AoPS - Problem

InterLoop

02.08.2024 13:53

my 100th post - and what a wonderful (?) one

EthanWYX2009

02.08.2024 14:05

Let $T=BD\cap XZYW,S=AC\cap BD.$ By $ZT/YT=AS/SC=XT/WT$ we get $TX\cdot TY=TZ\cdot TW,$ so $B,T$ lies on the root axis of $XYB$ and $ZTB,$ so $BD$ is the root axis, done.$\Box$

math_comb01

02.08.2024 14:06

BY DIT, $(X,Y);(Z,T);(BD \cap \ell, \infty_{\ell})$ are pairs of involution, as this can not be reflection over midpoints therefore it must an inversion so there exists point $P$ such that $PX \cdot PY = PZ \cdot PT = \overline{P BD\cap\ell} \cdot \overline{P \infty_{\ell}}$ implying $P$ lies on $BD$ and radical axes of both circles finishing our proof.

iamnotgentle

02.08.2024 14:38

Let second intersection of $AX\cap(XBY) =P$ and $CT\cap (BZT) =Q$ then observe that $(APCB) $ and $(AQCB) $ is cyclic this implies $(XQPT) $ is cyclic hence $D$ lie on the radical of $(BZT) $ and $(BXY) $ as desired.

sami1618

13.08.2024 22:48

Let $BD$ meet $l$ and $AC$ at $P$ and $Q$. Then $$\frac{PX}{PT}=\frac{QA}{QC}=\frac{PZ}{PY}$$Thus $PX\cdot PY=PZ\cdot PT$ so $P$ lies on the radical axis of $(XYB)$ and $(ZTB)$. As $B$ also lies on their radical axis, the result follows.

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SomeonesPenguin

14.08.2024 00:30

Cool problem!

