Radical center lies on Euler line

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Source: 2017 Taiwan TST Round 2, Quiz 2, Problem 2

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2312 posts

#1 h Apr 26, 2017, 3:24 PM • 15 Y

Y by v_Enhance, Ankoganit, guangzhou-2015, anantmudgal09, baopbc, Gluncho, 62861, mihajlon, buratinogigle, ThisIsASentence, Generic_Username, FRaelya, enhanced, Gaussian_cyber, Adventure10

Given a $\triangle ABC$ and three points $D, E, F$ such that $DB = DC,$ $EC = EA,$ $FA = FB,$ $\measuredangle BDC = \measuredangle CEA = \measuredangle AFB.$ Let $\Omega_D$ be the circle with center $D$ passing through $B, C$ and similarly for $\Omega_E, \Omega_F.$ Prove that the radical center of $\Omega_D, \Omega_E, \Omega_F$ lies on the Euler line of $\triangle DEF.$

Proposed by Telv Cohl

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andria

#2 h Apr 26, 2017, 5:05 PM • 9 Y

Y by nikolapavlovic, Gluncho, Ankoganit, Lsway, FRaelya, mhq, doxuanlong15052000, Adventure10, Mango247

Lemma:

Let $D,E,F$ be three points in the plain of $\triangle ABC$ such that $\triangle DBC,\triangle EAC,\triangle FAB$ are isosceles similar triangles then $\triangle ABC,\triangle DEF$ have the same centroid.
Proof of the lemma

Back to the problem:

Let $\Omega_E\cap \Omega_F=X\ ,\ \Omega_E\cap \Omega_D=Y\ ,\ \Omega_D\cap \Omega_F=Z$ . Let $AX\cap \odot(BXC)=D'$ and $\angle BDC=\theta$ ; since $E,F$ are circumcenters of $AXC,AXB\Longrightarrow \angle CXD'=\angle BXD'=\frac{\theta}{2}\Longrightarrow D'C=D'B$ and $\angle BD'C=180^{\circ}-\theta$ hence $D'$ is orthocenter of $\triangle BDC$ . Let $M$ be the midpoint of $BC$ , $H$ the orthocenter of $DEF$ and $G$ be the centroid of $\triangle ABC$ . From the Lemma $G$ is also the centroid of $\triangle BDC$ so if $AM\cap DH=W$ :

$\frac{GH}{GR}=\frac{GW}{GA}=\frac{MW-MG}{GA}=\frac{3}{2}.\frac{MW}{MA}-\frac{1}{2}=\frac{3}{2}.\frac{MD}{MD'}-\frac{1}{2}=\frac{3\cot^2(\frac{\theta}{2})}{2}-\frac{1}{2}=\text{constant}$
Hence $R$ is fixed point which lies on $GH$ .
Similarly $BY$ and $CZ$ cut the Euler line of $DEF$ with the same ratio hence $R$ is radical center of $\Omega_D,\Omega_E,\Omega_F$ .

Q.E.D

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#4 h Apr 27, 2017, 4:16 AM • 2 Y

Y by doxuanlong15052000, Adventure10

My solution.

We see triangle $ABC$ and $DEF$ are orthologic so the perpendicular through $A,B,C$ to $EF,FD,DE$ , reps, are concurrent at $K$ which is the radical center of the pair of circles $\Omega_D, \Omega_E, \Omega_F$ . We shall prove that $K$ lies on Euler line of $DEF$ by using tripolar coordinates of Euler line, this is equivalent to

$(DE^2-DF^2)KD^2+(EF^2-ED^2)KE^2+(FD^2-FE^2)KF^2=0$

$\iff DE^2(KD^2-KE^2)+EF^2(KE^2-KF^2)+FD^2(KF^2-KD^2)=0$

$\iff DE^2(CD^2-CE^2)+EF^2(AE^2-AF^2)+FD^2(BF^2-BD^2)=0$

$\iff DE^2(x^2-y^2)+ EF^2(y^2-z^2)+FD^2(z^2-x^2)=0$

Let $\angle DBC=\angle DCB=\angle ECA=\angle EAC=\angle FAB=\angle FBA=\theta$ and $BC=a,CA=b,AB=c,$ $DB=DC=a/(2\cos\theta)=x,EC=EA=b/(2\cos\theta)=y,FA=FB=c/(2\cos\theta)=z$ apply law of cosine, $EF^2=y^2+z^2+2yz\cos(A+2\theta)$ . We need to prove that

