Vietnam MO 2019 - pro 6_prove BS, CT, XY concurent

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Vietnam MO 2019 - pro 6_prove BS, CT, XY concurent

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Source: Vietnam MO 2019

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#1 h Jan 15, 2019, 7:22 AM • 4 Y

Y by nguyendangkhoa17112003, anantmudgal09, Adventure10, Mango247

Given an acute triangle $ABC$ and $(O)$ be its circumcircle, and $H$ is its orthocenter. Let $M, N, P$ be midpoints of $BC, CA, AB$ , respectively. $D, E, F$ are the feet of the altitudes from $A, B$ and $C$ , respectively. Let $K$ symmetry with $H$ through $BC$ . $DE$ intersects $MP$ at $X$ , $DF$ intersects $MN$ at $Y$ .
a) $XY$ intersects smaller arc $BC$ of $(O)$ at $Z$ . Prove that $K, Z, E, F$ are concyclic.
b) $KE, KF$ intersect $(O)$ at $S, T$ ( $S, T\ne K$ ), respectively. Prove that $BS, CT, XY$ are concurent.

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1614 posts

yayups

#2 h Jan 15, 2019, 9:01 AM • 5 Y

Y by enthusiast101, stroller, Vietnam06727, Adventure10, Mango247

Solved

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(a) We make a very important structural claim first.

Claim: $XY$ is the $A$ -symmedian.

Proof of Claim 1: It suffices to show that $AX$ is the isogonal of $AM$ in $\angle A$ . To do this, let $U$ be on $MP$ such that $AU$ tangent to $(ABC)$ . Then, we have
$\angle AUX=\angle(AU,AC)=\pi-\angle CAU=\angle ABC$ and
$\angle EXU = \angle(ED,MP)=\angle(ED,AC)=\angle CED=\angle ABC.$ Thus, $\angle AUX=\angle EXU$ , so $AUXE$ is an iscoleces trapezoid. But we have that $AEDB$ cyclic with center $P$ , so $PA=PE$ , implying that $P$ is the midpoinit of $UX$ . Thus, $(UX;P\infty_{AC})=-1$ . Projecting through $A$ , we get that
$(AU,AX;AB,AC)=-1,$ which is certainly enough to imply that $AX$ is the $\angle A$ -symmedian. It similarly follows that $AY$ is the $\angle A$ -symmedian, so we are done. $\blacksquare$

Now, we have that $Z$ is the harmonic conjugate of $A$ wrt $BC$ in $(ABC)$ . Let $T=EF\cap BC$ and $G=AT\cap (ABC)$ . It is well known that if we define an inversion $\phi$ at $T$ with power $TB\cdot TC$ , then $\phi(G)=A$ . Also, we have that $\phi$ swaps $(E,F)$ .

Now, we know $(BC;TD)=-1$ , so projecting from $A$ , we get $(GK;BC)=-1$ . Inverting this statement, we get $(A\phi(K);CB)=-1$ , which readiliy implies $\phi(K)=Z$ . Thus, $\phi(K)=Z$ and $\phi(E)=F$ , so $EFKZ$ cyclic, as desired.

