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Source: Iranian TST 2019, third exam day 2, problem 4

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164 posts

#1 h Apr 15, 2019, 12:01 PM • 5 Y

Y by anantmudgal09, Mathuzb, Number_Ninjas_2003, Adventure10, Mango247

Given an acute-angled triangle $ABC$ with orthocenter $H$ . Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$ . Prove that $AH$ is the external bisector of $\angle XHY$ .

Proposed by Mohammad Javad Shabani

This post has been edited 1 time. Last edited by Dadgarnia, Apr 15, 2019, 7:42 PM

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3042 posts

#2 h Apr 15, 2019, 1:17 PM • 3 Y

Y by Number_Ninjas_2003, Adventure10, Mango247

Dadgarnia wrote:

Given a triangle $ABC$ with orthocenter $H$ . Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$ . Prove that $AH$ is the external bisector of $\angle XHY$ .

Proposed by Mohammad Javad Shabani

Hopefully correct.

Solution

This post has been edited 1 time. Last edited by TheDarkPrince, Apr 15, 2019, 1:31 PM

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2125 posts

#3 h Apr 15, 2019, 3:18 PM • 2 Y

Y by Number_Ninjas_2003, Adventure10

Iran TST #3 2019 P4 wrote:

Given a triangle $ABC$ with orthocenter $H$ . Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$ . Prove that $AH$ is the external bisector of $\angle XHY$ .

Solution: Denote, $\omega_{N}$ as the nine-point circle WRT $\Delta ABC$ , $\omega_{N}'$ as its reflection over $AH$ and $X', Y'$ as the reflections of $X, Y$ over $AH$ . Hence, $X', Y'$ $\in$ $\omega_{N}$ . and $X, X' , Y, Y'$ are concyclic points. Also reflection of $H$ over $Y'$ be $H_{1}'$ lies on $\odot (ABC)$

Now, Let $HH_1' \cap \odot (ABC)=T_1$ and $HY \cap \odot (ABC)=T_2$ . Let $M'$ be the midpoint of $HT_2$ $\implies$ $M'$ $\in$ $\omega_{N}$

$HY \cdot HT_2 = HH_1' \cdot HT_1 \implies HY \cdot 2 HM' = 2 HY' \cdot HT_1 \implies HM'=HT_1$ Also, $HY \cdot HM' = HY' \cdot HT_1 \implies Y, Y', M', T_1 \text{ concyclic}$ $HD$ bisects $\angle YHY'$ $\implies$ $HD$ bisects $\angle T_1HM'$ , Hence, $T_1$ is the reflection of $M'$ over $AH$ $\implies$ $T_1$ $\equiv$ $\omega_{N}'$ $\cap$ $\odot (ABC)$ $\equiv X$

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102 posts

#5 h Apr 15, 2019, 3:41 PM • 9 Y

Y by Number_Ninjas_2003, ultralako, Aryan-23, tapir1729, AlastorMoody, Idio-logy, amar_04, Adventure10, Mango247

Let $\Gamma$ be the circumcircle and $\omega_9$ be the $9$ -point circle, and let $\Gamma'$ , $\omega_9'$ be their reflection in $AH$ . Considering negative inversion with center $H$ taking $\Gamma$ to $\omega_9$ , by symmetry it also takes $\Gamma’$ to $\omega_9'$ , so $X$ goes to one of the intersections of $\omega_9$ and $\Gamma'$ , say $Y'$ , which is symmetric to $Y$ in $AH$ . Now the conclusion is obvious.

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#6 h Apr 15, 2019, 9:58 PM • 3 Y

Y by Number_Ninjas_2003, Adventure10, Mango247

Dadgarnia wrote:

Given an acute-angled triangle $ABC$ with orthocenter $H$ . Reflection of nine-point circle about $AH$ intersects circumcircle at points $X$ and $Y$ . Prove that $AH$ is the external bisector of $\angle XHY$ .

Proposed by Mohammad Javad Shabani

Redefine $X$ and $Y$ by reflecting $\odot(ABC)$ in $A$ -altitude and intersecting with the nine-point circle; noting that it still suffices to show $AH$ bisects angle $XHY$ . Suppose $G$ lies on $AD$ with $\frac{AG}{DG}=2$ , and $N$ is the center of the nine-point circle, $O$ the circumcenter and $M$ the centroid of $\triangle ABC$ . Observe that $(HM;NO)=-1$ so $\angle HGM=90^{\circ}$ yields that $GH$ bisects angle $OGN$ ; so $G$ lies on the perpendicular bisector of $XY$ . Let $XY$ meet $AH$ at $Z$ and $\odot(DEF)$ meet $AH$ at $K$ . Reflect $H$ in $BC$ to get $H_A$ .

Note that $r^2=ZX \cdot ZY=ZA \cdot ZH_A=ZK \cdot ZD$ . So it is sufficient to show that $r^2=ZG \cdot ZH$ , or equivalently, $(AG;HD)=(H_AH;GK)$ . This is a direct computation, notably using $HD=DH_A, AG=2DG, AH=2AK$ , so I'll leave it at that.

