It's Problem 7 not p5

The conclusion is true not only for tangent points $M,N$ but for arbitrary points $M,N$ as well,since we have Disargues Involution Theorem

Indeed in Johnson's Advanced Euclidean Geometry,the author proved this problem in order to show Poncelet porism theorem...and he proved for the coxial pencils case.

See here

Again a chinese problem destroyed by Desargues' Involution :p Let $l \equiv \overline{MN}$. Denote $l\cap \omega_1 = {X,Y}$. By Desargues' Involution Theorem we have that $(S,T); (X,Y); (M,N)$ are pairs of involution, so respectively there exists a point $P$ on $l$ such that $PM\cdot PN=PX\cdot PY=PT\cdot PS$, thus $P$ is on the radical axis of circles $\omega_1$ and $\omega_2$, which is the line $XY$, so $PM \cdot PN= PR \cdot PQ=PS \cdot PT$.

Well, there is another way to do it short (and even prove much stronger result: $\omega_2$ can replace by any circle as long as: $EM=EN$, also, I don't know whether this is Desargues' Involution ) Consider the ratio of power of point to two circles (I don't know how to $LaTeX$ it ) $\frac{P_{S\rightarrow\omega_1}}{P_{S\rightarrow\omega_2}}=\frac{SB.SC}{SM.SN}=\frac{SB}{SM}.\frac{SC}{SN}=\frac{\text{sin}\widehat{EMN}}{\text{sin}\widehat{DBC}}.\frac{\text{sin}\widehat{ENM}}{\text{sin}\widehat{ACB}}$, this is symmetric for $(A,C),(B,D)$ so doing the same with $T$, we have: $\frac{P_{S\rightarrow\omega_1}}{P_{S\rightarrow\omega_2}}=\frac{P_{T\rightarrow\omega_1}}{P_{T\rightarrow\omega_2}}$, so: $Q,S,R,T$ are concyclic. $\blacksquare$

rmtf1111 wrote: ...so respectively there exists a point $P$ on $l$ such that $PM\cdot PN=PX\cdot PY=PT\cdot PS$... Is this a well-known result about Desargues' Involution? In particular, for any three pairs of points which are mapped to each other by an involution, does this result hold? And is this well-known enough to be stated without proof? Please enlighten me, as I am new to involutions. Thank you very much!

ManuelKahayon wrote: rmtf1111 wrote: ...so respectively there exists a point $P$ on $l$ such that $PM\cdot PN=PX\cdot PY=PT\cdot PS$... Is this a well-known result about Desargues' Involution? In particular, for any three pairs of points which are mapped to each other by an involution, does this result hold? And is this well-known enough to be stated without proof? Please enlighten me, as I am new to involutions. Thank you very much! Yes it is a well-know result about involutions, so for any $n$ pairs of an involution this result holds . In other words, any involution on a line is an inversion. And even here this is presented as the definiton of involution on a line. Im not too sure if it can be stated without a proof, but in competitions like IMO it should be pretty safe, unless it somehow trivializes the problem.

Vietnamisalwaysinmyheart wrote: Well, there is another way to do it short (and even prove much stronger result: $\omega_2$ can replace by any circle as long as: $EM=EN$, also, I don't know whether this is Desargues' Involution ) Consider the ratio of power of point to two circles (I don't know how to $LaTeX$ it ) $\frac{P_{S\rightarrow\omega_1}}{P_{S\rightarrow\omega_2}}=\frac{SB.SC}{SM.SN}=\frac{SB}{SM}.\frac{SC}{SN}=\frac{\text{sin}\widehat{EMN}}{\text{sin}\widehat{DBC}}.\frac{\text{sin}\widehat{ENM}}{\text{sin}\widehat{ACB}}$, this is symmetric for $(A,C),(B,D)$ so doing the same with $T$, we have: $\frac{P_{S\rightarrow\omega_1}}{P_{S\rightarrow\omega_2}}=\frac{P_{T\rightarrow\omega_1}}{P_{T\rightarrow\omega_2}}$, so: $Q,S,R,T$ are concyclic. $\blacksquare$ Actually Desargues Involution is much stronger than this problem,and... Hermitianism wrote: The conclusion is true not only for tangent points $M,N$ but for arbitrary points $M,N$ as wel

Shouldn't this be problem six?

What is desargues

smy2012 wrote: It's Problem 7 not p5 That's what the post said originally right?