(b) We claim that they are concurrent on the midpoint of $EF$ which we'll denote $L$ . Note that $AL$ is the $A$ symmedian of $ABC$ since $AEF$ and $ABC$ are inversely similar, so this in fact suffices. So it suffices to show that $T,L,C$ collinear in the following diagram:
$[asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.84, xmax = 16.96, ymin = -10.51, ymax = 9.99; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-5.22,6.21)--(-7.28,-3.47), linewidth(2) + wrwrwr); draw((-7.28,-3.47)--(4.6,-2.97), linewidth(2) + wrwrwr); draw((4.6,-2.97)--(-5.22,6.21), linewidth(2) + wrwrwr); draw(circle((-1.4905531357814978,0.35714250616837606), 6.9400803133775195), linewidth(2) + wrwrwr); draw((-6.66349327180283,-0.573016927694852)--(-1.4902815641864529,2.723358936785299), linewidth(2) + wrwrwr); draw(circle((-3.2047234321092506,-0.29357125308418885), 3.4700401566887606), linewidth(2) + wrwrwr); draw((-5.22,6.21)--(-4.715017339545034,-5.788388012409966), linewidth(2) + wrwrwr); draw((-4.715017339545034,-5.788388012409966)--(-7.954668182394771,2.882995118708958), linewidth(2) + wrwrwr); draw((-6.66349327180283,-0.573016927694852)--(-4.918893728437006,-0.944285012336753), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(4.6,-2.97), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(-1.4902815641864529,2.723358936785299), linewidth(2) + wrwrwr); draw((-4.918893728437006,-0.944285012336753)--(-7.28,-3.47), linewidth(2) + wrwrwr); draw((-7.954668182394771,2.882995118708958)--(-4.076887417994643,1.0751710045452236), linewidth(2) + wrwrwr); draw((-4.076887417994643,1.0751710045452236)--(4.6,-2.97), linewidth(2) + wrwrwr); /* dots and labels */ dot((-5.22,6.21),dotstyle); label("$A$", (-5.14,6.41), NE * labelscalefactor); dot((-7.28,-3.47),dotstyle); label("$B$", (-7.2,-3.27), NE * labelscalefactor); dot((4.6,-2.97),dotstyle); label("$C$", (4.68,-2.77), NE * labelscalefactor); dot((-4.81695553399102,-3.36633651237336),linewidth(4pt) + dotstyle); label("$D$", (-4.74,-3.21), NE * labelscalefactor); dot((-1.4902815641864529,2.723358936785299),linewidth(4pt) + dotstyle); label("$E$", (-1.42,2.89), NE * labelscalefactor); dot((-6.66349327180283,-0.573016927694852),linewidth(4pt) + dotstyle); label("$F$", (-6.58,-0.41), NE * labelscalefactor); dot((-4.715017339545034,-5.788388012409966),linewidth(4pt) + dotstyle); label("$K$", (-4.64,-5.63), NE * labelscalefactor); dot((-7.954668182394771,2.882995118708958),linewidth(4pt) + dotstyle); label("$T$", (-7.88,3.05), NE * labelscalefactor); dot((-4.076887417994643,1.0751710045452236),linewidth(4pt) + dotstyle); label("$L$", (-4,1.23), NE * labelscalefactor); dot((-4.918893728437006,-0.944285012336753),linewidth(4pt) + dotstyle); label("$H$", (-4.84,-0.79), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
We have $\angle FHD=\angle FEC=\pi-B$ and $\angle DFH=\angle CFE$ , so $\triangle DHF\sim \triangle CEF$ . But then by SAS, we have $\triangle KHF\sim\triangle CEL$ , so $\angle FKH=\angle LCE$ , so $\angle TKA=\angle LCA$ , implying that $LC$ passes through $T$ , as desired.

This post has been edited 2 times. Last edited by yayups, Jan 16, 2019, 12:01 AM

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Desimathematics

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Desimathematics

#3 h Jan 15, 2019, 10:08 AM • 1 Y

Y by Adventure10

@above

can you kindly explain the meaning of $(BC;TD)=-1$

Thanks

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1614 posts

yayups

#4 h Jan 15, 2019, 10:14 AM • 1 Y

Y by Adventure10

https://en.wikipedia.org/wiki/Cross-ratio

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894 posts

#5 h Jan 17, 2019, 8:54 PM • 4 Y

Y by yayups, Vietnam06727, Adventure10, Mango247

probably an alternative approach

another (trivial) property in the configuration

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#6 h Jan 18, 2019, 3:51 AM • 2 Y

Y by Vietnam06727, Adventure10

yayups wrote:

Claim: $XY$ is the $A$ -symmedian.

I thought this looked familiar- it's actually just Problem 4 from here.

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RC.