$(x^2+y^2+2xy\cos(C+2\theta))(x^2-y^2)+(y^2+z^2+2yz\cos(A+2\theta))(y^2-z^2)+(z^2+x^2+2zx\cos(B+2\theta))(z^2-x^2)=0$

$\iff xy(x^2-y^2)\cos(C+2\theta)+yz(y^2-z^2)\cos(A+2\theta)+zx(z^2-x^2)\cos(B+2\theta)=0$

$\iff ab(a^2-b^2)\frac{\cos(C+2\theta)}{\cos^4\theta}+bc(b^2-c^2)\frac{\cos(A+2\theta)}{\cos^4\theta}+ca(c^2-a^2)\frac{\cos(C+2\theta)}{\cos^4\theta}=0$

$\iff ab(a^2-b^2)(\cos C\cos 2\theta-\sin C\sin 2\theta)+bc(b^2-c^2)(\cos A\cos 2\theta-\sin A\sin 2\theta)+ca(c^2-a^2)(\cos B\cos 2\theta-\sin B\sin 2\theta)=0$

$\iff \cos 2\theta((a^2-b^2)(a^2+b^2-c^2)+(b^2-c^2)(b^2+c^2-a^2)+(c^2-a^2)(c^2+a^2-b^2))-\sin 2\theta(\frac{abc}{R}(a^2-b^2+b^2-c^2+c^2-a^2))=0$

The last equality is true, we are done.

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Lsway

#5 h Apr 27, 2017, 11:44 AM • 2 Y

Y by Leooooo, Adventure10

My proof:
Lemma 1: Rt $\triangle ABC$ $\overset{-}{\sim}$ Rt $\triangle ADE$ . $AB,AD$ are hypotenuses. $M$ is the midpoint of $BD$ ,then $M$ is on the perpendicular bisector of $CE$ .
Proof for lemma 1: Let $F,G$ be the midpoint of $AB,AD$ , respectively. $MF= \frac{1}{2} DA=EG,MG=\frac{1}{2} BA=CF,\measuredangle (MG,CF)=2 \measuredangle ABC=2\measuredangle AED=\measuredangle (GE,MF)$ $\Longrightarrow \triangle MCF \overset{+}{\sim} \triangle EMG \Longrightarrow MC=ME.$
Lemma 2(Obviously): $\triangle ABC$ and $\triangle A_1B_1C_1$ are both orthologic with $\triangle DEF$ .Let $P$ be the orthologic center of $\triangle ABC$ WRT $\triangle DEF$ , $P_1$ be the orthologic center of $\triangle A_1B_1C_1$ WRT $\triangle DEF$ . Let $A_2\in AA_1,B_2\in BB_1,C_2\in CC_1$ such that $\frac{AA_2}{AA_1}=\frac{BB_2}{BB_1}=\frac{CC_2}{CC_1}.$
Then $\triangle A_2B_2C_2$ and $\triangle DEF$ are orthologic, the orthologic center $P_2$ of $\triangle A_2B_2C_2$ WRT $\triangle DEF$ is on $PP_1,$ meanwhile $\frac{PP_2}{P_1P_2} = \frac{AA_2}{A_1A_2}$ .
Lemma 3: the same as the lemma mentioned in post #2.