$\frac{P_{S\rightarrow\omega_1}}{P_{S\rightarrow\omega_2}}=\frac{P_{T\rightarrow\omega_1}}{P_{T\rightarrow\omega_2}}$, so: $Q,S,R,T$ are concyclic. $\blacksquare$ Why is this always correct?

It's just showing that $\omega_1$, $\omega_2$, $(QRS)$ are coaxial. So we use the coaxiality lemma. Length ratios turn out to be easily turned into ratios of sines, and the end.

Incinerated using Desargues Involution. Let the line $MNST$ meet $(ABCD)$ again at $G,H$. Then by Desargues Involution theorem, the pairs $\{ M,N \}, \{ S,T \}, \{ G,H \}$ are the pairs of an Involution on $MNST$, so by the definition of an involution (as inversion on a line), there exists a point $X$, inversion at which these pairs are inverses. So the (possibly imaginary) circle of inversion is orthogonal to both the circles in the problem, hence $X$ is on the radical axis, and we are done by PoP.

Killed by Desargue involution: [asy][asy] unitsize(0.35inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.456542402457, xmax = 20.470327262609462, ymin = -16.969702588868746, ymax = 9.73463231698739; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-2.8,1.76), 6.190460263673148), linewidth(2) + wrwrwr); draw((2.8143543782521503,4.367838797072613)--(-5.402665858950897,-3.856754285418547), linewidth(2) + wrwrwr); draw((-5.402665858950897,-3.856754285418547)--(-8.658402308727117,-0.2402301530616422), linewidth(2) + wrwrwr); draw((-8.658402308727117,-0.2402301530616422)--(-6.9318735348908005,6.369709249810146), linewidth(2) + wrwrwr); draw((-6.9318735348908005,6.369709249810146)--(2.8143543782521503,4.367838797072613), linewidth(2) + wrwrwr); draw((-6.9318735348908005,6.369709249810146)--(-5.402665858950897,-3.856754285418547), linewidth(2) + wrwrwr); draw((2.8143543782521503,4.367838797072613)--(-8.658402308727117,-0.2402301530616422), linewidth(2) + wrwrwr); draw(circle((-2.474471034936861,-1.2868700523168353), 3.2760561429522506), linewidth(2) + wrwrwr); draw((-11.86092667588629,-12.500953706545447)--(-8.658402308727117,-0.2402301530616422), linewidth(2) + wrwrwr); draw((-8.933756085612032,-7.391098781780007)--(-5.402665858950897,-3.856754285418547), linewidth(2) + wrwrwr); draw((-0.41833024016745846,7.473969498624876)--(-11.86092667588629,-12.500953706545447), linewidth(2) + wrwrwr); draw((-7.452530952792372,-4.805378098204903)--(-0.7836421186393819,-4.092871019456275), linewidth(2) + wrwrwr); /* dots and labels */ dot((2.8143543782521503,4.367838797072613),dotstyle); label("$B$", (2.918060533226349,4.62710418450811), NE * labelscalefactor); dot((-5.402665858950897,-3.856754285418547),dotstyle); label("$C$", (-5.171019554761142,-3.902727062119724), NE * labelscalefactor); dot((-8.658402308727117,-0.2402301530616422),dotstyle); label("$D$", (-9.241486137498438,-0.013746250587276958), NE * labelscalefactor); dot((-6.9318735348908005,6.369709249810146),dotstyle); label("$A$", (-6.83031803434832,6.623447667761433), NE * labelscalefactor); dot((-6.0972836905649626,0.7884512270078892),linewidth(4pt) + dotstyle); label("$E$", (-6.0006687945547315,0.9973887604111593), NE * labelscalefactor); dot((-3.695498928749375,1.7531356605572304),linewidth(4pt) + dotstyle); label("$M$", (-4.082104927532057,2.1640830038708936), NE * labelscalefactor); dot((-5.714503036428075,-1.7713661628151642),linewidth(4pt) + dotstyle); label("$N$", (-6.337713798220877,-1.7248978076615538), NE * labelscalefactor); dot((-8.933756085612032,-7.391098781780007),linewidth(4pt) + dotstyle); label("$S$", (-9.319265753729086,-6.936132095115033), NE * labelscalefactor); dot((-11.86092667588629,-12.500953706545447),linewidth(4pt) + dotstyle); label("$T$", (-12.430450402955044,-12.458484847491109), NE * labelscalefactor); dot((-3.5343125661860584,-4.386753885691005),linewidth(4pt) + dotstyle); label("$Q$", (-3.4339414589433157,-4.187918988298771), NE * labelscalefactor); dot((-0.7836421186393819,-4.092871019456275),linewidth(4pt) + dotstyle); label("$R$", (-0.6857283521270525,-3.8768005233761746), NE * labelscalefactor); dot((-6.524234077677377,-3.1848841048888854),linewidth(4pt) + dotstyle); label("$X$", (-7.271069192988664,-3.3064166710180825), NE * labelscalefactor); dot((-0.41833024016745846,7.473969498624876),linewidth(4pt) + dotstyle); label("$Y$", (-0.32275680971735743,7.6864357562469685), NE * labelscalefactor); dot((-7.452530952792372,-4.805378098204903),linewidth(4pt) + dotstyle); label("$P$", (-7.867379584090306,-4.576817069452015), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that by Desargue involution, there is an involution on line $MN$ swapping the pairs $(X,Y),(M,N),(S,T)$ where $\{X,Y\}=MN\cap\omega_1$. Therefore, there is a point $P\in MN$ such that $PX\cdot PY=PM\cdot PN=PS\cdot PT$. This implies that $P$ is on the radical axis of $\omega_1$ and $\omega_2$, so on $QR$. Therefore, $PS\cdot PT=PQ\cdot PR$, implying the result.

