Invert about $K$, the configuration will be like this: $A,M,C,K$ are collinear points with $AM=MC$, $D,M,B,K$ are concyclic, $AD$ and $BC$ meet at $L$, $N$ is the point such that $NDMB$ is harmonic quadrilateral, and we want to prove that $L,M,N$ are collinear. It suffices to prove that if $KL$ intersect circle $DMBK$ at $N$, then $NDMB$ is harmonic. Let $KD$ meets line $LB$ at $P$. Lines $(KN,KD,KM,KB)$ meet line $LB$ at $L,P,C,B$, so it suffices to prove that $(LC,PB)=-1$. Let $DL$ and $DC$ meet $(DMBK)$ at $Q$ and $R$. It suffices to prove that $QKRB$ is harmonic. Let line $BD$ meets $AK$ at $X$. Lines $(QD,KD,RD,BD)$ meet $AK$ at $A,X,C,K$. So it suffices to prove that $(AC,XK)=-1$. Let $MX=a,XC=b,AM=a+b$. We have $AX\cdot XC=DX\cdot XB=MX\cdot XK$. So $(2a+b)b=a(b+CK)$, which gives $CK=\frac{b^2+ab}a$. Therefore $\frac{AX}{XC}=\frac{2a+b}b=\frac{2a}b+1$ and $\frac{AK}{KC}=\frac{2a+2b+CK}{CK}=\frac{2a+2b}{CK}+1=\frac{(2a+2b)a}{b^2+ab}+1=\frac{2a}{b}+1$. So $\frac{AX}{XC}=\frac{AK}{KC}$ and our proof is complete.

Another solution. Let $DA$ intersect $CB$ at $E$. Then $E,M,L$ are collinear by radical axis theorem. Let the circumcircles of $\triangle ADK,BCK$ intersect again at $F$. Then again $E,K,F$ are collinear. Since $EM\cdot EL=EB\cdot EC=EK\cdot EF$, $MLFK$ is cyclic. By angle chasing, it is easy to show $\triangle FBC\sim\triangle FDA$, so by spiral symmetry, $\triangle FBD\sim\triangle FCA$. Since $M,N$ are midpoints of $CA,BD$, we have $\angle NFM=\angle DFA=\angle DKA$, so $NFKM$ is cyclic. Thus $KLNM$ is cyclic.

oneplusone wrote: Another solution. Let $DA$ intersect $CB$ at $E$. Then $E,M,L$ are collinear by radical axis theorem. Let the circumcircles of $\triangle ADK,BCK$ intersect again at $F$. Then again $E,K,F$ are collinear. Since $EM\cdot EL=EB\cdot EC=EK\cdot EF$, $MLFK$ is cyclic. By angle chasing, it is easy to show $\triangle FBC\sim\triangle FDA$, so by spiral symmetry, $\triangle FBD\sim\triangle FCA$. Since $M,N$ are midpoints of $CA,BD$, we have $\angle NFM=\angle DFA=\angle DKA$, so $NFKM$ is cyclic. Thus $KLNM$ is cyclic. Really awesome!

Dear oneplusone and Mathlinkers, are you sure with your proof? I made a figure and I didn't find the collinearity in question... Sincerely Jean-Louis

Sorry, I omit to consider that the quadrilateral is cyclic... O.K. Sincerely Jean-Louis

By radical axis theorem, three lines $BC, AD$ and $LM$ intersect at one point. Call this point $T$. Consider a line passing through $B$ and parallel to $AC$, which intersect $TM$ at $X$ and $AD$ at $Y$. As $\frac{CM}{BX}=\frac{MT}{TX}=\frac{AM}{XY}$, from which $BX=XY$. Hence, the segment $XN$ is parallel to $BD$, as in triangle $YBD$, $X$ is the midpoint of $BY$ and $N$ is the midpoint of $BD$. Hence, $\angle{XNB}=\angle{ADB}=\angle{ACB}=\angle{MCB}=\angle{MLB}=\angle{XLB}$, from which $XBLN$ is concylic, but $\angle{KML}=\angle{YXL}=180-\angle{BXL}=180-\angle{BNL}=180-\angle{KNL}$, from which $\angle{KNL}+\angle{KML}=180$, which immediately leads to the fact that $K,L,M,N$ are on the same circle, as desired.

This problem is not new one. The problem is from 2007 scool 239's olympiad in Saint Petersburg. The author pf the problem is P.Sahipov.

