that if $M$ is the midpoint of $AB$ then $MX \parallel BC$. For this notice that by lemma 1.45 of egmo we have that if $K$ is the midpoint of $BC$ and $BI \cap EF=G$ then $\overline{(M,K,G)}$ are collinear. Now by a homothety we are essentially done $\square$. Now by our crucial claim and feuberach's theorem the problem is solved $\blacksquare$.

The point of tangency is the Anti-Steiner point of $O$ wrt $\triangle ABC$. One can show with angle chasing that points $O_A,A_C,B_C,O_B$ are collinear, and similarly other collinearities. So basically, $\triangle O_aO_bO_c$ is just the triangle formed by lines $$ A_CB_C,B_AC_A,C_B,A_B $$ Now our problem is simply equivalent to Problem 3.1 of this handout (also linked before). $\blacksquare$

Let the tangential triangle be $XYZ$. By Feuerbach theorem, it is sufficient to prove that $\Omega:=(O_aO_bO_c)$ is the nine-point circle of $\bigtriangleup XYZ$. Let $(AA_bA_c)\cap(ABC)=T\neq A$ and let the $A$-antopode be $A’$. Note that $\angle AA_bA_c=90^{\circ}-\angle B= 90^{\circ}-1/2\cdot\angle AOC=\angle OAA_c$ so AO is tangent to $(AA_bA_c)$. From this, it follows that $O_a$ is on $YZ$ and that $(ABC)$ and $(AA_bA_c)$ are orthogonal. We shall prove that O_a is the midpoint of YZ which finished the problem. Let $B’$, $C’$, and $T’$ be the midpoints of $AB$, $AC$, and $AT$, respectively. Let $P_{\infty}$ be the point at infinity along $YZ$. Indeed, \begin{align*} -1&=(A,T;A_b,A_c)\\ &= (AA,AT;AA_b,AA_c)\\ &=(A’,T;B,C)\\ &=(O,T’;B’,C’)\\ &=(P_{\infty},O_a;Y,Z) \end{align*}where the two last equalities follow from homothety in $A$ and inversion in $(ABC)$.