Nice Problem Blastzit wrote: Two circles $\Gamma$ and $\Omega$ intersect at two distinct points $A$ and $B$. Let $P$ be a point on $\Gamma$. The tangent at $P$ to $\Gamma$ meets $\Omega$ at the points $C$ and $D$, where $D$ lies between $P$ and $C$, and $ABCD$ is a convex quadrilateral. The lines $CA$ and $CB$ meet $\Gamma$ again at $E$ and $F$ respectively. The lines $DA$ and $DB$ meet $\Gamma$ again at $S$ and $T$ respectively. Suppose the points $P,E,S,F,B,T,A$ lie on $\Gamma$ in this order. Prove that $PC,ET,SF$ are parallel. Note that it suffices to show $PT=PE$ and analogously $PS=PF \;$.We'll show only the first part now.By similar triangles we get $PE=\dfrac{PA.PC}{AC}\; ,PT=\dfrac{PB.PD}{BD}$.So we have to show $\mathbb{P}=\left(\dfrac{PA\cdot PC}{PB\cdot PD}\cdot\dfrac{BD}{AC}\right)=1$. Invert at $P$ with arbitary radius.We get the following equivalent problem:- Inverted Problem wrote: Let $P$ be a point.Let $\ell$ be a line through $P$.Let $\chi \parallel \ell$ be another line.Let $A^*,B^*\in \chi$.Let a circle $\Gamma$ through $A^*,B^*$ intersect $\ell$ at $D^*,C^*$.Then $P=1$ Note that \[\mathbb{P}=\left(\dfrac{PA\cdot PC\cdot \tfrac{B^*D^*\cdot PB\cdot PD}{r^2}}{PB\cdot PD\cdot \tfrac{A^*C^*\cdot PA\cdot PC}{r^2}}\right)=\dfrac{B^*D^*}{A^*C^*}\].But now note that $A^*B^*C^*D^*$ is an isosceles trapezoid hence $B^*D^*=A^*C^*\implies \mathbb{P}=1$ as desired.$\blacksquare$ [asy][asy] import geometry; import olympiad; size(5cm); pair O=origin; pair A,B,C,D,P; B=dir(0); C=dir(40); D=dir(130); A=dir(170); dot(A^^B^^C^^D); draw(unitcircle); draw("$A$",A,SW); draw("$D$",D,W); draw("$C$",C,E); draw("$B$",B,SE); draw(A--B--C--D--A); P=(3D+C)/4; dot(P); draw("$P$",P,N); draw(segment(B,D),dashed); draw(segment(A,C),dashed); [/asy][/asy]

This was a really trivial problem. (or have i made some error coz seeing @above's solution makes me think like that). Point $P$ and the tangency was a total giveaway. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -6.011206066471126, xmax = 30.93982320423769, ymin = -16.77235533990069, ymax = 7.629720483677792; /* image dimensions */ /* draw figures */ draw(circle((4.116767655088003,-3.1240144748596026), 8), linewidth(2)); draw(circle((15.257808278160047,-2.587666845988283), 5), linewidth(2)); draw((-3.218368917261432,0.06906733821405941)--(19.076561899003014,0.6398896265306337), linewidth(1.2)); draw((-1.8437213223505866,-8.459984035289374)--(6.04295210612313,-10.888666009899055), linewidth(1.2)); draw((-1.8437213223505866,-8.459984035289374)--(13.91625205743987,2.22899453993527), linewidth(1.2)); draw((6.04295210612313,-10.888666009899055)--(19.076561899003014,0.6398896265306337), linewidth(1.2)); draw((13.91625205743987,2.22899453993527)--(11.588157913300714,-5.983792020313833), linewidth(1.2)); draw((-3.218368917261432,0.06906733821405941)--(11.97746127686943,-4.610451753887808), linewidth(1.2)); draw((6.471237295238069,4.521668798309095)--(19.076561899003014,0.6398896265306337), linewidth(1.2)); /* dots and labels */ dot((11.278894472152317,0.44024413757134956),linewidth(4pt) + dotstyle); label("$A$", (11.384570700583348,0.6234214271537831), NE * labelscalefactor); dot((11.588157913300714,-5.983792020313833),linewidth(4pt) + dotstyle); label("$B$", (11.672500798796664,-5.783023258092485), NE * labelscalefactor); dot((6.471237295238069,4.521668798309095),dotstyle); label("$P$", (6.561741555510315,4.750419501544638), NE * labelscalefactor); dot((13.91625205743987,2.22899453993527),linewidth(4pt) + dotstyle); label("$D$", (14.023929934205407,2.4229845409870046), NE * labelscalefactor); dot((-1.8437213223505866,-8.459984035289374),linewidth(4pt) + dotstyle); label("$S$", (-1.740242942973614,-8.278417442607886), NE * labelscalefactor); dot((11.97746127686943,-4.610451753887808),linewidth(4pt) + dotstyle); label("$T$", (12.080401771265528,-4.415355291579237), NE * labelscalefactor); dot((19.076561899003014,0.6398896265306337),linewidth(4pt) + dotstyle); label("$C$", (19.182677527193974,0.8393690008137696), NE * labelscalefactor); dot((-3.218368917261432,0.06906733821405941),linewidth(4pt) + dotstyle); label("E", (-3.131905084337972,0.26350880438713875), NE * labelscalefactor); dot((6.04295210612313,-10.888666009899055),linewidth(4pt) + dotstyle); label("F", (6.129846408190342,-10.701829102569958), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By reims theorem we get that $CD \parallel SF$. Again by reims theorem we get that $CD \parallel ET$. So we have $CD \parallel SF \parallel ET$ which implies the result. $\blacksquare$. @below ig those are the indication of being too pr0.

@above those are called side-effects of not sleeping

GeoMetrix wrote: Point $P$ and the tangency was a total giveaway. But this condition doesn't even necessary. Line $CD$ would be enough. Strange problem. I mean, for TST.