Consider an inversion of arbitrary about $K$, and for any point $P$, let $P'$ denote the image of $P$ under this inversion. $M'$ is sent to the harmonic conjugate of $K$ with respect to $A'$ and $C'$, and $N'$ is sent to the harmonic conjugate of $K$ with respect to $B'$ and $D'$. Hence, $M'$ and $N'$ lies on the polar of $K$ with respect to $(A'B'C'D')$. Let $E = A'D' \cap B'C'$. We observe that $E$ also lies on the polar of $K$. By the radical axis theorem on $(M'A'D')$, $(M'B'C')$, and $(A'B'C'D')$, we find that $M'$, $L'$, and $E'$ are collinear, so $M'$, $L'$, and $N'$ are collinear, so $KMLN$ is cyclic, as desired.

It helps while inverting the figure to see that it suffices to show that $KLMO$ is cyclic, where $O$ is the centre of the circle $\cdot (ABCD)$, and after inversion about $K$, this becomes the polar of $K$.

Mr, Y. Balkash, Бисйор нагз кардаен инчо пост карда!

A different approach: Let $E$, $F$ be points of intersection of $BD$ with $(BCM)$ and $(AMD)$ respectively. We have to prove that the cirrcumcircles of $AMD$, $BCM$, $KMN$ are coaxal. But circle $KMN$ is concyclic with circles $AMD$ and $BCM$ iff $\frac{P_{(AMD)}^K}{P_{(BMC)}^K}=\frac{P_{(AMD)}^N}{P_{(BMC)}^N}$ where $P_{(O)}^T$ is power of point $T$ with respect to circle $O$. But it's obvious that $\frac{P_{(AMD)}^K}{P_{(BMC)}^K}=\frac{KM.KA}{KM.KC}=\frac{KA}{KC}$ and $\frac{P_{(AMD)}^N}{P_{(BMC)}^N}=\frac{NF.ND}{NE.NA}=\frac{NF}{NE}$, so it suffices to prove $\frac{KA}{KC}=\frac{NF}{NE}$. But we have $KM=\frac{KC-KA}{2}$ and $KN=\frac{KD-KB}{2}$. We also have $KM.KC=KE.KB \Rightarrow NE-NK=KE=\frac{(KC-KA)KC}{2KB} \Rightarrow NE=\frac{(KC-KA)KC+KB(KD-KB)}{2KB}=\frac{KC^2-KB^2}{2KB}$. With the same calculations $NF=\frac{KD^2-KA^2}{2KD}$, so if suffices to prove that $\frac{KA}{KC}=\frac{\frac{KD^2-KA^2}{2KD}}{\frac{KC^2-KB^2}{2KB}}=\frac{KB.(KD-KA)(KD+KA)}{KD.(KC-KB)(KC+KB)} $ but $B\overset{\Delta}KC \sim A\overset{\Delta}KD\Rightarrow \frac{KD-KA}{KC-KB}=\frac{KA}{KB}=\frac{KD+KA}{KB+KC}\Rightarrow \frac{KB.(KD-KA)(KD+KA)}{KD.(KC-KB)(KC+KB)}=\frac{KB}{KD}.\frac{KA^2}{KB^2}=\frac{KA^2}{KB.KD}$ so it suffices to prove that $\frac{KA^2}{KB.KD}=\frac{KA}{KC} \Leftrightarrow KA.(KB.KD)=(KA^2)(KC) \Leftrightarrow KB.KD=KA.KC$ which is trivial.

Without of using $harmonic ratio$, nice work Owerlord!

I think it's easy for P6,isn't it? Here is my approach. Let $O$ be center of $(ABCD)$ $\implies$ $\angle ONK=\angle OMK=90$ $\implies$ $OMKN$ is cyclic. Let $AD\cap BC=T$. So, $T$ is radical center of $(ABCD),(ADLM),(BCML)$ $\implies$ $T\in LM$. Let $\ell$ be polar of $T$ wrt $(ABCD)$, $S=TO\cap \ell$,and $\ell\cap (ABCD)=E,F$. It's well-known that $K\in \ell$ and $\angle TSE=\angle OET=90$ $\implies$ $TS\cdot TO=TE^2=TC\cdot TB=TD\cdot TA$ $\implies$ $ADSO$ and $BCSO$ are cyclic. Also since $\angle OSK=90$ we have $OMSKN$ is cyclic. So $TS\cdot TO=TC\cdot TB=TL\cdot TM$ $\implies$ $SOLM$ is cyclic $\implies$ $OLMSKN$ is cyclic $\implies$ $LMKN$ is cyclic. So we are done!

