Orthopole - Anti-Steiner point related problem

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Orthopole - Anti-Steiner point related problem

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#1 h Aug 17, 2020, 8:55 AM • 1 Y

Y by amar_04

Let $ABC$ is a triangle with circumcircle $(O)$ and Anti-median triangle $A'B'C'$ ( $A$ is the midpoint of segment $B'C'$ , and similar to $B, C$ ). Let $A_1B_1C_1$ is the triangle created from three tangents of circle $(O)$ at $A, B, C$ .
a) Prove that $B'B_1, C'C_1, BC$ are concurrent at point $X$ . Similarly denote $Y, Z$ .
b) Prove that the orthopole of the Euler line of triangle $ABC$ with respect to triangle $XYZ$ lies on the circumcircle of triangle $A'B'C'$ .

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#2 h Aug 18, 2020, 10:26 PM • 5 Y

Y by Kar-98k, RudraRockstar, andrenguyen, Abhaysingh2003, Mango247

What a beautiful result! Hatsoff to you. Here is the Original Diagram

andrenguyen wrote:

Let $ABC$ is a triangle with circumcircle $(O)$ and Anti-median triangle $A'B'C'$ ( $A$ is the midpoint of segment $B'C'$ , and similar to $B, C$ ). Let $A_1B_1C_1$ is the triangle created from three tangents of circle $(O)$ at $A, B, C$ .
a) Prove that $B'B_1, C'C_1, BC$ are concurrent at point $X$ . Similarly denote $Y, Z$ .
b) Prove that the orthopole of the Euler line of triangle $ABC$ with respect to triangle $XYZ$ lies on the circumcircle of triangle $A'B'C'$ .

$\textbf{LEMMA:-}$ $ABC$ be a triangle and let $\Delta A'B'C'$ be the anticomplementary triangle of $\Delta ABC$ . Let $T$ be the Pole of $\overline{AC}$ WRT $\odot(ABC)$ and let $\overline{TB'}\cap\overline{CB}=\{K\}$ . Then $\overline{AK}$ is Perpendicular to the Euler Line $(\mathcal E)$ of $\Delta ABC$ . Diagram

Proof:- Let $O,H,G$ denote the Circumcenter , Orthocenter and Centroid of $\Delta ABC$ . Let $\overline{B'C'}\cap\odot(ABC)=Q$ and $\overline{A'B'}\cap\odot(ABC)=\{P\}$ . Let $\overline{AP}\cap\overline{CQ}=\{X\}$ and $\overline{AC}\cap\overline{PQ}=\{Z\}$ and let $\overline{B'X}\cap\overline{AC}=\{R\}$ . Let $\overline{CX}\cap\overline{AB}=\{M\}$ and $\overline{AX}\cap\overline{BC}=\{N\}$ and let $\Delta M_AM_BM_C$ be the Complementary Triangle of $\Delta ABC$ . So, by Pappu's Theorem on $\overline{M_C-M-A}$ and $\overline{M_A-N-C}$ we get that $X\in\overline{OG}\implies\{X\}\in\mathcal E$ and by Pascal's Theorem on $QAAPCC$ we get $X\in\overline{B'T}$ . Now by Brocard's Theorem we get $\overline{OX}\perp\overline{B'Z}$ . Now let $\overline{KC}\cap\overline{B'Z}=K^*$ . Then $-1=(A,C;R,Z)\overset{B'}{=}(K,K^*;C,\infty_{\overline{BC}})\implies CK=CK^*$ but $CA'=CB'$ . Hence, $\overline{AK}\|\overline{B'Z}\perp\mathcal E$ . $(\bigstar)$
$\text{[math]}$ $\blacksquare$ .