#7 h Jan 18, 2019, 7:30 AM • 4 Y

Y by stroller, Vietnam06727, Adventure10, Mango247

(a) Applying Pascal's on cyclic hexagon $MPFDEN$ , we have $\overline{A,X,Z,Y}$ . Let $M'$ be the midpoint of $EF$ , by Menelaus' in $\triangle DFE$ we have $\frac{DM'}{M'F}\times \frac{FY}{DY} \times \frac{DX}{XE}= 1$ , due to $||$ conditions. $\Rightarrow \overline{A,M,'X,Z,Y}$ . Now $M'FBZ, M'ECZ$ are concyclic. Also, $\triangle M'BF \sim \triangle EKH , \triangle M'CF \sim \triangle FKH$ $\Longrightarrow \angle EKF= \angle EZF$ .

(b) $\angle FBM' = \angle HKE \Rightarrow \angle ABS= \angle AKS \Rightarrow M' \in BS$ similarly for other one. Thus $BS\cap CT \cap XY = M$ done

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PROF65

#8 h Jan 18, 2019, 11:15 AM • 3 Y

Y by stroller, Vietnam06727, Adventure10

$a)$ $X$ is on the $A$ -symmedian ideed $\frac{d(X,AB)}{d(X,AC)}= \frac{XP \sin A}{XE\sin B},\frac{XP}{XE}=\frac{\sin A}{\sin C}$ thus $\frac{d(X,AB)}{d(X,AC)}=\frac{\sin C}{\sin B}=\frac{AB}{AC}$ idem for $Y$ ,
Now consider $U$ the symmetric of $Z$ in $BC$ then $U\in (BHC)$ and $U\in AM$ since the $MZ$ and the median are symmetric about the $BC$ -bisector ; more $PUMD$ is cyclic by radical axis theorem we conclude that $EF,HU,BC$ are concurent so $KZ,EF,BC$ are concurrent more by symmetry $KZMD$ is also cyclic then by radical axis theorem we get $KZEF$ is cyclic
$b)$ since $E,F,KZ\cap BC$ are collinear applying Pascal's to the hexagons $BSKTCA,ZKSBCA,ZKTCBA$ yields the result.

This post has been edited 1 time. Last edited by PROF65, Jan 18, 2019, 1:31 PM

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698 posts

#9 h Jan 18, 2019, 3:38 PM • 4 Y

Y by AlastorMoody, Vietnam06727, Adventure10, Mango247

(a) Let $MP$ intersect the tangent at $A$ to $(O)$ at $T$ . Note that $\triangle{XPE} \equiv \triangle{TPA}$ , thus $A(TX,BC)=K(AZ,BC)-1$ , this means $KZ, FE$ and $BC$ are concurrent, and we are done by radical axis.
(b) Let $MH\cap (O) = L, BE\cap (O)=B'$ and $LE\cap (O)=S'$ . Clearly, $(LK,BC)=-1 \implies$
$-1=E(LK,BC)=(S'S,B'A)=B(S'S,B'A)=B(S'S,FE)$ By Reim's $BS' \mid \mid FE$ , thus $BS$ bisects $EF$ , which means that the $A$ -symmedian, $BS$ and $EF$ are concurrent.

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1218 posts

#10 h Jan 25, 2019, 3:17 PM • 3 Y

Y by AlastorMoody, Vietnam06727, Adventure10

Here's my solution: Let $B'$ be the reflection of $B$ in $X$ , which lies on $AC$ (as $X$ lies on $MP$ ). Now, by Pascal on $DENMPF$ and $DEFPMM$ (along with the well known fact that $ME=MF$ ), we get that $A,X,Y$ are collinear and $BX \parallel EF$ . As $AX$ bisects $BB'$ , which is antiparallel wrt $\angle BAC$ (because $BB' \parallel EF$ ), so $AX$ is the $A$ -symmedian of $\triangle ABC$ . This means that $ABZC$ is harmonic. Note that showing that $KZEF$ is cyclic is equivalent to showing that $KZ,BC,EF$ are concurrent (since $BCEF$ is also cyclic) at the $A$ -Ex point $X_A$ of $\triangle ABC$ . Now, by APMO 2012 P4, we know that $ZD$ passes through the $A$ -Queue point $Q_A$ of $\triangle ABC$ . This gives $-1=(A,Z;B,C) \overset{D}{=} (K,Q_A;C,B) \overset{X_A}{=} (KX_A \cap (O),A;B,C) \Rightarrow Z=KX_A \cap (O)$ Thus, we get that $Z,K,X_A$ are collinear, proving part (a).