I guess complex bash wasn't so popular back then. Quick sketch: With $|a| = |b| = |c| = |d| = 1$ we have $k = \frac{ac(b+d)-bd(a+c)}{ac-bd}$, as well as $m = \frac{a+c}{2}$, $n=\frac{b+d}{2}$. The hardest part is to compute $\ell$. For this, note that $\frac{\ell - m}{d-m}/\frac{\ell - a}{d-a} = \frac{2\ell - a-c}{2d-a-c}\frac{d-a}{\ell - a}$ and $\frac{\ell - m}{c-m}/\frac{\ell - b}{c-b} = \frac{2\ell - a - c}{c-a}\frac{c-b}{\ell-b}$ are real, thus giving the equations $$ \frac{2\ell - a-c}{2d-a-c}\frac{d-a}{\ell - a} = \frac{2\bar{\ell} - \frac{1}{a}-\frac{1}{c}}{\frac{2}{d}-\frac{1}{a}-\frac{1}{c}}\frac{\frac{1}{d}-\frac{1}{a}}{\bar{\ell} - \frac{1}{a}} = \frac{2\bar{\ell}ac - a - c}{2ac - cd - ad}\frac{a-d}{a\bar{\ell} - 1} \mbox{ and } \frac{2\ell - a - c}{c-a}\frac{c-b}{\ell-b} = \frac{2\bar{\ell}-\frac{1}{a}-\frac{1}{c}}{\frac{1}{c}-\frac{1}{a}}\frac{\frac{1}{c}-\frac{1}{b}}{\bar{\ell}-\frac{1}{b}} = \frac{2\bar{\ell} ac-a-c}{a-c}\frac{b-c}{b\bar{\ell}-1}.$$From here solve with respect to $\ell$ and $\bar{\ell}$ to obtain an expression for $\ell$ and finally check that $\frac{\ell - k}{m-k}\frac{m-n}{\ell-n}$ is equal to its conjugate, thus giving the result.

Consider an inversion arround $\odot(ABC)$.This inversion send $M,N$ to $R=AA\cap CC$ and $BB\cap DD$ respectively.Also $L$ goes to $\odot(BCR)\cap\odot(ADR)-\{R\}=T$. Applying Pascal theorem on cyclic hexagon $CDDABB$ and $DAABCC$ we get $BB\cap DD$ and $AA\cap CC$ lies on the polar of $K$. It is enough to show $T$ lies on the polar of $K$ wrt $\odot(ABC)$,becoz it will show that $K,M,N,L$ will lie on a circle passing through the center of $ABCD$. Let $T'\ne R$ be the point where $\odot(BCR)$ cut the polar of $K$ for the second time.Let $X=AD\cap BC$.Since $X$ lies on polar of $K$ we have, $$XR\times XT'=XC\times XB=XD\times XA$$So $T'$ lies on $\odot(ADR)$,which implies $T\equiv T'$.

jgnr wrote: Invert about $K$, the configuration will be like this: $A,M,C,K$ are collinear points with $AM=MC$, $D,M,B,K$ are concyclic, $AD$ and $BC$ meet at $L$, $N$ is the point such that $NDMB$ is harmonic quadrilateral, and we want to prove that $L,M,N$ are collinear. It suffices to prove that if $KL$ intersect circle $DMBK$ at $N$, then $NDMB$ is harmonic. Let $KD$ meets line $LB$ at $P$. Lines $(KN,KD,KM,KB)$ meet line $LB$ at $L,P,C,B$, so it suffices to prove that $(LC,PB)=-1$. Let $DL$ and $DC$ meet $(DMBK)$ at $Q$ and $R$. It suffices to prove that $QKRB$ is harmonic. Let line $BD$ meets $AK$ at $X$. Lines $(QD,KD,RD,BD)$ meet $AK$ at $A,X,C,K$. So it suffices to prove that $(AC,XK)=-1$. Let $MX=a,XC=b,AM=a+b$. We have $AX\cdot XC=DX\cdot XB=MX\cdot XK$. So $(2a+b)b=a(b+CK)$, which gives $CK=\frac{b^2+ab}a$. Therefore $\frac{AX}{XC}=\frac{2a+b}b=\frac{2a}b+1$ and $\frac{AK}{KC}=\frac{2a+2b+CK}{CK}=\frac{2a+2b}{CK}+1=\frac{(2a+2b)a}{b^2+ab}+1=\frac{2a}{b}+1$. So $\frac{AX}{XC}=\frac{AK}{KC}$ and our proof is complete. bokum gibi çözüm