A Problem with some Nice Lemma

by TelvCohl, Aug 9, 2016, 10:20 PM

Problem

Problem : Given a $\triangle ABC$ with circumcenter $O.$ Let $P,$ $Q$ be the isogonal conjugate WRT $\triangle ABC$ and let $M$ be the midpoint of $PQ.$ Let $G_P,$ $G_Q$ be the centroid of the pedal triangle of $P,$ $Q$ WRT $\triangle ABC,$ respectively and let $N$ be the midpoint of $G_PG_Q.$ Prove that if $PQ$ is parallel to $G_PG_Q,$ then $O$ lies on $MN.$

Preliminaries

Lemma 1 : Given a $\triangle ABC$ and the points $E$ $\in$ $CA,$ $F$ $\in$ $AB.$ Let $M,$ $N$ be the intersection of $EF,$ $\odot (ABC)$ and let $P,$ $Q,$ $R,$ $S$ be the midpoint of $BE,$ $CF,$ $EF,$ $MN,$ respectively. Then $P,$ $Q,$ $R,$ $S$ are concyclic.

Proof : Let $O$ be the circumcenter of $\triangle ABC$ and let $A^*$ be the antipode of $A$ in $\odot (O).$ Let $H$ be the orthocenter of $\triangle AEF$ and let $J$ be the midpoint of $A^*H.$ From $OJ$ $\parallel$ $AH$ we get $OJ$ $\perp$ $EF,$ so $O,$ $J,$ $S$ are collinear. On the other hand, notice $PJ$ $\parallel$ $A^*B$ $\parallel$ $EH$ and $PR$ $\parallel$ $AB$ we get $\angle JPR$ $=$ $90^{\circ},$ so $J,$ $P,$ $R,$ $S$ are concyclic. Similarly, we can prove $Q$ lies on this circle, so we conclude that $P,$ $Q,$ $R,$ $S$ are concyclic. $\qquad \blacksquare$

Lemma 2 : Given a $\triangle ABC$ and a point $P.$ Let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle ABC$ and let $\omega$ be the pedal circle of $P$ WRT $\triangle DEF.$ Then the orthotransversal of $P$ WRT $\triangle ABC$ is the polar of $P$ WRT $\omega.$

Proof : Let $\triangle XYZ$ be the pedal triangle of $P$ WRT $\triangle DEF$ and let the perpendicular from $P$ to $AP$ cuts $BC,$ $FD,$ $DE$ at $A_1,$ $U,$ $V,$ respectively. From symmetry, it suffices to prove $A_1$ lies on the polar of $P$ WRT $\odot (XYZ).$ Obviously, $D,$ $P,$ $Y,$ $Z$ lie on the circle with diameter $DP,$ so from $UV$ $\perp$ $DP$ we get $UV,$ $YZ$ are antiparallel WRT $\angle EDF,$ hence $U,$ $V,$ $Y,$ $Z$ are concyclic. Let $J$ $\equiv$ $UV$ $\cap$ $YZ.$ Since ${JP}^2$ $=$ $JY$ $\cdot$ $JZ$ $=$ $JU$ $\cdot$ $JV,$ so combining $(A_1,P;U,V)$ $=$ $-1$ we get $J$ is the midpoint of $A_1P$ and $J$ lies on the radical axis of $\odot (XYZ)$ and the degenerate circle $P,$ hence we conclude that $A_1$ lies on the polar of $P$ WRT $\odot (XYZ).$ $\qquad \blacksquare$

Lemma 3 : Given a $\triangle ABC$ and a point $P.$ Let $\Omega$ be the pedal circle of $P$ WRT $\triangle ABC.$ Then the trilinear polar of $P$ WRT $\triangle ABC,$ the orthotransversal of $P$ WRT $\triangle ABC$ and the polar of $P$ WRT $\Omega$ are concurrent.

