isogonal cente of a special quadrilateral

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Lsway

#1 h Feb 2, 2017, 5:01 AM • 1 Y

Y by Adventure10

(2017.1.16)Given $\triangle ABC$ .Let $T\in\odot(ABC)$ , $X\in BC$ , $Y\in CA$ , $Z\in AB$ and $AX//BY//CZ\perp$ the Simson line of $T$ WRT $\triangle ABC$ .The isogonal center of $TXYZ$ is $R$ .
Prove that the midpoint of $RT \in\odot(ABC)$ .

This post has been edited 2 times. Last edited by Lsway, Feb 8, 2017, 11:08 AM

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#2 h Feb 7, 2017, 4:52 PM • 6 Y

Y by WizardMath, Saro00, Leooooo, enhanced, Adventure10, Mango247

Lemma 1 : Given a $\triangle ABC$ and a point $T$ at infinity. Let $P$ be a point lying on the Pivotal Isogonal cubic of $\triangle ABC$ with pivot $T.$ Then the midpoint of $P$ and the cevian quotient $(T/P)$ of $P,$ $T$ WRT $\triangle ABC$ lies on $\odot (ABC).$

Proof : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ and $\triangle Q_AQ_BQ_C$ be the circumcevian triangle of $Q$ WRT $\triangle ABC.$ Let $\triangle P^{a}P^{b}P^{c}$ be the anticevian triangle of $P$ WRT $\triangle ABC$ and let $M_a$ be the midpoint of $PP^{a}.$ Let $U,$ $V$ be the intersection of $BC$ with $AP,$ $AQ,$ respectively, then we get $(A,V;Q,Q_A) = \underbrace{(U,A;P, \infty ) = (A,U;P,M_a)}_{\text{harmonic conjugate WRT} \ P \ \text{and} \ P^{a}} \ ,$ so $BC,$ $PQ,$ $M_aQ_A$ are concurrent at $X.$ Let $PQ$ cuts $CA,$ $AB$ at $Y,$ $Z,$ respectively, then by Pascal's theorem we get $Q_AX,$ $Q_BY,$ $Q_CZ$ are concurrent at $K$ $\in$ $\odot (ABC).$ From $\tfrac{UA}{UP}$ $=$ $\tfrac{UP^{a}}{UM_a}$ we get $KM_a$ $\parallel$ $(T/P)P^{a},$ so the reflection $\widetilde{P}$ of $P$ in $K$ lies on $P^{a} (T/P).$ Similarly, we can prove $\widetilde{P}$ lies on $P^{b}(T/P),$ $P^{c}(T/P),$ so we conclude that $\widetilde{P}$ $\equiv$ $(T/P).$
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Lemma 2 : Let $T$ be the isogonal conjugate of the complement of $P$ WRT $\triangle ABC$ WRT $\triangle ABC.$ Then the Miquel associate of $P$ WRT $\triangle ABC$ is the cevian quotient $(P/T)$ of $P,$ $T$ WRT $\triangle ABC.$

Proof : Let $Q$ be the isotomic conjugate of $P$ WRT $\triangle ABC$ and let $\triangle P_aP_bP_c,$ $\triangle Q_aQ_bQ_c$ be the cevian triangle of $P,$ $Q$ WRT $\triangle ABC,$ respectively. Let $\triangle DEF$ be the anticevian triangle of the isotomcomplement of $Q$ WRT $\triangle ABC$ WRT $\triangle ABC.$ Since $\triangle DEF$ and $\triangle Q_aQ_bQ_c$ are homothetic, so the isogonal conjugate $D^*$ of $D$ WRT $\triangle ABC$ lies on the tangent of $\odot (BQ_cQ_a),$ $\odot (CQ_aQ_b)$ at $B,$ $C,$ respectively $\Longrightarrow$ $D^*$ is the radical center of $\odot (ABC),$ $\odot (BQ_cQ_a),$ $\odot (CQ_aQ_b).$

Note that $T$ is the isogonal conjugate (WRT $\triangle ABC$ ) of the isotomcomplement of $Q$ WRT $\triangle ABC,$ so $D^*$ is the A-vertex of the anticevian triangle of $T$ WRT $\triangle ABC$ $\Longrightarrow$ the Miquel associate $M_P$ of $P$ WRT $\triangle ABC$ lies on $P_a(P/T).$ Analogously, we can prove $M_P$ lies on $P_b(P/T),$ $P_c(P/T),$ so we conclude that $M_P$ $\equiv$ $(P/T).$
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Lemma 3 : Given a $\triangle ABC$ and a point $P.$ Let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle ABC$ and let $T$ be the Miquel point of $D,$ $E,$ $F$ WRT $\triangle ABC.$ Then $\odot (ADT),$ $\odot (BET),$ $\odot (CFT)$ pass through the Isogonal center $R$ of $ABCP.$