Im gonna look

This actually holds for arbitrary $MN.$ Let $X$ and $Y$ denote the intersection of $MN$ with $\omega_1.$ By Desargue's Involution Theorem, there is an involution swapping pairs $(X, Y),(M, N)$ and $(S, T).$ Thus, there exists point $P \in MN$ satisfying $$PX \cdot PY = PS \cdot PT = PM \cdot PN,$$so $P$ is on the radical axis of $\omega_1$ and $\omega_2,$ which is $QR$. Thus, $P = QR \cap MN,$ and moreover, $PR \cdot PQ = PS \cdot PT$ and we are done.

wait, this problem can be done with desargues involution?!?!?!

Kagebaka wrote: wait, this problem can be done with desargues involution?!?!?! Internet explorer?

Asuboptimal wrote: Kagebaka wrote: wait, this problem can be done with desargues involution?!?!?! Internet explorer? Nooooooooooo! See posts #19 and #20. Also, I always wanted to say that the title when read in the feed looks really cringy at first

Here is a different Solution China Girls MO 2017 P7 wrote: Let the $ABCD$ be a cyclic quadrilateral with circumcircle $\omega_1$.Lines $AC$ and $BD$ intersect at point $E$,and lines $AD$,$BC$ intersect at point $F$.Circle $\omega_2$ is tangent to segments $EB,EC$ at points $M,N$ respectively,and intersects with circle $\omega_1$ at points $Q,R$.Lines $BC,AD$ intersect line $MN$ at $S,T$ respectively.Show that $Q,R,S,T$ are concyclic. By Menelaus Theorem we get these equations$$\begin{cases} \left(\frac{BE}{EM}\cdot\frac{MN}{NS}\cdot\frac{SC}{CB}\right)=1\implies\frac{SC}{SN}=\left(\frac{EM\cdot CB}{MN\cdot BE}\right) \\ \left(\frac{TM}{MN}\cdot\frac{NE}{EA}\cdot\frac{AD}{DT}\right)=1\implies \frac{TM}{TD}=\left(\frac{AE\cdot MN}{NE\cdot AD}\right) \\ \left(\frac{SB}{BC}\cdot\frac{CE}{EN}\cdot\frac{NM}{MS}\right)=1\implies\frac{SB}{SM}=\left(\frac{BC\cdot EN}{EC\cdot NM}\right) \\ \left(\frac{TA}{AD}\cdot\frac{DE}{EM}\cdot\frac{MN}{NT}\right)=1\implies\frac{TA}{TN}=\left(\frac{EM\cdot AD}{DE\cdot MN}\right) \end{cases}$$So, $\frac{SC}{SN}\cdot\frac{TM}{TD}=1$ and similarly $\frac{SM}{SB}\cdot\frac{TA}{TN}=1$. Hence, $\frac{SC\cdot SB}{SN\cdot SM}=\frac{TD\cdot TA}{TN\cdot TM}\implies \frac{Pow_{\odot(\omega_1)}S}{Pow_{\odot(\omega_2)}S}=\frac{Pow_{\odot(\omega_1)}T}{Pow_{\odot(\omega_2)}T}$. Hence, by Coaxility Lemma $Q,R,S,T$ are concyclic. $\blacksquare$