Proof : Let $\triangle DEF$ be the anticevian triangle of $P$ WRT $\triangle ABC$ and let $B_1$ $\equiv$ $CA$ $\cap$ $FD,$ $C_1$ $\equiv$ $AB$ $\cap$ $DE.$ Let the perpendicular from $P$ to $BP$ cuts $CA,$ $FD$ at $Y_1,$ $E_1,$ respectively and let the perpendicular from $P$ to $CP$ cuts $AB,$ $DE$ at $Z_1,$ $F_1,$ respectively. By Lemma 2 $\Longrightarrow$ $E_1F_1$ is the polar of $P$ WRT $\Omega,$ so from Desargues's theorem ( $\triangle B_1E_1Y_1,$ $\triangle C_1F_1Z_1$ ) we conclude that the trilinear polar $B_1C_1$ of $P$ WRT $\triangle ABC,$ the orthotransversal $Y_1Z_1$ of $P$ WRT $\triangle ABC,$ polar $E_1F_1$ of $P$ WRT $\Omega$ are concurrent. $\qquad \blacksquare$

Corollary 4 : Given a $\triangle ABC$ and a point $P$ $\in$ $\odot (ABC).$ Then the trilinear polar of $P$ WRT $\triangle ABC,$ the orthotransversal of $P$ WRT $\triangle ABC$ and the Steiner line of $P$ WRT $\triangle ABC$ are concurrent.

Lemma 5 : Given a $\triangle ABC$ and a point $P$ $\in$ $\odot (ABC).$ Then the intersection $Q$ of the Steiner line of $P$ WRT $\triangle ABC$ and the orthotransversal of $P$ WRT $\triangle ABC$ lies on the Jerabek hyperbola of $\triangle ABC.$

Proof : Let the perpendicular from $P$ to $BP$ cuts $CA,$ $\odot (ABC)$ at $E,$ $B^*,$ respectively and let the perpendicular from $P$ to $CP$ cuts $AB,$ $\odot (ABC)$ at $F,$ $C^*,$ respectively. From Pascal's theorem ( $ABB^*PC^*C$ ) we get $EF$ passes through the circumcenter $O$ of $\triangle ABC.$ Let $H$ be the orthocenter of $\triangle ABC$ and let $T$ $\equiv$ $AH$ $\cap$ $\odot (O).$ Let $L$ be the reflection of $P$ in $BC.$ Since $H,$ $T$ are symmetry WRT $BC,$ so $BC,$ $PT$ and the Steiner line $HL$ of $P$ WRT $\triangle ABC$ are concurrent at $U.$ From Pascal's theorem ( $ATPC^*CB$ ) $\Longrightarrow$ $F,$ $U,$ $V$ $\equiv$ $AH$ $\cap$ $CC^*$ are collinear, so by Pascal's theorem ( $AHQOCB$ ) we conclude that $H,$ $O,$ $Q$ lie on a circumconic of $\triangle ABC.$ i.e. $Q$ lies on the Jerabek hyperbola of $\triangle ABC.$ $\qquad \blacksquare$

Corollary 6 (Corollary 4 $+$ Lemma 5) : Given a $\triangle ABC$ and a point $P$ $\in$ $\odot (ABC).$ Let $\mathcal{O}_P,$ $\mathcal{S}_P,$ $\mathcal{T}_P$ be the orthotransversal of $P$ WRT $\triangle ABC,$ the Steiner line of $P$ WRT $\triangle ABC,$ the trilinear polar of $P$ WRT $\triangle ABC,$ respectively. Then $\mathcal{S}_P$ $\parallel$ $\mathcal{T}_P$ $\Longleftrightarrow$ $\mathcal{T}_P$ $\parallel$ $\mathcal{O}_P$ $\Longleftrightarrow$ $\mathcal{O}_P$ $\parallel$ $\mathcal{S}_P$ $\Longleftrightarrow$ one of $\mathcal{O}_P,$ $\mathcal{S}_P,$ $\mathcal{T}_P$ is parallel to one of the asymptote of the Jerabek hyperbola of $\triangle ABC.$

Lemma 7 : Given a $\triangle ABC$ and a point $P.$ Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ and let $\triangle Q_aQ_bQ_c$ be the pedal triangle of $Q$ WRT $\triangle ABC.$ Then the line passing through $Q$ and the centroid $G_Q$ of $\triangle Q_aQ_bQ_c$ is perpendicular to the trilinear polar of $P$ WRT $\triangle ABC.$

