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\begin{document}
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\title[]
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{CIRCUMBILLIARD GEOMETRY}
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\date{}
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\author{Dominique Laurain}
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\address{Mathematics hobbyist, At his home, Toulouse, France}
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\email{dominique.laurain31@orange.fr}
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\begin{abstract}
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Mathematical billiards are at crossroads of various research works, but mainly covered in dynamicals systems. Dan Reznik found geometric invariants for the inner elliptic billiard. We propose a parametrization of a 3-orbit, a "circumbilliard", using euclidean plane geometry.
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\end{abstract}
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\maketitle
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Introduction}
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In this note we present analytical and geometrical results about 3-orbits in an elliptic billiard. They are deduced from Dan Reznik \cite{reznik} observations in the flavor of plane euclidean geometry.
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In physical or optical terms, we restrict the results to "specular reflection" on the ellipse, a perfect elastic reflection inducing a Snell law.
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\section{Definition in triangle geometry}
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We define a \textit{circumbilliard} as a not oriented 3-orbit ABC in an elliptical billiard.
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Points A,B,C are reflections points of a modelized blue "ball" on the elliptic boundary.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.4\linewidth]{Circumbilliard}
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\caption{Circumbilliard for an ellipse.}
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\label{circumbilliard}
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\end{figure}
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The 3-orbit ABC is choosen as the \textit{reference triangle} in triangle geometry.
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Common notation is used for triangle vertices (A,B,C) and triangle edge lengths ($a = BC$, $b = CA$, $c = AB$).
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The ellipse is named \textit{circumbilliard} by analogy with the ABC \textit{circumcircle,} the unique circle going through A,B and C.
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Point of view is changed versus representing orbits in a fixed ellipse : here we set a triangle ABC, and draw the ellipse centered at Mittenpunkt $X_9$.
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\section{Parametrization}
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One invariant for n-orbits in elliptic billiards is total length of the orbit, an extremal value (\textit{Lemma 2.3} in \cite{AST}) .For a 3-orbit ABC it is twice the semi-perimeter $s$.
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Another invariant discovered by Dan Reznik is the ratio between inradius and circumradius, or equivalently the sum of cosines at internal angles at A,B and C vertices (\textit{Theorem 1} in \cite{AST}))
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We use these invariants to find ABC edges lengths parametrization.
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\begin{theorem} \label{thm:parametrization}
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Edges lengths of a 3-orbits ABC can be defined as :
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$$a = \frac{2(1 - t) s}{k - 2t + 2}$$
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$$b = \frac{(k t - t^2 + k + 1 - w) s}{(1 + t)(k - 2t + 2)}$$
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$$c = \frac{(k t - t^2 + k + 1 + w) s}{(1 + t)(k - 2t + 2)}$$
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with
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$$ w = \pm \sqrt{ (1 - t^2)( h^2 - (t - k)^2) } $$
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$$ h = \sqrt{ 1 - 2k } $$
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and parameter $t$ in [$k - h$,$k + h$ ] is cosine internal angle A.
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\end{theorem}
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\begin{proof}
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We use Ravi substitution : $a = y + z$, $b = z + x$, $c = x + y$ to compute some geometric values for triangle ABC. The next formulas are from triangle geometry.
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Formula for semi-perimeter :
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$$s = \frac{1}{2}(a + b + c) = x + y + z$$
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Formula for circumradius :
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$$ R = \frac{abc}{4 S} = \frac{(x + y) (x + z) (y + z)}{4 S}$$
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\begin{flushleft}
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where S is ABC area.
