My solution: Let $KR\cap LS=M$.Let $MK\cap \odot (BKS)=X$,$ML\cap \odot (CLR) =Y $. By angle-chasing we can get that $SR\parallel KL $.On the other hand $\angle RXS=\angle MXS=\angle KBS=60=\angle MYR=\angle SYR $ so $XSRY $ is cyclic($A $ also lies on that circle since $\angle A=60$) then Reim's theorem implies that $KXYL $ is cyclic $\implies$ $MX\cdot MK=MY\cdot ML $ which means $M $ lies on radical axis of $\odot (BKS) $ and $\odot(CLR) $. Let $KB\cap CL=H $.Angle-chasing yields that $H\in \odot(BCT) $ $\textbf {Claim 1} $:$M$ lies on $A $-median. $\textbf {Proof}:$Let $\angle MSR=x $,$\angle MRS=y $. Sine Law in triangles $\triangle AKR,ALS $: $\frac {SL}{RK}=\frac {AL}{AK}\cdot \frac {sin (60-y)}{sin (60-x)}=\frac {AC}{AB}\cdot \frac {sin (60-y)}{sin (60-x)}=\frac {sin B}{sin C}\cdot \frac {sin (60-y)}{sin (60-x)}$ but since $KL\parallel SR $ $\frac {sin y}{sin x}=\frac {MS}{MR}=\frac {SL}{RK}=\frac {sin B}{sin C}\cdot \frac {sin (60-y)}{sin (60-x)}$ but from Trig Ceva in triangle $\triangle ASR $ we have $\frac {sin MAS}{sin MAR}\cdot \frac {sin (60-y)}{sin (60-x)}\cdot \frac {sin x}{sin y} =1$ so $\frac {sin B}{sin C}=\frac {sin MAS}{sin MAR}$ then Ratio Lemma yields that $M $ lies on $A $-median. $\textbf {Claim 2}:$ $H $ lies on $A $-median. $\textbf {Proof}:$ Sine Law in triangles $\triangle HAB,HAC,HBC $: $\frac {sin B}{sin C}=\frac{sin (60+C)}{sin (60+B)}=\frac{HB}{HC}=\frac {HB}{HA}\cdot \frac {HA}{HC}=\frac {sin HAB}{sin HAC}$ again Ratio Lemma yields that $H $ lies on $A $-median. So $A,M,H $ are collinear.Now let $AH\cap \odot (ASRXY)=G$ By easy angle-chasing we get that $G\in \odot (BTC) $ again by easy angle-chasing we get that $KXGH $ is cyclic which means $MX\cdot MK=MG\cdot MH $ so $M $ is radical center of $\odot (BSK),\odot (CLR),\odot (BTC) $.done.

tenplusten wrote: Let $KB\cap CL=H $.Angle-chasing yields that $H\in \odot(BCT) $ I think you are wrong

abbosjon2002 wrote: tenplusten wrote: Let $KB\cap CL=H $.Angle-chasing yields that $H\in \odot(BCT) $ I think you are wrong No it is true. Just do some angle chasing.

Do the same as: tenplusten wrote: Let $KR\cap LS=M$. Let $MK\cap \odot (BKS)=X$ and $ML\cap \odot (CLR) =Y$. By angle-chasing we can get that $SR\parallel KL $. On the other hand, $\angle RXS=\angle MXS=\angle KBS=60^{\circ}=\angle MYR=\angle SYR $ so $XSRY $ is cyclic ($A $ also lies on that circle since $\angle A=60^{\circ}$) Then Reim implies that $KXYL $ is cyclic so $MX\cdot MK=MY\cdot ML $ and thus $M $ lies on radical axis of $\odot (BKS) $ and $\odot(CLR) $. Let $KB\cap CL=H $. Angle-chasing yields that $H\in \odot(BCT).$ Let $\odot(ABK)$ and $\odot(ACL)$ meet again at $T'$. Easily we see $B, T', L$ collinear and $C, T', K$ collinear so $T=T'$. Then $\angle STR = 120^{\circ}$ so $T \in \odot(ASRXY)$. Also, by $AB \parallel CL$ and $AC \parallel BK$ we have $ABHC$ parallelogram so $\angle BHC = 60^{\circ} \implies H \in \odot(BTC)$ and $H$ is on the median. Next we have \[\angle ABT = \angle BLC = \angle TAC \quad \text{and} \quad \angle ACT = \angle CKB = \angle TAB,\]so $T$ is the spiral center sending $BA \mapsto AC$ i.e. $T$ is the A-Dumpty point and lies on the symmedian. Then by DDIT or isogonal line lemma whichever you like on $\{S, R\}$ and $\{K, L\}$ wrt $\angle A$ gives $T, M$ isogonal, so $M$ lies on the median. Finally tenplusten wrote: Let $AM\cap \odot (ASRXY)=G$. By easy angle-chasing we get that $KXGH $ is cyclic, which means $MX\cdot MK=MG\cdot MH $, so $M $ is radical center of $\odot (BSK),\odot (CLR),\odot (BTC) $ and we are done.