Proof : Let $D,$ $E,$ $F$ be the second intersection of $Q_aG_Q,$ $Q_bG_Q,$ $Q_cG_Q$ with $\odot (Q_aQ_bQ_c),$ respectively. Let $P_A,$ $P_B,$ $P_C$ be the intersection of the trilinear polar of $P$ WRT $\triangle ABC$ with $BC,$ $CA,$ $AB,$ respectively. Clearly, $AP,$ $BP,$ $CP$ is perpendicular to $Q_bQ_c,$ $Q_cQ_a,$ $Q_aQ_b,$ respectively, so from $P(P_A,A;B,C)$ $=$ $-1$ $=$ $(Q_aG_Q, Q_bQ_c; Q_cQ_a, Q_aQ_b)$ we get $PP_A$ $\perp$ $Q_aG_Q,$ hence $P_A$ is the antipode of $Q$ in $\odot (DQQ_a).$ Similarly, we can prove $P_B,$ $P_C$ is the antipode of $Q$ in $\odot (EQQ_b),$ $\odot (FQQ_c),$ respectively. From $Q_aG_Q$ $\cdot$ $G_QD$ $=$ $Q_bG_Q$ $\cdot$ $G_QE$ $=$ $Q_cG_Q$ $\cdot$ $G_QF$ $\Longrightarrow$ $\odot (DQQ_a),$ $\odot (EQQ_b),$ $\odot (FQQ_c)$ are coaxial with common radical axis $QG_Q,$ so we conclude that $QG_Q$ passes through the projection of $Q$ on the trilinear polar of $P$ WRT $\triangle ABC$ $\Longrightarrow$ $QG_Q$ is perpendicular to the trilinear polar of $P$ WRT $\triangle ABC.$ $\qquad \blacksquare$

Lemma 8 : Given a $\triangle ABC,$ a fixed point $M$ and a fixed angle $\theta.$ Let $P,$ $P^*$ be the isogonal conjugate WRT $\triangle ABC$ and let $\tau_P^*$ be the trilinear polar of $P^*$ WRT $\triangle ABC.$ Then $\measuredangle (\tau_P^*, MP)$ $=$ $\theta$ $\Longrightarrow$ $P$ lies on a fixed conic $\Gamma_{(M,\theta)}$ passing through $M$ and the symmedian point $K$ of $\triangle ABC.$

Proof : Actually, this lemma simplified follows from the construction of $P$ (for the given point $M$ and angle $\theta$ ). Let $\ell$ be a line passing through $M$ and let $\ell^*$ be the line passing through $M$ such that $\measuredangle (\ell^*, \ell)$ $=$ $\theta.$ If $P$ is the point on $\ell$ such that $\measuredangle (\tau_P^*, MP)$ $=$ $\theta,$ then $\tau_P^*$ $\parallel$ $\ell^*$ $\Longrightarrow$ $\tau_P^*,$ $\ell^*$ and the trilinear polar of the centroid $G$ of $\triangle ABC$ WRT $\triangle ABC$ are concurrent, hence the trilinear pole $Q^*$ of $\ell^*$ WRT $\triangle ABC$ lies on the circumconic of $\triangle ABC$ passing through $G,$ $P^*$ $\Longrightarrow$ $KP$ passes through the isogonal conjugate $Q$ of $Q^*$ WRT $\triangle ABC.$ Conversely, if $KQ$ cuts $\ell$ at $P,$ then $\tau_P^*$ is parallel to $\ell^*.$

Since the trilinear pole of the pencil with center $M$ lies on the circumconic $\mathcal{C}$ of $\triangle ABC$ with perspector $M,$ so $Q$ varies on a line (isogonal conjugate of $\mathcal{C}$ WRT $\triangle ABC$ ) when $\ell$ varies around $M.$ Since the intersection of $\ell^*,$ $AQ^*$ with $BC$ are harmonic conjugate WRT $B,$ $C,$ so pencil $\ell^*$ $\mapsto$ pencil $AQ^*$ is a homography when $\ell$ varies around $M.$ On the other hand, it's obvious that pencil $\ell$ $\mapsto$ pencil $\ell^*$ and pencil $AQ^*$ $\mapsto$ pencil $AQ$ are homography when $\ell$ varies around $M,$ so pencil $\ell$ $\mapsto$ pencil $AQ$ is a homography $\Longrightarrow$ pencil $\ell$ $\mapsto$ pencil $KQ$ is a homography when $\ell$ varies around $M,$ hence the intersection $P$ of $\ell,$ $KQ$ lies on a fixed conic passing through $M,$ $K.$ $\qquad \blacksquare$

Corollary 9 (Corollary 6 $+$ Lemma 8) : Given a $\triangle ABC$ and a point $M.$ Let $\Gamma_{(M, 90^{\circ})}$ be the conic defined at Lemma 8. Then $\Gamma_{(M, 90^{\circ})}$ and the Jerabek hyperbola of $\triangle ABC$ are homothetic.

