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\title{Rectangular Circumhyperbolae}
\author{Gaurav Goel \\ gmgoel@gmail.com \\ gauravgoel@college.harvard.edu}
\date{April 30, 2019}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}[theorem]
\theoremstyle{remark}
\newtheorem*{remark}{Remark}
\newtheorem{numremark}{Remark}[theorem]
\begin{document}
\maketitle
\begin{abstract}
This paper deals with the Euclidean properties of rectangular circumhyperbolae with respect to a triangle using as little analytic treatment as possible. Familiarity with projective geometry, specifically ideal points, conic sections and Pascal's Theorem, is assumed.
\end{abstract}
\tableofcontents
\newpage
\section{Introduction to Circumconics}
In this paper, the symbol $\measuredangle$ represents a directed angle modulo $\pi$.
\begin{theorem} The isogonal conjugate $l^*$ of a line $l$ in the plane of $\triangle ABC$ is a circumconic of $\triangle ABC$.
\end{theorem}
\begin{proof}
We use homogenous barycentric coordinates. Let the line be $l \equiv ux+vy+wz=0$. Isogonal conjugation maps $P(x:y:z) \mapsto P^*(\frac{a^2}{x}:\frac{b^2}{y}:\frac{c^2}{z})$. Therefore the line $l$ is mapped to \( l^* \equiv \frac{ua^2}{x}+\frac{vb^2}{y}+\frac{wc^2}{z} = 0 \equiv ua^2yz+vb^2zx+wc^2xy = 0,\) which is a second-degree curve and hence a conic. The reason it passes through the vertices is because a sequence of points on $l$ converging to $l\cap BC$ have isogonal conjugates converging to $A$. Because isogonal conjugation is a continuous mapping, continuity ensures that $A \in l^*$. The other vertices also lie on $l^*$ by symmetry.
\end{proof}
\begin{theorem}
The isogonal conjugate of the ideal line is the circumcircle $\Omega$ of $\triangle ABC$.
\end{theorem}
\begin{proof}
The ideal line $l_\infty \equiv x+y+z=0$ is mapped to $a^2yz+b^2zx+c^2xy = 0$, which is nothing but $\Omega \equiv (ABC)$.
\end{proof}
\begin{proof}[Aliter]
This theorem can also be proved by angle chasing. We show that the isogonal conjugate of a point $P$ is an ideal point iff $P \in \Omega$. For that, let $r_a$, $r_b$ and $r_c$ be the lines isogonal to $AP$, $BP$ and $CP$ with to respect the corresponding vertices.\\
First assume that $P \in \Omega$. Then $\measuredangle(AB,r_a) = \measuredangle PAC =\measuredangle PBC =\measuredangle(AB, r_b)$, hence $r_a \parallel r_b$. By symmetry, $r_c$ is also parallel to these lines and hence these concur at a point at infinity. For the converse, assume that $r_a\parallel r_b\parallel r_c$. Using essentially the same argument, $\measuredangle PAC = \measuredangle (AB, r_a) =\measuredangle(AB, r_b)=\measuredangle PBC\implies P\in \Omega$.
\begin{figure}[H]
\centering
{\includegraphics[width=0.50\textwidth]{linfstarisomega.eps}}
\caption{$\Omega^* \equiv l_\infty$}
\label{IsogonalCircumcircle}
\end{figure}
\end{proof}
\begin{corollary}
The nature of the circumconic $l^*$ may be determined by counting the number of intersections of $l$ with $\Omega\equiv (ABC)$. In particular, $l^*$ is an ellipse, parabola or hyperbola according to whether $l$ meets $\Omega$ in 0, 1 or 2 points respectively.
\end{corollary}
\begin{theorem}
$l^*$ is a rectangular hyperbola iff $l$ is a diameter of $\Omega$. Equivalently, $l^*$ is a rectangular hyperbola iff $H\equiv O^* \in l^*$.
\label{Rect}
\end{theorem}
\begin{proof}
Suppose that $l\cap \Omega = \{X_1,X_2\}$. Then $l^*$ is a hyperbola with points at infinity $Y_1=X_1^*$ and $Y_2=X_2^*$. By isogonal conjugates, $\measuredangle Y_1AY_2=-\measuredangle X_1AX_2$ and hence the angle between the asymptotes of $l^*$ is the angle subtended by $X_1X_2$ at $\Omega$. In particular, the asymptotes are perpendicular iff $X_1X_2$ is a diameter of $\Omega$.