Let $\odot (BKS)\cap \odot (BTC)=P ,\odot (BLR)\cap \odot (BTC)=Q $ ,$BP\cap AC=X ,CQ\cap AB=Y $ and $BP\cap CQ=Z $ be the radical center.By angle-chasing $CPSX,BQRY $ are cyclic hence $XS\parallel BR $,$YR\parallel CS $.So by Thales $\frac {AX}{AC}=\frac{AS\cdot AR}{AB\cdot AC}$ Analogously $\frac {AY}{AB}=\frac{AS\cdot AR}{AB\cdot AC}$ so $\frac {AY}{AB}=\frac {AX}{AC}$.Hence Ceva's theorem implies that $AZ$ bisects $BC $ and we are done.

Here is my solution for this problem. I have the same idea as Systematicworker, but a little bit different Solution Let $D$ be radical center of $(BCT)$, $(KBS)$, $(LCR)$; $E$ $\equiv$ $BD$ $\cap$ $AC$; $F$ $\equiv$ $CD$ $\cap$ $AB$, $V$ $\equiv$ $BD$ $\cap$ $CS$, $U$ $\equiv$ $CD$ $\cap$ $BR$ We have: $(ACER) = (SCVT)$ $\Leftrightarrow$ $\dfrac{\overline{EA}}{\overline{EC}} : \dfrac{\overline{RA}}{\overline{RC}} = \dfrac{\overline{VS}}{\overline{VC}} : \dfrac{\overline{TS}}{\overline{TC}}$ $\Leftrightarrow$ $\dfrac{\overline{EA}}{\overline{EC}} = \dfrac{\overline{RA}}{\overline{RC}} . \dfrac{\overline{VS}}{\overline{VC}} . \dfrac{\overline{TC}}{\overline{TS}}$ $\Leftrightarrow$ $\dfrac{\overline{EA}}{\overline{EC}} = \dfrac{\overline{VS}}{\overline{VC}} . \dfrac{\overline{BA}}{\overline{BS}}$ Similarly: $\dfrac{\overline{FA}}{\overline{FB}} = \dfrac{\overline{UR}}{\overline{UB}} . \dfrac{\overline{CA}}{\overline{CR}}$ Then: $\dfrac{\overline{EA}}{\overline{EC}} = \dfrac{\overline{FA}}{\overline{FB}}$ $\Leftrightarrow$ $\dfrac{\overline{VS}}{\overline{VC}} . \dfrac{\overline{BA}}{\overline{BS}} = \dfrac{\overline{UR}}{\overline{UB}} . \dfrac{\overline{CA}}{\overline{CR}}$ $\Leftrightarrow$ $\dfrac{\overline{ST}}{\overline{CK}} . \dfrac{\overline{BA}}{\overline{BS}} = \dfrac{\overline{RT}}{\overline{BL}} . \dfrac{\overline{CA}} {\overline{CR}}$ $\Leftrightarrow$ $\dfrac{ST . AB}{BS} = \dfrac{RT . AC}{CR}$ $\Leftrightarrow$ $\dfrac{AS . AK}{KS} = \dfrac{AR . AL}{LR}$ $\Leftrightarrow$ $\dfrac{AT . AK}{AK} = \dfrac{AT . AL}{AL}$ $\Leftrightarrow$ $AT = AT$ which is alway true So: $EF$ $\parallel$ $BC$ or $AD$ passes through midpoint of $BC$ Hence: radical center of $(BCT)$, $(KBS)$, $(LCR)$ lies on $A$ - median of $\triangle ABC$