Let $l\cap BD = \{O\}$. Case 1. $O\neq B$. By DIT, there is an involution swapping $(X,Y)$, $(Z,T)$ and $(O,\infty_{AC})$. But we know that a involution on a line is an inversion and since $O$ gets swapped to the point of infinity, the inversion must be centered at $O$. So we get $OX\cdot OY=OZ\cdot OT$ hence it lies on the radical axis of $(ZTB)$ and $(XYB)$ so $BD$ is the radical axis. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -3.22, xmax = 22.86, ymin = -11.56, ymax = 5.94; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); pen qqzzqq = rgb(0.,0.6,0.); draw((8.,-6.)--(18.,-6.)--(16.,0.)--(11.,2.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((8.,-6.)--(18.,-6.), linewidth(0.8) + qqzzcc); draw((18.,-6.)--(16.,0.), linewidth(0.8) + qqzzcc); draw((16.,0.)--(11.,2.), linewidth(0.8) + qqzzcc); draw((11.,2.)--(8.,-6.), linewidth(0.8) + qqzzcc); draw((8.,-6.)--(16.,0.), linewidth(0.8)); draw((2.546666666666666,-6.)--(8.,-6.), linewidth(0.8)); draw((16.,0.)--(14.909333333333333,3.272), linewidth(0.8)); draw(circle((15.01536231884058,-1.8437681159420267), 5.116866771947141), linewidth(0.8) + qqzzqq); draw(circle((10.273333333333335,-5.99304347826087), 7.726669798236012), linewidth(0.8) + qqzzqq); draw((11.,2.)--(18.,-6.), linewidth(0.8)); draw((2.546666666666666,-6.)--(14.909333333333333,3.272), linewidth(0.8)); /* dots and labels */ dot((8.,-6.),dotstyle); label("$A$", (7.7,-6.58), NE * labelscalefactor); dot((18.,-6.),dotstyle); label("$B$", (18.28,-6.48), NE * labelscalefactor); dot((16.,0.),dotstyle); label("$C$", (16.3,0.08), NE * labelscalefactor); dot((11.,2.),dotstyle); label("$D$", (10.86,2.36), NE * labelscalefactor); dot((2.546666666666666,-6.),linewidth(4.pt) + dotstyle); label("$Z$", (1.94,-6.1), NE * labelscalefactor); dot((10.13391304347826,-0.30956521739130455),linewidth(4.pt) + dotstyle); label("$X$", (9.56,-0.18), NE * labelscalefactor); dot((12.443478260869565,1.422608695652174),linewidth(4.pt) + dotstyle); label("$T$", (12.34,1.8), NE * labelscalefactor); dot((14.909333333333333,3.272),linewidth(4.pt) + dotstyle); label("$Y$", (14.92,3.7), NE * labelscalefactor); dot((11.876981132075471,0.9977358490566037),linewidth(4.pt) + dotstyle); label("$O$", (11.72,0.36), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Case 2. $O=B$ ($l$ is the parallel through $B$ to $AC$) We get $Y=Z=B$ and the circle $(XYB)$ becomes the circle passing through $B$ and $X$ and tangent to $BC$ and $(ZTB)$ is defined similarly. Let $(BBT)$ intersect $DC$ at $H$ and $(BBX)$ intersect $AD$ at $G$. We have $\angle ABH =\angle BTH=\angle ACH$ so $H$ lies on $(ABC)$ and similiary $G$ lies on $(ABC)$ so we have $DH\cdot DC=DG\cdot DA$ and since $AC$ is parallel to $XT$ we have $DH\cdot DT=DG\cdot DX$ so $GHTX$ is cyclic so $D$ lies on the radical axis of the circles. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -13.187030149311312, xmax = 42.369817808628284, ymin = -25.00999481626015, ymax = 12.269331842633335; /* image dimensions */ pen qqzzff = rgb(0.,0.6,1.); pen qqccqq = rgb(0.,0.8,0.); pen ffwwqq = rgb(1.,0.4,0.); draw((8.,-7.)--(18.,-7.)--(16.,0.)--(9.819640017320117,2.2571698256733708)--cycle, linewidth(0.8) + qqzzff); /* draw figures */ draw((8.,-7.)--(18.,-7.), linewidth(0.8) + qqzzff); draw((18.,-7.)--(16.,0.), linewidth(0.8) + qqzzff); draw((16.,0.)--(9.819640017320117,2.2571698256733708), linewidth(0.8) + qqzzff); draw((9.819640017320117,2.2571698256733708)--(8.,-7.), linewidth(0.8) + qqzzff); draw((8.,-7.)--(16.,0.), linewidth(0.8)); draw((8.,-7.)--(5.922781251999055,-17.567566404500827), linewidth(0.8)); draw((5.922781251999055,-17.567566404500827)--(23.05521944077962,-2.5766829893178325), linewidth(0.8)); draw((23.05521944077962,-2.5766829893178325)--(16.,0.), linewidth(0.8)); draw(circle((18.,-1.8996446713562773), 5.100355328643722), linewidth(0.8) + qqccqq); draw(circle((9.470464259224334,-9.437010211650191), 8.87085107105608), linewidth(0.8) + qqccqq); draw((9.264199574312139,-0.5685574940328231)--(13.698304072318887,0.8406174738035777), linewidth(0.8)); draw(circle((13.,-4.642857142857144), 5.527759261127386), linewidth(0.8) + ffwwqq); draw((13.698304072318887,0.8406174738035777)--(8.,-7.), linewidth(0.8)); draw((13.698304072318887,0.8406174738035777)--(18.,-7.), linewidth(0.8)); /* dots and labels */ dot((8.,-7.),dotstyle); label("$A$", (7.258047171952292,-6.962053738202001), NE * labelscalefactor); dot((18.,-7.),dotstyle); label("$B$", (18.370209188912267,-7.762053738202001), NE * labelscalefactor); dot((16.,0.),dotstyle); label("$C$", (16.16777678552027,0.42515720129346235), NE * labelscalefactor); dot((9.819640017320117,2.2571698256733708),dotstyle); label("$D$", (8.990059796332202,2.683219273203582), NE * labelscalefactor); dot((23.05521944077962,-2.5766829893178325),linewidth(4.pt) + dotstyle); label("$T$", (23.240197614521783,-2.216349373393847), NE * labelscalefactor); dot((5.922781251999055,-17.567566404500827),linewidth(4.pt) + dotstyle); label("$X$", (5.413009823807276,-18.413289926457282), NE * labelscalefactor); dot((9.264199574312139,-0.5685574940328231),linewidth(4.pt) + dotstyle); label("$G$", (8.636195514542927,-0.2139169700018545), NE * labelscalefactor); dot((13.698304072318887,0.8406174738035777),linewidth(4.pt) + dotstyle); label("$H$", (13.467109768857128,1.1920462068478426), NE * labelscalefactor); dot((12.938395907762672,-1.2721189811485274),linewidth(4.pt) + dotstyle); label("$I$", (13.100220763302747,-0.9382010308032136), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Mahdi_Mashayekhi

19.08.2024 15:51

Let $R$ be the intersection of $BXY$ and $BZT$. Let $K$ be such that $KX \parallel AB$ and $KT \parallel BC$. Note that since $XT \parallel AC$ then $B,K,D$ are collinear. Claim $: XRTK$ is cyclic. Proof $:$ Note that $\angle XRT = \angle XRB + \angle BRT = \angle XYB + \angle BZT = 180 - \angle B = 180 - \angle XKT$. Now we have $\angle RKX = \angle RTX = \angle RTZ = \angle RBZ$ so $R$ lies on $BD$ as wanted.