Back to the problem:
Define $O,G$ as the circumcenter and centroid of $\triangle DEF$ ,respectively.
Let $E_A \in AC,F_A \in AB$ such that $E_AE \perp AE,F_AF \perp AF$ , $A_2$ be the midpoint of $E_AF_A$ ,similarly define $B_2,C_2$ .From lemma 1 we can know $A_2$ is on the perpendicular bisector of $EF$ ,similarly for $B_2,C_2.$
Hence $\triangle A_2B_2C_2$ and $\triangle DEF$ are orthologic, $O$ is the the orthologic center of $\triangle A_2B_2C_2$ WRT $\triangle DEF$ .
Let $P$ be the orthologic center of $\triangle ABC$ WRT $\triangle DEF$ (scilicet the radical center of $\Omega_D, \Omega_E, \Omega_F$ ), $P_1$ be the orthologic center of $\triangle A_1B_1C_1$ WRT $\triangle DEF$ .
Obviously $A_2\in AA_1,B_2\in BB_1,C_2\in CC_1$ ,meanwhile $\frac{AA_2}{AA_1}=\frac{BB_2}{BB_1}=\frac{CC_2}{CC_1}.$
From lemma 2 we can know $O,P,P_1,$ are colinear,and obviouly $P,P_1,G$ are colinear $\Longrightarrow$ $P,P_1 \in$ the Euler line of $\triangle DEF.$

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#6 h Apr 28, 2017, 11:49 PM • 8 Y

Y by FRaelya, Ankoganit, doxuanlong15052000, mhq, enhanced, Gaussian_cyber, Adventure10, Mango247

Official solution :

Let $T$ be the second intersection of $\Omega_E, \Omega_F$ and let $H_D$ be the second intersection of $AT$ with $\odot (BTC).$ From $\measuredangle BCH_D = \measuredangle BTA = \tfrac{1}{2} \measuredangle BFA = \tfrac{1}{2} \measuredangle CDB$ we get $BH_D \perp CD.$ Similarly, $CH_D \perp BD,$ so $H_D$ is the orthocenter of $\triangle BDC.$ Analogously, if $H_E, H_F$ is the orthocenter of $\triangle CEA, \triangle AFB,$ respectively, $BH_E, CH_F$ is the radical axis of $(\Omega_F, \Omega_D),$ $(\Omega_D, \Omega_E),$ respectively, so the radical center $P$ of $\Omega_D, \Omega_E, \Omega_F$ lies on $AH_D, BH_E, CH_F.$

Let $X, Y, Z$ be the reflection of $D, E, F$ in $BC, CA, AB,$ respectively. From $\triangle BXC \stackrel{+}{\sim} \triangle BFA$ $\Longrightarrow$ $\triangle BCA \stackrel{+}{\sim} \triangle BXF,$ so $\tfrac{AE}{BF} = \tfrac{CA}{AB} = \tfrac{FX}{BF}$ $\Longrightarrow$ $AE = FX.$ Similarly, we get $AF$ $=$ $EX,$ so $AEXF$ is a parallelogram. Let $O_T$ be the circumcenter of $\triangle DEF$ and let $Q$ be the reflection of $P$ in $O_T,$ then from the discussion above we get $XQ \perp EF.$ Analogously, $YQ, ZQ$ is perpendicular to $FD, DE,$ respectively. Let $H_T$ be the orthocenter of $\triangle DEF.$ Notice $\tfrac{DX}{DH_D} = \tfrac{EY}{EH_E} = \tfrac{FZ}{FH_F},$ so $H_T, P, Q$ are collinear. i.e. $P$ lies on the Euler line $O_TH_T$ of $\triangle DEF.$ $\blacksquare$

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#7 h Sep 23, 2017, 8:29 AM • 1 Y

Y by Adventure10

$\textbf{Proof :}$

$\textbf{Lemma :}$

Given a $\triangle ABC$ and three points $D, E, F$ such that $DB = DC,$ $EC = EA,$ $FA = FB,$ $\measuredangle BDC = \measuredangle CEA = \measuredangle AFB$ . Let $D'$ be reflection of $D$ wrt $BC$ and $D_1$ be the circumcenter of $(D'BC)$ . Then $D_1$ lies on the perpendicular bisector of $EF$ .

