collinear wanted, starting with a cyclic quadr.

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

collinear wanted, starting with a cyclic quadr.

G H J

Reveal topic tags

cyclic quadrilateral

L

G H BBookmark kLocked kLocked NReply

Source: 2021 Nordic MC p4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30629 posts

#1 h May 15, 2021, 11:35 AM

Y by

Let $A, B, C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$ . Let $X \ne D$ be the point on $\omega$ such that $DX$ and $EF$ are parallel. Let $Y$ be the reflection of $D$ through $EF$ and suppose that $Y$ is inside the circle $\omega$ .
Show that $A, X$ , and $Y$ are collinear.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

386 posts

#3 h May 15, 2021, 12:27 PM • 1 Y

Y by Mango247

$[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 = -12.503988125868464, xmax = 9.00006180677948, ymin = -6.927953789716436, ymax = 6.475915899243479; /* image dimensions */ pen wwzzqq = rgb(0.4,0.6,0); /* draw figures */ draw(circle((0,0), 4), linewidth(0.4) + red); draw((-8.154058564822963,-0.9219273465879328)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); draw((-1.612329064547513,3.660654994343955)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); draw((-1.612329064547513,3.660654994343955)--(-8.154058564822963,-0.9219273465879328), linewidth(0.4) + blue); draw((-2.889584353828225,2.7659179781965912)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); draw((-1.8968541003492212,-2.3923980697013247)--(1.6565885029890675,3.6408398113298586), linewidth(0.4) + blue); draw((-1.612329064547513,3.660654994343955)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); draw((-2.889584353828225,2.7659179781965912)--(-0.47249042781922146,-3.9719960719541514), linewidth(0.4) + blue); draw((-8.154058564822963,-0.9219273465879328)--(-1.992057807278974,0.2639644283969095), linewidth(0.4) + blue); draw((-2.889584353828225,2.7659179781965912)--(1.6565885029890675,3.6408398113298586), linewidth(0.4) + blue); draw((-2.889584353828225,2.7659179781965912)--(-1.8968541003492212,-2.3923980697013247), linewidth(0.4) + blue); draw((-8.154058564822963,-0.9219273465879328)--(-1.8968541003492212,-2.3923980697013247), linewidth(0.4) + blue); draw((-1.8968541003492212,-2.3923980697013247)--(-1.992057807278974,0.2639644283969095), linewidth(0.4) + blue); draw(circle((-4.91109805798729,-1.1705409107014284), 3.252476218691984), linewidth(0.4) + linetype("2 2") + wwzzqq); draw((-1.8968541003492212,-2.3923980697013247)--(-2.380898965077216,-3.214237097367623), linewidth(0.4) + blue); /* dots and labels */ dot((-8.154058564822963,-0.9219273465879328),dotstyle); label("$E$", (-8.068175135277407,-0.6840648518735976), NE * labelscalefactor); dot((-2.380898965077216,-3.214237097367623),dotstyle); label("$A$", (-2.282332104071683,-2.9742943850591947), NE * labelscalefactor); dot((-0.47249042781922146,-3.9719960719541514),linewidth(4pt) + dotstyle); label("$B$", (-0.37782543963313225,-3.769847801849981), NE * labelscalefactor); dot((-1.612329064547513,3.660654994343955),dotstyle); label("$C$", (-1.5108863665775867,3.8963942144975965), NE * labelscalefactor); dot((-2.889584353828225,2.7659179781965912),linewidth(4pt) + dotstyle); label("$D$", (-2.788593369302184,2.9561947219266673), NE * labelscalefactor); dot((-1.992057807278974,0.2639644283969095),linewidth(4pt) + dotstyle); label("$F$", (-1.8966092353246349,0.4489960750708557), NE * labelscalefactor); dot((1.6565885029890675,3.6408398113298586),linewidth(4pt) + dotstyle); label("$X$", (1.7436503384756334,3.824071176607525), NE * labelscalefactor); dot((-1.8968541003492212,-2.3923980697013247),linewidth(4pt) + dotstyle); label("$Y$", (-1.8001785181378729,-2.202848647565099), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Classical geometry at its' finest!

Essentially we want to show that $EYAF$ is a cyclic quad.
Since if that is true we have that following:
$\angle YAF = \angle YEF = \angle FED = \angle XDC = \angle XAC = \angle XAF$
But showing that $EYAF$ is cyclic is pretty easy, since we have that:
$\angle EAF = \angle EAD + \angle DAF = \angle DCB +\angle DBC = \angle EDB = \angle EDF = \angle EYF$ therefore proving that $EYAF$ is cyclic and in turn proving that $A,Y$ and $X$ are collinear.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

520 posts

zuss77

#4 h May 15, 2021, 2:52 PM

Y by

Let $AD\cap BC=G$ and let $GX$ meet $(ABC)$ at $L$ . Since $EF$ is a polar of $G$ and $DX\parallel EF$ , we have $AL\parallel DX$ . If $Z$ - center of this trapezoid, then $Z\in EF$ and $ZE$ bisects $\angle AZD.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

92 posts

#5 h May 31, 2021, 11:39 PM

Y by

Inversion on the circumference with center $E$ and ratio $\sqrt{EA\cdot EB}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#7 h Apr 11, 2022, 9:25 AM

Y by

$\angle BAX = \angle BDX = \angle DFE = \angle YFE$ so if $\angle BAY = \angle BAX$ then $AYFE$ is cyclic so we'll prove it in order to solve the Problem. $\angle EAF = \angle 180 - \angle BAC = \angle 180 - \angle BDC = \angle FDE = \angle FYE$ so $AYFE$ is cyclic.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a