Collinearity

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

Collinearity

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Turkey JBMO TST 2016 P4

The post below has been deleted. Click to close.

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

1345 posts

#1 h May 22, 2016, 10:11 PM • 2 Y

Y by Adventure10, Mango247

In a trapezoid $ABCD$ with $AB<CD$ and $AB \parallel CD$ , the diagonals intersect each other at $E$ . Let $F$ be the midpoint of the arc $BC$ (not containing the point $E$ ) of the circumcircle of the triangle $EBC$ . The lines $EF$ and $BC$ intersect at $G$ . The circumcircle of the triangle $BFD$ intersects the ray $[DA$ at $H$ such that $A \in [HD]$ . The circumcircle of the triangle $AHB$ intersects the lines $AC$ and $BD$ at $M$ and $N$ , respectively. $BM$ intersects $GH$ at $P$ , $GN$ intersects $AC$ at $Q$ . Prove that the points $P, Q, D$ are collinear.

Z K Y

The post below has been deleted. Click to close.

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

339 posts

rmrf

#2 h May 23, 2016, 3:29 AM • 2 Y

Y by Adventure10, Mango247

After many diagrams by hand and computer I do not see the collinearity mentioned..

Please check the wording very carefully!

Z K Y

The post below has been deleted. Click to close.

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

69 posts

#3 h May 23, 2016, 4:15 AM • 1 Y

Y by Adventure10

The wording seems all correct to me. Tough one to proof..

Regards
David

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

241 posts

#5 h May 23, 2016, 1:59 PM • 1 Y

Y by Adventure10

Since $EBCF$ is cyclic we get $\triangle EBG\sim \triangle EFC$ $\Longrightarrow$ $\tfrac{EB}{EG}=\tfrac{EF}{EC}..(\star)$ , so from $AB\parallel CD$ we get $EA.ED$ $=$ $EB.EC...(\star\star)$ $\Longrightarrow$ combining $(\star)$ and $(\star\star)$ and $\measuredangle DEF=\measuredangle GEA$ we get $\triangle EAG\sim \triangle EFD$ $\Longrightarrow$ $\measuredangle EGA$ $=$ $\measuredangle EDF$ $=$ $\beta-\theta$ and $\measuredangle EAG$ $=$ $\measuredangle EFD$ $=$ $\theta$ and let $\measuredangle EFB=\alpha$ $\Longrightarrow$ by angle-chasing we get $\measuredangle DBF=\measuredangle AGB=\alpha+\theta...(\blacksquare)$ $\Longrightarrow$ by $(\blacksquare)$ we get $\odot (GAD)$ $\cap$ $\odot (BFD)$ $\in$ $AD$ hence $G\in \odot (AHBMN)$ $\Longrightarrow$ by Pascal theorem in $(BGANMH)$ we get $D=AH\cap BN$ , $Q=GN\cap AM$ and $P$ $=$ $GH$ $\cap$ $BM$ are collinear hence $P$ , $Q$ and $D$ are collinear.

Z K Y

The post below has been deleted. Click to close.

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

1345 posts

#6 h May 23, 2016, 5:03 PM • 2 Y

Y by Adventure10, Mango247

FabrizioFelen wrote:

Since $EBCF$ is cyclic we get $\triangle EBG\sim \triangle EFC$ $\Longrightarrow$ $\tfrac{EB}{EG}=\tfrac{EF}{EC}..(\star)$ , so from $AB\parallel CD$ we get $EA.ED$ $=$ $EB.EC...(\star\star)$ $\Longrightarrow$ combining $(\star)$ and $(\star\star)$ and $\measuredangle DEF=\measuredangle GEA$ we get $\triangle EAG\sim \triangle EFD$ $\Longrightarrow$ $\measuredangle EGA$ $=$ $\measuredangle EDF$ $=$ $\beta-\theta$ and $\measuredangle EAG$ $=$ $\measuredangle EFD$ $=$ $\theta$ and let $\measuredangle EFB=\alpha$ $\Longrightarrow$ by angle-chasing we get $\measuredangle DBF=\measuredangle AGB=\alpha+\theta...(\blacksquare)$ $\Longrightarrow$ by $(\blacksquare)$ we get $\odot (GAD)$ $\cap$ $\odot (BFD)$ $\in$ $AD$ hence $G\in \odot (AHBMN)$ $\Longrightarrow$ by Pascal theorem in $(BGANMH)$ we get $D=AH\cap BN$ , $Q=GN\cap AM$ and $P$ $=$ $GH$ $\cap$ $BM$ are collinear hence $P$ , $Q$ and $D$ are collinear.

