I don't see the definition of $P$ and $Q$.

amogususususus wrote: Given a concyclic quadrilateral $ABCD$ with circumcenter $O$. Let $E$ be the intersection of $AD$ and $BC$, while $F$ be the intersection of $AC$ and $BD$. A circle $w$ are tangent to $BP$ and $CP$ such that $F$ is the orthocenter of $\triangle QEP$. Prove that $EO$ passes through the center of $w$. Are you sure? What I wrote down was $w$ tangent to $BD$ and $AC$ Oh, also $PQ$ is the diameter of $w$

My apologies, I'm sure that the new one is correct. If there are something wrong just tell me.

It's a very nice problem! Me and my friends used an entire day to solve it [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.05417022997491cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.1) + fontsize(5); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.901874083657221, xmax = 37.31480683624242, ymin = -23.59939481593597, ymax = 0.3682910821441535; /* image dimensions */ /* draw figures */ draw(circle((6.110108296557107,-9.391635118745668), 9.24893830610868), linewidth(1.2)); draw((0.5121720494348282,-2.0291647749440007)--(7.07952027869446,-18.589629476163655), linewidth(1.2)); draw((1.7636255736374915,-17.555638286895828)--(14.072002935584258,-14.09812994475315), linewidth(1.2)); draw((0.5121720494348282,-2.0291647749440007)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw((1.7636255736374915,-17.555638286895828)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw((6.110108296557107,-9.391635118745668)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw(circle((8.664067765760477,-11.293588067362759), 4.162555330363527), linewidth(1.2)); draw((6.177745551650191,-16.315681451124615)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw((xmin, 3.094086685976248*xmin-38.100964787798226)--(xmax, 3.094086685976248*xmax-38.100964787798226), linewidth(1.2)); /* line */ draw((9.944195346446698,-7.332762363610539)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw((4.7946698156678025,-12.828067256585111)--(22.48499691950008,-21.5861386240802), linewidth(1.2) + linetype("4 4")); draw((8.664067765760477,-11.293588067362759)--(6.177745551650191,-16.315681451124615), linewidth(1.2)); draw((2.104535589111946,-21.785224302561588)--(22.48499691950008,-21.5861386240802), linewidth(1.2)); draw((0.5121720494348282,-2.0291647749440007)--(2.104535589111946,-21.785224302561588), linewidth(1.2)); draw((2.104535589111946,-21.785224302561588)--(14.072002935584258,-14.09812994475315), linewidth(1.2)); draw((7.7028051621324165,-14.267817734978168)--(2.104535589111946,-21.785224302561588), linewidth(1.2)); draw((9.470498564816785,-11.89414318760482)--(7.7028051621324165,-14.267817734978168), linewidth(1.2)); draw((5.2300637601480515,-18.229892902884323)--(6.177745551650191,-16.315681451124615), linewidth(1.2)); draw((6.177745551650191,-16.315681451124615)--(13.15968393329803,-19.772278117075587), linewidth(1.2)); draw(circle((7.420906741668152,-13.804634647840464), 2.801928668105427), linewidth(1.2)); /* dots and labels */ dot((0.5121720494348282,-2.0291647749440007),dotstyle); label("$A$", (0.6560974642191724,-1.6576855567685205), NE * labelscalefactor); dot((1.7636255736374915,-17.555638286895828),dotstyle); label("$B$", (1.9175546167496909,-17.177431130325985), NE * labelscalefactor); dot((7.07952027869446,-18.589629476163655),dotstyle); label("$C$", (7.230965047105511,-18.209532436941874), NE * labelscalefactor); dot((14.072002935584258,-14.09812994475315),dotstyle); label("$D$", (14.226318347502023,-13.698867467287243), NE * labelscalefactor); dot((22.48499691950008,-21.5861386240802),linewidth(4.pt) + dotstyle); label("$E$", (22.636032697705478,-21.26761038247044), NE * labelscalefactor); dot((6.177745551650191,-16.315681451124615),linewidth(4.pt) + dotstyle); label("$F$", (6.313541663446952,-15.992425926433667), NE * labelscalefactor); dot((6.110108296557107,-9.391635118745668),linewidth(4.pt) + dotstyle); label("$O$", (6.275315689127845,-9.07352457467529), NE * labelscalefactor); dot((8.664067765760477,-11.293588067362759),linewidth(4.pt) + dotstyle); label("$G$", (8.798229994188882,-10.984823290630642), NE * labelscalefactor); dot((9.944195346446698,-7.332762363610539),linewidth(4.pt) + dotstyle); label("$P$", (10.097913121038507,-7.009321961443509), NE * labelscalefactor); dot((5.2300637601480515,-18.229892902884323),linewidth(4.pt) + dotstyle); label("$Q$", (5.396118279788393,-17.941950616708127), NE * labelscalefactor); dot((12.755638531716768,-10.528136863490365),linewidth(4.pt) + dotstyle); label("$X$", (12.926635220652397,-10.2203038042485), NE * labelscalefactor); dot((4.7946698156678025,-12.828067256585111),linewidth(4.pt) + dotstyle); label("$Y$", (4.937406587959114,-12.513862263394925), NE * labelscalefactor); dot((9.789787051374512,-15.301034008150287),linewidth(4.pt) + dotstyle); label("$Z$", (9.945009223762082,-14.998550594136884), NE * labelscalefactor); dot((2.104535589111946,-21.785224302561588),linewidth(4.pt) + dotstyle); label("$K$", (2.2615883856216503,-21.496966228385084), NE * labelscalefactor); dot((6.959302836278751,-16.568278538391407),linewidth(4.pt) + dotstyle); label("$H$", (7.116287124148191,-16.260007746667416), NE * labelscalefactor); dot((9.470498564816785,-11.89414318760482),linewidth(4.pt) + dotstyle); label("$L$", (9.639201429209228,-11.596438879736356), NE * labelscalefactor); dot((13.15968393329803,-19.772278117075587),linewidth(4.pt) + dotstyle); label("$S$", (13.308894963843462,-19.47098958947241), NE * labelscalefactor); dot((4.812677299922594,-18.148707336514423),linewidth(4.pt) + dotstyle); label("$R$", (4.9756325622782205,-17.827272693750807), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] First, flip the condition. $K$ is defined as $AB \cap CD$. Let $G$ be the center of the circle, which is $EO$ intersect the angle bisector of $\angle AFD$. $P$ is defined as the perpendicular line through $G$ to $EF$ intersect $G$, wlog it's at 'outside'. I redefined the point $Q$ here (see below). $H$ is the foot of perpendicular from $P$ to $EF$. $X$ is the second intersection of $EP$ and $(G)$. We want to show $\angle FXP = 90^{\circ}$, which is equivalent to $PXFH$ cyclic. Let $Y$, $Z$ be the points of tangencies at $(G)$ to $AC$, $BD$ respectively. Let $L=KF\cap EO$ which $KL bot EO$ by Brokard's. Clearly, $GLZHFY$ is cyclic. Let $S$ be a point on $BC$ such that $FS$ is tangent to $(GLZHFY)$. $R=LF \cap BC$, $Q=GF \cap BC$. Now we are ready to solve the problem. Claim: $Y$, $Z$, $E$ collinear. Proof: it suffices to show $\{(Y,Z),(G,L),(F,H)\}$ are reciprocal pairs. Projecting through $F$ onto $BC$, we need to show that $\{(B,C),(Q,R),(S,E)\}$ are reciprocal pairs. But $(B,C;R,S)=(B,C;Q,E)=-1$ and hence an inversive center exists as inversion preserves cross ratios. Now by radical center on $(YZHF)$ and $(PXZY)$, $(PXHF)$ is cyclic as desired. :>

Iranian tst 2023 https://artofproblemsolving.com/community/q2h3032970p27294059