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Source: Iran TST 2015,third exam,second day,problem 6

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228 posts

ATimo

#1 h May 29, 2015, 7:14 PM • 3 Y

Y by Rmasters, Tawan, Adventure10

$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$ . If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$ , intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$ , intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$ .

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2312 posts

#2 h May 29, 2015, 8:24 PM • 11 Y

Y by mihajlon, Rmasters, DreamComeTrue, Tawan, AlastorMoody, enhanced, Smkh, Adventure10, L.M., XX-math-XX, Rounak_iitr

My solution :

Let $P, Q$ be the projection of $X$ on $AC, AB$ , respectively .
Let $M$ be the midpoint of $BC$ and $T$ be a point such that $TB \parallel AC, TC \parallel AB$ .

Since $\{ BT, BX \}, \{ CT, CX \}$ are isogonal conjugate WRT $\angle CBA, \angle ACB$ , respectively ,
so $X, T$ are the isogonal conjugate of $\triangle ABC$ $\Longrightarrow P, Q, M, H'$ are concyclic ( pedal circle of $\{ X, T \}$ ) .

Since $B, M, X, Q$ are concyclic and $C, M, X, P$ are concyclic ,
so $\angle QH'P=\angle QMP=\angle QBX+\angle PCX=\angle ACB+\angle CBA$ . ... $(\star)$

Since $H', X, Q, Y$ are concyclic and $H', X, P, Z$ are concyclic ,
so $\angle ZXY=\angle ZXH'+\angle YXH'=\angle APH'+\angle AQH'=\angle QH'P-\angle BAC$ ,
hence combine $(\star)$ we get $\angle ZXY=180^{\circ}-2\angle BAC=\angle CXB$ $\Longrightarrow \angle ZXC=\angle YXB$ .

Q.E.D

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95 posts

#3 h May 29, 2015, 9:45 PM • 2 Y

Y by Adventure10, Mango247

beautifull prove
thanks

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158 posts

THVSH

#4 h May 30, 2015, 3:54 PM • 4 Y

Y by buratinogigle, Mathuzb, Adventure10, Mango247

Another solution:

Let $M$ be the midpoint of $BC$ . $U, V$ are the projection of $X$ on $AB, AC$ , respectively.
$T$ is the projection of $A$ on $OX$ . Then $AMH'T$ is the parallelogram.

We have $\angle AUM=\angle BXM=90^{\circ}-\angle BAC \Longrightarrow UM\perp AC$ . Similarly, $VM\perp AB$
$\Longrightarrow M$ is the orthocenter of $\triangle AUV$ .
On the other hand, $A, T, U, V, X$ are concyclic in the circle $\odot (AX)$ , $AMH'T$ is the parallelogram.
$\Longrightarrow H'$ is the orthocenter of $\triangle TUV$ . $\Longrightarrow \angle UH'V=180^{\circ}-\angle BAC$
$\Longrightarrow \angle YXZ=180^{\circ}-\angle XYZ-\angle XZY=180^{\circ}-\angle H'UX-\angle H'VX=\angle UH'V+\angle UXV-180^{\circ}=180^{\circ}-2.\angle BAC=\angle BXC$
$\Longrightarrow \angle ZXC=\angle YXB$ . Q.E.D

This post has been edited 1 time. Last edited by THVSH, May 30, 2015, 3:55 PM

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824 posts

andria

#5 h May 30, 2015, 6:34 PM • 7 Y

Y by seyyed_khandan, Tawan, AlastorMoody, HolyMath, Gaussian_cyber, Adventure10, Mango247