\end{proof}
\section{Revisiting Wallace-Simson Lines}
Since we will need the discussion of Wallace-Simson lines in the following article, it is worth revising their properties.
\newpage
\begin{theorem}
Let $l_P$ denote the Simson line of $P\in \Omega \equiv (ABC)$ with respect to $\triangle ABC$, and let $H$ denote the orthocenter of $\triangle ABC$. Then $l_P$ bisects $PH$, and this point of bisection lies on the nine-point circle $\Omega_9$ of $\triangle ABC$.
\label{ImpTheo1}
\end{theorem}
\begin{proof}
This proof can be found in \cite{EGMO}.\\
Let $X$, $Y$ and $Z$ be the feet of perpendiculars from $P$ to $BC$, $CA$ and $AB$ respectively. By definition, $X, Y, Z \in l_P$. Let $AH$ meet $\Omega$ again in $H'\neq A$ and let $PX$ meet $\Omega$ again in $K' \neq P$. Let $K$ be the orthocenter of $\triangle PBC$. Then, we know that $K'H'$ is the image of $KH$ in $BC$. Let $L\in l_P\cap AH$.
Then $LA \parallel XK'$ and $\measuredangle AK'P=\measuredangle ABP=\measuredangle ZBP = \measuredangle ZXP$ where the last equality follows because $PZXB$ is cyclic with diameter $PB$. $\therefore AK' \parallel l_P \equiv LX \implies ALXK'$ is a parallelogram.\\
Because $AH \parallel PK$ and $AH=PK=2R\cos A$, $AHKP$ is also a parallelogram. Consequently, $LH \parallel PX$ along with $LH = LA + AH = XK'+PK=KX+PK=PX$ implies that $PLHX$ is also a parallelogram. Therefore, $LX \equiv l_P$ bisects $PH$. Moreover, because the homothety $\mathbb{H}(H,\frac{1}{2})$ maps $\Omega$ to $\Omega_9$, the midpoint of $PH$ also lies on $\Omega_9$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{SimsonLine1.eps}
\caption{Simson line bisects the segment PH}
\label{SimsonLine}
\end{figure}
\begin{proof}[Aliter]
This proof is due to Ross Honsberger, and was taken from \cite{Lemmas}.\\
With notation as in the previous proof, let $D$ be the foot of altitude from $A$ and let $E \in PH' \cap BC$. Further, let $M$ be the midpoint of $PE$. Since the triangles $\triangle HEH'$ and $\triangle XME$ are isosceles, $\measuredangle MXE =\measuredangle XEM = \measuredangle CEH' =\measuredangle HEC$, we get that $MX \parallel HE$. But on the other hand, becuase $PYCX$ is cyclic, $\measuredangle YXC = \measuredangle YPC = \frac{\pi}{2} - \measuredangle PCA = \frac{\pi}{2}-\measuredangle PH'A = \measuredangle CEH' = \measuredangle HEC$ and hence $l_P \parallel HE$. This means that $M \in l_P$ and that $l_P$ is the $P$-midline of $\triangle PEH$. Therefore, it passes through the midpoint of $PH$. We can finish with $\mathbb{H}(H,\frac{1}{2})$ as before.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{SimsonLine2.eps}
\caption{Another proof that Simson line bisects PH}
\label{SimsonLine2}
\end{figure}
\begin{theorem}
Let $P$ and $P'$ be antipodes of $\Omega$. Then $P'^*$, the isogonal conjugate of $P'$, is the ideal point of $l_P$.
\label{IsogSimson}
\end{theorem}
\begin{proof}
With the same notation as before, because $AK' \parallel l_P$ and because $PYAZ$ is cyclic, $\measuredangle BAK'=\measuredangle AZY=\measuredangle APY$. Further, because $AP \perp AP'$ and $PY\perp AC$, we get $\measuredangle APY = \frac{\pi}{2}-\measuredangle YAP =\measuredangle P'AC$. Therefore, finally, $\measuredangle BAK' = \measuredangle P'AC$, which means that the isogonal conjugate of $P'$ lies on $AK'$ and hence on $l_P$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{SimsonLine3.eps}
\caption{$P'^* \in l_P$}
\label{SimsonLine3}
\end{figure}
\begin{theorem}
If $P, Q \in \Omega$, the $\measuredangle(l_P,l_Q)$ is negative of the angle subtended by arc $PQ$ in $\Omega$.