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\end{flushleft}
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Inradius :
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$$ r = \frac{S}{s} = \frac{2S}{x + y + z} $$
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Ratio between inradius and circumradius :
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$$ k = \frac{r}{R} = \frac{4 x y z}{(x + y)(x + z)(y + z)} = k $$
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Cosines of internal angles using cosines law :
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$$\cos(A) = \frac{b^2 + c^2 - a^2}{2bc} = \frac{x^2 + x(y + z) - yz }{(x + y)(x + z)} = t$$
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$$\cos(B) = \frac{c^2 + a^2 - b^2}{2ca} = \frac{y^2 + y(z + x) - zx }{(y + z)(y + x)} $$
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$$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab} = \frac{z^2 + z(x + y) - xy }{(z + x)(z + y)} $$
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Adding all cosines and factoring :
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$$\cos(A) + cos(B) + cos(C) = \frac{x^2 (y + z) + y^2 (z + x) + z^2 (x + y) + 6xyz }{(x + y)(x + z)(y + z)} $$
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or
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$$\cos(A) + cos(B) + cos(C) = 1 + \frac{4xyz }{(x + y)(x + z)(y + z)} = 1 + k$$
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From previous formulas we deduce variables $x$,$y$,$z$ are roots of :
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\begin{enumerate}
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\item $x + y + z - s = 0$
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\medskip
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\item $k(x + y)(x + z)(y + z) - 4xyz = 0$
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\medskip
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\item $ x^2 + x y + x z - y z - t(x + y)(x + z) = 0 $
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\end{enumerate}
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\begin{flushleft}
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We choose the couple of first and third equations, expressing y and z from x,k,s and t. Putting back expression into second equation we get :
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\end{flushleft}
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$$ x = \frac{k s}{k - 2t + 2}$$
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The first and third equations give sum and product of y and z :
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$$ \sigma = y + z = s - x = \frac{2 (1 - t) s}{k - 2t + 2}$$
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$$ \mu = y z = \frac{1 - t)x ( x + y + z) }{1 + t} = \frac{(1 - t) k s^2}{(1 + t)(k - 2t + 2)}$$
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Hence y,z are roots of the binomial equation : $U^2 - \sigma U + \mu = 0$.
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$$ y = \frac{(1 - t^2 + w)s}{(1 + t)(k - 2t + 2)}$$
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$$ z = \frac{(1 - t^2 - w)s}{(1 + t)(k - 2t + 2)}$$
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with
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$$w^2 = (1 - t^2)(1 - 2k - (t - k)^2) $$
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Substituting x,y,z into a,b,c we get a,b,c parametrization in k,s,t.
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\end{proof}
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The common part in a,b,c parametrization ($\frac{s}{k - 2t + 2}$ ) is a scaling factor very useful in trilinear coordinates computations. Changing sign of $w$ is simply swapping $b$ and $c$ or equivalenty reflecting the 3-orbit on minor axis.
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\section{Equation in trilinear coordinates}
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Something is missing in the definition of the \textit{circumbilliard ellipse} because we need five points, no three aligned, to define an ellipse and here we have only three of them : A,B,C.
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Dan Reznik found two more points which are on the ellipe for all 3-orbits, points $X_{88}$ (isogonal conjugate of $X_{44}$) and $X_{100}$ (Feuerbach anticomplement point):
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$$X_{88} = \frac{1}{b+c-2a} : \frac{1}{c+a-2b} : \frac{1}{a+b-2c} $$
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$$X_{100} = \frac{1}{b-c} : \frac{1}{c-a} : \frac{1}{a-b} $$
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Property for these points : the sum of denominators of trilinear coordinates is 0.
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These points are sometimes not well defined (for example, for isosceles ABC), that's why, we prefer the reflection of B and C with respect to Mittenpunkt $X_9$ center of the ellipse in the proof of the next theorem.
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\begin{theorem} \label{thm:equation}
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Circumbilliard ellipse $(\Phi)$ is locus of points $M = \alpha : \beta : \gamma$ whose trilinear coordinates are satisfying equation :
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$$ \alpha \beta + \alpha \gamma + \beta \gamma = 0$$
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which is the isogonal conjugate of the line $(\Phi^*)$ with equation
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$$ \alpha + \beta + \gamma = 0$$
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\end{theorem}
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\begin{proof}
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Computing trilinear coordinates of B, and C reflections with respect to Mittenpunkt we get :
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$$D = 2(-a+b+c)b : (a+b-c)(a-b-c) : 2(a+b-c)b$$
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$$E = 2(-a+b+c)c : 2(a-b+c)c : (a-b+c)(a-b-c)$$
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Isogonal conjugates of these points are :
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$$D^* = \frac{1}{(-a+b+c)b} : \frac{2}{(a+b-c)(a-b-c)} : \frac{1}{(a+b-c)b}$$
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$$E^* = \frac{1}{(-a+b+c)c} : \frac{1}{(a-b+c)c} : \frac{2}{(a- b+c)(a-b- c)}$$
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Trilinear coordinates $\alpha : \beta : \gamma$ of $D^*$ and $E^*$ are satisfying the line equation $\alpha + \beta + \gamma = 0$.