$T$ is just the Fermat point of $\triangle ABC$, i.e. $\angle BTC = \angle CTA = \angle ATB = 120^\circ$. Let $M = \overline{KB} \cap \overline{LC}$ ; $X = \odot(BKS) \cap \odot(BTC) \ne B$ ; $Y= \odot(CLR) \cap \odot(BTC) \ne C$ ; $\overline{BX} \cap \overline{CY}$ ; $P = \overline{SX} \cap \overline{LY}$ ; $Q = \overline{RY} \cap \overline{KX}$. [asy][asy] size(250); pair M=dir(-80),L=dir(40),K=dir(160),A=0.47*L+0.53*K,B=extension(A,foot(A,K,(L+M)/2),K,M),C=extension(A,foot(A,L,(K+M)/2),L,M),T=extension(C,K,B,L),S=extension(K,T,A,B),R=extension(L,T,A,C),X=2*foot(B,circumcenter(B,S,K),circumcenter(B,T,C))-B,Y=2*foot(C,circumcenter(C,R,L),circumcenter(B,T,C))-C,N=extension(B,X,C,Y),P=extension(S,X,L,Y),Q=extension(K,X,R,Y); dot("$A$",A,dir(A)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$B$",B,dir(180)); dot("$C$",C,dir(30)); dot("$T$",T,dir(-90)); dot("$S$",S,dir(220)); dot("$R$",R,dir(0)); dot("$X$",X,dir(30)); dot("$Y$",Y,dir(200)); dot("$N$",N,dir(70)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); draw(A--B--C--A^^R--S,orange); draw(K--L--M--K,red); draw(circumcircle(K,S,B)^^circumcircle(L,R,C),blue); draw(K--R^^L--S^^A--M^^B--N^^C--N,brown); draw(B--L^^C--K^^A--T,fuchsia); draw(circumcircle(B,T,C),green); draw(L--P--S^^K--Q--R,purple); [/asy][/asy] We want to show $N$ lies on $A$-median of $\triangle ABC$. Angle chase shows $P,Q \in \odot(BTC)$. Pascal on $BXQYCT$ wrt $\odot(BTC)$ gives points $K,N,R$ are collinear. Similarly points $L,N,S$ are collinear. Thus $N = \overline{KR} \cap \overline{LS}$. Now from cyclic quad $ASTR$ we obtain $\triangle ASR$ is equilateral. Then $\triangle ASR$ and $\triangle MLK$ are homothetic, i.e. $\overline{AM},\overline{SL},\overline{RK}$ concur. So $N \in \overline{AM}$. Now $ABMC$ is a parallelogram, i.e. $AM$ is the $A$-median of $\triangle ABC$, and we are done. $\blacksquare$

Let $X,Y = (BSK) , (CRL) \cap (BTC)$ respectively and let $M$ be the midpoint of $BC$. By a simple angle chasing we get that $\angle RYT , \angle RYL = \angle B , 60^{\circ}$ respectively. Also note that $\frac{TR}{RL} = \frac{sin \angle TCA}{sin 60} . \frac{CT}{CL}$ and we also have $CL=CA$ and since $T$ is $A-$dumpty point(coz $\angle BTC =120 = 2\angle A$ and $AT$ is symmedian since it passes through the point $D$ st $BCD$ is equilateral and $DB,DC$ would be both tangent to $(ABC)$) , $\angle TCA = \angle MAC$. So : $$\frac{sin \angle YCB}{sin \angle YCA} = \frac{sin \angle YLR}{sin \angle YTR} = \frac{YT}{YL} = \frac{RT}{RL} . \frac{sin 60}{sin \angle B} = \frac{sin \angle TCA}{sin 60} . \frac{CT}{CA} . \frac{sin 60}{sin \angle B} = \frac{sin \angle MAC}{sin 60} . \frac{CT}{CA} . \frac{sin 60}{sin \angle B}$$Also note that $\frac{CT}{BT} = \frac{AB}{AC} ^2$(a well-known property of Dumpty point). So multiplying everything , since $\frac{sin \angle MAC}{sin \angle MAB} = \frac{AB}{AC}$ , we're done by sine ceva.

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Let $J$ be the humpty point, $H = (KSD) \cap (ATJ)$, similarly define $I$. Claim 1: $B \in (KHSD),C \in (IREL)$ Claim 2:$K-H-G-R$, $S-G-I-L$ Claim 3: $DHGJ,GIEJ$ are cyclic By some angle chase and radical axes we are done.