Posting for Storage. Suppose $MNST$ intersects $\omega_1$ at $X$ and $Y$. Therefore, By Desargues Involution Theorem, there exists an involution on the same line swapping $(M,N); (X,Y); (S,T)$. Let it be point $P$. Therefore, we have \[ PX \cdot PY = PM \cdot PN = PS \cdot PT \]The first equality says that $P$ lies on the radical axis of $\omega_1$ and $\omega_2$, which means $P \in QR$ as well. Therefore, \[ PQ \cdot PR = PM \cdot PN = PS \cdot PT \]Therefore, $QRST$ is cyclic.

First, define $P = \overline{RQ} \cap \overline{MN}$, and let the intersections of $\overline{MN}$ with $(ABCD)$ be $Y_1$ and $Y_2$. By Desargues Involution on $\overline{Y_1NMY_2PST}$ on quadrilateral $ABCD$, we get the following involuted pairs: $$Y_1 \longleftrightarrow Y_2, N \longleftrightarrow M, S \longleftrightarrow T.$$Since $PQ \cdot PR = PY_2 \cdot PY_1 = PM \cdot PN$ by power of a point, however, we find that the desired involution is an inversion about $P$ swapping $M$ and $N$. Hence, $PS \cdot PT = PQ \cdot PR$ aswell and we are done.

From DIT we know that there exist a point $P$ in the line MN such that $PM*PN=PT*PS=PX*PY$ (Where $X$ and $Y$ are the intersection of line $MN$ with the circumcircle of $ABCD$) so we know that $P$ must lie on the radical axis of (ABCD) and (MNQR) but we know that also $P$ lies on line MN ,which means that p is the intersection of $MN$ and $QR$ ,which means that $PM*PN=PQ*PR=PT*PS$,and we are done from power of a point.

haha funie Let $MN$ intersect $\omega_1$ at $X$ and $Y$. By DIT, we have that on line $MN$ there is an involution swapping pairs $(M,N),(X,Y),$ and $(S,T)$. Letting $P$ be the center of this inversion, we have \[PM\cdot PN=PX\cdot PY\iff P\in QR\text{, hence }PQ\cdot PR=PM\cdot PN=PS\cdot PT\]which implies the conclusion.

Name the line passing through MN, $l$. Now we introduce a function on $l$: Let A and B be any two fixed points not on $l$, for any point $X$ on $l$ let $f(X)$ be the second intersection of the circumcircle of triangle ABX with line $l$. Note that $f$ is an inversion(and hence an involution on $l$), it's center is the intersection of $AB$ and $l$. Let F,G be the intersections of $l$ and $\omega_1$ Now apply the function $f$, for fixed points $Q$,$R$ and line $l$. We know that $f$ mapps $M$ to $N$, and $F$ to $G$, so by DIT we can tell that $f$ aslo maps $S$ to $T$ which means $QRST$ is cyclic.