\label{anglesimson}
\end{theorem}
\begin{proof}
Let perpendiculars from $P$ and $Q$ to $BC$ meet $\Omega$ again in $P_1$ and $Q_1$ other than $P$ and $Q$ respectively. Then $PP_1QQ_1$ (not necessarily in that order) is an isosceles trapezium. Moreover, from the first proof of Theorem \ref{ImpTheo1}, we know that $l_P\parallel AP_1$ and $l_Q\parallel AQ_1$. Hence, \(\measuredangle(l_P,l_Q) = \measuredangle P_1AQ_1 = \measuredangle P_1PQ_1 = - \measuredangle PP_1Q.\) Consequently, $\measuredangle(l_P,l_Q) = -\measuredangle PSQ$ for $S \in \Omega$ as required.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{SimsonLine4.eps}
\caption{$\measuredangle(l_P,l_Q)=-\measuredangle PSQ$ for $S \in \Omega$. The negative angles are highlighted in a different shade.}
\label{SimsonLine4}
\end{figure}
\begin{corollary}
Simson lines of antipodal points are perpendicular.
\label{Antipodal}
\end{corollary}
\begin{corollary}
Because isogonal lines of antipodal points are perpendicular, Theorem \ref{IsogSimson} means that the Simson line of a point is perpendicular to its isogonal line.
\end{corollary}
\begin{theorem}
Simson lines of antipodal points $P$ and $P'$ of $\Omega$ intersect on $\Omega_9$.
\end{theorem}
\begin{proof}
Let $M$ and $M'$ be the midpoints of $PH$ and $PH'$ respectively. Because $\mathbb{H}(H,\frac{1}{2}): PP' \mapsto MM'$, $M$ and $M'$ are antipodal on $\Omega_9$. Further, by Corollary \ref{Antipodal}, $l_P \perp l_{P'}$. If $X_P \in l_P \cap l_{P'}$ then $\measuredangle MX_PM' = \frac{\pi}{2}$, because of which $X_P \in \Omega_9$.
\end{proof}
This theorem, as it turns out, links very beautifully with the concept of the asymptotes of rectangular circumhyperbolae, as the following sections will develop.
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{SimsonLine5.eps}
\caption{Simson lines of antipodal points meet on $\Omega_9$}
\label{SimsonLine5}
\end{figure}
\section{Two Useful Theorems on Conics}
\subsection{Brocard's Theorem}
\begin{theorem}[Brocard's Theorem]
Let $ABCD$ be a quadrilateral inscribed in a conic $\mathcal{C}$. Let $M \in AD\cap BC$, $N \in AB \cap CD$, $P \in AA \cap CC$, $Q \in BB \cap DD$ and $J \in AC \cap BD$. (Here $AA$ means the tangent to $\mathcal{C}$ at $A$ and so on.) Then $M, N, P$ and $Q$ are all collinear on the polar $j$ of $J$ with respect to $\mathcal{C}$.
\label{Brocard}
\end{theorem}
\begin{proof}
By Pascal's Theorem on $AABCCD$, $P \in MN$. Similarly, by Pascal's Theorem on $ABBCDD$, $Q \in MN$. Moreover, since the $J$ belongs to $AC$, the polar of $P$ and to $BD$, the polar of $Q$, La Hire's Theorem tells us that $PQ\equiv j$ as needed.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.50\textwidth]{Brocard.eps}
\caption{Brocard's Theorem}
\label{BrocardPic}
\end{figure}
\subsection{8 Points on a Conic}
\begin{theorem}
For a quadrilateral $ABCD$, assign $M$, $N$ and $J$ as before, i.e. let $M \in AD\cap BC$, $N\in AB\cap CD$ and $J \in AC\cap BD$. Suppose that quadrilaterals $ABCD$ and $A'B'C'D'$ are assigned the same $M$, $N$ and $J$. Then the 8 points $A, B, C, D, A', B', C', D'$ lie on a conic.
