Ptolemy triangle and QA-Orthopole-circle

by TelvCohl, Feb 28, 2018, 11:08 AM

Notation

For two lines $\tau, \eta$ and a points $P,$ let $\mathbb{T}_{\tau,\eta}(P)$ be the midpoint between the reflection of $P$ in $\tau$ and the reflection of $P$ in $\eta.$

Preliminaries

Lemma 1 : Given two lines $\ell_1, \ell_2$ and two points $P,Q.$ Let $X$ be the intersection of $\ell_1, \ell_2.$ Then $\triangle XPQ \stackrel{-}{\sim} \triangle X\mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q).$

Proof : Let $P_i, Q_i$ be the projection of $P,Q$ on $\ell_i,$ respectively where $i \in \{1,2 \}$ and let $Y \equiv \mathbb{T}_{\ell_1,\ell_2} (P), Z \equiv \mathbb{T}_{\ell_1,\ell_2} (Q),$ then $Y, Z$ is the orthocenter of $\triangle XP_1P_2, \triangle XQ_1Q_2,$ respectively, so $\frac{XP}{XQ} = \frac{XY}{XZ}.$ Since $XY, XZ$ is the isogonal conjugate of $XP, XQ$ WRT $\angle (\ell_1, \ell_2),$ respectively, so we conclude that $\triangle XPQ \stackrel{-}{\sim} \triangle XYZ.$ $\qquad \blacksquare$

Corollary 1.1 : Given two lines $\ell_1, \ell_2$ and two points $P,Q.$ Then $PQ$ and $\mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q)$ are antiparallel WRT $\angle (\ell_1,\ell_2).$

Corollary 1.2 : Given two lines $\ell_1, \ell_2$ and three points $P,Q, R.$ Then $\triangle PQR \stackrel{-}{\sim} \triangle \mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q)\mathbb{T}_{\ell_1,\ell_2} (R).$

Corollary 1.3 : Given a $\triangle ABC$ and a point $P \in BC.$ Let $E, F$ be the projection of $B, C$ on $CA, AB,$ respectively. Then $\mathbb{T}_{AB,AC} (P)$ lies on $EF.$

Lemma 2 : Given a quadrangle $P_1P_2P_3P_4$ with Euler-Poncelet point $E$ and Isogonal center $I.$ Then $\mathbb{T}_{P_1P_2,P_3P_4} (I) = \mathbb{T}_{P_1P_3,P_2P_4} (I) = \mathbb{T}_{P_1P_4,P_2P_3} (I) = E.$

Proof : It suffices to prove $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E.$ Let $O_4$ be the circumcenter of $\triangle P_1P_2P_3$ and let the perpendicular from $E$ to $P_2P_3$ cuts $P_1P_4, IO_4$ at $U, V,$ respectively. By Miquel triangle and Isogonal center (Property 11), $I$ is the image of the isogonal conjugate $Q_4$ of $P_4$ WRT $\triangle P_1P_2P_3$ under the inversion WRT $\odot (O_4),$ so $\measuredangle VIP_1 = \measuredangle Q_4P_1O_4 = \measuredangle (\perp P_2P_3, P_1P_4) = \measuredangle VUP_1$ $\Longrightarrow$ $P_1, I, U, V$ are concyclic. Notice that $E$ is the orthopole of $O_4Q_4$ WRT $\triangle P_1P_2P_3$ we get $P_1V \perp O_4Q_4,$ so $U$ is the projection of $I$ on $P_1P_4.$ Similarly, the intersection of the perpendicular from $I,E$ to $P_2P_3, P_1P_4,$ respectively lies on $P_2P_3,$ so $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E.$ $\qquad \blacksquare$

Lemma 3 : Given a quadrilateral $P_1P_2P_3P_4$ and a point $T.$ Then the pedal circle of $T$ WRT $\triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3$ are concurrent.