Corollary 9.1 : Let $H$ be the orthocenter of $\triangle ABC.$ Then $\Gamma_{(H, 90^{\circ})}$ is the Jerabek hyperbola of $\triangle ABC.$

Corollary 9.2 : Let $O$ be the circumcenter of $\triangle ABC.$ Then $\Gamma_{(O, 90^{\circ})}$ passes through the incenter and the excenter of $\triangle ABC.$

Lemma 10 : Given a $\triangle ABC$ and the points $D_1,$ $D_2$ $\in$ $BC,$ $E_1,$ $E_2$ $\in$ $CA,$ $F_1,$ $F_2$ $\in$ $AB.$ Let $G_1,$ $G_2$ be the centroid of $\triangle D_1E_1F_1,$ $\triangle D_2E_2F_2,$ respectively. Let $P$ be the point such that $\text{dist}(P,BC) : \text{dist}(P,CA) : \text{dist}(P,AB) = \frac{1}{D_1D_2} : \frac{1}{E_1E_2} : \frac{1}{F_1F_2}.$ Then $G_1G_2$ is parallel to the trilinear polar $\tau_P$ of $P$ WRT $\triangle ABC.$

Proof : First, we prove the following lemma.

Lemma 10.1 : Given a $\triangle ABC$ and the points $E$ $\in$ $CA,$ $F$ $\in$ $AB.$ Let $P$ be the point such that $\text{dist}(P,BC) : \text{dist}(P,CA) : \text{dist}(P,AB) = \frac{1}{BC} : \frac{1}{CE} : \frac{1}{FB}.$ Then $EF$ is parallel to the trilinear polar $\varsigma_P$ of $P$ WRT $\triangle ABC.$

Proof : Let $Y,$ $Z$ be the intersection of $\varsigma_P$ with $CA,$ $AB,$ respectively. From $[\triangle PCB]$ $=$ $[\triangle PBF]$ $\Longrightarrow$ $BP$ passes through the midpoint of $CF,$ so $(BC,BF;BP,CF)$ $=$ $-1$ $=$ $B(C,F;P,Y)$ $\Longrightarrow$ $CF$ $\parallel$ $BY.$ Similarly, we can prove $BE$ $\parallel$ $CZ,$ so from little Pappus's theorem we conclude that $EF$ $\parallel$ $YZ$ $\equiv$ $\varsigma_P .$ $\qquad \square$

Back to the main proof :

Let $TD_1,$ $TD_2$ be the parallel from $D_1,$ $D_2$ to $AB,$ $AC,$ respectively. Let $E$ $\in$ $TD_2$ be the point such that $EE_2$ $\parallel$ $D_2E_1$ and let $F$ $\in$ $TD_1$ be the point such that $FF_1$ $\parallel$ $D_1F_2.$ Since $E_1G_1$ passes through the midpoint $U$ of $F_1D_1,$ $FF_2,$ so combining $\tfrac{G_1U}{E_1G_1}$ $=$ $\tfrac{1}{2}$ we get $G_1$ is the centroid of $\triangle FE_1F_2,$ hence $FG_1$ passes through the midpoint $V$ of $E_1F_2$ and $\tfrac{G_1V}{FG_1}$ $=$ $\tfrac{1}{2}.$ Similarly, we can prove $V$ lies on $EG_2$ and $\tfrac{G_2V}{EG_2}$ $=$ $\tfrac{1}{2},$ so we get $EF$ $\parallel$ $G_1G_2.$ Let $P^*$ be the point such that $\triangle ABC$ $\cup$ $P$ $\sim$ $\triangle TD_1D_2$ $\cup$ $P^*.$ By Lemma 10.1 $\Longrightarrow$ $EF$ is parallel to the trilinear polar $\tau_P^*$ of $P^*$ WRT $\triangle TD_1D_2,$ so we conclude that $G_1G_2$ $\parallel$ $\tau_P^*$ $\Longrightarrow$ $G_1G_2$ $\parallel$ $\tau_P.$ $\qquad \blacksquare$

Corollary 11 : Given a $\triangle ABC$ and two points $P,$ $Q.$ Let $\triangle P_aP_bP_c,$ $\triangle Q_aQ_bQ_c$ be the pedal triangle of $P,$ $Q$ WRT $\triangle ABC,$ respectively. Let $J$ $\in$ $\odot (ABC)$ be the point such that the Steiner line of $J$ WRT $\triangle ABC$ is parallel to $PQ.$ Then the line connecting the centroid of $\triangle P_aP_bP_c,$ $\triangle Q_aQ_bQ_c$ is parallel to the trilinear polar of $J$ WRT $\triangle ABC.$