Yes, this is also the official solution.

Z K Y

The post below has been deleted. Click to close.

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

233 posts

#7 h Jun 3, 2016, 7:24 PM • 1 Y

Y by Adventure10

Hello.

My approach:

I will use the following lemma:

Lemma:Given is a triangle $\triangle{ABC}$ and a point $D\in BC$ .Then $\frac{\sin \hat{BAD}}{\sin \hat{CAD}}=\frac{BD}{CD}\cdot \frac{AC}{AB}$

Proof:See here.

We will next prove that the points $C,D,H,G$ are concyclic.Consider $T\equiv FH\cap BC$ and $S\equiv FG\cap DA$ .

Obviously $\overline{FGES}$ is the angle bisector of $\angle{BEC}$ .

We have $\angle{SHT}=\angle{SHF}=\angle{DBF}=\angle{EBC}+\angle{CBF}=\angle{EBC}+\frac{\angle{BEG}}{2}=\angle{EGC}$ ,thus $SHTG$ is cyclic.

Hence,it suffices to show that $ST\parallel AB$ ,because then $\angle{SHG}=\angle{STG}=180^{\circ}-\angle{GCD}\Rightarrow CDHG$ cyclic,q.e.d.

Notice that $\angle{TFC}=\angle{EFC}=\angle{BFC}-\angle{BFH}=180^{\circ}-\angle{BEC}-\angle{BDH}=180^{\circ}-\angle{DEA}-\angle{EDA}=\angle{DAE}$ .

The lemma gives $\frac{\sin \hat{BFT}}{\sin \hat{TFC}}=\frac{BT}{CT}\cdot \frac{CF}{BF}=\frac{BT}{CT}\Rightarrow \frac{BT}{CT}=\frac{\sin \hat{EDA}}{\sin \hat{DAE}}=\frac{AE}{ED}$ .

From the angle bisector theorem we get $\frac{AE}{ED}=\frac{AS}{SD}$ ,thus,finally $\frac{BT}{CT}=\frac{AS}{SD}\Rightarrow ST\parallel AB$ ,and we are done.

From the above,we also get that $AHBG$ is cyclic (simple angle-chasing),thus $A,H,B,G,M,N$ are concyclic.

From Menelaus' theorem,it suffices to show that $\frac{DE}{DB}\cdot \frac{BP}{MP}\cdot \frac{MQ}{EQ}=1$ .

$\bullet$ From the lemma,we have $\frac{\sin \hat{MNG}}{\sin \hat{BNG}}=\frac{MQ}{EQ}\cdot \frac{EN}{MN}\Rightarrow \frac{MQ}{EQ}=\frac{\sin \hat{MBG}}{\sin \hat{BMG}}\cdot \frac{MN}{EN}=\frac{MG}{BG}\cdot \frac{MN}{EN}$ .

$\bullet$ Similarly, $\frac{BP}{MP}=\frac{\sin \hat{BGP}}{\sin \hat{PGM}}\cdot \frac{BG}{GM}=\frac{\sin \hat{ADC}}{\sin \hat{DAC}}\cdot \frac{BG}{GM}=\frac{AC}{CD}\cdot \frac{BG}{GM}$ .

Thus,it suffices to show that $\frac{DE}{DB}\cdot \frac{MN}{NE}\cdot \frac{AC}{CD}=1$ .We have $\angle{EMN}=\angle{ABD}=\angle{EDC}\Rightarrow MNDC$ cyclic.

Thus $\frac{MN}{CD}=\frac{EN}{EC}$ ,and the above relation becomes $\frac{DE}{DB}\cdot \frac{EN}{CE}\cdot \frac{AC}{EN}=1\Leftrightarrow \frac{DE}{DB}=\frac{CE}{AC}$ ,which is obvious,and we are done.

