Hard problem involving circumcenter and concurrent lines

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Hard problem involving circumcenter and concurrent lines

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Source: AQGO 2020 Problem 3

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#1 h Jun 20, 2020, 6:31 PM • 4 Y

Y by amar_04, AmirKhusrau, mueller.25, mijail

Let $\triangle{ABC}$ be a triangle with circumcenter $O$ . Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$ . Let $D$ be the projection of $A$ on $\overline{BC}$ . Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$ . Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$ . Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$ . Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix

This post has been edited 2 times. Last edited by GeoMetrix, Jun 20, 2020, 6:32 PM

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#2 h Jun 21, 2020, 1:09 AM • 6 Y

Y by amar_04, AmirKhusrau, mijail, Mango247, Mango247, Mango247

Applying inversion with centre $A,\mathrm {power}\left(AB\cdot\frac {AC}{2}\right),$ and reflection through the angle bisector of $BAC .$
We obtain

New Problem wrote:

Let $ABC$ be a triangle. Let $O$ be its projection. Let $M$ be the midpoint of $CA, N$ is the midpoint of $AB.$ $D$ is the projection of $A$ onto $BC.$ $E$ lies on $MN$ such that $AE$ is tangent to $(O).$ $(E,EA)$ ( is the image of $OT$ in the problem) intersects $\odot (ABC)$ again at $T, \odot (AOT)$ intersects $\odot (DMN)$ (image of $\odot (BOC)$ ) at $U,V$ respectively. $UV$ intersects $(O)$ at $X,Y, AX$ cuts $BC$ at $R, AY$ cuts $BC$ at $S.$ Then $\odot (ABC),AD,\odot (ARS)$ are concurrent.

Proof- Let $I$ be the midpoint of $BC\implies I$ lies on $\odot (DMN).$ $AD$ intersects $\odot(ABC)$ at $F ,ED$ cuts $OF$ at $H.$ $\widehat {AEH}=2\widehat {AED}=180^{\circ}-\widehat {AOF}\implies H$ lies on $\odot (AOT).$ But $\widehat{OID}=\widehat {OHD}=90^{\circ}.$ If $OF$ intersects $BC$ at $G.$ Then $GD\cdot GI=GH\cdot GO$ or $G$ lies on the radical axis of $\odot (DMN),$ which is $=UV.$ $G$ lies on $UV.$ Now, $\widehat {FGS}=\tfrac 12\widehat {AOF}=\widehat {AYF}=\widehat {SYF}\implies FYGS$ is a concyclic quadrilateral. $F$ is the Miquel point of the quadrilateral formed by $AY,XY,AX,BC.$ We conclude that $\odot (ASR)$ passes through the Miquel point of the quadrilateral.

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#3 h Jun 21, 2020, 7:47 AM • 2 Y

Y by amar_04, AmirKhusrau

GeoMetrix wrote:

Let $\triangle{ABC}$ be a triangle with circumcenter $O$ . Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$ . Let $D$ be the projection of $A$ on $\overline{BC}$ . Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$ . Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$ . Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$ . Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix

Solution. Let $A'$ be the point on $(ABC)$ such that $AA'\parallel BC,$ $A''$ be the midpoint of $AA',$ $RS\cap MN=J,$ tangent at $A$ to $(ABC)$ intersect $BC$ at $L_A.$ Let $AD$ intersect $(ABC)$ again at $H_A,$ $L_AH_A$ intersect $(ABC)$ again at $Z.$ Let $P$ be the midpoint of $AZ.$ By homothety at the centroid, it is clear that the centroid lies on $DT,$ so by IMO Shortlist 2011/G4, we have that $D,T,A'$ are collinear.
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Claim 1. $P,H_A\in(AUV).$
Proof. We prove that radical axis of $(AH_AP)$ and $(BOC)$ is $A'D,$ which would imply that $P,H_A\in(AUV).$ It is clear that $D$ lies on the radical axis. It suffices to show that $A'$ lies on the radical axis. Invert around $A,$ we obtain the following equivalent problem:

Quote:

Let $ABC$ be a triangle with circumcenter $O$ and the antipode of $A$ be $A_1,$ Let $\omega$ be the reflection of $(ABC)$ on $BC.$ $A'$ be the point on $(ABC)$ such that $AA'\parallel BC.$ Tangent at $A$ to $(ABC)$ cuts $BC$ at $L_A, BC\cap AO=F,BC\cap A'O=E,$ $Q$ be the reflection of $A$ on $E.$ Prove that $FQ$ is the radical axis of $\omega$ and $(L_AAA_1).$

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We work on the above problem, see that $F$ lies on the radical axis of both the circles. So it suffices to show that $Q$ lies on the radical axis. Without loss of generality assume that $\angle B\ge \angle C.$ Define the map $f:\mathbb{R}^2\to \mathbb{R}$ by $f(P)\overset{\text{def}}{=} \operatorname{Pow}_{\omega}(P)-\operatorname{Pow}_{(L_AAA_1)}(P).$ It is well known and easy to see that $f$ is linear. Also $f(F)=0,$ and it is clear that the midpoints of $BC$ and $EF$ is common, let it be $M,$ so, $f(E)=f(E)+f(F)=2f(M)=f(B)+f(C).$ Observe that
$f(Q)=2f(E)-f(A)=2f(B)+2f(C)-f(A).$ Let $G$ be the foot of $A_1$ on $BC.$ It is a simple exercise to prove that $f(A)=2bc\cos A.$ Note that $f(B)=BL_A\cdot BG$ and $f(C)=-CL_A\cdot CG.$ By Steiner ratio theorem we have that $\tfrac{BL_A}{CL_A}=\tfrac{c^2}{b^2},$ let $x\overset{\text{def}}{=}\tfrac{a}{b^2-c^2},$ then it follows that $BL_A=xc^2$ and $CL_A=xb^2.$ Therefore $f(B)=BL_A\cdot BG=xbc^2\cos C$ and similarly $f(C)=-xb^2c\cos B.$ We wish to prove that $f(Q)=0,$ so we want
$xbc(c\cos C-b\cos B)=bc\cos A\iff a(b\cos B-c\cos C)=(c^2-b^2)\cos A$ which is just a simple computation using cosine formula. $~~\square$

Claim 2. $AO,A'Z,BC$ are concurrent.
Proof. Let $A_1$ be the $A-$ antipode with respect to $(ABC).$ Let $AO\cap BC=K.$ Note that $\angle L_AZA=\angle AZH_A=\angle AKH_A=\angle AKD=\angle AKL_A,$ so $AL_AZK$ is cyclic, since $\angle A'ZH_A=90^{\circ}$ hence $A',K,Z$ are collinear. $~~\square$

Claim 3. $A'',J,P$ are collinear.
Proof. Notice that $P$ is the Miquel point of $ARJY$ since $P=(ARS)\cap (AXY).$ Hence $\angle YJP=\angle YSP=\angle AA''P.~\square$

Because of the above two claims and homothety at $A$ with scale $0.5,$ we have that $A,O,J$ are collinear, which is what we wanted. $~~\blacksquare$

This post has been edited 3 times. Last edited by ayan.nmath, Jun 21, 2020, 8:05 AM

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#4 h Jun 21, 2020, 8:18 AM

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I just have a small doubt , in which course will u study these type of problems , intermediate or inroductory ? I am currently in introductory to algebra B.

This post has been edited 2 times. Last edited by thin21, Jun 21, 2020, 8:41 AM
Reason: typo

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#5 h Jun 29, 2020, 6:44 PM

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This is a proof of the Claim 1 on "ayan.math" solution without inversion,Can someone check it?

Attachments:

Solution aqco.pdf (404kb)

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#6 h Aug 19, 2020, 7:09 PM • 1 Y

Y by GeoMetrix

Here goes the official solution by @enhanced

GeoMetrix wrote:

Let $\triangle{ABC}$ be a triangle with circumcenter $O$ . Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$ . Let $D$ be the projection of $A$ on $\overline{BC}$ . Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$ . Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$ . Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$ . Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix

Let $\Psi$ denote the homothety centered at $A$ with ratio $\frac{1}{2}$ . Now note that from 2011G4 and stuff we know that $\odot(KQR)$ is tangent to $\odot(ABC)$ now let the tangent to $\odot(ABC)$ at $A$ meet $BC$ at $T$ then by radical axis theorem on $\odot(ABC),\odot(AQR),\odot(KQR)$ we have that $\Psi(T)\in \odot(AKO)$ . Let $AO\cap BC=O_A$ note that $\Psi(O_A)A\cdot \Psi(O_A)O=\Psi(O_A)R\cdot \Psi(O_A)Q$ hence $\Psi(O_A)\in \ell$ . Now let $O_A^*$ be the reflection of $O_A$ in the midpoint of $BC$ , I claim that $O_A^*\in \ell$ . To prove this note that it suffices to show that line joining $N$ and the midpoint of $O\Psi(T)$ is perpendicular to $\Psi(O_A)O_A^*$ as $O\Psi(T)$ is the diameter of $\odot(AKO)$ . Now let $A^*$ be the antipode of $A$ in $\odot(ABC)$ and let $N^*$ be the reflection of $O$ in $BC$ and let $S$ be the midpoint of $TA^*$ now by a homothety at $A$ with ratio $2$ it suffices to show $SN^*\perp \Psi(O_A)O_A^*$ . Now finally let $S^*$ be the midpoint of $TO_A$ now since $\angle TAA^*=90$ and as $S^*\Psi(O_A)$ is the $T-$ midline of $\triangle O_ATA$ so we have $\angle S^*\Psi(O_A)O=90=\angle S^*PO$ hence $S^*\Psi(O_A)OP$ is cyclic . Again by similar mid-lines argument we have $S^*SO\Psi(O_A)$ is cyclic so $\Longrightarrow S^*SOP\Psi(O_A)$ is cyclic $\Longrightarrow \Psi(O_A)PS=90$ . Now $\Psi(O_A)(O_A,O_A^*;P,\infty_{BC})=-1=S(O,N^*;P,\infty_{\perp BC})$ now $\Psi(O_A)O_A\perp SO , \Psi(O_A)P\perp SP$ so this give $\Psi(O_A)O_A^*\perp SN^*$ so we have proven our claim .

Now we have shown that $O_A^*\in \ell$ that is $X,Y,O_A^*$ are collinear . Now let $K^*$ be the second intesection of $GD$ with $\odot(ABC)$ then $AK^*\parallel BC$ and the line $K^*H'$ is the reflection of $AO$ in the perpendicular bisector of $BC$ hence $K^* , H' , O_A^*$ are collinear hence there is an involution on $\odot(ABC)$ swapping $(B,C) ,(X,Y) , (K^* ,H')$ now projecting this involution on $BC$ through $A$ we have that $AH'$ is the radical axis of $\odot(ABC) ,\odot(AUV)$ hence $H'\in \odot(AUV)$ so we are done $\qquad \blacksquare$

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#7 h Yesterday at 1:21 PM

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$AD\cap (ABC)=F,W\in (ABC)$ where $AW\parallel BC$ . Let $A'$ be the antipode of $A$ and $H\in BC$ where $A'H\perp BC$ . Note that $F\in (AUV)$ since $DU.DV=DB.DC=DA.DF$ and $2OT=A'H=DF$ hence $D,T,W$ are collinear. It sufficies to show that $(\overline{AX},\overline{AY}),(\overline{AB},\overline{AC}),(\overline{AA},\overline{AO})$ is an involution. Apply $\sqrt{\frac{bc}{2}}$ inversion.

New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and $M,N$ are the midpoints of $AB,AC$ . Let $AD$ be the altitude which meets $(ABC)$ at $F$ . $W\in (ABC)$ such that $AW\parallel BC$ and $K\in MN$ such that $AK$ is tangent to $(ABC)$ . Let $(AKO)$ meet $(DMN)$ at $U,V$ and $UV$ intersects $(ABC)$ at $X,Y$ . Show that $XY,BC,FW$ concur.

Set $FW\cap BC=P$ .
$\measuredangle MKD+\measuredangle (PO,MN)=(\measuredangle B-\measuredangle C)+\measuredangle FPD=90$ Thus, $KD\perp OP$ . If $KD\cap OP=Z$ , then $Z$ lies on $(AOK)$ since $OK$ is diameter. So $D,Z,O,Q$ lie on the circle with diameter $DO$ . If $Q$ is the midpoint of $BC$ , we have $Pow(P,(DMN))=PD.PQ=PO.PZ=Pow(P,(AOK))$ so $P$ lies on the radical axis of these circles which is $UV$ as desired. $\blacksquare$

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