Complex bash is useless at the contest. Coordinates: $(A,B,C,Q)=(a^2,b^2,c^2,ab)$. Then $$m=\frac{b^2+c^2}{2},\ b_1=\frac{ac^2+bc^2-ab^2}{b},\ e=\frac{bc^4+ac^4-ab^4-ab^2c^2}{ac^2+2bc^2-2ab^2-b^3},$$intersection of line passing through $C$ and midpoint of $B_1E$ with circumcircle $ABC$ different than $C$: $$k=\frac{a(2a^2b^2-a^2c^2-2abc^2+b^2c^2)(b^3+2ab^2-2bc^2-ac^2)}{b(ac^2+2bc^2-2ab^2-b^3)(2c^2-b^2-2ab+a^2)}.$$We are left with verifying that $b,b_1,m,k$ are concyclic.

Let $L$ be the midpoint of arc $AB$, not containing $C$. Let $S$ be the intersection of $KC$ and $(BB_1M)$ and let $T=MS\cap AC$. We claim that $S$ is the midpoint of $BT\parallel B_1E$ from which we are done by the homothety. Observe that $B_1$ lies on the circle centered at $L$ with radius $LB$. Claim 01. $\measuredangle BB_1M=\measuredangle QB_1C$. Proof. Note that $\measuredangle BB_1L=90^\circ-\measuredangle CAB=\measuredangle (LM,AC)$, hence we get that \begin{align*} \measuredangle BMB_1=\measuredangle BB_1L-\measuredangle MB_1L=\measuredangle (LM,AC)-\measuredangle LQB_1=\measuredangle QB_1C, \end{align*}where the second equality comes from an inversion at $L$ with radius $LB$ swapping $M$ and $Q$. Claim 02. $SB\parallel QB_1$. Proof. Indeed, \begin{align*} \measuredangle SBM&=\measuredangle SKM=\measuredangle CKM\\&=\measuredangle CKB+\measuredangle BKM\\&=\measuredangle B_1AM+\measuredangle BB_1M\\&=\measuredangle B_1AM+\measuredangle QB_1C\\&=\measuredangle (QB_1,AB). \end{align*} Now, we easily obtain that $\measuredangle SMB=\measuredangle SKB=\measuredangle CKB=\measuredangle CAB=\measuredangle TAB\implies MS\parallel TA$ and as $M$ is the midpoint of $AB$, we conclude that $S$ is the midpoint of $BT$. [asy][asy]import olympiad; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,M,K,Q,E,S,T,L,B1; O=(0,0);C=dir(120);B=dir(200);A=dir(340);M=midpoint(A--B); L=intersectionpoints(circumcircle(A,B,C),C--100incenter(A,B,C)-99C)[1];Q=-L;B1=intersectionpoints(C--A,circle(C,abs(C-B)))[0];E=extension(C,B,Q,B1);K=intersectionpoints(circumcircle(A,B,C),circumcircle(M,B1,B))[0]; S=intersectionpoints(circumcircle(M,B,B1),K--100C-99K)[0];T=extension(S,B,C,A); draw(B1--E,grey+0.5);draw(C--E,grey+0.5);draw(B--T,grey+0.5);draw(C--T,grey+0.5);draw(K--S,grey+0.5);draw(Q--L,grey+0.5);draw(B--B1,grey+0.5); draw(circumcircle(M,B,B1),royalblue); draw(circumcircle(A,B,C),royalblue); draw(A--B--C--cycle,red);draw(arc(L,A,B),royalblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$L$",L,dir(L)); dot("$K$",K,dir(K)); dot("$Q$",Q,dir(Q)); dot("$B_1$",B1,dir(B1)); dot("$E$",E,dir(E)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); [/asy][/asy]