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Moreover these two points are distinct when the ellipse is not a circle ($k < \frac{1}{2}$) because the squared distance is not 0 :
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$$(D^*E^*)^2 = (1-2k)(\frac{2(k+4)(1-t)s}{4k^2t-6kt^2+5k^2-4kt+12t^2+10k-8t-4})^2 $$
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The ABC circumcircle has equation :
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$$a \beta \gamma + b \gamma \alpha + c \alpha \beta = 0$$
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The isogonal conjugate of a line is an ellipse through A,B,C if and only if the line doesn't intersect the ABC circumcircle (\cite {Wolf}).
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Substituting $\alpha$ with $-\beta -\gamma$, the circumcircle intersects the line $D^*E^*$ with equation $\alpha + \beta + \gamma = 0$ if only if it exists one real root $\beta$ of equation :
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$$c \beta^2 + (- a + b + c ) \beta \gamma + b \gamma^2 = 0$$
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The squared discriminant of this binominal equation is $\gamma^2$ times $\Delta$ :
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$$\Delta = a^2-2ab+b^2-2ac-2bc+c^2 = \frac{-4k(k + 4)(1-t)s^2}{(k-2t+2)^2(1+t)} $$
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is always negative, and there is no real root.
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The isogonal conjugate of the line $D^*E^*$ is the circumbilliard ellipse with equation : $\alpha \beta + \alpha \gamma + \beta \gamma = 0$.
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\end{proof}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.8\linewidth]{IsogonalConjugation.png}
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\caption{Isogonal conjugation}
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\label{isogonalconjugation}
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\end{figure}
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We can choose many couple of points ($P^*$,$Q^*$) instead of ($D^*$,$E^*$) to define the line $(\Phi^*)$.
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A simple choice is $P^* = 1 : -2 : 1$ and $Q^* = 1 : 1 : -2$ with :
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$$(P^*Q^*)^2 = \frac{36(1-2k)k^2(1-t)^2s^2}{(4kt-2t^2-5k+2)^2(k-2t+2)^2} $$
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The line $(\Phi^*)$ is the \textit{center line} $X_{44}X_{513}$ and \textit{triangle centers} on that line have isogonal conjugates on the ellipse $(\Phi)$ : triangle centers $X_{88}$, $X_{100}$, $X_{190}$, $X_{651}$, $X_{162}$, $X_{660}$, $X_{662}$, $X_{673}$, $X_{799}$, $X_{823}$, $X_{897}$, $X_{31002}$ and so on.
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\section{Ellipse axis}
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It is easy to compute the two intersection points of a line : $ l = l_a : l_b : l_c $ equation $ l_a \alpha + l_b \beta + l_c \gamma = 0 $ with the ellipse equation $ \alpha \beta + \alpha \gamma + \beta \gamma = 0 $.
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We set a generic point $ P = 0 : (1 - m)c : mb $ on the BC line and intersect line $X_9P$ with the ellipse to get two points $M_i$ and $M_j$.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.8\linewidth]{CircumbilliardAxis.png}
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\caption{Axis}
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\label{axis}
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\end{figure}
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We maximize or minimize function $q_{ij}(m) = (M_iM_j)^2$ the squared distance between $M_i$ and $M_j$ to get the optimal values for m : $m = m_1$ for major axis $M_1M_2$ and $m = m_2$ for minor axis $M_3M_4$.
