Generalization of Parry Reflection Point

by TelvCohl, Sep 14, 2017, 4:42 AM

Preliminaries

Lemma : Given a $\triangle ABC$ and a point $P.$ Then $P$ lies on the Pivotal Isogonal cubic with pivot $U$ if and only if $P,$ the isogonal conjugate $V$ of $U$ WRT $\triangle ABC,$ the cevian quotient $\left( U/P \right)$ WRT $\triangle ABC$ are collinear.

Proof : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC,$ $\mathcal{C}_U$ be the circumconic of $\triangle ABC$ passing through $P,$ $U,$ and $\mathcal{C}_V$ be the circumconic of $\triangle ABC$ passing through $P,$ $V.$ It's well-known that the tangent of $\mathcal{C}_U$ at $P$ passes through $\left( U/P \right) ,$ so $P, V, \left( U/P \right)$ are collinear if and only if $(P,A;B,C)_{\mathcal{C}_U}$ $=$ $(V,A;B,C)_{\mathcal{C}_V}.$ Let $K$ be the intersection of $BC$ with $QU,$ then notice $\mathcal{C}_V$ is the isogonal conjugate of $QU$ WRT $\triangle ABC$ we know $AK$ and the tangent of $\mathcal{C}_V$ at $A$ are isogonal conjugate WRT $\angle A,$ so $(V,A;B,C)_{\mathcal{C}_V}$ $=$ $A(U,K;C,B)$ $=$ $U(A,Q;C,B)$ $=$ $U(Q,A;B,C),$ hence we conclude that $U \in PQ \Longleftrightarrow (V,A;B,C)_{\mathcal{C}_V} = U(P,A;B,C) = (P,A;B,C)_{\mathcal{C}_U} \Longleftrightarrow \left( U/P \right) \in PV \ . \qquad \blacksquare$
Corollary : Given a $\triangle ABC$ and a point $U.$ Then a point $P$ lies on the Pivotal Isogonal cubic with pivot $U$ if and only if the cevian quotient $\left( U/P \right)$ of $P$ and $U$ WRT $\triangle ABC$ lies on it.

Proof : Let $P^*,$ $U^*,$ $\left( U/P \right)^*$ be the isogonal conjugate of $P,$ $U,$ $\left( U/P \right)$ WRT $\triangle ABC,$ respectively. Then $P, P^*, U \text{ are collinear }. \stackrel{\text{Lemma}}{\Longleftrightarrow} P, U^*, \left( U/P \right) \text{ are collinear }. \stackrel{\text{Lemma}}{\Longleftrightarrow} U, \left( U/P \right) , \left( U/P \right)^* \text{ are collinear }. \qquad \blacksquare$
Main result

Notation

Given a $\triangle ABC,$ a point $P$ and a point $W$ lying on the line $\mathcal{L}_{\infty}$ at infinity. Let $\triangle DEF$ be the cevian triangle of $W$ WRT $\triangle ABC$ and let $\triangle XYZ$ be the circlecevian triangle of $P$ WRT $\triangle ABC.$

Property 1 : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC.$ Then $DX, EY, FZ$ are concurrent $\Longleftrightarrow$ $W \in PQ.$

Proof :

( $\Longrightarrow$ ) Clearly, $A$ is the center of the spiral similarity of $BY \mapsto ZC,$ so $\measuredangle EAY$ $=$ $\measuredangle ZAF$ and $AY \cdot AZ$ $=$ $AB \cdot AC$ $=$ $AE \cdot AF$ $\Longrightarrow$ $\triangle AEY$ $\stackrel{+}{\sim}$ $\triangle AZF,$ hence $\measuredangle (EY, FZ)$ is independent of the position of $W$ $\Longrightarrow$ the intersection of $EY, FZ$ lies on a fixed circle $\mathbf{C}_X$ passing through $Y, Z$ when $W$ varies on $\mathcal{L}_{\infty}.$ Similarly, the intersection of $(FZ, DX),$ $(DX, EY)$ lie on a fixed circle $\mathbf{C}_Y,$ $\mathbf{C}_Z,$ respectively. Let $\bar{W}$ be the second intersection of $\mathbf{C}_Y,$ $\mathbf{C}_Z.$ From $\measuredangle Y\bar{W}Z$ $=$ $\measuredangle Y\bar{W}X$ $+$ $\measuredangle X\bar{W}Z$ $=$ $\measuredangle ACB$ $+$ $\measuredangle BPA$ $+$ $\measuredangle CBA$ $+$ $\measuredangle APC$ $=$ $\measuredangle CAB$ $+$ $\measuredangle BPC$ we get $\bar{W} \in \mathbf{C}_X,$ so $\mathbf{C}_X, \mathbf{C}_Y, \mathbf{C}_Z$ are concurrent at $\bar{W}.$ This establish the uniqueness of the position of $W$ such that $DX, EY, FZ$ are concurrent.