Now as we know triangles $ABC$ and $DEF$ have same centroid, so it is enough to prove that the lines through the midpoints of $BC, CA, AB$ which are perpendicular to sides of $DEF$ meet at point $X$ on the Euler line of $DEF$ . Consider point $D_1$ as in lemma. Similarly define $E_1, F_1$ . Let $D'$ be midpoint of $BC$ , similarly define $E', F'$ . We get that $\frac{DD'}{DD_1} = \frac{EE'}{EE_1} = \frac{FF'}{FF_1}$ , so $X$ lie on the line between the orthocenter of $DEF$ and the circumcenter of $DEF$ . Done

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#8 h Sep 23, 2017, 10:25 AM • 2 Y

Y by Adventure10, Mango247

andria wrote:

Lemma:

Let $D,E,F$ be three points in the plain of $\triangle ABC$ such that $\triangle DBC,\triangle EAC,\triangle FAB$ are isosceles similar triangles then $\triangle ABC,\triangle DEF$ have the same centroid.
Proof of the lemma

Back to the problem:

Let $\Omega_E\cap \Omega_F=X\ ,\ \Omega_E\cap \Omega_D=Y\ ,\ \Omega_D\cap \Omega_F=Z$ . Let $AX\cap \odot(BXC)=D'$ and $\angle BDC=\theta$ ; since $E,F$ are circumcenters of $AXC,AXB\Longrightarrow \angle CXD'=\angle BXD'=\frac{\theta}{2}\Longrightarrow D'C=D'B$ and $\angle BD'C=180^{\circ}-\theta$ hence $D'$ is orthocenter of $\triangle BDC$ . Let $M$ be the midpoint of $BC$ , $H$ the orthocenter of $DEF$ and $G$ be the centroid of $\triangle ABC$ . From the Lemma $G$ is also the centroid of $\triangle BDC$ so if $AM\cap DH=W$ :

$\frac{GH}{GR}=\frac{GW}{GA}=\frac{MW-MG}{GA}=\frac{3}{2}.\frac{MW}{MA}-\frac{1}{2}=\frac{3}{2}.\frac{MD}{MD'}-\frac{1}{2}=\frac{3\cot^2(\frac{\theta}{2})}{2}-\frac{1}{2}=\text{constant}$
Hence $R$ is fixed point which lies on $GH$ .
Similarly $BY$ and $CZ$ cut the Euler line of $DEF$ with the same ratio hence $R$ is radical center of $\Omega_D,\Omega_E,\Omega_F$ .

Q.E.D

I think the diagram is wrong... $AERF$ in diagram doesn't make parallelogram

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#11 h Sep 5, 2020, 8:59 PM • 3 Y

Y by Gaussian_cyber, Abhaysingh2003, Mango247

Here is a Different Proof.

TelvCohl wrote:

Given a $\triangle ABC$ and three points $D, E, F$ such that $DB = DC,$ $EC = EA,$ $FA = FB,$ $\measuredangle BDC = \measuredangle CEA = \measuredangle AFB.$ Let $\Omega_D$ be the circle with center $D$ passing through $B, C$ and similarly for $\Omega_E, \Omega_F.$ Prove that the radical center of $\Omega_D, \Omega_E, \Omega_F$ lies on the Euler line of $\triangle DEF.$

Proposed by Telv Cohl

$\textbf{LEMMA 1:-}(\text{Well Known})$ $ABC$ be a triangle and $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$ . Then $\Delta ABC$ and $\Delta DEF$ have the same centroid $G$ .

Proof:- Notice that $\Delta BDC\stackrel{-}{\sim}\Delta CEA\stackrel{-}{\sim}\Delta AFB$ . Let $D^*$ be the reflection of $D$ over $BC$ . Then $\Delta CD^*B\stackrel{+}{\sim}\Delta CEA\implies\Delta CED^*\stackrel{+}{\sim}\Delta CAB$ . So, $\measuredangle FAE=A+2\measuredangle CAE$ also $\measuredangle AED^*=\pi-2\measuredangle CAE-A\implies\overline{AF}\|\overline{D^*E}$ . Similarly we get $\overline{AE}\|\overline{D^*F}$ . Hence, $D^*$ is the reflection of $A$ over midpoint of $EF$ . Let $\{T,K\}$ be the midpoints of $BC,EF$ repectively, So clearly if $\overline{DK}\cap\overline{AT}=G$ . Then $\frac{GD}{GK}=\frac{GA}{GT}=2\implies\{G\}$ is the Centroid of $\Delta ABC$ and $\Delta DEF$ . $\qquad$ $\blacksquare$