Proof : Let $H_{ij}$ be the projection of $T$ on $P_iP_j$ where $i<j$ and $i, j \in \{ 1,2,3,4 \}.$ Consider an inversion with center $T$ and denote the image of $\Box$ as ${\Box}^*,$ then $H^*_{ij}, H^*_{ik}, H^*_{il}$ are collinear where $\{ i, j, k, l \} = \{ 1,2,3,4 \},$ so $\odot (H^*_{23}H^*_{24}H^*_{34}), \odot (H^*_{13}H^*_{14}H^*_{34}), \odot (H^*_{12}H^*_{14}H^*_{24}), \odot (H^*_{12}H^*_{13}H^*_{23})$ are concurrent $\Longrightarrow$ $\odot (H_{23}H_{24}H_{34}), \odot (H_{13}H_{14}H_{34}), \odot (H_{12}H_{14}H_{24}), \odot (H_{12}H_{13}H_{23})$ are concurrent. $\qquad \blacksquare$

Lemma 4 : Given a $\triangle ABC$ with orthocenter $H$ and a point $P.$ Let $Y, Z$ be the reflection of $P$ in $CA, AB,$ respectively. Then the anti-Steiner point $T$ of $HP$ WRT $\triangle ABC$ lies on $\odot (AYZ).$

Proof : Let $E, F$ be the reflection of $H$ in $CA, AB,$ respectively, then $T$ lies on $EY, FZ$ and $E, F$ lie on $\odot (ABC),$ so $\measuredangle YAZ = \measuredangle EAF = \measuredangle ETF = \measuredangle YTZ$ $\Longrightarrow$ $T \in \odot (AYZ).$ $\qquad \blacksquare$

Main result

In this section, I'll give a brief introduction to Ptolemy triangle, QA-Orthopole-circle and related transformation WRT a quadrangle. All symbols in this section bear the same meaning.

Notation

Given a quadrangle $P_1P_2P_3P_4$ with Euler-Poncelet point $E,$ Isogonal center $I$ and a point $T.$ Let $T_X \equiv \mathbb{T}_{P_1P_4, P_2P_3} (T), T_Y \equiv \mathbb{T}_{P_1P_3, P_2P_4} (T), T_Z \equiv \mathbb{T}_{P_1P_2, P_3P_4} (T).$

Property 1 : $\triangle T_XT_YT_Z$ is the image of the pedal triangle of $P_i$ WRT $\triangle P_jP_kP_l$ under the spiral similarity with center $E,$ ratio $\frac{IT}{IP_i}$ and angle $\measuredangle TIP_i$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$

Proof : It suffices to prove the case when $i=4.$ Let $\triangle A_4B_4C_4$ be the pedal triangle of $P_4$ WRT $\triangle P_1P_2P_3.$ From Lemma 2 $\Longrightarrow$ $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E,$ so combining Corollary 1.2 for lines $P_1P_4, P_2P_3$ and points $P_4, I, T$ we get $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_XA_4.$ Similarly, we can prove $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_YB_4,$ $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_ZC_4,$ so $\triangle T_XT_YT_Z$ is the image of $\triangle A_4B_4C_4$ under the spiral similarity with center $E,$ ratio $\frac{IT}{IP_4}$ and angle $\measuredangle TIP_4.$ $\qquad \blacksquare$

Corollary 2 : $E$ lies on $\odot (T_XT_YT_Z).$

Corollary 3 : Let $O_T$ be the circumcenter of $\triangle T_XT_YT_Z.$ Then the mapping $T \mapsto O_T$ is linear.

Proof : Let $M_4$ be the circumcenter $\triangle A_4B_4C_4,$ then from Property 1 we get $\triangle ITP_4 \stackrel{-}{\sim} \triangle EO_TM_4,$ so the mapping $T \mapsto O_T$ is linear. $\qquad \blacksquare$

Remark : $\triangle T_XT_YT_Z$ is called the Ptolemy triangle of $T$ and the mapping $T \mapsto O_T$ is called the QA-Orthopole transformation WRT the quadrangle $P_1P_2P_3P_4.$

