let $P=AO\cap k$, $Q=BO\cap k$ and let midpoint of $UV$ is $T$ and let symetry of $A,B$ to $T$ are $R,S$ and let symetry of $O$ is $Z$. we have $OP=AO$ and $OQ=BO$ so $ABPQ$ parallelogram. Similarly $ASRB$ parallelogram. So $\angle BAQ=180^{\circ}-\angle ABP$ and $\angle BAS=180^{\circ}-\angle ABR$. So we have $\angle QAS=\angle BAS-\angle BAQ=180^{\circ}-\angle ABR-180^{\circ}+\angle ABP=\angle PBR$. So we have $BP=AQ$ and $BR=AS$ and $\angle QAS=\angle PBR$. Then we have $PBR$ and $QAS$ congruent triangles and from there we have $PR=QS$ and $PR\parallel QS$. because of $BO=OQ$ and $BT=TS$ we have $OT\parallel QS$ and because of $O$ center of $k$ and $T$ is midpoint of $UV$ we have $OT$ is perpendicular to $UV$ so $QS$ is perpendicular to $UV$ and we have $OT=\frac{QS}{2}$ so we have $OZ=2OT=QS$ and $OT\parallel QS$. Similarly we have $OZ=PR$ and $OT\parallel PR$. So we have $OZ=PR=QS$ and $OT\parallel QS\parallel PR$. From there we have $OQP$ and $ZSR$ congruent triangles. So $OP=ZR=ZS$ and $\angle RZS=\angle POQ$. Because of $ABPQ$ parallelogram we have $\angle POQ=\angle AOB=2\angle ACB$ so $\angle RZS=2\angle ACB$. Also $O$ is center so $OV=PO$ after that we have because of $O,Z$ symetric $OV=ZV$. From there we have $ZR=ZV=OP$. Similarly we have $ZV=ZR=ZS=ZU$. So we have $VRSU$ cyclic with center $Z$. Because of $AS\parallel BR$ we have $\angle SAC+\angle RBC=\angle ACB$. let $L=RM\cap NS$ and $G=LC\cap (Z)$ and $K=RM\cap (Z)$. Because of $GN \cdot NS = UN \cdot NV$ and $UN \cdot NV =AN \cdot NC$ (if we write power of point $N$ in circles $k$ and $(Z)$ we have these equations) we have $GN \cdot NS = AN \cdot NC$ so $GCSA$ cyclic. From there we have $\angle CGS=\angle CAS$. $G$ on $(Z)$ so $\angle RGS=\frac{\angle RZS}{2}=\angle ACB$ and we have $\angle SAC+\angle RBC=\angle ACB$. From there we have $\angle CGR=\angle RGS-\angle SGC=\angle ACB-\angle CAS=\angle RBC$. So we have $\angle CGS=\angle CAS$ and $\angle CGR=\angle RBC$. Similarly we have $\angle CKS=\angle CAS$ and $\angle CKR=\angle RBC$. So we have $\angle CGS=\angle CKS=\angle CAS$ and $\angle CGR=\angle CKR=\angle RBC$. So $G,K,C$ on intersection of circles $(ASC),(BRC)$. This is impossible so $K=G$. İf $K=G$ we have, intersection of $SN,RM$ on $(Z)$. So $\angle RLS=\angle CKR=\angle RBC$ Because of $QS\parallel PR$ ,$\angle QSN+\angle PRM=\angle RLS=\frac{\angle RZS}{2}=\angle ACB$. let midpoints of $AM,BN$ are $F,Y$. İf we apply homotety with center $A$ and ratio $1/2$ we have points $Y,T,O$ goes to $M,R,P$. So we have $\angle YTO=\angle PRM$. Similarly $\angle FTO=\angle QSN$. So $\angle FTY=\angle YTO+\angle FTO=\angle QSN+\angle PRM=\angle ACB$. let midpoint of $MN$ is $W$. Because of $Y,W$ midpoints of $AM,MN$ we have $YW\parallel AC$ and $W,F$ midpoints of $AN,BC$ we have $FW\parallel BC$. Because of $YW\parallel AC$ and $FW\parallel BC$ ,$\angle FWY=\angle ACB$. from there we have $\angle FTY=\angle FWY=\angle ACB$. So we have $\angle FWY=\angle FTY$ this shows that $WTFY$ cyclic so we get result.

Let $P,Q,R,S$ be the midpoints of $BM,AN,MN,UV$ respectively and let $X=k\cap (CMN)\neq A$. Define $B'=XM\cap k\neq X$ and $C'=XN\cap k\neq X$. We have $\angle XA'A=\angle XBA=\angle XCA=\angle XNM$, so $AA'\parallel UV$. Similarly $BB'\parallel UV$. Let the action $\Phi$ be the reflection through the perpendicular bisector of $UV$. Then $\Phi(A)=A'$ and $\Phi(B)=B'$. Also let $M',N',X'$ be $\Phi(M),\Phi(N),\Phi(X)$ respectively. Since $B'M,A'N$ meet at point $X$ on $k$, then $ BM',AN'$ meet at point $X'$ on $k$. Also $MS=SM'$ and $NS=SN'$. Since $P,S$ are the midpoints of $BM,NN'$ respectively, then $PS\parallel BX'$. Similarly $QS\parallel AX'$. Thus $\angle QSP=\angle AX'B=\angle ACB$. Since $P,R$ are the midpoints of $BM,MN$ respectively, then $PR\parallel BC$. Similarly $QR\parallel AC$. Thus $\angle QRP=\angle ACB$. Combining the last two results, we have $\angle QRP=\angle ACB=\angle QSP$, so $P,Q,R,S$ are concyclic, as desired.

ELMO SL 2010 G6

Cries to be complexed.