For part (b), apply Pascal on $KSBACT$ and $AZKSBC$ , and use the fact that $KZ,BC,EF$ are concurrent (proved in part (a)), to get that $BS,CT,AZ$ concur on $EF$ .

This post has been edited 1 time. Last edited by math_pi_rate, Jan 25, 2019, 3:17 PM

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121 posts

#11 h Mar 20, 2019, 5:09 PM • 5 Y

Y by RAMUGAUSS, AlastorMoody, Vietnam06727, Adventure10, Mango247

Special Thanks to Alastormoody.
solution(a)
solution(b).

Attachments:

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Generic_Username

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Generic_Username

#12 h Mar 21, 2019, 2:28 AM • 2 Y

Y by Vietnam06727, Adventure10

a)
By Pascal on $NMPFDE,$ we see $A,X,Y$ are collinear. Now angle chasing shows that $BDXP,CYDN$ are cyclic, which implies that $BX\parallel CY.$ Now Law of Sines in $\triangle BDX,\triangle CDY$ shows
$\frac{BX}{\sin A}=\frac{BD}{\sin 2B},\frac{CY}{\sin A}=\frac{CD}{\sin 2C},$ and substituting $\frac{BD}{CD}=\frac{c\cos B}{b\cos C}$ shows $\frac{BX}{CY}=\frac{c^2}{b^2}.$ Finally if $XY\cap BC=U,$ this and $BX\parallel CY$ implies $\frac{BU}{CU}=\frac{c^2}{b^2},$ so $XY$ is a symmedian.

Invert at $A$ fixing $BFEC.$ Then $K$ maps to $AD\cap EF$ and $Z$ maps to $AZ\cap EF.$ But symmedians intersect antiparallels at their midpoint, so $Z'$ is the midpoint of $EF.$ This along with $(Z,K';F,E)=-1$ shows that $BK'N'C$ is cyclic, and inverting back finishes.

b) is surprisingly easy. By Converse Pascal $BS,CT$ concur on $EF.$ Now the previous problem implies that $EF,KZ,BC$ concur by radical axis, so Pascal on $BCAZKS$ wins.

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#13 h Apr 10, 2019, 10:49 AM • 5 Y

