Symmedian perpendicular to Euler line

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Symmedian perpendicular to Euler line

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Source: 2019 CWMI P2

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179 posts

#1 h Aug 13, 2019, 6:15 AM • 2 Y

Y by rashah76, Adventure10

Let $O,H$ be the circumcenter and orthocenter of acute triangle $ABC$ with $AB\neq AC$ , respectively. Let $M$ be the midpoint of $BC$ and $K$ be the intersection of $AM$ and the circumcircle of $\triangle BHC$ , such that $M$ lies between $A$ and $K$ . Let $N$ be the intersection of $HK$ and $BC$ . Show that if $\angle BAM=\angle CAN$ , then $AN\perp OH$ .

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#2 h Aug 13, 2019, 8:03 AM • 1 Y

Y by Adventure10

Let $AD$ be diameter in the circumcircle of $ABC$ and let $AO \cap BC=E, AO=R, \angle A=\alpha$ . Let the tangents to the circumcircle of $ABC$ at points $B, C$ meet at $T$ . Because $AO$ and $AH$ , as well as $AM$ and $AN$ are isogonals we deduce that $\angle HAN= \angle MAO$ . On the other hand $OM \cdot OT=OC^2=OA^2$ and also $KH \| AD$ thus $\angle OMA= \angle OAT= \angle HNA$ . This means that the triangles $AOM$ and $AHM$ are similar so $NH/AH=MO/AO=\cos \alpha$ and $NH=AH \cos \alpha =2R \cos^2 \alpha$ . Note that by symmetry $ON=OE=OD-DE=OD-HN=R-2R \cos^2 \alpha$ . Now it remains to check that $OA^2+NH^2=ON^2+AH^2 \iff R^2+(2R \cos^2 \alpha)^2=(R-2R \cos^2 \alpha)^2+(2R \cos \alpha)^2$ , which is equivalent to the perpendicularity of $AN$ and $OH$ .

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670 posts

#3 h Aug 13, 2019, 11:46 AM • 2 Y

Y by Adventure10, Mango247

Here is a 100% synthetic solution.

We start by proving a lemma.

Lemma: The only points $K$ on line $AH$ with the property that $\angle{ABK}=\angle{ACK}$ are $K=A$ and $K=H.$
Proof: Let $B'$ be the symmetric of $B$ in line $AH.$ Then the angle condition translates as $K\in (AB'C).$ Hence $K$ takes at most two values. Since $A$ and $H$ clearly work, the lemma is proved.

Note that $K$ is the symmetric of $A$ wrt $M$ . Let $P=AO\cap BC$ , $T$ the intersection of the tangent in $K$ at $(BKC)$ with line $BC$ and $H'=AP\cap KT.$

Since the reflection of $H$ in $M$ lies on $AO$ , it's easy to see that $N$ and $P$ are symmetric wrt $M.$ $AN$ is the $A$ -symmedian in $\triangle{ABC}$ , so $KP$ is the $K$ -symmedian in $\triangle{KBC}.$ Therefore $(H'P,H'T ; H'B,H'C)=-1.$ But $\angle{PH'T}=90^0$ (if $O'$ is the circumcenter of $\triangle{KBC}$ , then $KT\perp KO'\parallel AO$ ), so $\angle{BH'A}=\angle{CH'A}.$

Consider the inversion in $A$ of radius $\sqrt{AB\cdot AC}$ composed with a reflection in the bisector of $\angle{BAC}.$ Denote by $H^*,K^*,M^*$ the images of $H',K,M$ .
$\angle{BH'A}=\angle{CH'A}$ gives $\angle{ABH^*}=\angle{ACH^*}$ and $H'\in AO$ gives $H^*\in AH.$ By the Lemma it follows that $H^*=H.$
Combining this with $\angle{AH'K}=90^0$ , we obtain $HK^*\perp AK^*.$ (1)
But $K^*$ is the midpoint of $\overline{AM^*}$ and $M^*$ is the second intersection of $AN$ with $(ABC)$ , meaning that $OK^*\perp AK^*$ . (2)

Finally, from (1) amd (2) we obtain $AN\perp OH.$ This ends our proof.

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241 posts

#4 h Aug 13, 2019, 12:14 PM • 5 Y

Y by Pluto1708, UzbekMathematician, Assassino9931, CabinetMatesha, Adventure10

I think it's much simpler

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Let $X$ be on $BC$ such that $AX$ touches $(ABC)$ . Clearly $OX \perp AN$ ( $AN$ is the symmedian and so the polar of $X$ ). The condition tells us $HN \parallel AO$ and so $\angle AHN= \angle MOA = 180^0- \angle AXN$ and $AH \perp XN$ , so $H$ is the orthocenter of the trinagle $\Delta AXN$ . Now we have $XH \perp AN$ , so $X,O,H$ are collinear and this line is perpendicular to $AN$ .

This post has been edited 1 time. Last edited by MilosMilicev, Aug 13, 2019, 12:18 PM

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#5 h Aug 13, 2019, 12:53 PM • 6 Y

Y by MilosMilicev, MilosMilicevonAoPS, AlastorMoody, amar_04, Adventure10, LLL2019

Wow, @MilosMilicev is so good at geo! No wonder he got all the girls at IMO.

