Collinearity of Kiepert perspectors

by TelvCohl, Sep 1, 2018, 7:38 AM

Preliminaries

Lemma 1 : Given a $\triangle ABC$ with isogonal conjugate $(P,P^*), (Q,Q^*).$ Then $S$ $\equiv$ $PQ$ $\cap$ $P^*Q^*$ and $T$ $\equiv$ $PQ^*$ $\cap$ $P^*Q$ are isogonal conjugate WRT $\triangle ABC.$

Proof : Let $X, Y$ be the intersection of $AQ$ with $PQ^*, P^*Q^*,$ respectively. Since $A(Q,Q^*;T,P) = (X,Q^*;T,P) \stackrel{Q}{=} (Y,Q^*;P^*,S) = A(Q,Q^*;P^*,S) = A(Q^*,Q;S,P^*),$ so $(AS, AT)$ are isogonal conjugate WRT $\angle A.$ Similarly, $(BS, BT), (CS, CT)$ are isogonal conjugate WRT $\angle B, \angle C,$ respectively, so we conclude that $S,$ $T$ are isogonal conjugate WRT $\triangle ABC.$ $\qquad \blacksquare$

Lemma 2 : Given a $\triangle ABC$ and a point $U$ lying on the perpendicular bisector of $BC.$ Let $V$ be the point such that $\measuredangle ABU = \measuredangle VCB, \measuredangle ACU = \measuredangle VBC.$ Then $A, U, V$ are collinear.

Proof : Let $W$ be the reflection of $A$ in the perpendicular bisector of $BC,$ then $\measuredangle UCW = \measuredangle VCB, \measuredangle UBW = \measuredangle VBC$ $\Longrightarrow$ $V,W$ are isogonal conjugate WRT $\triangle BUC,$ so we conclude that $A, U, V$ are collinear. $\qquad \blacksquare$

Lemma 3 : Given a $\triangle ABC$ with circumcircle $\odot (O, R)$ and 9-point center $N.$ Let $P,Q$ be the isogonal conjugate WRT $\triangle ABC$ and $M$ be the midpoint of $PQ.$ Then $MN = \frac{OP \cdot OQ}{2R}.$

Proof : Let $\triangle A_1B_1C_1$ be the circumcevian triangle of $P$ WRT $\triangle ABC$ and $A_2, B_2, C_2$ be the reflection of $A_1, B_1, C_1$ in $BC, CA, AB,$ respectively, then from Properties of Hagge circle (Property 1, Corollary 1.1, Property 3) we get $H \in \odot (A_2B_2C_2),$ the reflection $T$ of $Q$ in $N$ is the circumcenter of $\triangle A_2B_2C_2$ and $\triangle A_1B_1C_1 \cup P \stackrel{-}{\sim} \triangle A_2B_2C_2 \cup P$ where $H$ is the orthocenter of $\triangle ABC,$ so $\frac{OP}{2MN} = \frac{OP}{TP} = \frac{\text{circumradius of }\triangle A_1B_1C_1}{\text{circumradius of }\triangle A_2B_2C_2} = \frac{R}{HT} = \frac{R}{OQ}. \qquad \blacksquare$
Main result

Notation : Given a $\triangle ABC$ with centroid $G,$ circumcenter $O,$ orthocenter $H,$ 9-point center $N,$ symmedian point $K$ and $\theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ].$ Let $\triangle A_{\theta}B_{\theta}C_{\theta}$ be the Kiepert triangle of $\triangle ABC$ with angle $\theta,$ $K_{\theta}$ be the perspector of $\triangle ABC, \triangle A_{\theta}B_{\theta}C_{\theta},$ and let $A^*_{\theta}, B^*_{\theta}, C^*_{\theta}, K^*_{\theta}$ be the isogonal conjugate of $A_{\theta}, B_{\theta}, C_{\theta}, K_{\theta}$ WRT $\triangle ABC,$ respectively.