Let $\omega_1$ be the circle tangent to $BD$ at $M$ and tangent to $AC$ at $N$. Define $\omega_2$ as the circle tangent to $BF$ at $S$ and tangent to $AF$ at $T$. Redefine $Q,R=\omega_1\cap \omega_2$. We claim that $\dfrac{P(X,\omega_1)}{P(X,\omega_2)}$ is constant for $X={A,B,C,D}$, thus by coaxial circles lemma, this would imply that $A,B,C,D,Q,R$ are concyclic. We need $$\frac{AN}{AT}=\frac{BM}{BS}=\frac{CN}{CS}=\frac{DM}{DT}.$$But this is easy by sine law, \begin{align*} \frac{AN}{AT}&=\frac{\sin{\angle ATN}}{\sin{\angle ANT}}=\frac{\sin{\angle NSC}}{\sin{\angle CNS}}=\frac{CN}{CS}\\ \frac{AN}{AT}&=\frac{\sin{\angle ATN}}{\sin{\angle ANT}}=\frac{\sin{\angle MSB}}{\sin{\angle BMS}}=\frac{BM}{BS}\\ \frac{AN}{AT}&=\frac{\sin{\angle ATN}}{\sin{\angle ANT}}=\frac{\sin{\angle DTM}}{\sin{\angle DMT}}=\frac{DM}{DT}. \end{align*}We are done. [asy][asy] size(8cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen org=magenta;pen heavy=heavymagenta; pair O,A,B,C,D,E,F,G,M,N,S,T,Q,R; O=(0,0);D=dir(15);A=dir(110);B=dir(205);C=dir(330);path w=circumcircle(A,B,C); E=extension(A,C,B,D);F=extension(A,D,B,C);G=2.2*incenter(B,E,C)-1.2*E;N=foot(G,E,C);M=foot(G,E,B); S=extension(B,C,N,M);T=extension(A,D,N,M);Q=intersectionpoints(circle(G,abs(G-M)),w)[0];R=intersectionpoints(circle(G,abs(G-M)),w)[1]; draw(w,heavyblue+dashed);draw(A--B--C--D--cycle,heavygreen);draw(circle(G,abs(G-M)),heavyblue); draw(circumcircle(Q,R,S),heavyblue);draw(D--T,heavygreen);draw(C--F,heavygreen); draw(M--T,olive);draw(A--C,olive);draw(B--D,olive); clip((-1,1)--(3.5,1)--(3.5,-1.5)--(-1,-1.5)--cycle); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A$",A,dir(A)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); [/asy][/asy]

Hermitianism wrote: This is a very classical problem. Let the $ABCD$ be a cyclic quadrilateral with circumcircle $\omega_1$.Lines $AC$ and $BD$ intersect at point $E$,and lines $AD$,$BC$ intersect at point $F$.Circle $\omega_2$ is tangent to segments $EB,EC$ at points $M,N$ respectively,and intersects with circle $\omega_1$ at points $Q,R$.Lines $BC,AD$ intersect line $MN$ at $S,T$ respectively.Show that $Q,R,S,T$ are concyclic. Let $MN$ intersect $\omega _1$ at points $X$ and $Y$ and define $Z \equiv QR \cap MN$. We see that $ZM \cdot ZN = ZX \cdot ZY = ZQ \cdot ZR$ because $Z$ lies on radical axis of $\omega_1$ and $\omega_2$. Due to DIT in quadrilateral $ACBD$, we see that a unique involution swaps the points $(M, N), (X, Y), (S, T)$ which means that the involution is inversion about the circle with center $Z$, radius $\sqrt{ZQ \cdot ZR}$. This means that $ZQ \cdot ZR = ZS \cdot ZT$ as desired.

