Connection between Inconic and Circumconic

by TelvCohl, Aug 7, 2018, 2:57 PM

Notation : Given a $\triangle ABC$ and a point $T.$ Let $\tau (T, \triangle ABC),$ $\mathcal{C} (T, \triangle ABC),$ $\mathcal{K} (T, \triangle ABC)$ be the tripolar of $T,$ inconic of $\triangle ABC$ with perspector $T,$ circumconic of $\triangle ABC$ with perspector $T$ WRT $\triangle ABC,$ respectively. When no confusion can be caused, we abbreviate $\square (T, \triangle ABC)$ as $\square_{T}.$

Property : Given a $\triangle ABC$ with isogonal conjugate $(P,Q), (R,S).$ Then $\tau_R$ is tangent to the $\mathcal{K}_{P}$ if and only if $S\in \mathcal{C}_{Q}.$

Proof :

Lemma 1 : Given a $\triangle ABC$ with isogonal conjugate $(P,Q), (R,S).$ Then $P\in\tau_{R} \Longleftrightarrow R\in\mathcal{K}_{P} \Longleftrightarrow S \in \tau_{Q}.$

Proof : First, when a line $\ell$ passing through $P$ varies around $P,$ the mapping $\ell \cap AB \mapsto \ell \cap AC$ is a homography between $AB$ and $AC,$ so the mapping sending the harmonic conjugate of $\ell \cap AB$ WRT $(A,B)$ to the harmonic conjugate of $\ell \cap AC$ WRT $(A,C)$ is a homography, too $\Longrightarrow$ the locus of the tripole of $\ell$ WRT $\triangle ABC$ is a circumconic of $\triangle ABC,$ hence to prove the first part it suffices to prove $P\in\tau_{R}$ when $R \in \mathcal{K}_{P}.$ Let $\triangle P^aP^bP^c$ be the anticevian triangle of $P$ WRT $\triangle ABC,$ then $\mathcal{K}_{P}$ is tangent to $P^bP^c, P^cP^a, P^aP^b$ at $A, B, C,$ respectively. Let $\triangle R_aR_bR_c$ be the cevian triangle of $R$ WRT $\triangle ABC$ and assume $R\in\mathcal{K}_{P},$ then $P^a$ lie on the polar $R_bR_c$ of $R_a$ WRT $\mathcal{K}_{P},$ so note that $\tau_{R}$ passes through the harmonic conjugate of $R_b, R_c$ WRT $(A,C), (A,B),$ respectively we get $P\in\tau_{R}.$

For the second part, let $X_P, X_Q$ be the intersection of $BC$ with $\tau_{P}, \tau_{Q},$ respectively, then $AX_P, AX_Q$ are isogonal conjugate WRT $\angle A$ $\Longrightarrow$ the isogonal conjugate of $\tau_{Q}$ WRT $\triangle ABC$ is tangent to $P^bP^c$ at $A.$ Similarly, we can prove the isogonal conjugate of $\tau_{Q}$ WRT $\triangle ABC$ is tangent to $P^cP^a, P^aP^b$ at $B,C,$ respectively, so $\mathcal{K}_{P}$ is the isogonal conjugate of $\tau_{Q}$ WRT $\triangle ABC$ $\qquad \blacksquare$

Lemma 2 : Given a $\triangle ABC$ and a point $P.$ Then the perspector of the circumconic of $\triangle ABC$ with center $P$ is the isotomcomplement of the anticomplement of $P$ WRT $\triangle ABC.$

Proof : Let $Q$ be the anticomplement of $P$ WRT $\triangle ABC,$ $R$ be the isotomcomplement of $Q$ WRT $\triangle ABC$ and $A_P$ be the reflection of $A$ in $P.$ By Lemma 1, it suffices to prove that the isogonal conjugate $A_P^*$ of $A_P$ WRT $\triangle ABC$ lies on $\tau_S$ where $S$ is the isogonal conjugate $R$ WRT $\triangle ABC.$ Let $\triangle Q_aQ_bQ_c$ be the cevian triangle of $Q$ WRT $\triangle ABC,$ $\triangle R^aR^bR^c$ be the anticevian triangle of $R$ WRT $\triangle ABC$ and $B_s, C_s$ be the intersection of $CA, AB$ with $\tau_S,$ respectively. Note that $\triangle Q_aQ_bQ_c, \triangle R^aR^bR^c$ are homothetic and $BQ \parallel CA_P, CQ \parallel BA_P,$ so we conclude that $B(C,A_P^*;B_s,A) = B(A,A_P;R^a,C) = Q_c(A,C;Q_a,\parallel BC) = Q_b(B,A;Q_a,\parallel BC) = C(A,A_P;B, R^a) = C(B,A_P^*;A,C_s).$ i.e. $A_P^*\in\tau_S.$ $\qquad \blacksquare$

Corollary 2.1 : Given a $\triangle ABC$ and points $P, Q.$ Then $\tau_Q$ is tangent to $\mathcal{K}_{P}$ if and only if $P \in \mathcal{C}_{Q}.$

Proof : Consider the homology taking $Q$ to the centroid of $\triangle ABC,$ then the desired result simplified follows from Lemma 2. $\qquad \blacksquare$

Back to the main problem :

By Lemma 1, $\mathcal{K}_{P}, \mathcal{K}_{S}$ is the isogonal conjugate of $\tau_Q, \tau_R$ WRT $\triangle ABC,$ respectively, so $\tau_R$ is tangent to the $\mathcal{K}_{P}$ if and only if $\tau_Q$ is tangent to $\mathcal{K}_{S}.$ But from Corollary 2.1 we know $\tau_Q$ is tangent to $\mathcal{K}_{S}$ if and only if $S \in \mathcal{C}_{Q},$ so we conclude that $\tau_R$ is tangent to the $\mathcal{K}_{P}$ if and only if $S\in \mathcal{C}_{Q}.$ $\qquad \blacksquare$

This post has been edited 2 times. Last edited by TelvCohl, Feb 14, 2020, 10:37 AM

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Connection between Inconic and Circumconic

by TelvCohl, Aug 7, 2018, 2:57 PM

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