Coaxal circles

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Coaxal circles

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9775 posts

jayme

#1 h Oct 13, 2014, 12:24 PM • 3 Y

Y by Lsway, Adventure10, Mango247

Dear Mathlinkers,

1. ABC a triangle
2. P a point
3. A’B’C’ the P-cevian triangle
4. Q the pivot point wrt ABC and A’, B’, C’
5. (1), (2), (3) the circumcircles wrt to AA’Q, BB’Q, CC’Q.

Prove : (1), (2) and (3) are coaxal.

Sincerely
Jean-Louis

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#2 h Oct 13, 2014, 12:40 PM • 2 Y

Y by Adventure10, Mango247

Can you explain please, what is exactly Q?

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9775 posts

jayme

#3 h Oct 13, 2014, 12:44 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
Q is the point of concurs of the circles (AB'C'), (BC'A') and (CA'B')...

Sincerely
Jean-Louis

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4146 posts

#4 h Oct 28, 2014, 4:49 AM • 4 Y

Y by shapi, Lsway, Adventure10, Mango247

Let $\tau_a,\tau_b,\tau_c$ denote the radical axes of the circumcircle $(O)$ of $\triangle ABC$ with $\odot(AB'C'),$ $\odot(BC'A'),$ $\odot(CA'B'),$ resp. $\tau_b,\tau_c$ and $QA'$ are pairwise radical axes of $(O),\odot(BC'A')$ and $\odot(CA'B')$ concurring at their radical center $X$ and similarly we have $Y \equiv \tau_c \cap \tau_a \cap QB'$ and $Z \equiv \tau_a \cap \tau_b\cap QC'.$ Hence if $AX,BY,CZ$ cut $(O)$ again at $U,V,W,$ we have $XU \cdot XA=XQ \cdot XA'$ $\Longrightarrow$ $U \in \odot(AQA')$ and similarly $V \in \odot(BQB')$ and $W \in \odot(CQC').$

Since $AA',BB',CC'$ concur at $P$ and $XA',YB',ZC'$ concur at $Q,$ then by Cevian Nest Theorem, $AX \equiv AU,$ $BY \equiv BV$ and $CZ \equiv CW$ concur at a point $R.$ Thus, $\odot(AQA'),$ $\odot(BQB')$ and $\odot(CQC')$ are coaxal with common radical axis $QR.$

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#5 h Oct 20, 2015, 8:49 PM • 6 Y

Y by mineiraojose, baladin, Lsway, enhanced, Adventure10, Mango247

We prove the stronger result as following :

Given a $\triangle ABC$ and a point $P$ . Let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle ABC$ and let $Q$ be the Miquel point of $D,$ $E,$ $F$ WRT $\triangle ABC$ . Then $\odot (AQD),$ $\odot (BQE),$ $\odot (CQF)$ are coaxial and the second intersection of these three circles is the isogonal conjugate (WRT $\triangle ABC$ ) of the antigonal conjugate of $P$ WRT $\triangle ABC$ .

Proof : Let $X$ be the Miquel point of the complete quadrilateral $\{CA, AB, BP, CP\}$ (define $Y,$ $Z$ similarly). Since $X$ is the center of the spiral similarity of $BF$ $\mapsto$ $EC$ and $FA$ $\mapsto$ $PE$ , so we get $XB$ $\cdot$ $XC$ $=$ $XE$ $\cdot$ $XF$ $=$ $XA$ $\cdot$ $XP$ and $\angle CXB,$ $\angle EXF,$ $\angle PXA$ share the same angle bisector $\ell$ , hence if $\Psi_X$ is the Inversion with center $X$ and factor $XB$ $\cdot$ $XC$ followed by the reflection in $\ell$ then $(B,C),$ $(E,F),$ $(A,P)$ is the image of each other under $\Psi_X$ . From $Y$ $\in$ $\odot (APF)$ and $\odot (BCF)$ we get $\Psi_X(Y)$ $\in$ $\odot (PAE)$ and $\odot (CBE)$ , so $\Psi_X(Y)$ coincide with $Z$ . i.e. $\Psi_X$ also maps $Y,$ $Z$ to each other