Property 1 : $\color{blue} K \in K_{\theta}K_{-\theta}.$

Proof : Let $\triangle T_aT_bT_c$ be the tangential triangle of $\triangle ABC.$ Note that $\measuredangle CBC_{-\theta} = \measuredangle T_bCB_{\theta}, \measuredangle BCB_{-\theta} = \measuredangle T_cBC_{\theta},$ so we conclude that $B(C,T_b; B_{\theta},B_{-\theta}) = C(B,T_b; B_{\theta},B_{-\theta}) = B(C,T_c;C_{\theta},C_{-\theta}) = C(B,T_c;C_{\theta},C_{-\theta}).$ i.e. $K \equiv BT_b \cap CT_c, K_{\theta} \equiv BB_{\theta} \cap CC_{\theta}, K_{-\theta} \equiv BB_{-\theta} \cap CC_{-\theta}$ are collinear. $\qquad \blacksquare$

Corollary 1.1 : $\color{blue} G \in K_{\theta}K^*_{-\theta}.$

Proof : From Property 1 we get $K$ is the intersection of $K_{\theta}K_{-\theta}, K^*_{\theta}K^*_{-\theta},$ so by Lemma 1 we conclude that $G$ lies on $K_{\theta}K^*_{-\theta}, K_{-\theta}K^*_{\theta}.$ $\qquad \blacksquare$

Property 2 : $\color{blue} H \in A_{\theta}A^*_{\frac{\pi}{2}-\theta}, B_{\theta}B^*_{\frac{\pi}{2}-\theta}, C_{\theta}C^*_{\frac{\pi}{2}-\theta}, K_{\theta}K^*_{\frac{\pi}{2}-\theta}.$

Proof : Simple angle chasing yields $\measuredangle HBA_{\theta} = \measuredangle A^*_{\frac{\pi}{2} - \theta}CB, \measuredangle HCA_{\theta} = \measuredangle A^*_{\frac{\pi}{2} - \theta}BC,$ so by Lemma 2 we get $H$ lies on $A_{\theta}A^*_{\frac{\pi}{2}-\theta}.$ Similarly, we can prove $H \in B_{\theta}B^*_{\frac{\pi}{2}-\theta}, C_{\theta}C^*_{\frac{\pi}{2}-\theta}.$ By Desargues's theorem for $\triangle BC_{\theta}C^*_{\frac{\pi}{2}-\theta}$ and $\triangle CB_{\theta}B^*_{\frac{\pi}{2}-\theta}$ $\Longrightarrow$ $BC, B_{\theta}C_{\theta}, B^*_{\frac{\pi}{2}-\theta}C^*_{\frac{\pi}{2}-\theta}$ are concurrent, so by Desargues's theorem for $\triangle BB_{\theta}B^*_{\frac{\pi}{2}-\theta}$ and $\triangle CC_{\theta}C^*_{\frac{\pi}{2}-\theta}$ we conclude that $H$ lies on $K_{\theta}K^*_{\frac{\pi}{2}-\theta}.$ $\qquad \blacksquare$

Corollary 2.1 : $\color{blue} O \in A^*_{\theta}A^*_{\frac{\pi}{2}-\theta}, B^*_{\theta}B^*_{\frac{\pi}{2}-\theta}, C^*_{\theta}C^*_{\frac{\pi}{2}-\theta}, K_{\theta}K_{\frac{\pi}{2}-\theta}.$

Proof : From Property 2 we get $H$ is the intersection of $K_{\theta}K^*_{\frac{\pi}{2}-\theta}, K_{\frac{\pi}{2}-\theta}K^*_{\theta},$ so by Lemma 1 $\Longrightarrow$ $O$ lies on $K_{\theta}K_{\frac{\pi}{2}-\theta}.$ Analogously, we can prove $O \in A^*_{\theta}A^*_{\frac{\pi}{2}-\theta}, B^*_{\theta}B^*_{\frac{\pi}{2}-\theta}, C^*_{\theta}C^*_{\frac{\pi}{2}-\theta}.$ $\qquad \blacksquare$

Property 3 : Given $\color{blue} \phi, \sigma \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ].$ Then $\color{blue} K^*_{-(\phi + \sigma)} \in K_{\phi} K_{\sigma}.$