\label{8points}
\end{theorem}
\begin{proof}
Consider the projective transformation that maps $MN$ to the ideal line. Then $ABCD$ and $A'B'C'D'$ are mapped to concentric parallelograms with parallel sides which clearly determine the degenerate conic $AC\cup BD$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{8pointsonconic.eps}
\caption{8 Points on a Conic}
\label{8pointsonconic}
\end{figure}
\section{Onto Rectangular Circumhyperbolae}
Since a conic is completely determined by five points, Theorem \ref{Rect} tells us that a rectangular circumhyperbola of $\triangle ABC$ is characterized completely by the fifth point $P$ it contains. Let $\mathcal{H}(P)$ denote the rectangular circumhyperbola containing $P$. Further, let $Z$ be the center of $\mathcal{H}(P)$. Then $Z$ is called the Poncelet Point of $P$ with respect to $\triangle ABC$, or in a more symmetric formulation, the Poncelet Point of the quadrilateral $ABCP$.
\begin{theorem}
$Z$ lies on the nine-point circle $\Omega_9$ of the orthocentric system $ABCH$.
\label{zonnine}
\end{theorem}
\begin{proof}
This proof was taken from the online blog linked in \cite{RectBlog}. Let $D$ be the fourth intersection of $\mathcal{H}(P)$ with $\Omega \equiv (ABC)$ and let $H'$ be the orthocenter of $\triangle DBC$. By Theorem \ref{Rect}, $H' \in \mathcal{H}(P)$. Moreover, $AHH'D$ is a parallelogram inscribed in a hyperbola $\mathcal{H}(P)$. Applying Brocard's Theorem to $AHH'D$, the center of the parallelogram $AHH'D$ must be the pole of the ideal line $l_\infty$ with respect to $\mathcal{H}(P)$, which is none other than \emph{its} center $Z$. Hence, $Z$ is the midpoint of $HD$ and once again using $\mathbb{H}(H,\frac{1}{2}):D\mapsto Z$, we get that $Z \in \Omega_9$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{RectHype1.eps}
\caption{$Z \in \Omega_9$}
\label{RectHype1}
\end{figure}
\begin{corollary}
Given any four points $A$, $B$, $C$ and $D$ in a plane, the nine-point circles of $\triangle ABC$, $\triangle BCD$, $\triangle CDA$ and $\triangle DAB$ concur.
\label{corr}
\end{corollary}
\begin{proof}
Consider the rectangular hyperbola $\mathcal{H}$ that passes through $A$, $B$, $C$ and $D$. Then its center $Z$, the Poncelet Point of quadrilateral $ABCD$, is the desired point of concurrency.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{RectHype2.eps}
\caption{$Z$ as the point of concurrency. Here $\Omega_A$ denotes the nine-point circle of $\triangle BCD$ and so on.}
\label{RectHype1}
\end{figure}
\begin{remark}
An elementary proof of this result and further reading about the Poncelet Point can be found in \cite{Poncelet}.
\end{remark}
\begin{theorem}[Main Theorem]
Let $PQ$ be a diameter of $\Omega \equiv (ABC)$ and let $\mathcal{H}$ denote the rectangular circumhyperbola that is the isogonal conjugate of the line $PQ$ with respect to $\triangle ABC$. Then the asymptotes of $\mathcal{H}$ are the Simson lines $l_P$ and $l_Q$ of $P$ and $Q$ respectively with respect to $\triangle ABC$.
\end{theorem}
\begin{proof}
As previously, let $D$ be the fourth intersection of $\mathcal{H}$ with $\Omega$ and let $Z$ be the center of $\mathcal{H}$. Let $O$ denote the circumcenter of $\triangle ABC$ and let $A'$ denote the antipode of $A$ in $\Omega$. Let $F \in PD\cap BC$ and $E \in l_P\cap BC$. Let $P'$ and $Q'$ be the midpoints of $PH$ and $QH$ respectively. Finally, let the line parallel to $PQ$ through $A$ meet $\Omega$ again in $G$.
We know from Theorem \ref{ImpTheo1} that $P' \in l_P$ and from Theorem \ref{IsogSimson} that $Q^* \in l_P$. Because $Q^*$ is one of the points at infinity of $\mathcal{H}$, $l_P$ is parallel to one of the asymptotes of $\mathcal{H}$. Hence, to show that it \emph{is} one of the asymptotes, it suffices to show that $Z \in l_P$. The fact that $l_Q$ is the other asymptote follows by symmetry.