Let $W$ $\equiv$ $\odot (CAY)$ $\cap$ $\odot (ABZ)$ . Since $\odot (CAY),$ $\odot (ABZ)$ is the image of $\odot (BPZ),$ $\odot (PCY)$ , respectively under $\Psi_X$ , so $W$ is the image of $D$ under $\Psi_X$ $\Longrightarrow$ $\triangle XDC$ $\stackrel{+}{\sim}$ $\triangle XBW$ , hence we get $\measuredangle XCB$ $=$ $\measuredangle XWB$ $\Longrightarrow$ $W$ lies on $\odot (BCX)$ . Furthermore, from $\triangle XDP$ $\stackrel{+}{\sim}$ $\triangle XAW$ we get $\measuredangle XPA$ $=$ $\measuredangle XWA$ , so $W$ $\in$ $\odot (APX)$ . Analogously, we can prove $W$ lies on $\odot (BPY)$ and $\odot (CPZ)$ , so we conclude that $\odot (BCX),$ $\odot (CAY),$ $\odot (ABZ),$ $\odot (APX),$ $\odot (BPY),$ $\odot (CPZ)$ have a common point $W$ . ...... $(\bigstar)$

Let $\Phi_W$ be the Inversion with center $W$ . From $(\bigstar)$ we get $\triangle \Phi_W(X)\Phi_W(Y)\Phi_W(Z)$ is the cevian triangle of $\Phi_W(P)$ WRT $\triangle \Phi_W(A)\Phi_W(B)\Phi_W(C)$ . Since $D$ lies on $\odot (CAZ)$ and $\odot (ABY)$ , so $\Phi_W(D)$ lies on $\odot (\Phi_W(C)\Phi_W(A)\Phi_W(Z))$ and $\odot (\Phi_W(A)\Phi_W(B)\Phi_W(Y))$ , hence $\Phi_W(D)$ is the Miquel point of the complete quadrilateral with the sides $\Phi_W(C)\Phi_W(A),$ $\Phi_W(A)\Phi_W(B),$ $\Phi_W(B)\Phi_W(P),$ $\Phi_W(C)\Phi_W(P)$ (similar discussion for $\Phi_W(E),$ $\Phi_W(F)$ ).

From Three concurrent radical axes $\Longrightarrow$ $\Phi_W(B)\Phi_W(F),$ $\Phi_W(C)\Phi_W(E),$ $\Phi_W(P)\Phi_W(D)$ are concurrent (at the isogonal conjugate of the complement (WRT $\triangle \Phi_W(B)\Phi_W(P)\Phi_W(C)$ ) of $\Phi_W(A)$ WRT $\triangle \Phi_W(B)\Phi_W(P)\Phi_W(C)$ ), so $\odot (WBF),$ $\odot (WCE),$ $\odot (WPD)$ are coaxial, hence notice $\odot (WBF),$ $\odot (WCE)$ is the image of $\odot (DCE),$ $\odot (DBF)$ , respectively under $\Psi_X$ we get $W$ $\in$ $\odot (AQD)$ . Analogously, we can prove $W$ lies on $\odot (BQE)$ and $\odot (CQF)$ , so $\odot (AQD),$ $\odot (BQE),$ $\odot (CQF)$ are coaxial with common points $Q,$ $W$ .

Finally, from $\measuredangle BWC$ $+$ $\measuredangle CPB$ $=$ $\measuredangle BXC$ $+$ $\measuredangle CPB$ $=$ $\measuredangle BXP$ $+$ $\measuredangle PXC$ $+$ $\measuredangle CPB$ $=$ $\measuredangle (AB, CP)$ $+$ $\measuredangle (BP, CA)$ $+$ $\measuredangle (CP, BP)$ $=$ $\measuredangle BAC$ and similarly we get $\measuredangle CWA$ $+$ $\measuredangle APC$ $=$ $\measuredangle CBA,$ $\measuredangle AWB$ $+$ $\measuredangle BPA$ $=$ $\measuredangle ACB$ we conclude that $W$ is the isogonal conjugate (WRT $\triangle ABC$ ) of the antigonal conjugate of $P$ WRT $\triangle ABC$ .
____________________________________________________________
See also here : isogonal cente of a special quadrilateral (Lemma 3 at post #2)

This post has been edited 1 time. Last edited by TelvCohl, Feb 25, 2017, 2:09 PM
Reason: Add another proof

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#6 h Nov 4, 2015, 2:44 AM • 3 Y

Y by Omez, SerdarBozdag, Adventure10

TelvCohl wrote:

We prove the stronger result as following :

Given a $\triangle ABC$ and a point $P .$ Let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle ABC$ and let $Q$ be the Miquel point of $D,$ $E,$ $F$ WRT $\triangle ABC .$ Then $\odot (AQD),$ $\odot (BQE),$ $\odot (CQF)$ are coaxial and the second intersection of these three circles is the isogonal conjugate (WRT $\triangle ABC$ ) of the antigonal conjugate of $P$ WRT $\triangle ABC .$

Here's another approach. We use directed angles mod 180º throughout the proof.