Proof : Note that for a fixed $\tau \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ the mapping $\mathbb{I}: K_{\theta} \mapsto K_{\tau - \theta}$ is an involution on the Kiepert hyperbola of $\triangle ABC,$ so $K_{\theta} K_{\tau - \theta}$ passes through the pole $X$ of $\mathbb{I}$ for all $\theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ].$ Consider the case when $\theta = 0$ and $\theta = \frac{\pi}{2}$ we conclude that $X = K^*_{-\tau}$ by Corollary 1.1 and Property 2 $\Longrightarrow$ $K^*_{-\tau} \in K_{\theta} K_{\tau - \theta}.$ $\qquad \blacksquare$

Property 4 : Let $\color{blue} T$ be the intersection of $\color{blue} OH$ with $\color{blue} A_{\theta}A^*_{\theta}.$ Then $\color{blue} \frac{OH}{OT} = 4\cos^2\theta-1.$

Proof : Let $D$ be the intersection of $AA^*_{\theta}$ with the perpendicular bisector of $BC$ and let $BD = CD = \eta,$ then from $\measuredangle ADO = \measuredangle DAH = \measuredangle OAA_{\theta}$ we get $OA^2 = OD \cdot OA_{\theta}$ $\Longrightarrow$ $D$ is the image of $A_{\theta}$ under the inversion WRT $\odot(O),$ so $\eta = \frac{BC \cdot R}{2 \cdot OA_{\theta} \cdot \cos \theta}.$ On the other hand, by Lemma 2 we get $\measuredangle A^*_{\theta}CD = \measuredangle ABC, \measuredangle A^*_{\theta}BD = \measuredangle ACB,$ so $\frac{HA^*_{\theta}}{A_{\frac{\pi}{2}-\theta}A^*_{\theta}} = \frac{AA^*_{\theta}}{DA^*_{\theta}} = \frac{2R \cdot \sin\theta}{\eta} = \frac{4 \cdot OA_{\theta} \cdot \sin \theta \cdot \cos \theta}{BC},$ hence by Menelaus' theorem for $\triangle OHA_{\frac{\pi}{2}-\theta}$ and $\overline{A^*_{\theta}A_{\theta}T}$ we conclude that $\frac{HT}{OT} = \frac{HA^*_{\theta}}{A_{\frac{\pi}{2}-\theta}A^*_{\theta}} \cdot \frac{A_{\frac{\pi}{2}-\theta}A_{\theta}}{OA_{\theta}} = -2\cos 2\theta = -2\left(2\cos^2\theta -1 \right ) \Longrightarrow \frac{OH}{OT} = 4\cos^2\theta-1. \qquad \blacksquare$
Corollary 4.1 : $\color{blue} A_{\theta}A^*_{\theta}, A_{-\theta}A^*_{-\theta}, B_{\theta}B^*_{\theta}, B_{-\theta}B^*_{-\theta}, C_{\theta}C^*_{\theta}, C_{-\theta}C^*_{-\theta}, K_{\theta}K^*_{\theta}, K_{-\theta}K^*_{-\theta}$ are concurrent on $\color{blue} OH.$

Proof : From Property 4, $A_{\theta}A^*_{\theta}, A_{-\theta}A^*_{-\theta}, B_{\theta}B^*_{\theta}, B_{-\theta}B^*_{-\theta}, C_{\theta}C^*_{\theta}, C_{-\theta}C^*_{-\theta}$ are concurrent at $V.$ By Desargues's theorem for $\triangle BC_{\theta}C^*_{\theta}$ and $\triangle CB_{\theta}B^*_{\theta}$ we get $BC, B_{\theta}C_{\theta}, B^*_{\theta}C^*_{\theta}$ are concurrent, so by Desargues's theorem for $\triangle BB_{\theta}B^*_{\theta}$ and $\triangle CC_{\theta}C^*_{\theta}$ we conclude that $V$ lies on $K_{\theta}K^*_{\theta}.$ Analogously, we can prove $V \in K_{-\theta}K^*_{-\theta}.$ $\qquad \blacksquare$