$\textbf{LEMMA 2:-}$ Given two (not homothetic) triangles $\Delta ABC$ and $\Delta DEF$ . Let $P$ be a point such that the line passing through $D,$ $E,$ $F$ and parallel to $\overline{AP},$ $\overline{BP},$ $\overline{CP}$ , respectively are concurrent. Then $P$ lies on a circumconic of $\triangle ABC$ or the line at infinity.
Proof:- See Lemma 1 here(#9) $\qquad$ $\blacksquare$

$\textbf{LEMMA 3:-}$ . $ABC$ be a triangle and $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$ . If $\{H_A\}$ is the orthocenter of $\Delta BDC$ . Then $\overline{AH_A}\perp\overline{EF}$ .

Proof- Notice that $\{\Delta DEF,\Delta ABC\}$ are Orthologic Triangles. Let $X$ be the Orthology Center of $\Delta ABC,\Delta DEF$ except $O$ where $O$ is the circumcenter of $\Delta ABC$ . Let $Y$ be the Perspector of $\{\Delta ABC,\Delta DEF\}$ . Then from $\textbf{LEMMA 2}$ we get that $\{A,B,C,X,Y\}$ lie on a Conic and it's well known that $X\in\mathcal{K}$ which is the Kiepert Hyperbola of $\Delta ABC$ . Let $\measuredangle DBC=\theta$ . Now by Sondat's Theorem we get that $O,X,Y$ are collinear. So from (2) here(#5) we get that $Y$ is $K(90^\circ-\theta)$ of $\Delta ABC\implies\overline{AY}$ passes through $H_A$ . Hence, $\overline{AH_A}\perp\overline{EF}$ . $\qquad$ $\blacksquare$
____________________________________________________________________________________________

Coming back to the Problem. Let $H$ be the Orthocenter of $\Delta DEF$ . So from $\textbf{LEMMA 1}$ we get that $\overline{HG}$ is the Euler Line $(\mathcal E)$ of $\Delta DEF$ and define $\{H_B,H_C\}$ as the Orthocenters of $\Delta ACE,ABF$ respectively. Then $\Delta BDC\cup H_A\stackrel{-}{\sim}\Delta CEA\cup H_B\stackrel{-}{\sim}\Delta AFB\cup H_C (\bigstar)$ . Let $\overline{HD}\cap\overline{AG}=\{V\}$ and $\overline{AH_A}\cap\mathcal E=\{U\}$ . So, $\frac{GV}{GA}=\frac{GT-TV}{\frac{2}{3}TA}=\frac{1}{2}-\frac{TV}{TA}=\frac{1}{2}-\frac{TD}{TH_A}$ and clearly this is a Constant $\mathcal P$ from $(\bigstar)$ . Also clearly the Radical Center of $\Omega_D,\Omega_E,\Omega_F$ is the Orthology Center $X$ of $\Delta ABC$ . So, $\overline{AH_A}\cap\overline{BH_B}\cap\overline{CH_C}=X\in\mathcal{E}$ . $\qquad$ $\blacksquare$

Remark:- Here goes a more General Result, which I have not proved yet.

GENERALIZATION:- $ABC$ be a triangle and $P$ be an arbitary point on the Kiepert Hyperbola $\mathcal K$ of $\Delta ABC$ . Let $\Delta DEF$ be a Kiepert Triangle of $\Delta ABC$ . Then the Parallel Lines from $D,E,F$ to $\overline{AP},\overline{BP},\overline{CP}$ councurs at a point $Q$ and also $Q$ lies on the Kiepert Hyperbola $\mathcal{K}_1$ of $\Delta DEF$ and $\overline{PQ}$ passes through the Centroid $G$ of $\Delta ABC$ .

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