Y by Vietnam06727, Adventure10, Mango247, Mango247, Mango247

Here is my solution for this problem
Solution
a) Let $G$ $\equiv$ $EF$ $\cap$ $BC$
Denote $d(X; YZ)$ is distance from point $X$ to line $YZ$
Since: $(BCDG) = - 1$ and $M$ is midpoint of $BC$ , we have: $MB^2 = MC^2 = \overline{MD} . \overline{MG}$
Then: $\dfrac{\overline{XD}}{\overline{XE}} = \dfrac{\overline{MD}}{\overline{MC}} = \dfrac{\overline{DM}}{\overline{MB}} = \dfrac{\overline{MB}}{\overline{GM}} = \dfrac{\overline{MB} + \overline{DM}}{\overline{GM} + \overline{MB}} = \dfrac{\overline{DB}}{\overline{GB}}$ or $BX$ $\parallel$ $EG$
So: $(BX; BP) \equiv (FE; FA) \equiv (CA; CB) \equiv - (CB; CA)$ (mod $\pi$ )
But: $(PB; PX) \equiv (AB; AC) \equiv - (AC; AB)$ (mod $\pi$ ) then: $\triangle PXB$ $\stackrel{-}{\sim}$ $\triangle ABC$
Hence: $\dfrac{d(X; AB)}{d(X; AC)} = \dfrac{d(X; AB)}{d(B; AC)} . \dfrac{d(B; AC)}{d(X; AC)} = \dfrac{2 PB}{AC} = \dfrac{AB}{AC}$ or $AX$ is $A$ - symmedian of $\triangle ABC$
Similarly: $AY$ is $A$ - symmedian of $\triangle ABC$
Hence: $A$ , $X$ , $Y$ lie on $A$ - symmedian of $\triangle ABC$
We have: $\dfrac{KB}{KC} . \dfrac{ZB}{ZC} = \dfrac{BD}{CD} . \dfrac{AC}{AB} . \dfrac{AB}{AC} = \dfrac{BG}{CG}$
Then: $K$ , $Z$ , $G$ are collinear
So: $\overline{GK} . \overline{GZ} = \overline{GB} . \overline{GC} = \overline{GE} . \overline{GF}$ or $E$ , $F$ , $K$ , $Z$ lie on a circle
b) We have: $\dfrac{ZB}{ZC} . \dfrac{SC}{SA} . \dfrac{TA}{TB} = \dfrac{AB}{AC} . \dfrac{EC}{EA} . \dfrac{KA}{KC} . \dfrac{FA}{FB} . \dfrac{KB}{KA} = \dfrac{AB}{AC} . \dfrac{DC}{DB} . \dfrac{KB}{KC} = \dfrac{AB}{AC} . \dfrac{KC}{KB} . \dfrac{AC}{AB} . \dfrac{KB}{KC} = 1$
Then: $AZ$ , $BS$ , $CT$ concurrent or $XY$ , $BS$ , $CT$ concurrent

This post has been edited 6 times. Last edited by khanhnx, May 29, 2019, 8:12 AM

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#14 h Jan 14, 2020, 11:26 AM • 3 Y

Y by GeoMetrix, mueller.25, Adventure10

VMO 2019 P6 wrote:

Given an acute triangle $ABC$ and $(O)$ be its circumcircle, and $H$ is its orthocenter. Let $M, N, P$ be midpoints of $BC, CA, AB$ , respectively. $D, E, F$ are the feet of the altitudes from $A, B$ and $C$ , respectively. Let $K$ symmetry with $H$ through $BC$ . $DE$ intersects $MP$ at $X$ , $DF$ intersects $MN$ at $Y$ .
a) $XY$ intersects smaller arc $BC$ of $(O)$ at $Z$ . Prove that $K, Z, E, F$ are concyclic.
b) $KE, KF$ intersect $(O)$ at $S, T$ ( $S, T\ne K$ ), respectively. Prove that $BS, CT, XY$ are concurent.

$(a)$ By Pascal's Theorem on $DFPMNE$ we get that $\overline{A-X-Y}$ . So, redifine $Z$ as $AX\cap\odot(ABC)$ .

Claim:- $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$ .
By Three Tangents Lemma we get that $MF=ME$ . Hence, $MM\|EF$ WRT $\odot(X_5)$ . So, by Pascal's Theorem on $DEFPMM$ we get that $BX\|FE$ . Let $BX\cap AC=L$ . Now as $MP\|AC$ we get that $BX=XL$ . Hence, $AZ$ bisects $FE$ . Now as $FE$ and $BC$ are antiparallel, we get that $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$ .

Let $KZ\cap BC=X_A'$ also it's well known that $K\in\odot(ABC)$ . Then $-1=(AZ;BC)\overset{K}{=}(DX_A';BC)$ Now if $EF\cap BC=X_A$ then $(X_AD;BC)=-1\implies\boxed{X_A'\equiv X_A}$ . Hence, $EF,BC,KZ$ are concurrent at a point $X_A$ .So, $X_AF\cdot X_AE=X_AB\cdot X_AC=X_AK\cdot X_AZ\implies E,F,K,X\text{ are concyclic}.\blacksquare$
$(b)$ From ELMO Shortlist 2019 G1 we get that $CT$ bisects $EF$ and $BS$ bisects $EF$ . Hence, $CT,BS,XY$ are concurrent at the midpoint of $XY$ . $\blacksquare$

This post has been edited 11 times. Last edited by amar_04, Jan 14, 2020, 1:40 PM

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