Property 4 : Let $\varrho$ be a line passing through $T,$ $R_{ij}$ be the intersection of $\varrho$ with $P_iP_j$ and $H_{kl}$ be the projection of $R_{ij}$ on $P_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Then $H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23}$ are concurrent on $\odot (O_T).$

Proof : From Corollary 1.3 $\Longrightarrow$ $T_Y \in H_{13}H_{24},$ $T_Z \in H_{12}H_{34},$ so note that $\{ \varrho, H_{13}H_{24} \}, \{ \varrho, H_{12}H_{34}\}$ are antiparallel WRT $\angle (P_1P_3, P_2P_4), \angle (P_1P_2, P_3P_4),$ respectively we get $\measuredangle (H_{13}H_{24}, H_{14}H_{23}) = \measuredangle (H_{13}H_{24}, P_2P_4) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (P_3P_4, H_{12}H_{34}) = \measuredangle (P_1P_3, \varrho) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (\varrho, P_1P_2) = \measuredangle (P_1P_3, P_1P_2) + \measuredangle (P_2P_4, P_3P_4) = \measuredangle T_YT_XT_Z,$ hence the intersection of $H_{13}H_{24}, H_{12}H_{34}$ lies on $\odot (O_T).$ Similarly, the intersection of $\{ H_{12}H_{34}, H_{14}H_{23} \}, \{ H_{14}H_{23}, H_{13}H_{24}\}$ lies on $\odot (O_T),$ so $H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23}, \odot (O_T)$ are concurrent at $T_{\varrho}.$ $\qquad \blacksquare$

Remark : The mapping $\varrho \mapsto T_{\varrho}$ is called the Quang Duong’s transformation WRT the quadrangle $P_1P_2P_3P_4.$

Property 5 : Let $O_i$ be the circumcenter of $\triangle P_jP_kP_l$ and $V_i$ be the orthopole of $O_iT$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Then $V_i$ lies on $\odot (O_T)$ for any $i \in \{ 1,2,3,4 \}.$

Proof : Let $U$ be the midpoint of $P_2P_3,$ $Q$ be the projection of $T$ on $P_2P_3$ and $E_1, E_4$ be the Poncelet point of $P_2P_3P_4T, P_1P_2P_3T,$ respectively, then $E, U, E_1, V_1$ lie on the 9-point circle of $\triangle P_2P_3P_4$ and $E, U, E_4, V_4$ lie on the 9-point circle of $\triangle P_1P_2P_3.$ From Lemma 3 $\Longrightarrow$ the pedal circle of $T$ WRT $\triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3$ are concurrent at $R,$ so note that $Q, U, E_1, E_4$ lie on the 9-point circle of $\triangle TP_2P_3$ we get $\measuredangle EV_1R + \measuredangle RV_4E = \measuredangle EV_1E_1 + \measuredangle E_1V_1R + \measuredangle RV_4E_4 + \measuredangle E_4V_4E = \measuredangle EUE_1 + \measuredangle E_1QR + \measuredangle RQE_4 + \measuredangle E_4UE = \measuredangle E_1QE_4 + \measuredangle E_4UE_1 = 0,$ hence $E, U, V_1, V_4$ are concyclic. Similarly, we can prove $V_2, V_3$ lie on $\odot (EUV_4),$ so $E, U, V_1, V_2, V_3, V_4$ lie on a circle $\odot (O_T^*).$

Let $N_i$ be the 9-point center of $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Since $O_1O_4, O_1T$ is the Steiner line of $U, V_1$ WRT the medial triangle of $\triangle P_2P_3P_4,$ respectively, so $\measuredangle O_T^*N_1N_4 = \measuredangle V_1EU = \measuredangle O_4O_1T.$ Similarly, we can prove $\measuredangle O_T^*N_4N_1 = \measuredangle O_1O_4T,$ so $\triangle O_T^*N_1N_4 \stackrel{-}{\sim} \triangle TO_1O_4.$ Analogously, we can prove $\triangle O_T^*N_2N_4 \stackrel{-}{\sim} \triangle TO_2O_4,$ $\triangle O_T^*N_3N_4 \stackrel{-}{\sim} \triangle TO_3O_4,$ so $\triangle N_1N_2N_3 \cup O_T^* \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T.$