( $\Longleftarrow$ ) Let $\triangle P_aP_bP_c, \triangle Q_aQ_bQ_c$ be the cevian triangle of $P, Q$ WRT $\triangle ABC,$ respectively and let $\triangle P_AP_BP_C$ be the circumcevian triangle of $P$ WRT $\triangle ABC.$ It is well-known that $BC, QP_A$ and the parallel from $P$ to $AQ$ are concurrent at $R,$ so combining $AP_a \cdot P_AP_a$ $=$ $PP_a \cdot XP_a$ we get $\tfrac{XP_a}{P_AP_a}$ $=$ $\tfrac{AP_a}{PP_a}$ $=$ $\tfrac{Q_aP_a}{RP_a},$ hence $\triangle PRP_A$ and $\triangle AQ_aX$ are homothetic with center $P_a.$ Let $PQ$ cuts $BC, CA, AB$ at $A^*,$ $B^*,$ $C^*,$ respectively and let $J$ $\equiv$ $DX \cap QP_A,$ $K$ $\equiv$ $AD \cap P_AA^*.$ Obviously, $P_AA^*$ and $DX$ are parallel and $\triangle KP_AQ, \triangle DXQ_a$ are homothetic with center $A,$ so $QK$ $\parallel$ $BC$ $\Longrightarrow$ $\tfrac{QP_A}{RP_A}$ $=$ $\tfrac{QK}{A^*R}$ $=$ $\tfrac{A^*D}{A^*R}$ $=$ $\tfrac{JP_A}{RP_A},$ hence $J$ is the reflection of $Q$ in $P_A$ $\Longrightarrow$ $DX$ is the image of $P_AA^*$ under the homothety $\mathbf{H}^{Q}_{2} .$ Similarly, $EY,$ $FZ$ is the image of $P_BB^*,$ $P_CC^*,$ under the homothety $\mathbf{H}^{Q}_{2},$ respectively. From Pascal's theorem we get $P_AA^*, P_BB^*P_CC^*$ are concurrent at $S \in \odot (ABC),$ so we conclude that $DX,EY,FZ$ are concurrent at the image $T$ of $S$ under the homothety $\mathbf{H}^{Q}_{2}.$ $\qquad \blacksquare$

Property 2 : Let $V$ be the second intersection of $QT$ with $\odot (ABC).$ Then the Steiner line of $V$ WRT $\triangle ABC$ is perpendicular to $PQ.$

Proof : Let $\triangle Q_AQ_BQ_C$ be the circumcevian triangle of $Q$ WRT $\triangle ABC.$ Notice $\triangle ASQ_A$ $\stackrel{-}{\sim}$ $\triangle A^*P_aP_A,$ so from $\tfrac{AQ}{Q_AQ}$ $=$ $\tfrac{PP_a}{P_AP_a}$ we know $\triangle QSQ_A$ $\stackrel{-}{\sim}$ $\triangle A^*PP_A$ $\Longrightarrow$ $\measuredangle (AV,AQ)$ $=$ $\measuredangle QSQ_A$ $=$ $\measuredangle P_APA^*$ $=$ $\measuredangle (AP,PQ),$ hence the isogonal conjugate of $AV$ WRT $\angle A$ is parallel to $PQ.$ i.e. the Steiner line of $V$ WRT $\triangle ABC$ is perpendicular to $PQ.$ $\qquad \blacksquare$

Corollary 1 : $T$ lies on $\odot (ADQ),$ $\odot (BEQ),$ $\odot (CFQ).$

Proof : Since $AD$ and $AV$ are isogonal conjugate WRT $\angle A,$ so $\measuredangle DAQ$ $=$ $\measuredangle PAV$ $=$ $\measuredangle P_ASV$ $=$ $\measuredangle DTQ$ $\Longrightarrow$ $A,D,Q,T$ are concyclic. $\qquad \blacksquare$

Property 3 : Let $I, I_a, I_b, I_c$ be the incenter, A-excenter, B-excenter, C-excenter of $\triangle ABC,$ respectively. Then $T$ lies on $\odot (II_aX),$ $\odot (II_bY),$ $\odot (II_cZ).$