We will apply DIT on $ABCD$ and line $MN$. Therefore we have \begin{align*} (S, T), (M, N) \end{align*}are the involution pair. It is known that an involution on a line is either reflection WRT a fixed point or an inversion. Clearly they are not reflection. So the pairs are inversion. Suppose the center of inversion is $O$ and we get $$OS\cdot OT=OM\cdot ON$$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12., xmax = 6., ymin = -7., ymax = 5.4; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqffff = rgb(0.,1.,1.); /* draw figures */ draw(circle((0.4787705612490665,-0.42923049660257007), 5.), linewidth(0.8)); draw((1.6321475408029935,4.435923308061272)--(3.9883459309953677,-3.990533730542375), linewidth(0.8) + blue); draw((-3.004854840122384,-4.0159217382512855)--(-4.509038568390384,-0.7781718760423679), linewidth(0.8) + blue); draw((-4.509038568390384,-0.7781718760423679)--(3.9883459309953677,-3.990533730542375), linewidth(0.8)); draw((-3.004854840122384,-4.0159217382512855)--(1.6321475408029935,4.435923308061272), linewidth(0.8)); draw(circle((-1.0544133919328547,-3.802538896381981), 1.607352935425707), linewidth(0.8)); draw((xmin, 0.36932104999553933*xmin-2.1195359640545486)--(xmax, 0.36932104999553933*xmax-2.1195359640545486), linewidth(0.8) + ffxfqq); /* line */ draw(circle((-5.124229257636485,-12.756939975809772), 8.733266542825747), linewidth(0.8) + qqffff); draw(circle((-4.18751358472585,-3.666072878036084), 2.6928342565447796), linewidth(0.8) + red); draw((-4.18751358472585,-3.666072878036084)--(-0.5381481223222374,-5.324726017078879), linewidth(0.8)); draw((-1.7920310306334626,-1.8053128007951837)--(-1.9375196098602507,-5.145560440095857), linewidth(0.8)); draw((-10.776593302393856,-6.099558717869544)--(-4.509038568390384,-0.7781718760423679), linewidth(0.8)); draw((-4.509038568390384,-0.7781718760423679)--(1.6321475408029935,4.435923308061272), linewidth(0.8) + blue); draw((-3.004854840122384,-4.0159217382512855)--(3.9883459309953677,-3.990533730542375), linewidth(0.8) + blue); draw((-5.155934164228222,-4.02373098329519)--(-3.004854840122384,-4.0159217382512855), linewidth(0.8)); /* dots and labels */ dot((-3.004854840122384,-4.0159217382512855),dotstyle); label("$B$", (-3.5511385043475667,-4.783152351392899), NE * labelscalefactor); dot((3.9883459309953677,-3.990533730542375),dotstyle); label("$C$", (4.083412841028966,-4.580644358146837), NE * labelscalefactor); dot((1.6321475408029935,4.435923308061272),dotstyle); label("$D$", (1.7748217180238601,4.896729725768863), NE * labelscalefactor); dot((-4.509038568390384,-0.7781718760423679),dotstyle); label("$A$", (-4.847189661122362,-0.5304844932255977), NE * labelscalefactor); dot((-1.7920310306334626,-1.8053128007951837),linewidth(4.pt) + dotstyle); label("$E$", (-2.072830153651315,-1.3810180648590578), NE * labelscalefactor); dot((-2.463611367549884,-3.029399501099019),dotstyle); label("$M$", (-2.943614524609381,-2.6568184223092484), NE * labelscalefactor); dot((-0.486027308419252,-2.299036079926453),dotstyle); label("$N$", (-0.6755250002534883,-1.7860340513511817), NE * labelscalefactor); dot((-2.54068728529934,-4.414562893394242),dotstyle); label("$Q$", (-2.8018589293371376,-4.985660344638961), NE * labelscalefactor); dot((-0.5381481223222374,-5.324726017078879),dotstyle); label("$R$", (-0.5135186056566388,-5.8969463142462395), NE * labelscalefactor); dot((-5.155934164228222,-4.02373098329519),dotstyle); label("$S$", (-5.576218436808185,-3.649107589214952), NE * labelscalefactor); dot((-10.776593302393856,-6.099558717869544),dotstyle); label("$T$", (-11.489451839593192,-5.653936722350966), NE * labelscalefactor); dot((-4.18751358472585,-3.666072878036084),dotstyle); label("$Z$", (-4.280167280033389,-3.3048440006966464), NE * labelscalefactor); dot((-1.8174028298105072,-2.387819639176234),dotstyle); label("$U$", (-1.7690681637822223,-2.8795772148799164), NE * labelscalefactor); dot((-1.9375196098602507,-5.145560440095857),dotstyle); label("$V$", (-1.72856656513301,-5.8969463142462395), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Suppose the line $QR$ intersect $MN$ at $Z$. We claim $O\equiv Z$. For the proof Just take a circle with radius $\sqrt{ZM\cdot ZM}$. Therefore we get $M, N$ are inverses of each other. Now by the PoP we have \begin{align*} ZS\cdot ZT=ZN\cdot ZN=ZQ\cdot ZR \end{align*}Which means $S, T, R, Q$ are cyclic.$\square$

Nice problem! Solved with Dimath27 By Coaxiality Lemma it's enough to prove that: $\frac{P(S,\omega_1)}{P(S,\omega_2)} =\frac{P(T,\omega_1)}{P(T,\omega_2)}$, then we need to prove: $$\frac{SN.SM}{SC.SB} = \frac{TM.TN}{TD.TA}$$ Then see that $\angle NSC = \angle EMS - \angle EBS = \angle ENM - \angle EAT= \angle ATN$ and also $\angle ANT = \angle SMB$ so $\triangle SMB \sim \triangle TNA$, and $\triangle SNC \sim \triangle TMD$. Then we have: $\frac{SM}{TN}= \frac{SB}{TA}$ and $\frac{SN}{TM}=\frac{SC}{TD}$ this implies $\frac{SN.SM}{SC.SB} = \frac{TM.TN}{TD.TA}$ As desired. $\blacksquare$