Property 5 : $\color{blue} N \in K_{\theta}K_{\frac{\pi}{2}+\theta}.$

Proof : Let $D_{-\theta}, E_{-\theta}, F_{-\theta}$ be the midpoint of $AA_{-\theta}, BB_{-\theta}, CC_{-\theta},$ respectively. First, note that $AB_{\theta}A_{-\theta}C_{\theta}, BC_{\theta}B_{-\theta}A_{\theta}, CA_{\theta}C_{-\theta}B_{\theta}$ are parallelogram, so $\triangle D_{-\theta}E_{-\theta}F_{-\theta}$ is the medial triangle of $\triangle A_{\theta}B_{\theta}C_{\theta}.$ On the other hand, simple angle chasing yields the second intersection of $AA_{\frac{\pi}{2}-\theta}$ with $\odot (A_{\frac{\pi}{2}-\theta}BC)$ lies on $\odot (B_{\theta}), \odot (C_{\theta})$ where $\odot (B_{\theta}), \odot (C_{\theta})$ is the circle with center $B_{\theta}, C_{\theta}$ and containing $A,$ so $AK_{\frac{\pi}{2}-\theta} \perp E_{\theta} F_{\theta}.$ Analogously, we can prove $BK_{\frac{\pi}{2}-\theta} \perp F_{\theta} D_{\theta}, CK_{\frac{\pi}{2}-\theta} \perp D_{\theta} E_{\theta},$ so by Sondat's theorem for $\triangle ABC$ and $\triangle D_{-\theta}E_{-\theta}F_{-\theta}$ we conclude that $N \in K_{-\theta} K_{\frac{\pi}{2}-\theta}.$ $\qquad \blacksquare$

Application

Notation : Let $G, O, N, K, F_1, F_2, S_1, S_2$ be the centroid, circumcenter, 9-point center, symmedian point, 1st Fermat point, 2nd Fermat point, 1st Isodynamic point, 2nd Isodynamic point of $\triangle ABC,$ respectively.

Theorem (Basic properties of Fermat points and Isodynamic points) :

(1) $\qquad$ $G \in F_1S_2, F_2S_1.$ $\qquad$ $K \in F_1F_2, S_1S_2.$ $\qquad$ $\overline{GON} \parallel F_1S_1 \parallel F_2S_2.$ $\qquad$ (2) $\qquad$ $GK$ is the G-symmedian of $\triangle GF_1F_2.$ $\qquad$ (3) (Lester circle) $O, N, F_1, F_2$ are concyclic.

Proof : (1) is a consequence of the results in the previous section. Let $T$ be the intersection of $GO, F_1F_2$ and $\triangle^{1}_{\mathbf{N}}, \triangle^{2}_{\mathbf{N}}$ be the 1st Napoleon triangle, 2nd Napoleon triangle of $\triangle ABC,$ respectively. First, note that $O$ lies on $S_1S_2,$ so $T$ is the reflection of $O$ in $G.$ Since $G$ is the center of $\triangle^{i}_{\mathbf{N}},$ $F_{j}$ lies on the circumcircle of $\triangle^{i}_{\mathbf{N}}$ and $O,F_i$ are isogonal conjugate of $\triangle^{i}_{\mathbf{N}}$ for $\{ i, j \} = \{ 1, 2 \},$ so by Lemma 3 we get $\frac{KF_1}{KF_2} = \frac{TF_1}{TF_2} = \frac{GO \cdot GF_1}{GO \cdot GF_2} \cdot \frac{\text{circumradius of }\triangle^{2}_{\mathbf{N}}}{\text{circumradius of }\triangle^{1}_{\mathbf{N}}} = \left ( \frac{GF_1}{GF_2} \right )^2,$ hence $GK$ is the G-symmedian of $\triangle GF_1F_2$ and (2) is proved. Finally, from the previous conclusion and $(T,K; F_1, F_2) = -1$ we get $\overline{GON}$ is tangent to $\odot (GF_1F_2)$ at $G,$ so $TO \cdot TN = {TG}^2 = TF_1 \cdot TF_2$ and hence we conclude that $O, N, F_1, F_2$ are concyclic. i.e. (3) is proved. $\qquad \blacksquare$

Remark : It is possible to prove (2) or (3) directly.