Proof : Let $M$ be the midpoint of arc $BC$ in $\odot (ABC)$ and let $N$ be image of $M$ under the homothety $\mathbf{H}^{Q}_{2}.$ Since $AB \cdot AC$ $=$ $AQ \cdot AX$ $=$ $AI \cdot AI_a,$ so note that $\measuredangle (AM, AX) = \measuredangle MSP_A = \measuredangle NTX$ we get $\left\{\begin{array}{cc} \triangle AIX \stackrel{+}{\sim} \triangle AQI_a \Longrightarrow \measuredangle (AI_a,AX) = \measuredangle (QI_a,IX) = \measuredangle NIX \\\\ \triangle AIQ \stackrel{+}{\sim} \triangle AXI_a \Longrightarrow \measuredangle (AI,AX) = \measuredangle (IQ,XI_a) = \measuredangle NI_aX \end{array}\right\| \Longrightarrow T \in \odot (II_aX) \ . \qquad \blacksquare$
Corollary 2 : $T$ lies on $\odot (XI_bI_c),$ $\odot (YI_cI_a),$ $\odot (ZI_aI_b).$

Proof : By the same reason, as in the proof of Property 3, we get

$\left\{\begin{array}{cc} \triangle BIQ \stackrel{+}{\sim} \triangle BYI_b, \ \triangle BQI_b \stackrel{+}{\sim} \triangle BIY \Longrightarrow \measuredangle I_bYI = \measuredangle I_bYB + \measuredangle BYI = \measuredangle QIB + \measuredangle BI_bQ = \measuredangle IQI_b \\\\ \triangle CIQ \stackrel{+}{\sim} \triangle CZI_c, \ \triangle CQI_c \stackrel{+}{\sim} \triangle CIZ \Longrightarrow \measuredangle IZI_c = \measuredangle IZC + \measuredangle CZI_c = \measuredangle QI_cC + \measuredangle CIQ = \measuredangle I_cQI \end{array}\right\| \stackrel{\text{Prop. 3}}{\Longrightarrow} \measuredangle I_bTI_c = \measuredangle I_cQI_b \ .$ On the other hand, from $\triangle AXI_b$ $\stackrel{+}{\sim}$ $\triangle AI_cQ,$ $\triangle AXI_c$ $\stackrel{+}{\sim}$ $\triangle AI_bQ$ we get $\measuredangle I_bXI_c$ $=$ $\measuredangle I_bXA$ $+$ $\measuredangle AXI_c$ $=$ $\measuredangle QI_cA$ $+$ $\measuredangle AI_bQ$ $=$ $\measuredangle I_cQI_b,$ so $\measuredangle I_bTI_c$ $=$ $\measuredangle I_bXI_c$ $\Longrightarrow$ $T \in \odot (XI_bI_c).$ $\qquad \blacksquare$

Corollary 3 : $Q$ and $T$ are antigonal conjugate WRT $\triangle I_aI_bI_c.$

Corollary 4 : $Q$ and $T$ are antigonal conjugate WRT the anticevian triangle $\triangle Q_a^{-1}Q_b^{-1}Q_c^{-1}$ of $Q$ WRT $\triangle ABC.$

Proof : Note that $I, I_a, I_b, I_c, Q, Q_a^{-1}, Q_b^{-1}, Q_c^{-1}$ lie on a conic (well-known), so $Q$ and $T$ are antigonal conjugate WRT $\triangle Q_a^{-1}Q_b^{-1}Q_c^{-1}.$ $\qquad \blacksquare$

Property 4 : $T$ lies on the Pivotal Isogonal Circular Cubic of $\triangle ABC$ passing through $P,$ $Q.$

Proof : Since the reflection of $Q$ in $A^*$ lies on $DX,$ so $D(A,Q_a;QX,)$ $=$ $-1$ $\Longrightarrow$ $Q_a^{-1} \in DX.$ Similarly, $Q_b^{-1} \in EY,$ $Q_c^{-1} \in FZ,$ so from the Corollary of Lemma we conclude that $T,$ the cevian quotient $\left( U/Q \right)$ of $Q$ and $U$ WRT $\triangle ABC$ where $U$ is the point at infinity on $PQ,$ lies on the Pivotal Isogonal Circular Cubic of $\triangle ABC$ passing through $P,$ $Q.$ $\qquad \blacksquare$

This post has been edited 1 time. Last edited by TelvCohl, Jan 27, 2021, 3:12 PM

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Parry reflection point

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Generalization of Parry Reflection Point

by TelvCohl, Sep 14, 2017, 4:42 AM

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