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1595 posts

#7 h Jan 6, 2020, 12:31 PM • 3 Y

Y by Idio-logy, Adventure10, Mango247

Let $D,E,F$ denote foot of altitude from $A,B,C$ to $BC$ , $CA$ , $AB$ . Let $P_A = EF\cap BC$ and similarly $P_B = DF\cap AC$ , $P_C=DE\cap AB$ . Let $G$ be the centroid, $N$ be the nine-point center, $O$ be the circumcenter of $\triangle ABC$ . Finally, let $M$ be the midpoint of $BC$ . We first prove the following lemma.

Lemma: $\odot(ABC)$ , $\odot(DEF)$ and $\odot(HG)$ are coaxal with radical axis $\overline{P_AP_BP_C}$ .

Proof: It suffices to show that $P_A$ lie on the radical axis. First, note that $P_AB\cdot P_AC = P_AE\cdot P_AF$ so the first two circles are done. To see the third circle, recall that the projection $K_A$ of $H$ on to $AM$ lie on both $\odot(HG)$ and $\odot(BHC)$ . Thus this power is also equal to $PK_A\cdot PH$ , done. $\blacksquare$

Back to the main problem. Let $P$ be the projection of $G$ onto $AH$ and let $O'$ be the reflection of $O$ across $AH$ . Notice that $P$ is the centroid of $\triangle HOO'$ . Therefore if $N'$ is the reflection of $N$ across $AH$ , then $P\in ON'$ . Therefore $PX=PY$ . However, if $T=\overline{P_AP_BP_C}\cap AH$ , then by the lemma, $TX\cdot TY = TH\cdot TP$ so $X,Y,H,P$ are concyclic. Hence we are done.

This post has been edited 1 time. Last edited by MarkBcc168, Jan 6, 2020, 2:37 PM

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994 posts

khina

#8 h Feb 28, 2020, 6:04 PM • 2 Y

Y by AlastorMoody, Mango247

whoa these solutions are so overcomplicated

different solution

This post has been edited 1 time. Last edited by khina, Feb 28, 2020, 6:05 PM
Reason: Uwu

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1740 posts

#9 h Mar 21, 2020, 8:24 PM • 2 Y

Y by rashah76, SPHS1234

Solved with eisirrational, goodbear, Th3Numb3rThr33.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ size(7cm); defaultpen(fontsize(10pt)); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-2.7610213408027877,5.933356659746497)--(-5.28,1.09)--(2.46,1.09)--cycle, rvwvcq); /* draw figures */ draw((-2.7610213408027877,5.933356659746497)--(-5.28,1.09), rvwvcq); draw((-5.28,1.09)--(2.46,1.09), rvwvcq); draw((2.46,1.09)--(-2.7610213408027877,5.933356659746497), rvwvcq); draw(circle((-1.41,2.15397929378948), 4.0135958861864465), wrwrwr); draw(circle((-2.085510670401394,2.979688682978509), 2.0067979430932237), wrwrwr); draw(circle((-4.112042681605576,2.15397929378948), 4.0135958861864465), wrwrwr); draw(circle((-3.4365320112041817,2.979688682978509), 2.0067979430932237), wrwrwr); draw((-4.5578028381004945,4.64401729150342)--(-0.12037089975989268,2.5729164294431177), wrwrwr); draw((-5.4016717818456845,2.572916429443112)--(-0.9642398435050779,4.644017291503414), wrwrwr); draw(circle((-2.3842613779589934,3.5116783298732486), 2.45081088682496), linetype("2 2") + wrwrwr); draw(circle((-3.1377813036465847,3.5116783298732486), 2.4508108868249603), linetype("2 2") + wrwrwr); draw((-2.7610213408027877,5.933356659746497)--(-2.7610213408027877,-1.6253980721675363), wrwrwr); /* dots and labels */ dot("$A$",(-2.7610213408027877,5.933356659746497),N); dot("$B$",(-5.28,1.09),SW); dot("$C$",(2.46,1.09),SE); dot("$H$",(-2.7610213408027877,3.8053980721675362),dir(300)); dot("$D$",(-2.7610213408027877,1.09),dir(-60)); dot("$Y'$",(-0.9642398435050779,4.644017291503414),NE); dot("$X'$",(-0.12037089975989268,2.5729164294431177),SE); dot("$X$",(-4.5578028381004945,4.64401729150342),NW); dot("$Y$",(-5.4016717818456845,2.572916429443112),W); dot("$H_A$",(-2.7610213408027877,-1.6253980721675363),S); /* end of picture */ [/asy]$

Let $\Gamma$ be the circumcircle, $\omega$ the nine-point circle, and $\Gamma'$ and $\omega'$ their respective reflections across $\overline{AH}$ . Also let $\Psi$ denote negative inversion at $H$ with radius $\sqrt{AH\cdot HD}$ .

Note that:

$\Psi$ swaps $\Gamma$ and $\omega$ ;
$\Psi$ swaps $\Gamma'$ and $\omega'$ .