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Coming back to the Problem. Let $\overline{CC_1}\cap\overline{BC'}=\{M\}$ ; $\overline{B_1C}\cap\overline{B'B}=\{N\}$ . Let $\overline{CN}\cap\overline{BM}=\{T\}$ ; $\overline{CM}\cap\odot(ABC)=\{L\}$ and $\overline{BN}\cap\overline{AC}=F$ . Now notice that $-1=(C,L;A,B)\overset{C}{=}(T,M;B,\infty_{\overline{MT}})\implies MB=MT$ and as $FA=FC$ . So we get $A,M,N$ are collinear. So, $\{\Delta C'B'B,\Delta C_1B_1B\}$ are Perspective triangles, hence by Desargues Theorem we get that $\overline{C'C_1},\overline{B'B_1},\overline{BC}$ are concurrent at $X$ and we are done with the first part. Now from $\textbf{LEMMA}$ we get that $\overline{A'X}\|\overline{B'Y}\|\overline{C'Z}\perp\mathcal E$ . Now it suffices to prove the following Proposition, we will further Prove that the Concurrency Point is the Anti-Steiner Point of $\mathcal E$ WRT $\Delta A'B'C'$ .

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$\textbf{PROPOSITION:-}$ $ABC$ be a triangle. Let $\mathcal E$ be the Euler Line of $\Delta ABC$ and $D\in\mathcal E$ such that $\overline{AD}\perp\mathcal{E}$ . Let $\Delta M_AM_BM_C$ be the Complementary Triangle of $\Delta ABC$ and $\overline{AD}\cap\odot(ABC)=Y$ and $\overline{AD}\cap\overline{M_BM_C}=K$ . Let $\{X\}$ be the Euler Reflection Point of $\Delta ABC$ . Let $\overline{CK}\cap\overline{M_AM_C}=\{Q\{$ and $\overline{M_AM_B}\cap\overline{BK}=\{P\}$ . Then $\overline{XD}\perp\overline{PQ}$ .

By Pappu's Theorem on $\overline{M_C-K-M_B}$ and $\overline{B-M_A-C}$ we have $A,P,Q$ collinear. It's well known that the Simson Line of $X$ is at $\infty^{\perp\overline{OH}}\implies\{AX,AY\}$ are Isogonal Conjugates WRT $\Delta ABC\implies\overline{XY}\|\overline{BC}\implies M_AX=M_AY$ . Let $\overline{AM_A}\cap\mathcal E=\{G\}$ where $G$ is the Centroid of $\Delta ABC$ . Let $\{V\}\in\overline{AY}$ such that $\overline{M_AV}\perp\overline{AY}$ , then $\mathcal{E}\|\overline{M_AV}\implies AG:GM_A=AD:DV=2:1\implies DV=YV\implies M_AD=M_AY=M_AX$ . So if $\mathcal E\cap\overline{BC}=\{T\}$ then $TDM_AX$ is a cyclic quad as $\mathcal E$ and $\overline{XT}$ are reflections of each other over $\overline{BC}$ . Now from $(\bigstar)$ we get that $\overline{QB}\|\overline{AY}\implies\measuredangle ABQ=\measuredangle BAY=\measuredangle BCY=\measuredangle XBC$ . $(\bigstar\bigstar)$ . Now let $\Psi$ denote the Compostion of Inversion around $\odot(B,\sqrt{BA\cdot BC}/2)$ with a reflection around the angle bisector of $\angle ABC$ . So, $\Psi(M_A)=$ and $\Psi(X)=Q$ from $(\bigstar\bigstar)$ . So, $BX\cdot BQ=BM_A\cdot BA$ but from $(\bigstar\bigstar)$ we get $\measuredangle ABQ=\measuredangle XBM_A$ . So, $\Delta BM_AX\stackrel{+}{\sim}\Delta BQA\implies\measuredangle(AP,XD)=\pi-\measuredangle(AD,XD)-\measuredangle DAP=\pi-(\pi/2-\measuredangle BM_AX)-\measuredangle BQA=\pi/2\implies\overline{XD}\perp\overline{PQ}$ . So in original problem's terms we get that the Orthopole of $\mathcal E$ WRT $\Delta XYZ$ is the Euler Reflection Point of $\odot(A'B'C')\in\odot(A'B'C')$ . $\qquad$ $\blacksquare$

This post has been edited 3 times. Last edited by amar_04, Aug 18, 2020, 11:21 PM

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