We will solve the problem without the tangency condition, redefining $\omega_2=(MNP),$ where $P$ is some point on $\omega_1.$ We also define $\omega_3=(STP),$ noting that it suffices to prove $\omega_1,\omega_2,$ and $\omega_3$ are coaxial. By DIT, there exists an involution mapping pairs $(S,T),(M,N),$ and the two intersections of $\overline{MN}$ and $\omega_1.$ If this involution is a reflection, we are done, so assume there exists $X$ on $\overline{MN}$ such that $X$ is the center of the inversion that maps these pairs to each other. Assume FTSOC that the circles $\omega_i$ are not coaxial. Then, $X$ is the radial center of $\omega_i$ so it must lie at the point of infinity along the line perpendicular to $\overline{MN},$ which is absurd. $\square$

What is Desargues Involution Theorem?

hef4875 wrote: What is Desargues Involution Theorem? here

Algebraic geometry for toddlers solution: Let $C_1$ be the circle tangent at $M,N$ to $AC,BD$ and $C_2$ the circle tangent at $S,T$ to $BC, AD$, and let $C$ be the circle $(ABCD)$. Our goal is to show $C$ is a linear combination of $C_1,C_2$ (and then $P,Q\in C_2$). Indeed, $C$ is a linear combination of $\ell_{AC}\ell_{BD}$ and $\ell_{AD}\ell_{BC}$ because it belongs to the pencil formed by those two conics. Let $\ell$ be the line $MNST$. Then $\ell_{AC}\ell_{BD}$ is a linear combination of $C_1, \ell^2$ because it is in their pencil, and $\ell_{AD}\ell_{BC}$ is similarly a combination of $C_2, \ell^2$. Thus, \[C\in \text{span}(C_1, C_2, \ell^2)\]But the coefficient of $\ell^2$ in the combination must be 0, because all of $C,C_1,C_2$ vanish at the point $(1:i:0)$ in the complex projective plane, while $\ell^2$ doesn't. This finishes the proof.

[asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-92.77340,43.99502); pair B = (-112.59209,-63.08388); pair C = (48.23124,-63.08388); pair D = (-34.72002,67.40149); pair F = (-358.35356,-63.08388); pair E = (-62.43713,20.95766); pair N = (-11.51705,-17.71103); pair M = (-95.20340,-33.94673); pair R = (-14.41727,-110.85427); pair Q = (-100.31038,-81.44765); pair T = (-463.12222,-105.32542); pair Y = (56.03498,-4.60548); pair K = (-187.02322,-51.76038); pair X = (-120.72246,-38.89760); pair S = (-245.38977,-63.08388); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(A--B, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-32.18042,-22.59334), 90.03066), linewidth(0.5) + blue); draw(C--D, linewidth(0.5)); draw(D--B, linewidth(0.5)); draw(D--F, linewidth(0.5)); draw(F--C, linewidth(0.5)); draw(circle((-46.10009,-63.25098), 57.18281), linewidth(0.5) + blue); draw(F--T, linewidth(0.5)); draw(T--Y, linewidth(0.5) + blue); draw(R--K, linewidth(0.5)); dot("$A$", A, NW); dot("$B$", B, dir(140)); dot("$C$", C, NE); dot("$D$", D, NE); dot("$F$", F, NW); dot("$E$", E, 2*dir(95)); dot("$N$", N, NE); dot("$M$", M, dir(110)); dot("$R$", R, SE); dot("$Q$", Q, SW); dot("$T$", T, NW); dot("$Y$", Y, NE); dot("$K$", K, dir(90)); dot("$X$", X, NW); dot("$S$", S, NW); [/asy][/asy] Let $K=QR\cap MN$, and $X$, $Y$ be the points of second intersections of $MN$ with $\odot(ABCD)$. Now we apply DIT on quadrangle $ABCD$ w.r.t. line $MN$. This gives that $(S,T)$, $(M,N)$ and $(X,Y)$ are reciprocal pairs under some Involution. Now if this Involution turns out to be a reflection, then we get that $MN$ and $XY$ share the common midpoints, and thus they also share the same perpendicular bisectors too. Also note that $\odot(MNQR)$ and $\odot(ABCD)$ share $QR$ as a common chord. So the center of $\odot(MNQR)$ lies on the perpendicular bisector of $QR$ and $MN$, and the center of $\odot(ABCD)$ lies on the perpendicular bisector of $QR$ and $XY$. This however forces that $\odot(MNQR)$ and $\odot(ABCD)$ share the same centers which further forces $\odot(MNQR)\equiv\odot(ABCD)$ which is a crazy contradiction. So this Involution must indeed be an Inversion w.r.t. some point on the line $MN$. Now note that $K$ lies on the radical axis of $\left\{\odot(ABCD),\odot(MNQR)\right\}$. This gives that $KX\cdot KY=KQ\cdot KR=KM\cdot KN$ which means that the Inversion $\mathbf{I}(\odot(K,\sqrt{KX\cdot KY}))$ swaps $(X,Y)$ and $(M,N)$. However we previously had that $(S,T)$, $(M,N)$ and $(X,Y)$ are reciprocal pairs under some Involution. So by DIT, this must be the same Involution, that is the center of our Inversion is indeed $K$. So now we can finally deduce that $KS\cdot KT=KX\cdot KY=KQ\cdot KR$ due to our Involution which clearly finishes.