Let $\Psi$ send $X$ and $Y$ to $X'$ and $Y'$ , respectively. Then $X'$ and $Y'$ are the intersections of $\Gamma'$ and $\omega$ . Thus $X$ and $Y$ are the reflections of $Y'$ and $X'$ across $\overline{AH}$ , so $XYX'Y'$ is an isosceles trapezoid and $\overline{AH}$ bisects $\angle XHY$ . End proof.

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626 posts

#11 h Jun 12, 2020, 3:32 PM

Y by

The problem is just a shadow of the following:

Claim: Let $\ell$ be a line through the exsimilicenter $H$ of circles $\omega, \gamma$ . Let the reflection $\gamma'$ of $\gamma$ across $\ell$ intersect $\omega$ at $X,Y$ ; then $\ell$ is the external angle bisector of $\angle XHY$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.643470659725064, xmax = 3.651868867601753, ymin = -2.191103221604837, ymax = 1.9984781363025583; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(2)); draw((xmin, -3.4494843194342395*xmin-1.743302900012535)--(xmax, -3.4494843194342395*xmax-1.743302900012535), linewidth(2)); /* line */ draw(circle((-0.30322843739700656,0), 0.4), linewidth(2)); draw(circle((-0.6761890095104617,-0.10812067473744177), 0.4), linewidth(2)); draw(circle((-0.9324014302836379,-0.2703016868436044), 1), linewidth(2)); draw((-0.9901987764884226,0.13966525352008985)--(0.06186946814305705,-0.16341211512693504), linewidth(2)); draw((-0.17043406540790262,0.3773137351966124)--(-0.8972773630680387,-0.44146725102284434), linewidth(2)); /* dots and labels */ label("$\omega$", (-0.19772028586812995,1.0552739562501081), NE * labelscalefactor); dot((-0.505380728995011,0),dotstyle); label("$H$", (-0.48287503797701015,0.05723232386902697), NE * labelscalefactor); dot((-0.7096344581992715,0.7045700360760642),dotstyle); label("$\gamma$", (-0.5102937641413255,0.4301269997037166), NE * labelscalefactor); dot((-0.9901987764884226,0.13966525352008985),linewidth(4pt) + dotstyle); label("$X$", (-1.0860870135919491,0.13948850236197322), NE * labelscalefactor); dot((-0.8972773630680387,-0.44146725102284434),linewidth(4pt) + dotstyle); label("$Y$", (-0.9928633446332767,-0.5569471422116383), NE * labelscalefactor); dot((-0.17043406540790262,0.3773137351966124),linewidth(4pt) + dotstyle); label("$X'$", (-0.1483665787723622,0.41915950923799045), NE * labelscalefactor); dot((0.06186946814305705,-0.16341211512693504),linewidth(4pt) + dotstyle); label("$Y'$", (0.10388570193933952,-0.20598744730840102), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Proof: Let the reflections of $\omega, X,Y$ across $\ell$ be $\omega', X', Y'$ . Consider the inversion at $H$ swapping $\omega$ and $\gamma$ ; then this inversion must also swap $\omega'$ and $\gamma'$ . It follows that $X$ is swapped with $Y'$ , and the desired result is immediate.

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Eka01

#12 h Nov 21, 2024, 11:09 AM

Y by

Let $X'$ and $Y'$ be the reflection of $X$ and $Y$ about $AH$ . By $\sqrt{-HA.HD}$ inversion, $X$ is sent to $Y'$ and $Y$ is sent to $X'$ and oobviously $XX'Y'Y$ is an isoceles trapezoid so $H$ is the intersection of diagonals and $AH$ is the perpendicular bisector of parallel sides so the conclusion follows.

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CatinoBarbaraCombinatoric

107 posts

CatinoBarbaraCombinatoric

#13 h Mar 8, 2025, 5:13 PM

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Let $\omega$ the circumcircle of $ABC$ , $\omega_1$ the reflection of the nine-point circle respect $AH$ . Let $O$ and $O_1$ the center of this circle.
Notice that the equation $pow_{\omega}(P)=4 pow_{\omega_1}(P)$ describe a circle. $X,Y,H$ all satisfy the equation. So this circle is $(XYH)$ .
Let $Z$ be the midpoint of minor arc $XY$ of $(XYH)$ . $Z$ is on line $OO_1$ and is such that $pow_{\omega}(Z)=4 pow_{\omega_1}(Z)$
$OZ^2-R^2=4(O_1Z^2-\frac{R^2}{4})$
Because the radius of $\omega_1$ is half of $\omega$ .
So $ZO=2ZO_1$
$Z$ is the reflection of $O$ respect $O_1$ .
Let $P'$ be the projection of $P$ on $AH$ for every $P$ .
$O_1'$ is the midpoint of $O'H$ because it's the projection also of the center of nine-point circle which is midpoint $OH$ .
So $Z'=H$ and $HZ$ is perpendicular to $AH$ . But $HZ$ is angle bisector of $XHY$ .

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