We know that $BC$ is the Simson line of $A'$ with respect to $\triangle ABC$. Using Theorem \ref{anglesimson}, $\measuredangle (BC,l_P)=\measuredangle(l_{A'},l_P)=-\measuredangle A'QP =\measuredangle PQA'$. But because of the homothety $\mathbb{H}(O,-1): \triangle A'QP \mapsto \triangle APQ$, $\measuredangle PQA'=\measuredangle QPA$, which in turn equals $\measuredangle GAP$ because of our stipulation that $AG \parallel PQ$. Therefore, $\measuredangle (BC,l_P) =\measuredangle GAP =\measuredangle GAC +\measuredangle CAP$.
Because $D$ is the isogonal conjugate of the ideal point of $PQ$ with respect to $\triangle ABC$, $AD$ and $AG$ are isogonal with respect to $\angle BAC$ implying that $\measuredangle GAC = \measuredangle BAD$. Consequently, $\measuredangle (BC, l_P) = \measuredangle BAD + \measuredangle CAP = \measuredangle BAP + \measuredangle CAD$. However, $\measuredangle BAP = \measuredangle BDP = \measuredangle BDF$ and $\measuredangle CAD = \measuredangle CBD = \measuredangle FBD$. Summing up, $\measuredangle (BC,l_P) = \measuredangle BDF +\measuredangle FBD = \measuredangle BFD = \measuredangle (BC,PD) \implies l_P \parallel PD$.
This means that $l_P$ is the $H$-midline of $\triangle PDH$ and contains the midpoint of $DH$, which we know from the proof of the previous theorem to be $Z$. This concludes the proof.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=0.5\textwidth]{TurkishDelightFinal.eps}
\caption{The Turkish Delight!}
\label{TurkishDelight}
\end{figure}
\begin{numremark}
The proof of this theorem that is presented here is due entirely to the author.
\end{numremark}
\section{The Circles $Z$ Belongs To}
Let $Z$ be the Poncelet Point of $P$ with respect to $\triangle ABC$. This section develops two cute results taken from \cite{RectBlog}.
\subsection{The Pedal Circle}
\begin{theorem}
$Z$ lies on the pedal circle of $P$ with respect to $\triangle ABC$.
\label{pedal}
\end{theorem}
\begin{proof}
Let $D, E, F$ be the feet of perpendiculars from $P$ to $BC, CA, AB$ respectively. Let $K, L, M$ denote the midpoints of $AC, AB, AP$ respectively.
Then $\measuredangle EZF = \measuredangle EZM + \measuredangle MZF$. Because of Corollary \ref{corr}, $Z \in (MKE)\cap(MLF)$. Hence, $\measuredangle EZM = \measuredangle EKM$ and $\measuredangle MZF = \measuredangle MLF$. Therefore, $\measuredangle EZF = \measuredangle EKM + \measuredangle MLF = \measuredangle AKM +\measuredangle MLA$.
By midlines, it is evident that $\measuredangle AKM = \measuredangle ACP =\measuredangle ECP$. But $ECDP$ is cyclic, so $\measuredangle ECP = \measuredangle EDP$. Consequently, $\measuredangle AKM = \measuredangle EDP$. Similarly, $\measuredangle MLA = \measuredangle PDF$, so that $\measuredangle EZF = \measuredangle AKM + \measuredangle MLA = \measuredangle EDP +\measuredangle PDF = \measuredangle EDF \implies Z \in (DEF)$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{Pedal.eps}
\caption{$Z$ lies on the Pedal Circle $(DEF)$}
\label{PedCirc}
\end{figure}
\subsection{The Cevian Circle}
\begin{theorem}
$Z$ lies on the cevian circle of $P$ with respect to $\triangle ABC$.
\label{cevian}
\end{theorem}
\begin{proof}
Let $U, V, W$ be the feet of cevians in $\triangle ABC$. Let $J_U, J_V, J_W$ and $I$ be the corresponding excenters and incenter of $\triangle UVW$.
Applying Theorem \ref{8points} to quadrilaterals $ABCP$ and $J_UJ_VJ_WI$ we get that these 8 points lie on a conic. However, $I$ is the orthocenter of $\triangle J_UJ_VJ_W$, and therefore this conic must be a rectangular hyperbola. Consequently, this conic is nothing but $\mathcal{H}(P)$. To conclude, by Theorem \ref{zonnine}, the center $Z$ of this conic lies on the nine-point circle of the orthocentric system $J_UJ_VJ_WI$, which is nothing but $(UVW)$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{Cevian.eps}
\caption{$Z$ lies on the Cevian Circle $(UVW)$}
\label{CevCirc}
\end{figure}
\section{Applications}
We end with nice consequences of the theory developed above, which are rather difficult to prove using elementary synthetic geometry.
