International Zhautykov Olympiad 2011 - Problem 6

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International Zhautykov Olympiad 2011 - Problem 6

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cyclic quadrilateral

power of a point

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#1 h Jan 17, 2011, 7:00 AM • 2 Y

Y by Adventure10, Mango247

Diagonals of a cyclic quadrilateral $ABCD$ intersect at point $K.$ The midpoints of diagonals $AC$ and $BD$ are $M$ and $N,$ respectively. The circumscribed circles $ADM$ and $BCM$ intersect at points $M$ and $L.$ Prove that the points $K ,L ,M,$ and $N$ lie on a circle. (all points are supposed to be different.)

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jgnr

#2 h Jan 17, 2011, 10:17 AM • 5 Y

Y by Tawan, Adventure10, Mango247, and 2 other users

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.

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1459 posts

#3 h Jan 17, 2011, 12:28 PM • 14 Y

Y by nima-amini, dizzy, Tawan, tenplusten, TRYTOSOLVE, lahmacun, tiendung2006, Assassino9931, seyyedmohammadamin_taheri, Adventure10, Mango247, and 3 other users

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.

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jgnr

#4 h Jan 17, 2011, 12:39 PM • 2 Y

Y by Adventure10, Mango247

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!

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jayme

#5 h Jan 18, 2011, 5:28 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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

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jayme

#6 h Jan 18, 2011, 2:18 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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

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lasha

#7 h Jan 18, 2011, 9:34 PM • 2 Y

Y by Tawan, Adventure10

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.

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baysa

#8 h Feb 20, 2011, 8:47 AM • 2 Y

Y by Adventure10, Mango247

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.

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Zhero

#9 h Apr 9, 2011, 2:49 AM • 3 Y

Y by Tawan, Adventure10, Mango247

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.

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kunalsinghal1994

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kunalsinghal1994

#10 h Apr 16, 2011, 9:35 PM • 2 Y

Y by Adventure10, Mango247

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 $K$ .

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Itama

#11 h Jan 4, 2015, 3:03 PM • 2 Y

Y by Adventure10, Mango247

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

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#12 h Jan 4, 2015, 4:54 PM • 3 Y

Y by a-shi2213, Tawan, Adventure10

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.

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Itama

#13 h Jan 5, 2015, 4:06 PM • 2 Y

Y by Adventure10, Mango247

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

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#14 h Dec 14, 2021, 11:14 AM • 1 Y

Y by GuvercinciHoca

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! :-D

:-D

This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 14, 2021, 11:17 AM
Reason: not a bit easy, easy

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#15 h Jan 2, 2022, 4:32 PM

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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.

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Mathematicsislovely

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Mathematicsislovely

#16 h Jan 2, 2022, 6:35 PM

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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 $T\equiv T'$ .

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#17 h Jan 25, 2023, 3:49 AM • 1 Y

Y by Mango247

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

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