Lemma 12 : Let $I,$ $I_a,$ $I_b,$ $I_c$ be the incenter, A-excenter, B-excenter, C-excenter of $\triangle ABC,$ respectively. Let $(P,Q),$ $(R,S)$ be the isogonal conjugate WRT $\triangle ABC$ such that $PQ$ $\parallel$ $RS$ and let $M,$ $N$ be the midpoint of $PQ,$ $RS,$ respectively. Then $I,$ $I_a,$ $I_b,$ $I_c,$ $M,$ $N$ lie on a conic (rectangular hyperbola).

Proof : First, we prove the following lemma.

Lemma 12.1 : Given a $\triangle ABC$ and the points $E$ $\in$ $CA,$ $F$ $\in$ $AB.$ Let $Y$ be the isotomic conjugate of $E$ WRT $A,$ $C$ and let $Y_0$ be the anticomplement of $Y$ WRT $\triangle ABC.$ Let $Z$ be the isotomic conjugate of $F$ WRT $A,$ $B$ and let $Z_0$ be the anticomplement of $Z$ WRT $\triangle ABC.$ Let $\mathcal{P}$ be the parabola tangent to $BC,$ $CA,$ $AB,$ $EF$ and let $T$ be the tangency point of $\mathcal{P},$ $EF.$ Then $T,$ $Y_0,$ $Z_0$ are collinear.

Proof : Let $\triangle A_0B_0C_0$ be the antimedial triangle of $\triangle ABC$ and let $O_0$ be the circumcenter of $\triangle A_0B_0C_0.$ Let $M$ be the Miquel point of the complete quadrilateral $\mathcal{Q}$ with the sides $BC,$ $CA,$ $AB,$ $EF$ (focus of $\mathcal{P}$ ) and let $M^*$ be the reflection of $M$ in $EF.$ Since $M^*T$ passes through the tangency point of $\mathcal{P}$ and the line at infinity, so $M^*T$ is parallel to the Newton line of $\mathcal{Q}$ $\Longrightarrow$ $M^*T$ $\parallel$ $YZ$ $\parallel$ $Y_0Z_0,$ hence it suffices to prove $M^*$ $\in$ $Y_0Z_0.$

Let $M_1$ be the projection of $O_0$ on $Y_0Z_0.$ Since $M$ is the center of the spiral similarity of $BF$ $\mapsto$ $CE,$ so $\tfrac{MB}{MC}$ $=$ $\tfrac{BF}{CE}$ $=$ $\tfrac{AZ}{AY},$ hence combining $\measuredangle BMC$ $=$ $\measuredangle ZAY$ we get $\triangle AYZ$ $\stackrel{+}{\sim}$ $\triangle MCB.$ On the other hand, since $E,$ $F$ is the midpoint of $B_0Y_0,$ $C_0Z_0,$ respectively, so from Lemma 1 we get $\odot (EFM_1)$ passes through the midpoint $M_0$ of $Y_0Z_0,$ hence note that $M_0F$ $\parallel$ $AC$ we get $\measuredangle M_1EF$ $=$ $\measuredangle M_1M_0F$ $=$ $\measuredangle ZYA$ $=$ $\measuredangle FEM.$ Similarly, we can prove $\measuredangle M_1FE$ $=$ $\measuredangle EFM,$ so we conclude that $M,$ $M_1$ are symmetry WRT $EF$ $\Longrightarrow$ $M_1$ $\equiv$ $M^*.$ $\qquad \square$

Lemma 12.2 : Given a fixed conic $\mathcal{C}$ and two fixed lines $\ell,$ $\ell_1$ such that $\ell_1$ is tangent to $\mathcal{C}$ at $T.$ Let $V$ be a fixed point on $\ell_1$ and let $R$ be a point varies on $\ell.$ Let $S$ $\equiv$ $TR$ $\cap$ $\mathcal{C}$ and let the tangent of $\mathcal{C}$ at $S$ cuts $VR$ at $K.$ Then $K$ lies on a fixed conic when $R$ varies on $\ell.$ Furthermore, if $\ell$ cuts $\ell_1$ at $J,$ then this conic passes through $V$ and the harmonic conjugate $U$ of $V$ WRT $J,$ $T.$