Too easy problem for problem 6
My solution:
Lemma1: let $R,T$ projections of $X$ on $AB,AC$ and assume that $M$ is midpoint of $BC$ then $M$ is orthocenter of $\triangle ART$ for prove see Russia 2013 grade 9.
Lemma2: in cyclic quadrilateral $ABCD$ points $M,N$ are midpoints of $AB$ and $CD$ and $AC\cap CD=P$ also $H_1,H_2$ are projections of $P$ on $AD,BC$ then $\triangle H_1MN=\triangle H_2MN$ .(it's well known)
Lemma3: in triangle $ABC$ , $AD,BE,CF$ are altitudes and $M,N,S,T$ are midpoints of $BC,CE,BF,EF$ and $AL$ is altitude in $\triangle AEF$ and let $R$ be the reflection of $L$ WRT midpoint of $EF$ then quadrilateral $NLRS$ is cyclic.
Prove: note that $R$ is foot of $H$ (orthocenter of $\triangle ABC$ ) on $EF$ .then using lemma 2 for the cyclic quadrilateral $BFEC$ we get that $\angle SRN=\angle SMN$ note that $\angle ADM=\angle MSA=90\longrightarrow AMDS$ is cyclic. So $\angle SMN=\angle 180-A$ so $\angle SRN=180-A$ also using lemma 1 we get that $T$ is orthocenter of $ASN$ so $\angle STN=180-A$ and the lemma proved.
BACK TO THE MAIN PROBLEM:
Let $M$ midpoint of $BC$ then from the lemma3,1 $RMH'T$ is cyclic and $M$ is orthocenter of $ARZ$ so $\angle RMT=\angle RH'T=180-A$ but because $XH'ZT,XH'YR$ are cyclic we get that $180-A=\angle RH'T=\angle XZT+\angle XYR=\angle A+\angle TXY\longrightarrow \angle ZXY=180-2A=\angle CXB\longrightarrow \angle CXZ=\angle YXB$
DONE

Attachments:

This post has been edited 1 time. Last edited by andria, May 30, 2015, 6:35 PM

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137 posts

#6 h Jun 2, 2015, 9:47 AM • 2 Y

Y by Adventure10, Mango247

andria, can you provide a link for Lemma 2?

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824 posts

andria

#7 h Jun 2, 2015, 11:40 AM • 2 Y

Y by Achillys, Adventure10

Achillys wrote:

andria, can you provide a link for Lemma 2?

See http://www.artofproblemsolving.com/community/c6h296040
I also solved the problem 3 in second round exam using this lemma.

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2280 posts

#8 h Jan 5, 2016, 7:03 AM • 5 Y

Y by anantmudgal09, rashah76, ILOVEMYFAMILY, Adventure10, Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 9.5, ymin = -6.58, ymax = 6.3; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); pen wwzzqq = rgb(0.4,0.6,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((2.92,4.04)--(1.04,-0.08)--(6.24,-0.04)--cycle, zzttqq); draw(arc((3.680402725208175,-5.312354277062829),0.6,60.844980720518215,64.10457654225013)--(3.680402725208175,-5.312354277062829)--cycle, qqwuqq); draw(arc((3.680402725208175,-5.312354277062829),0.6,473.51728378153996,476.7768796032719)--(3.680402725208175,-5.312354277062829)--cycle, qqwuqq); draw((3.9073430096064508,-0.0028055971098801327)--(3.8554441513586597,-0.4238833910061812)--(4.27652194525496,-0.47578224925397233)--(4.328420803502752,-0.054704455357671315)--cycle, qqwuqq); draw((-0.09509926716624112,-3.589552397047317)--(0.08102693760250201,-3.203573692979646)--(-0.3049517664651692,-3.0274474882109033)--(-0.48107797123391227,-3.4134261922785742)--cycle, qqwuqq); /* draw figures */draw((1.04,-0.08)--(6.24,-0.04), zzttqq); draw(circle((3.6300996777336425,1.2270418946263961), 2.901202311956972)); draw((4.328420803502752,-0.05470445535767114)--(3.680402725208175,-5.312354277062829)); draw((2.92,4.04)--(6.46639011001663,-0.3182143520686293)); draw((6.46639011001663,-0.3182143520686293)--(1.2260256631944162,0.327673261894146)); draw((1.04,-0.08)--(3.680402725208175,-5.312354277062829)); draw((3.680402725208175,-5.312354277062829)--(6.24,-0.04)); draw((1.2260256631944162,0.327673261894146)--(3.680402725208175,-5.312354277062829)); draw((3.680402725208175,-5.312354277062829)--(6.46639011001663,-0.3182143520686293)); draw((2.92,4.04)--(-0.48107797123391227,-3.4134261922785742)); draw((-0.48107797123391227,-3.4134261922785742)--(3.680402725208175,-5.312354277062829)); draw((-0.48107797123391227,-3.4134261922785742)--(4.328420803502752,-0.05470445535767114), wwzzqq); draw((2.92,4.04)--(3.680402725208175,-5.312354277062829), wwzzqq); /* dots and labels */dot((2.92,4.04),dotstyle); label("$A$", (2.68,4.4), NE * labelscalefactor); dot((1.04,-0.08),dotstyle); label("$B$", (0.48,-0.28), NE * labelscalefactor); dot((6.24,-0.04),dotstyle); label("$C$", (6.44,-0.04), NE * labelscalefactor); dot((2.951579196497248,-0.06529554464232885),linewidth(3pt) + dotstyle); label("$H$", (2.84,-0.62), NE * labelscalefactor); dot((3.680402725208175,-5.312354277062829),linewidth(3pt) + dotstyle); label("$X$", (3.84,-5.72), NE * labelscalefactor); dot((4.328420803502752,-0.05470445535767114),linewidth(3pt) + dotstyle); label("$H'$", (4.32,0.18), NE * labelscalefactor); dot((1.2260256631944162,0.327673261894146),linewidth(3pt) + dotstyle); label("$Y$", (0.98,0.48), NE * labelscalefactor); dot((6.46639011001663,-0.3182143520686293),linewidth(3pt) + dotstyle); label("$Z$", (6.62,-0.76), NE * labelscalefactor); dot((-0.48107797123391227,-3.4134261922785742),linewidth(3pt) + dotstyle); label("$P$", (-0.8,-4.06), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Let the projection of $X$ on $AB$ be $P$ .