1. Proof of (2) without using (1) :

Since $F_1, F_2$ are antigonal conjugate WRT $\triangle ABC,$ so the center of the Kiepert hyperbola $\mathcal{K}$ of $\triangle ABC$ is the midpoint of $F_1F_2$ $\Longrightarrow$ the isogonal conjugate of $\mathcal{K}$ WRT $\triangle GF_1F_2$ is the perpendicular bisector of $F_1F_2,$ hence the tangent of $\mathcal{K}$ at $G$ is the G-symmedian of $\triangle GF_1F_2.$ On the other hand, $\mathcal{K}$ is the isotomic conjugate of $GK$ WRT $\triangle ABC,$ so $GK$ is tangent to $\mathcal{K}$ at $G$ and hence $GK$ is the G-symmedian of $\triangle GF_1F_2.$ $\qquad \blacksquare$

2. Proof of (3) without using (1) and (2) :

Lemma : Given a $\triangle ABC$ and a circle $\Omega$ passing through $B$ and $C.$ Let $E, F$ be the intersection of $\Omega$ with $CA, AB,$ respectively and let $Y \in BE, Z \in CF$ be the points such that $\frac{BE}{EY} = \frac{CF}{FZ} = \kappa.$ Suppose that $U, V$ is the intersection of the tangent of $\odot (AYZ)$ at $A$ with $BC, YZ,$ respectively, then $\frac{UA}{AV} = \kappa.$

Proof : Let $B^* \in CA, C^* \in AB$ be the points such that $\frac{BA}{AC^*} = \frac{CA}{AB^*} = \kappa$ and let $A^*$ be the intersection of $B^*Z, C^*Y.$ Clearly, $\triangle ABC, \triangle A^*B^*C^*$ are homothetic and $\triangle AEY \stackrel{-}{\sim} \triangle AFZ,$ so $\frac{VY}{VZ} = \left ( \frac{AY}{AZ} \right )^2 = \left ( \frac{AE}{AF} \right )^2 = \frac{AE}{AF} \cdot \frac{AB}{AC} = \frac{C^*Y}{B^*Z} \cdot \frac{A^*B^*}{A^*C^*},$ hence by Menelaus' theorem we conclude that $V \in B^*C^*$ $\Longrightarrow$ $\frac{UA}{AV} = \frac{\text{dist}(A,BC)}{\text{dist}(A,B^*C^*)} = \kappa.$ $\qquad \blacksquare$

Back to the main problem :

Let $\triangle N^{1}_aN^{1}_bN^{1}_c, \triangle N^{2}_aN^{2}_bN^{2}_c$ be the 1st Napoleon triangle, 2nd Napoleon triangle of $\triangle ABC,$ respectively and let $T^*$ be the intersection of $F_1F_2$ with the tangent of $\odot (GF_1F_2)$ at $G.$ Since $F_1, F_2$ is the reflection of $N^{2}_a, N^{1}_a$ in $GN^{1}_a, GN^{2}_a,$ respectively, so by Lemma for $\triangle GN^{1}_aN^{2}_a$ we get the reflection $\overline{T}$ of $T^*$ in $G$ lies on the perpendicular bisector of $BC.$ Similarly, we can prove $\overline{T}$ lies on the perpendicular bisector of $CA, AB,$ so $\overline{T} \equiv O$ and hence we conclude that $T^*N \cdot T^*O = T^*G^2 = T^*F_1 \cdot T^*F_2$ $\Longrightarrow$ $O, N, F_1, F_2$ are concyclic. $\qquad \blacksquare$

Kiepert perspector

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Telv Cohl's Geometry Blog

Collinearity of Kiepert perspectors

by TelvCohl, Sep 1, 2018, 7:38 AM

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