Proof : Let $\tau_R$ be the polar of $R$ WRT $\mathcal{C}$ and let $L$ $\in$ $\tau_R$ be the pole of $\ell$ WRT $\mathcal{C}.$ Since pencil $\tau_R$ $\mapsto$ pencil $VR$ is a homography when $R$ varies on $\ell,$ so $P$ $\equiv$ $\tau_R$ $\cap$ $VR$ lies on a fixed conic $\mathcal{H}_P$ passing through $L,$ $V.$ Since $P$ coincide with $T$ when $R$ coincide with $J,$ so $T$ $\in$ $\mathcal{H}_P.$ Let $Q$ $\equiv$ $\tau_R$ $\cap$ $ST,$ $X$ $\equiv$ $\ell$ $\cap$ $PT.$ Since $X(P,R;K,V)$ $=$ $(P,R;K,V)$ $=$ $(Q,R;S,T)$ $=$ $-1$ $=$ $X(T,J;U,V),$ so $K,$ $U,$ $X$ are collinear. Since the intersection of $TP,$ $UX$ lies on a fixed line $\ell,$ so pencil $TP$ $\mapsto$ pencil $UX$ is a homography $\Longrightarrow$ pencil $VP$ $\mapsto$ pencil $UX$ is a homography, hence we conclude that $K$ $\equiv$ $VP$ $\cap$ $UX$ lies on a fixed conic passing through $U,$ $V.$ $\qquad \square$

Back to the main proof :

We prove the following equivalence property of Lemma 12.

Given a $\triangle ABC$ and a point $L$ at infinity. Let $J,$ $K$ be the isogonal conjugate WRT $\triangle ABC$ such that $L$ $\in$ $JK$ and let $\mathcal{C}$ be the inconic of $\triangle ABC$ with focus $J,$ $K.$ Then the center $T$ of $\mathcal{C}$ lies on a fixed conic.

Let $V$ $\equiv$ $JK$ $\cap$ $BC$ and let the reflection of $BC$ in $JK$ cuts $CA,$ $AB$ at $E,$ $F,$ respectively. Let $A^*,$ $B^*,$ $C^*,$ $E^*,$ $F^*$ be the midpoint of $BC,$ $CA,$ $AB,$ $BE,$ $CF,$ respectively. Clearly, $EF$ is tangent to $\mathcal{C},$ so $T$ lies on the Newton line $E^*F^*$ of the complete quadrilateral formed by $\triangle ABC,$ $EF.$ Let $\mathcal{P}$ be the parabola tangent to $B^*C^*,$ $C^*A^*,$ $A^*B^*,$ $E^*F^*.$ By Lemma 12.1 $\Longrightarrow$ the intersection $G$ of $EF,$ $E^*F^*$ is the tangency point of $E^*F^*,$ $\mathcal{P},$ so note that $GV$ $\equiv$ $EF$ is parallel to the Newton line of the complete quadrilateral formed by $\triangle A^*B^*C^*,$ $E^*F^*$ we get $GV$ passes through the tangency point of $\mathcal{P}$ with the line $\mathcal{L}_{\infty}$ at infinity which is fixed when $J,$ $K$ varies. Finally, by Lemma 12.2 ( $\ell$ $\equiv$ $BC,$ $\ell_1$ $\equiv$ $\mathcal{L}_{\infty}$ ) we conclude that $T$ lies on a fixed conic passing through the points at infinity with the direction $\parallel JK,$ $\perp JK,$ respectively. $\qquad \blacksquare$

Main Proof

Proof of the original problem : Let $O$ be the circumcenter of $\triangle ABC.$ Clearly, $N$ is the centroid of the pedal triangle of $M$ WRT $\triangle ABC,$ so if $M^*$ is the isogonal conjugate of $M$ WRT $\triangle ABC,$ then from Lemma 7 we get $MN$ is perpendicular to the trilinear polar $\tau_M^*$ of $M^*$ WRT $\triangle ABC.$ On the other hand, from Corollary 6 and Corollary 11 we know if $PQ$ $\parallel$ $G_PG_Q,$ then they are parallel to one of the asymptote of the Jerabek hyperbola of $\triangle ABC,$ so combining Corollary 9.2 and Lemma 12 we get $OM$ is perpendicular to $\tau_M^*,$ hence we conclude that $M,$ $N,$ $O$ are collinear. $\qquad \blacksquare$

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Telv Cohl's Geometry Blog

A Problem with some Nice Lemma

by TelvCohl, Aug 9, 2016, 10:20 PM

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