Now we observe that $\frac{CA}{CX} = \frac{b}{CX} = \frac{b \cos{C}}{CX \cos{C}} = \frac{BH'}{BX \cos{C}} = \frac{BH'}{BP}$
Also obviously, $\angle H'BP = \angle ACX$

Thus we have $\triangle H'BP \sim \triangle ACX$

Since $Y,H',X,P$ are concyclic,
Thus angle $\angle CXA = \angle BPH' = \angle H'XY$

That is, $XC,XY$ and $XH',XA$ are isogonal lines with respect to $\angle CXY$

By our lemma here
( $AY \cap CH' = B$ , $YH' \cap AC = Z$ ),
$XB$ and $XZ$ are also a pair of isogonal lines with respect to $\angle CXY$

Thus $\angle BXY = \angle ZXC$

hence proved.

This post has been edited 6 times. Last edited by utkarshgupta, Jun 20, 2017, 5:35 PM

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33 posts

#9 h Aug 15, 2019, 7:37 AM • 3 Y

Y by k12byda5h, Adventure10, Mango247

ATimo wrote:

$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$ . If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$ , intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$ , intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$ .

Let $A’$ be the point such that $AHMA’$ form a rectangle. Let $P, Q$ be the projection of $X$ on $AB, AC$ .

Claim : $H’$ is the orthocentre of $\triangle A’PQ$

Proof : Note that $A’H’MA$ forms a parallelogram. So, $A’H’ \perp PQ$ . Note that $M$ is the orthocentre of $APQ$ . So, by the converse of a well-known lemma, we have $H’$ to be the orthocentre of $\triangle A’PQ$ . $\square$

So, we have $\measuredangle{PH’Q} = 180 - \measuredangle{BAC}$
$\measuredangle{ZXY} = \measuredangle{H’QA} + \measuredangle{H’PA} = \measuredangle{PH’Q} - \measuredangle{BAC} = 180 - 2 \measuredangle{BAC} = \measuredangle{BXC}$ Thus, we have $\measuredangle{ZXC} = \measuredangle{YXB}$ . $\blacksquare$

This post has been edited 5 times. Last edited by MathDelicacy12, Aug 20, 2019, 2:32 PM

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#10 h Mar 26, 2020, 9:31 PM • 1 Y

Y by itslumi

Direct application of Moving Points. ( This solution is wrong in many ways, as I did not understand Moving Points very well, but I didnt find a button to remove it, so just ignore.)

Let's animate A around the circle $\omega$ (contains $B, C$ and is tangent to $XB, XC$ ).
First, let's define some points, let $K = YX \cap BC, K'$ is the reflection of K wrt. the midpoint $M$ of BC, and $O$ the center of $\omega$

The first transformation is $A \rightarrow H \rightarrow H' \rightarrow H'X \rightarrow H'Y \rightarrow Y \rightarrow K \rightarrow K' \rightarrow XZ_1 \rightarrow Z_1$ ,
with $XZ_1$ being the reflection of $XK'$ wrt. $XC$ , $Z_1 \in AC$ .

The second is $A \rightarrow H \rightarrow H' \rightarrow H'X \rightarrow HZ \rightarrow Z$ , so it suffices to prove that these transformations coincide for 3 points, ie. $Z = Z_1$ for three choices of $A$ .

