The main claim: $\omega$ is the nine-point circle of $\triangle ABC$, then we notice the incircle satisfy $(1)$ all at once so use inversion at a vertice to construct the corresponding circle. The rest can be done by bashing.

Here is a proof of the main claim mentioned above. Let $M_a$ be the midpoint of $AH$. We will prove that $KM_a \perp DA_1$, which will imply that $A_2$ lies on the nine-point circle with diameter $A_1M_a$. We actually claim that $K$ is the orthocenter of $\triangle DM_aA_1$. Firstly, notice that $A_1M_a \perp QR$, as $QR$ is the radical axis of $(AH)$ and $(BC)$. So, we are left to prove that $DM_a \perp AA_1$. We will prove that if $AA \cap BC=E$, then $E \in DM_a$, which will be sufficient, as we know that $DE \perp AA_1$ by Poland 2018/5. Indeed, notice that $EM_a \perp AA_1$ as $M_a$ is the orthocenter of $\triangle AA_1E$ ($M_aA_1 \perp AA$ since $A_1M_a \parallel AO \perp AA$), so we are done!

Does this problem consisted of 2 independent problems (One is easy orthocenter and another is inversion which is completely different)

Wait am I tripping or after proving that $\omega$ is the nine-point circle the problem is obviously done by Feuerbach point?

nope A'B'C' must be different

nobodyknowswhoIam wrote: Wait am I tripping or after proving that $\omega$ is the nine-point circle the problem is obviously done by Feuerbach point? There's a condition lost in this version. This problem also requires $\omega$ and $\odot A',\odot B',\odot C'$ are internally tangent.

See here for an almost purely geometric proof: A bit disappointed that a problem that appeared previously showed up in China TST again ,but it is really beautiful (regardless of the meaningless part 1). LoloChen wrote: Sketch of my solution: By invertion and reflection at A where the 9. circle becomes itself, circle Oa becomes the incircle, so length chase (not hard) gets Oa is the orthocenter of the triangle formed by AB, AC and NbNc (where Na is midpoint of arc BAC and Nb Nc similarly defined) and Ob, Oc are similar. Since the Miquel point of any 4 of lines AB, BC, CA, NaNb, NbNc, NcNa is always X101, those quad's orthocenter lines are the same. Since orthocenter of NaNbNc is Bevan point and obviously H, Be and J, the Speiker point are concurrent, we are done. Below is the proof in simplified Chinese.

Solved with Mogmog8, Ritwin, and v4913- a new addition to DMW-anti ! [asy][asy] //certain anti prob //setup size(9cm); defaultpen(fontsize(10pt)); 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 int i=0; pair A,B,C,H,P,Q,R,D,A1,Ma,Ka,A2; A=(3,10); B=(0,0); C=(14,0); H=orthocenter(A,B,C); P=foot(A,B,C); Q=foot(B,C,A); R=foot(C,A,B); D=extension(A,A+B-C,Q,R); A1=(B+C)/2; Ma=(A+H)/2; Ka=extension(Q,R,A,A1); A2=foot(Ma,A1,D); //draw filldraw(A--B--C--cycle,blu1,blu); draw(P--A--A1^^R--D,blu); draw(A--D,purple+dashdotted); draw(D--A1,magenta); draw(Ma--A2,purple); draw(circumcircle(P,Q,R),blu+dashdotted); draw(circumcircle(A,Q,R),blu); //draw tangent circles pair I,Ia; I=incenter(A,B,C); Ia=2*circumcenter(B,I,C)-I; draw(incircle(A,B,C),red+dotted); draw(circle(Ia,distance(Ia,foot(Ia,B,C))),red+dotted); pair Ainvert(pair P,real pow) {return A+pow/(distance(P,A)*distance(P,A))*(P-A);} real r=distance(A,Q)*distance(A,C)/2; pair S,T; S=Ainvert(foot(Ia,A,C),r); T=Ainvert(foot(Ia,A,B),r); pair O=2*circumcenter(A,S,T)-A; draw(circle(O,distance(O,S)),red+dotted); draw(foot(Ia,A,B)--B^^C--foot(Ia,A,C),blu+dashed); clip(box((-4,-8),(18,12)) ); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.09),a,linewidth(.3)); label(s,P,v);} string labels[]={"$A$","$B$","$C$","$H$","$P$","$Q$","$R$", "$D$","$A_1$","$M_a$","$K_a$","$A_2$"}; pair points[]={A,B,C,H,P,Q,R, D,A1,Ma,Ka,A2}; //12 real dirs[]={ 90,160,50,-50,-110,80,180, 90,-80,-170,-100,-60}; pen colors[]={blu,blu,blu,blu,blu,blu,blu, purple,blu,purple,purple,magenta}; for (i=0; i<12; ++i) pt(labels[i],points[i],dir(dirs[i]),colors[i]); [/asy][/asy] Wasn't stated in the problem but $H$ is the orthocenter, and we'll call the midpoint of $\overline{AH}$ as $M_a$. Here is a slightly different proof of the claim that $A_2$ lies on the 9-point circle of $\triangle ABC$: Proof. Observe that \[(QR;DK_a)\overset A=(CB;\infty_{BC}A_1)=-1\]while $\overline{A_1Q}$, $\overline{A_1R}$ are tangent to $(AH)$. As a result, $\overline{A_1D}$ is the polar of $K_a=\overline{QR}\cap\overline{AA_1}$ wrt $(AH)$, meaning that $\overline{M_aK_a}\perp\overline{A_1D}$. But $M_a$ and $A_1$ are antipodes on the nine-point circle, so $\angle M_aA_2A_1=90^\circ$ means that $A_2$ lies on the nine-point circle too.$\qquad\qquad\square$ The problem in fact requires the centers to be different. So, instead, we have to take the incircle, the $A$-excircle, and the circle inside the triangle externally tangent to $\overline{AB},\overline{AC}$. This works because former two circles touch the 9-point circle by Feuerbach, while all three centers lie on the bisector of $\angle A$.

well... never mind the original problem mentions that all of the circles should be in $\omega$. as shown in the img below...

so actually the easiest (and probly only) way to solve this is to use incursion and vector. details see wechat program 数之谜