CIRCLE POINTS. Let $\omega_1$ intersect $MN$ at $X$ and $Y$. By DIT there exists an involution swapping $(X, Y), (M, N), (S, T)$. Since $(XYQRC_1C_2)$ and $(QRMNC_1C_2)$ are coconic, it follows that $(STQRC_1C_2)$ are coconic, where $C_1$ and $C_2$ are the circle points.

ok what By the Coaxial Lemma we just need to show \[\frac{SM\cdot SN}{SB\cdot SC}=\frac{TM\cdot TN}{TA\cdot TD}\]or \[\frac{SM}{TM}\cdot \frac{SN}{TN}=\frac{SB\cdot SC}{TA\cdot TD}\]but now by Menelaus on $\triangle FST$ with lines $\overline{ANC}$ and $\overline{BMD}$: \[\frac{SM}{TM}\cdot \frac{TD}{FD}\cdot \frac{FB}{SB}=1\]\[\frac{SN}{TN}\cdot \frac{TA}{FA}\cdot \frac{FC}{SC}=1\]so then \[\frac{SM}{TM}\cdot \frac{SN}{TN}=\frac{FA\cdot SC\cdot FD\cdot SB}{FB\cdot FC\cdot TD\cdot TA}=\frac{SB\cdot SC}{TA\cdot TD}\]and we are done. $\blacksquare$

Let $\overline{MN} = \ell$, and let $X, Y = \ell \cap \omega_1$. Then by DIT there is an involution swapping $(S, T)$, $(X, Y)$ and $(M, N)$. Clearly the involution must be an inversion around a point $P$. From this, we find that $PS \cdot PT = PX \cdot PY = PN \cdot PM \implies P \in \overline{QR}$ by radical axis. Then $PS \cdot PT = PM \cdot PN = PQ \cdot PR \implies$ $QRST$ is cyclic by PoP. also cool 1000 posts

By DIT, there exists a point $K$ on line $NM$ such that $$KS\cdot KT=KN\cdot KM=Pow_{\omega_1}(K).$$Since $KN\cdot KM=Pow_{\omega_2}(K)$, $K$ lies on the radical axis, which is $QR$. Thus, $$KQ\cdot KR=KN\cdot KM=KS\cdot KT,$$as desired.

Angle chasing gives $\triangle TAN \sim \triangle SBM$ and $\triangle TDM \sim \triangle SCN$, implying the ratios \[\frac{TA}{TN} = \frac{SB}{SM}, \quad \frac{TD}{TM} = \frac{SC}{SN}.\] We finish by Coaxiality Lemma on $S$, $T$ with respect to $\omega_1$, $\omega_2$, as \[\frac{\operatorname{pow}(S,\omega_1)}{\operatorname{pow}(S,\omega_2)} = \frac{SB \cdot SC}{SM \cdot SN} = \frac{TA \cdot TD}{TM \cdot TN} = \frac{\operatorname{pow}(T,\omega_1)}{\operatorname{pow}(T,\omega_2)}. \quad \blacksquare\]

Let $PQ\cap MN=X$. Let $MN\cap\omega_1=Z_1,Z_2$. DIT on $A,B,C,D$ and line $MN$ gives $(M,N),(S,T),(Z_1,Z_2)$ swapped. Since $XZ_1\cdot XZ_2=XP\cdot XQ=XM\cdot XN$ this involution is inversion at $X$, so $XS\cdot XT=XP\cdot XQ$ as well,