On the other hand, from Interesting Properties related to Four 9-point Centers (Property 1 and Corollary 11) we get $\triangle N_1N_2N_3 \cup M_4 \cup E \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup P_4 \cup I$ and $E, I$ is the image of $M_4, P_4$ under the inversion WRT $\odot (N_1N_2N_3), \odot (O_1O_2O_3),$ so combining $\triangle EO_TM_4 \stackrel{-}{\sim} \triangle ITP_4,$ by Property 1, we conclude that $\triangle N_1N_2N_3 \cup O_T \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T$ $\Longrightarrow$ $O_T \equiv O_T^*$ and $\odot (O_T) \equiv \odot (O_T^*).$ $\qquad \blacksquare$

Remark : $\odot (O_T)$ is called the QA-Orthopole-circle of $T$ WRT the quadrangle $P_1P_2P_3P_4.$

Property 6 : The image of $\varrho$ under the QA-Orthopole transformation WRT $P_1P_2P_3P_4$ is the perpendicular bisector of $ET_{\varrho}.$

Proof : Let $\gamma$ be the perpendicular bisector of $ET_{\varrho},$ then note that $ET_X$ and $IT$ are antiparallel WRT $\angle (P_1P_4, P_2P_3),$ by Corollary 1.1 and Lemma 2, we get $\measuredangle (EO_T, \gamma) = \measuredangle (ET_X, T_{\varrho}T_X) = \measuredangle (\varrho, IT),$ so $\gamma$ is the image of $\varrho$ under the QA-Orthopole transformation WRT $P_1P_2P_3P_4.$ $\qquad \blacksquare$

Let $\triangle XYZ$ be the diagonal triangle of the quadrangle $P_1P_2P_3P_4$ where $X, Y, Z$ is the intersection of $\{ P_1P_4, P_2P_3 \}, \{ P_1P_3, P_2P_4 \}, \{ P_1P_2, P_3P_4 \},$ respectively, then from New proof of the property of Poncelet point and Corollary 2 we know $E$ is one of the intersection of $\odot (O_T)$ and $\odot (XYZ).$ The following property gives the character of another intersection of these two circles.

Property 7 : Let $H^{\mathbf{D}}$ be the orthocenter of $\triangle XYZ.$ Then the anti-Steiner point $S$ of $H^{\mathbf{D}}T$ WRT $\triangle XYZ$ lies on $\odot (O_T).$

Proof : Let $X^T, Y^T. Z^T$ be the reflection of $T$ in $YZ, ZX, XY,$ respectively, then $T_X, T_Y, T_Z$ lie on $\odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T),$ respectively (well-known). On the other hand, from Lemma 4 we know $S$ lies on $\odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T),$ so $\measuredangle T_YST_Z = \measuredangle T_YSX^T + \measuredangle X^TST_Z = \measuredangle T_YYX^T + \measuredangle X^TZT_Z.$ Note that $\{ YT, YT_Y \}, \{ ZT, ZT_Z \}$ are isogonal conjugate WRT $\angle (T_1T_3, T_2T_4), \angle (T_1T_2, T_3T_4),$ respectively, so we get $\measuredangle T_YST_Z = \measuredangle T_YYP_1 + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle P_1ZT_Z = \measuredangle P_4YT + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle TZP_4 = \measuredangle P_4YZ + \measuredangle P_1YZ + \measuredangle YZP_1 + \measuredangle YZP_4 = \measuredangle P_2P_4P_3 + \measuredangle P_3P_1P_2,$ hence by Property 1 we conclude that $\measuredangle T_YST_Z = \measuredangle T_YT_XT_Z$ $\Longrightarrow$ $S \in \odot (O_T).$ $\qquad \blacksquare$

Telv Cohl's Geometry Blog

Ptolemy triangle and QA-Orthopole-circle

by TelvCohl, Feb 28, 2018, 11:08 AM

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