\subsection{The Big Picture}
\begin{theorem}[Nine Concurrent Circles]
Let $A, B, C$ and $D$ be any four points in a plane. Let $\Omega_A$ denote the nine-point circle of $\triangle BCD$, and define $\Omega_B, \Omega_C$ and $\Omega_D$ similarly. Let $\Gamma_A$ denote the pedal circle of $A$ with respect to $\triangle BCD$, and define $\Gamma_B$, $\Gamma_C$ and $\Gamma_D$ similarly. Finally, let $\Lambda$ denote the cevian circle $(MNJ)$ of quadrilateral $ABCD$, where $M \in AD\cap BC$, $N \in AB\cap CD$ and $J \in AC\cap BD$. Then the nine circles $\Omega_A, \Omega_B, \Omega_C, \Omega_D, \Gamma_A, \Gamma_B, \Gamma_C, \Gamma_D$ and $\Lambda$ concur.
\label{nineconcur}
\end{theorem}
\begin{proof}
This is merely a symmetric formulation of Corollary \ref{corr} and Theorems \ref{pedal} and \ref{cevian}. The concurrency point is none other than the Poncelet Point $Z$ of quadrilateral $ABCD$, the center of the rectangular hyperbola through $A, B, C$ and $D$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{NineCirclePoint.eps}
\caption{Nine Concurrent Circles}
\label{NineConcur}
\end{figure}
\begin{remark}
The other points of intersections include the midpoints of the six sides of $ABCD$ and the feet from one vertex to the segments determined by the other three.
\end{remark}
\begin{corollary}[The Anticenter]
Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Omega$. Let $\Omega_A$ denote the nine-point circle of $\triangle BCD$, and define $\Omega_B, \Omega_C$ and $\Omega_D$ similarly. Let $l_A$ denote the Simson Line of $A$ with respect to $\triangle BCD$, and define $l_B, l_C$ and $l_D$ similarly. Finally, let $H_A$ denote the orthocenter of $\triangle BCD$, and define $H_B, H_C$ and $H_D$ similarly. Then $l_A, l_B, l_C, l_D, \Omega_A, \Omega_B, \Omega_C$ and $\Omega_D$ concur at the common bisection point of $AH_A, BH_B, CH_C$ and $DH_D$.
\end{corollary}
This point of concurrency is called the anticenter of cyclic quadrilateral $ABCD$.
\begin{proof}
In the case when the four points are cyclic, the pedal circles degenerate to Simson Lines. We can use Theorems \ref{ImpTheo1}, \ref{zonnine} and \ref{nineconcur} to see that the anticenter is none other than the Poncelet Point $Z$ of quadrilateral $ABCD$.
\end{proof}
\begin{corollary}
With notation as before, quadrilateral $H_AH_BH_CH_D$ is cyclic and homothetic to $ABCD$, the homothety being $\mathbb{H}(Z,-1):ABCD\mapsto H_AH_BH_CH_D$.
\end{corollary}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{Anticenter.eps}
\caption{Anticenter}
\label{anticenter}
\end{figure}
\begin{remark}
The standard way to prove the existence of the anticenter is using complex numbers and setting the circumcircle $\Omega$ to be the unit circle $\{z: z \in \mathbb{C}, \lvert z \rvert = 1\} $. Then the complex number $z$ denoting the anticenter of points $A, B, C$ and $D$ given by $a, b, c$ and $d$ respectively is given by:
\[z=\frac{a+b+c+d}{2}\]
\end{remark}
\subsection{Feuerbach's Theorem and the Feuerbach Hyperbola}
\begin{theorem}[Feuerbach's Theorem]
The nine-point circle $\Omega_9$ of a triangle $\triangle ABC$ is tangent to its incircle $\omega$ and three excircles $\omega_A,\omega_B,\omega_C$.
\label{feu}
\end{theorem}
\begin{proof}
This proof was taken from \cite{RectBlog}.
Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$, and let $O$ denote the circumcenter of $\triangle ABC$. By the Six Point Circle Theorem, which can be found in \cite{Lemmas}, they share a common pedal circle; call this pedal circle $\omega_{PQ}$. Then from Theorems \ref{zonnine} and \ref{pedal}, $\omega_{PQ}$ meets $\Omega_9$ in the Poncelet Points $Z_P$ and $Z_Q$ of $P$ and $Q$ respectively, with respect to $\triangle ABC$.
Now $\mathcal{H}(P)$ and $\mathcal{H}(Q)$ are distinct lines iff $QO$ and $PO$ are distinct lines, as these are the isogonal conjugates of the hyperbolae. In this case, $Z_P \neq Z_Q$ and $\lvert \omega_{PQ} \cap \Omega_9 \rvert = 2$.
If we let the lines $PO$ and $QO$ move closer to each other, the $Z_P$ and $Z_Q$ move closer to each other on the nine-point circle. Consequently if $P, O, Q$ are collinear, then $\omega_{PQ}$ and $\Omega_9$ touch. In the particular case when $P\equiv Q$ is the incenter or one of the excenters, we get Feuerbach's Theorem.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{Feuerbach.eps}
\caption{Feuerbach's Theorem}
\label{Feuerbach}
\end{figure}
\begin{remark}
The point of tangency between $\omega$ and $\Omega_9$ is called the Feuerbach Point $F$ of $\triangle ABC$. It is the ETC center $X_{11}$. The points of tangency with the excircles, denoted by $F_A, F_B$ and $F_C$ respectively form the Feuerbach triangle of $\triangle ABC$.
\end{remark}
\begin{theorem}[The Feuerbach Hyperbola]
The isogonal conjugate of the line $OI$ of a triangle has its center at $F$. This hyperbola $\mathcal{H}(I)$, called the Feuerbach Hyperbola of $\triangle ABC$, passes through the Gergonne Point $G_E$, the Nagel Point $N$, the Mittenpunkt $M$ and the Schiffler Point $S$ of $\triangle ABC$.
\end{theorem}
\begin{proof}
The center of the hyperbola $\mathcal{H}(I)$ must lie on the pedal circle $\omega$ of $I$ by Theorem \ref{pedal}, and must lie on the nine-point circle $\Omega_9$ of $\triangle ABC$ by Theorem \ref{zonnine}. But we know from Theorem \ref{feu} that these circles are tangent at $F$ and hence $F$ must be the desired center of $\mathcal{H}(I)$.
It is well-known (see \cite{Lemmas}) that the isogonal conjugate of the Gergonne Point $G_E$ ($X_7$) is $X_{55}$, the in-similicenter of the incircle $\omega$ and circumcircle $\Omega$ of $\triangle ABC$, and that the isogonal conjugate of the Nagel Point $N$ ($X_8$) is $X_{56}$, the ex-similicenter of $\omega$ and $\Omega$. Both of these centers of similitude of $\omega$ and $\Omega$ obviously belong to the line $OI$ and hence their isogonal conjugates belong to $\mathcal{H}(I)$.
The isogonal conjugate of the Mittenpunkt $M$ ($X_9$) of $\triangle ABC$, called the Isogonal Mittenpunkt, is $X_{57}$, the homothetic center of the contact and excentral triangles of $\triangle ABC$. This result can be found in \cite{isogmittenpunkt}. Because this homothetic center takes $I$ to $V$, the Bevan Point of $\triangle ABC$, it lies on the line $VI \equiv OI$. This shows that the Mittenpunkt lies on $\mathcal{H}(I)$.
Finally, the isogonal conjugate of the Schiffler Point $S$ ($X_{21}$) of a triangle is the orthocenter of its intouch triangle, labelled $X_{65}$ in the ETC. This can be found in \cite{etcX65}. This point obviously belongs to the Euler Line of the intouch triangle, which is the $OI$ line of the reference triangle $\triangle ABC$. Thus, the Schiffler Point $S$ also lies on $\mathcal{H}(I)$.
\end{proof}
\begin{figure}[H]
\centering
\includegraphics[width=.5\textwidth]{FeuHyp.eps}
\caption{The Feuerbach Hyperbola $\mathcal{H}(I)$}
\label{FeuHyp}
\end{figure}
\begin{remark}
The line OI is tangent to its isogonal conjugate $\mathcal{H}(I)$. This can be seen by an obvious proof by contradiction.
\end{remark}
\bibliography{Bibliography}
\bibliographystyle{ieeetr}
\end{document}
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