Proof : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ and $\triangle Q_AQ_BQ_C$ be the circumcevian triangle of $Q$ WRT $\triangle ABC.$ Let $Y,$ $Z$ be the second intersection of $\odot (AEF)$ with $BP,$ $CP,$ respectively. From Pascal's theorem (for $YEAAFZ$ ) $\Longrightarrow$ the intersection $X$ of $BC$ and $YZ$ lies on the tangent of $\odot (AEF)$ at $A.$

Simple angle chasing yields $AQ_B,$ $AY$ are isogonal conjugate WRT $\angle BAX,$ so $Q_B,$ $Y$ are isogonal conjugate WRT $\triangle ABX$ $\Longrightarrow$ $XQ_B$ is the isogonal conjugate of $YZ$ WRT $\angle (AX,BC).$ Similarly, we can prove $XQ_C$ is the isogonal conjugate of $YZ$ WRT $\angle (AX,BC),$ so $Q_B,$ $Q_C,$ $X$ are collinear $\Longrightarrow$ $X$ lies on the polar of $Q$ WRT $\odot (ABC).$

Let $O$ be the circumcenter of $\triangle ABC,$ then we get $\begin{align*} \measuredangle (AD,AR) &= \measuredangle (AD,AO) + \measuredangle (AO,AR) \\ &= \measuredangle (\perp BC, AQ) + \measuredangle (AQ,OQ) \\ &= \measuredangle (BC, \perp OQ) \\ &= \measuredangle (XD,XR) \ , \end{align*}$ so $A,$ $D,$ $R,$ $X$ are concyclic $\Longrightarrow$ $R$ $\in$ $\odot (ADT).$ Analogously, we can prove $R$ lies on $\odot (BET),$ $\odot (CFT).$
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Back to the main problem :

Let $\widetilde{R}$ be the Miquel point of $X,$ $Y,$ $Z$ WRT $\triangle ABC$ and $V$ be the infinity point on $AX,$ $BY,$ $CZ.$ From Lemma 2 $\Longrightarrow$ $\widetilde{R}$ is the cevian quotient $(V/T)$ of $T,$ $V$ WRT $\triangle ABC,$ so from Lemma 1 we get the midpoint of $T\widetilde{R}$ lies on $\odot (ABC),$ hence it suffices to prove $\widetilde{R}$ is the Isogonal center of $TXYZ.$

From Lemma 3 $\Longrightarrow$ $\widetilde{R}$ lies on $\odot (ATX),$ $\odot (BTY),$ $\odot (CTZ),$ so $\begin{align*} \measuredangle (TY,TZ) + \measuredangle (XY,XZ) &= \left (\measuredangle (TY,T\widetilde{R}) + \measuredangle (T\widetilde{R},TZ) \right ) + \left ( \measuredangle (XY,X\widetilde{R}) + \measuredangle (X\widetilde{R},XZ) \right ) \\ &= \left (\measuredangle (BV,B\widetilde{R}) + \measuredangle (C\widetilde{R},CV) \right ) + \left ( \measuredangle (CA,C\widetilde{R}) + \measuredangle (B\widetilde{R},AB) \right ) \\ &= \measuredangle (CA,AB) \\ &= \measuredangle (\widetilde{R}Y,\widetilde{R}Z) \ . \end{align*}$ Similarly, we can prove $\measuredangle (TZ,TX)$ $+$ $\measuredangle (YZ,YX)$ $=$ $\measuredangle (\widetilde{R}Z,\widetilde{R}X)$ and $\measuredangle (TX,TY)$ $+$ $\measuredangle (ZX,ZY)$ $=$ $\measuredangle (\widetilde{R}X,\widetilde{R}Y),$ so we conclude that $\widetilde{R}$ is the Isogonal center of $TXYZ.$

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Lsway

#3 h Feb 8, 2017, 11:08 AM • 2 Y

Y by Adventure10, Mango247

Thank you for your attention dear Telv Cohl,the lemma 1 is a nice conclusion that I haven't seen before.
Here is my proof :Step1:Obviously the Miuqel points of complete quadrileteral $<TY,TZ,XY,XZ>$ ,
$<TZ,TX,YZ,XY>$ , $<TX,TY,XZ,YZ>$ are $A,B,C$ ,respectively.
Let $D= TX \cap YZ$ , $E= XA \cap YZ$ . $F$ is the midpoint of $XD$ .
Obviously $A$ is the midpoint of $XE$ $\Rightarrow AF//YZ\Rightarrow \measuredangle AFT=\measuredangle(YZ,XD)=\measuredangle YBX=\measuredangle ABT \Rightarrow F\in \odot(ABC)$
Step2:Let $S$ be midpoint of $RT$ . $\triangle ART\cup S \overset{+}{\sim} \triangle AXD \cup F \Rightarrow S,T,A,F$ are concyclic $\Rightarrow S\in \odot(ABC)$

This post has been edited 3 times. Last edited by Lsway, Feb 9, 2017, 5:52 AM

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