$[asy]import graph; size(13.953224465305587cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.6097843202905344, xmax = 11.343440145015052, ymin = -2.782553526622707, ymax = 5.112033999800181; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((3.5,3.)--(6.,3.)--(7.,0.)--(0.,0.)--cycle, aqaqaq); /* draw figures */ draw((3.5,3.)--(6.,3.), uququq); draw((6.,3.)--(7.,0.), uququq); draw((7.,0.)--(0.,0.), uququq); draw((0.,0.)--(3.5,3.), uququq); draw((0.,0.)--(6.,3.)); draw((7.,0.)--(3.5,3.)); draw(circle((5.8684210526315805,1.2894736842105268), 1.7155795803164857)); draw((4.421052631578948,2.210526315789474)--(7.495962746954337,1.8319875823181127)); draw(circle((4.183967403585045,-0.8679348071700893), 4.2730427172870105)); draw((3.5,3.)--(3.966144491165257,3.3995524209987917)); draw((3.966144491165257,3.3995524209987917)--(6.341983701792522,1.9740488946224288)); draw((3.5,3.)--(6.341983701792522,1.9740488946224288)); draw((7.495962746954337,1.8319875823181127)--(0.,0.)); draw((7.495962746954337,1.8319875823181127)--(6.,3.)); draw((7.495962746954337,1.8319875823181127)--(7.,0.)); draw((3.966144491165257,3.3995524209987917)--(7.495962746954337,1.8319875823181127)); draw((4.421052631578948,2.210526315789474)--(2.8102902624554438,2.4088202249618087)); draw((3.966144491165257,3.3995524209987917)--(6.,3.)); draw(circle((4.75,2.013360497012464), 1.5924689977690278)); draw((6.,3.)--(5.916208450011215,0.9289641857046727)); draw((6.341983701792522,1.9740488946224288)--(3.210688397609971,1.6053441988049855)); /* dots and labels */ dot((3.5,3.),linewidth(3.pt) + dotstyle); label("$A$", (3.2128018287655564,2.9744629585926545), NE * labelscalefactor); dot((6.,3.),linewidth(3.pt) + dotstyle); label("$B$", (6.05852523945288,3.079374420737809), NE * labelscalefactor); dot((7.,0.),linewidth(3.pt) + dotstyle); label("$C$", (7.146981659208861,-0.1728809057619851), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$D$", (-0.26239035479269585,-0.04174157808054176), NE * labelscalefactor); dot((4.421052631578948,2.210526315789474),linewidth(3.pt) + dotstyle); label("$E$", (4.419283643434836,1.8335508077640976), NE * labelscalefactor); dot((7.495962746954337,1.8319875823181127),linewidth(3.pt) + dotstyle); label("$F$", (7.553513575021336,1.9122344043729635), NE * labelscalefactor); dot((6.341983701792522,1.9740488946224288),linewidth(3.pt) + dotstyle); label("$G$", (6.05,1.7286393456189428), NE * labelscalefactor); dot((3.966144491165257,3.3995524209987917),linewidth(3.pt) + dotstyle); label("$H$", (3.907840265477207,3.525248134854716), NE * labelscalefactor); dot((6.197059925012732,2.408820224961808),linewidth(3.pt) + dotstyle); label("$T$", (6.294576029279479,2.4499056478668813), NE * labelscalefactor); dot((2.8102902624554438,2.4088202249618087),linewidth(3.pt) + dotstyle); label("$S$", (2.6882445180397827,2.554817110012036), NE * labelscalefactor); dot((3.210688397609971,1.6053441988049855),linewidth(3.pt) + dotstyle); label("$N$", (2.858725644025659,1.7024114800826542), NE * labelscalefactor); dot((5.916208450011215,0.9289641857046727),linewidth(3.pt) + dotstyle); label("$M$", (6.0847531049891685,0.8631197829214171), NE * labelscalefactor); dot((5.96758003189726,2.198691096559587),linewidth(3.pt) + dotstyle); label("$P$", (5.7,2.3056523874172936), NE * labelscalefactor); dot((4.895632572430932,1.8037435093449148),linewidth(3.pt) + dotstyle); label("$Q$", (4.799587693711023,1.492588555792345), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

This post has been edited 2 times. Last edited by gavrilos, Jun 3, 2016, 7:29 PM

Z K Y

The post below has been deleted. Click to close.

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

479 posts

#8 h Jun 4, 2020, 8:15 PM • 3 Y

Y by Mango247, Mango247, Mango247

I think this is a really hard problem and requires an effective solution. Here is motivation for it.

First notice that we are asked to prove collinearity. We can use Menelaus, radical axis, completing angles $180$ or "Pascal Theorem"

Secondly let's focus on the points which are asked to be proved and think on the relationships between them. We realize $AH \cap BN = \{D\}$ , $AM \cap NG =\{Q\}$ and $HG \cap BM = \{P\}$ . So it makes sense to use Pascal Theorem. Then we should prove that $AHBGMN$ is cyclic.

It is already given that $AHBMN$ is cyclic. Because of that we must try to prove $G$ is on the $(AHBMN)$ also!

Thanks,
Burak

This post has been edited 1 time. Last edited by electrovector, Jun 4, 2020, 8:15 PM
Reason: little typo

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a