From previous post #4 we known that $\odot(AQD),$ $\odot(BQE),$ $\odot(CQF)$ meet at a 2nd point $R.$ Let $\odot(BQE)$ and $\odot(CQF)$ cut $CA,AB$ again at $Y,Z,$ resp. Thus we have $\angle (RQ,RZ)=\angle (FQ,FA)=\angle (EQ,EY)=\angle (RQ,RY)$ $\Longrightarrow$ $R \in YZ.$ Hence since $\angle (RB,RZ)=\angle (EB,EC)$ and $\angle (RY,RC)=\angle (FB,FC)$ $\Longrightarrow$ $\angle (RB,RC)=\angle (EB,EC)+\angle (FB,FC)=\angle (AB,AC)+\angle (PB,PC).$ But if $K$ is the antigonal conjugate of $P$ and $K^*$ the isogonal conjugate of $K,$ we have $\angle (K^*B,K^*C)=\angle (AB,AC)+\angle (KC,KB)=\angle (AB+AC)+\angle (PB,PC)$ $\Longrightarrow$ $K^* \in \odot(RBC)$ and similarly $K^*$ also lies on $\odot(RCA)$ and $\odot(RAB)$ $\Longrightarrow$ $R \equiv K^*.$

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Lsway

#7 h Feb 25, 2017, 1:32 PM • 2 Y

Y by Adventure10, Mango247

Last year Telv Cohl shared this nice problem with me.At that time I got a proof from which I discoverd a new proof of the famous theorem “Poncelet point $\in$ Ceva circle”.Permit me to write down my solution here:
We only need to prove:
Let $R$ be the isogonal center of $ABCP$ , $PA\cap BC=D,PB\cap AC=E,PC\cap AB=F.$
Then $\measuredangle ARD=\measuredangle AEF-\measuredangle CBA$
Proof:Let $O$ be the circumceter of $\triangle ABC$ , $T$ be a point such that $\measuredangle TAQ=\measuredangle ADO$ and $QT \perp BC.$
Then $\measuredangle AQT=\measuredangle OAD \Rightarrow \measuredangle OQT=\measuredangle RAD$
$\measuredangle AQT=\measuredangle OAD,\measuredangle TAQ=\measuredangle ADO \Rightarrow \triangle QAT \overset{-}{\sim} \triangle ADO\Rightarrow \measuredangle ATQ= \measuredangle AOD, \frac{QT}{QA}=\frac{QA}{AD} \Rightarrow QT\cdot AD=OA \cdot QA=OQ \cdot RA \Rightarrow \frac{OQ}{QT}= \frac{AD}{RA},$ notice $\measuredangle OQT=\measuredangle RAD$ then $\triangle OQT \overset{-}{\sim} \triangle DAR \Rightarrow \measuredangle ARD=\measuredangle OTQ$
Let $OD\cap QT=N,$ then $AN \perp EF$ .
$\measuredangle ATQ=\measuredangle AOD \Rightarrow A,O,N,T$ are concyclic $\Rightarrow \measuredangle OTQ=\measuredangle OAN=\measuredangle OAC-\measuredangle NAC=\measuredangle AEF-\measuredangle CBA$

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This post has been edited 3 times. Last edited by Lsway, Mar 1, 2017, 11:53 PM
Reason: 分式

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#8 h May 4, 2017, 2:16 PM • 2 Y

Y by Adventure10, Mango247

Yesterday Lsway told me this beautiful problem and also told me it's related to the IMO shortlist 2012 G8. I also found an interesting problem which is equivalent to IMO shortlist 2012 G8 and similar to jayme's problem:

Given a triangle $\Delta ABC$ and a point $P$ , let $\Delta DEF$ be the cevian triangle of $P$ with respect to $\Delta ABC$ , $Q$ is the Miquel point of $D, E, F$ with respect to $\Delta ABC$ . $EF\cap BC=A_1$ , similarly define $B_1, C_1$ . Let $X$ be the pedal point of the circumcenter of $\Delta DEF$ on line $A_1B_1C_1$ . Then $(QDA_1), (QEB_1), (QFC_1)$ are coaxial and the second intersection point of these three circles is $X$ .

So is there any relationship between trilinear polar and isogonal center?

This post has been edited 3 times. Last edited by Leooooo, May 8, 2017, 3:51 PM

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#9 h Jul 29, 2018, 2:38 AM • 2 Y

Y by Adventure10, Mango247

how to prove AN is perpendicular to EF？

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