1) Let $A$ be the second intersection of $XO$ and $\omega$ , so $H = M = H' \rightarrow Y=B, Z=C \rightarrow K=B \rightarrow K' = C \rightarrow Z_1 = C = Z$

2) Let $A = B$ , this gives us that $H' = C$ , implying that the line perpendicular to $H'X$ is, in fact, line $OC \rightarrow Z = C$ .
As line $AB$ is $XB \rightarrow K'=C \rightarrow Z_1=C=Z$

3) The case $A=C$ is analogous, so we are done :-D

:-D

This post has been edited 2 times. Last edited by fcomoreira, Jun 29, 2020, 1:23 PM
Reason: Wrong solution

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104 posts

#11 h Dec 25, 2020, 4:10 PM • 2 Y

Y by MarkBcc168, R8kt

Let $A'$ is the reflection of $A$ over the perpendicular bisector of $BC$ . BY DDIT, $\square BYZC$ , $(XB,XZ) , (XY,XC) , (XA,XH')$ are reciprocal pairs. So, It suffices to show that $\angle A'XC =\angle AXB = \angle ZXH'$ . Which is true since $Z$ the isogonal conjugate of $A'$ wrt. $\triangle XH'C$ . ( $\angle XCZ = \angle ABC = \angle A'CB$ and $\angle XH'Z = \angle AH'C = 90^{\circ}$ )

This post has been edited 1 time. Last edited by k12byda5h, Feb 17, 2021, 1:18 AM
Reason: Typo

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662 posts

#12 h Dec 9, 2023, 8:23 PM • 1 Y

Y by Rounak_iitr

Felt somewhat easy for a P6
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Let $E$ and $F$ be the feet of perpendicular from $X$ to $AB$ and $AC$
CLaim 1: $\triangle EBH' \sim \triangle XCA$
Proof
Now, $\measuredangle BXH'=\measuredangle BEH' = \measuredangle AXC$ , now by Isogonal Lines lemma we're done.

This post has been edited 1 time. Last edited by math_comb01, Dec 9, 2023, 8:24 PM

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#13 h Mar 13, 2024, 3:04 PM

Y by

andria wrote:

Too easy problem for problem 6
My solution:
Lemma1: let $R,T$ projections of $X$ on $AB,AC$ and assume that $M$ is midpoint of $BC$ then $M$ is orthocenter of $\triangle ART$ for prove see Russia 2013 grade 9.
Lemma2: in cyclic quadrilateral $ABCD$ points $M,N$ are midpoints of $AB$ and $CD$ and $AC\cap CD=P$ also $H_1,H_2$ are projections of $P$ on $AD,BC$ then $\triangle H_1MN=\triangle H_2MN$ .(it's well known)
Lemma3: in triangle $ABC$ , $AD,BE,CF$ are altitudes and $M,N,S,T$ are midpoints of $BC,CE,BF,EF$ and $AL$ is altitude in $\triangle AEF$ and let $R$ be the reflection of $L$ WRT midpoint of $EF$ then quadrilateral $NLRS$ is cyclic.
Prove: note that $R$ is foot of $H$ (orthocenter of $\triangle ABC$ ) on $EF$ .then using lemma 2 for the cyclic quadrilateral $BFEC$ we get that $\angle SRN=\angle SMN$ note that $\angle ADM=\angle MSA=90\longrightarrow AMDS$ is cyclic. So $\angle SMN=\angle 180-A$ so $\angle SRN=180-A$ also using lemma 1 we get that $T$ is orthocenter of $ASN$ so $\angle STN=180-A$ and the lemma proved.
BACK TO THE MAIN PROBLEM:
Let $M$ midpoint of $BC$ then from the lemma3,1 $RMH'T$ is cyclic and $M$ is orthocenter of $ARZ$ so $\angle RMT=\angle RH'T=180-A$ but because $XH'ZT,XH'YR$ are cyclic we get that $180-A=\angle RH'T=\angle XZT+\angle XYR=\angle A+\angle TXY\longrightarrow \angle ZXY=180-2A=\angle CXB\longrightarrow \angle CXZ=\angle YXB$
DONE

Very interesting that $RMH'T$ is cyclic is just All Russian MO 2015, grade 10, problem 7 https://artofproblemsolving.com/community/c6h1126020p5204821

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