This post has been edited 1 time. Last edited by huricane, Aug 13, 2019, 1:08 PM

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MilosMilicevonAoPS

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MilosMilicevonAoPS

#6 h Aug 13, 2019, 3:23 PM • 7 Y

Y by 62861, huricane, Pluto1708, Aryan-23, Adventure10, Mango247, LLL2019

Can confirm, @MilosMilicev got me at the IMO.

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#7 h Nov 17, 2019, 1:28 PM • 4 Y

Y by amar_04, RAMUGAUSS, RudraRockstar, Adventure10

Let $L$ be the midpoint of $A$ symmedian chord of $\triangle ABC$ . It is sufficient to prove that $\angle ALH=90$ .Let $AO \cap \odot(ABC)=R, AM \cap \odot(ABC)=Q, BB \cap CC=P$ and $AO \cap \odot(BOC)=G$ an $AL \cap \odot(ABC)=E$ . Now do a $\sqrt{bc}$ inversion at $A$ followed by reflection along the $\angle A$ bisector. Let $D$ be the foot of the perpendicular from $A$ to $BC$ . Note that $L \mapsto K, H \mapsto G, D \mapsto R, N \mapsto Q$ . It suffice to prove that $\angle AGK=90$ . We have given that $K,N,H$ collinear $\iff A,G,Q,L$ lie on a circle. Note that $AN.NE=BN.CN=HN.HK$ $\implies A,H,E,K$ is conclyc. So $M=LG \cap AQ$ . Hence $\angle AGL=\angle OGL=\angle OPL$ and $\angle LGQ=\angle LAR$ . Hence $\angle RGQ=\angle OPL+\angle QAL=\angle OMA=\angle RQA \implies R,G,K,Q$ lie on a circle. So $\angle AGK=\angle AQR=90$ .

This post has been edited 2 times. Last edited by L-.Lawliet, Nov 19, 2019, 8:20 AM

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#8 h Dec 26, 2019, 10:15 PM • 4 Y

Y by AlastorMoody, rashah76, lahmacun, Adventure10

I guess this solution has not been posted yet...
Reflect $N$ in $M$ to get point $P$ .Observe that the reflection of $H$ in $M$ is $A$ antipode and reflection of $K$ in $M$ is $A$ ,so we have that $A$ , $O$ and $P$ are collinear.Therefore $AD$ and $AP$ are isogonal in $\angle A$ .Since $AP$ is parallel to $KH$ ,we have $\angle HAN$ = $\angle HKA$ . $(1)$
Now let $AN$ intersect $(ABC)$ in $Q$ . $NQ\times NA=NB\times NC=NH\times NK$ .
So $HAKQ$ is cyclic.
Therefore $\angle HKA=\angle HQA$ $(2)$ .
From $(1)$ and $(2)$ ,we conclude that $HA$ = $HQ$ so $OH\perp AN$

This post has been edited 2 times. Last edited by mmathss, Dec 26, 2019, 10:21 PM

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Mahdi_Mashayekhi

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#9 h Jun 13, 2022, 12:04 PM

Y by

Note that it's well known that $K$ is reflection of $A$ across $M$ so $ABKC$ is parallelogram. Note that $\angle CKH = \angle CBH = \angle CAH$ and $\angle CAN = \angle BAM = \angle CKM$ so $\angle HAN = \angle AKH$ . Let $HM$ meet $ABC$ at $S$ Note that $HM = MS$ so $AHKS$ is parallelogram. Note that $\angle AOH = \angle AMB = \angle CMK$ and $\angle NAO = \angle NAM + \angle MAO = \angle NAM + \angle MKH = \angle NAM + \angle NAH = \angle MAH = \angle MKS$ and $KS || AH \perp BC$ so $\angle CMK + \angle MKS = \angle 90 \implies AN \perp OH$ .

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#10 h Oct 26, 2023, 8:08 PM

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Consider $(ABC)$ , $\Omega=(BHC)$ , $\omega=(H,AH)$ . Note that $N$ lies on radical axis of the first two circles. Note that $\Omega$ and $\omega$ intersect on the lines $AB, AC$ in the points $P, Q$ such that $PC=AC, QB=BA$ . $\triangle ABC \equiv \triangle AQP$ , so $AN$ is its $A$ -median, and $HK$ is a perpendicular bisector of $QP$ , so $N$ lies on the segment $PQ$ , hence it's radical center of $(ABC),\Omega,\omega$ , so $OH \perp AN$ .

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#11 h Dec 19, 2023, 8:35 AM

Y by

As said above it is well known that $K$ is reflection of $A$ across $M$ and $ABKC$ is a parallelogram.
Realize that the intersection of $AM$ and $(BHC)$ is the $A-Humpty$ point and to prove the statement is enough to show that $AN$ $\cap$ $HO$
is the $A - Dumpty$ point. $\angle CKH = \angle CBH = \angle CAH$ and $\angle CAN = \angle BAM = \angle CKM$ so $\angle HAN = \angle AKH$ . Let $AN$ intersect $(ABC)$ at $T$ . Note that by PoP in $(BHC)$ and $(ABC)$ we get that : $AN.NT = HN.NK$ so $AHTK$ is cyclic so $\angle HAN= \angle HKT=\angle AKH$ . so $HA=HT$ .Its also well known that the line $OA-DUMPTY$ is the perpendicular bisector of $AT$ . So our statement is proved.

This post has been edited 1 time. Last edited by gvccimac_08, Dec 19, 2023, 8:42 AM
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