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$$ m = \frac{-c(a^2+b^2+c^2+ab-2ac-2bc)\pm \Delta}{(a^2+b^2+c^2-2ab-2ac-2bc)(b - c)} $$
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with
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$$ \Delta^2 = (a^3+b^3+c^3-a^2b-ab^2-a^2c-b^2c-ac^2-bc^2+3abc)abc $$.
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Using parametrization we deduce formulas for squared major and minor lengths :
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$$ p^2 = (X_9M_1)^2 = \frac{4(1+h)s^2}{(3-h)(3+h)^2} $$
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$$ q^2 = (X_9M_3)^2 = \frac{4(1-h)s^2}{(3+h)(3-h)^2} $$
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and focal distance : $f = \frac{4\sqrt{h}s}{9 - h^2}$.
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\section{Cartesian coordinates}
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Drawings have been done previously with the following layout : BC segment is set horizontal, A is drawn up BC and incenter $X_1$ is set at cartesian plane origin point with (0,0) as cartesian coordinates.
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It is convenient for dynamics in elliptic billiard to choose another layout where the center of ellipse $X_9$ is at (0,0) and major, minor axis are horizontal and vertical. The 3-orbits are then drawn as triangles ABC inscribed in the fixed ellipse.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.8\linewidth]{TriangleGeometryLayout.png}
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\caption{Triangle geometry layout}
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\label{triglay}
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\end{figure}
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We will morph drawing using a translation and a rotation.
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Translation vector from $(X_1,X_9) = (x_t,y_t)$ with :
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$$x_t = \frac{2(h^2t-5t+4)s}{(h^2+4t-5)(9-h^2)} \sqrt{ \frac{4h^2-(h^2+2t-1)^2}{1-t^2} } $$
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$$y_t = \frac{2(h^2t+2t^2-5t+2)(1-h^2)s}{(h^2+4t-5)(9-h^2)} \sqrt{ \frac{1}{1-t^2} } $$
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Cosine and sine of rotation angle $\theta_t = \theta$ are computed using edge lengths triangle $P_1$X_9$P_2$ :
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$$\cos \theta_t = \frac{1+h}{2} \sqrt{ \frac{2h-h^2-2t+1}{2h(1-t)} }$$
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$$\sin \theta_t = \frac{1-h}{2} \sqrt{ \frac{2h+h^2+2t-1}{2h(1-t)} }$$
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Ellipse equation in cartesian coordinates is given from $p^2$ and $q^2$ formulas :
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\begin{equation}\label{l}
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(3 + h)(1 - h)x^2 + (3 - h)(1 + h)y^2 = \frac{f^2(1 - h^2)(9 - h^2)}{4h}
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\end{equation}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.8\linewidth]{CartesianCoordinatesLayout.png}
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\caption{Cartesian coordinates layout}
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\label{cartlay}
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\end{figure}
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Using ellipse equation (\ref l) we can deduce the squared cartesian coordinates $x^2$ and $y^2$ for any point $M$ on the ellipse such as $(X_9M)^2 = x^2 + y^2$. For A,B,C vertices we get :
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$$x_A^2 = \frac{(h^2+2h+2t-1)(1+h)^2s^2}{2(3+h)^2h(1+t)}$$
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$$y_A^2 = \frac{(-h^2+2h-2t+1)(1-h)^2s^2}{2(3-h)^2h(1+t)}$$
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$$x_B^2 = \frac{(u+v-2(h^2+2h-3)w)(1+h)^2s^2}{2(h^2+4t-5)^2(3+h)^2h}$$
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$$y_B^2 = \frac{(-u+v+2(h^2-2h-3)w)(1-h)^2s^2}{2(h^2+4t-5)^2(3-h)^2h}$$
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$$x_C^2 = \frac{(u+v+2(h^2+2h-3)w)(1+h)^2s^2}{2(h^2+4t-5)^2(3+h)^2h}$$
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$$y_C^2 = \frac{(-u+v-2(h^2-2h-3)w)(1-h)^2s^2}{2(h^2+4t-5)^2(3-h)^2h}$$
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with
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$$u = 2h^4t+4h^2t^2-8h^2t+20t^2-26t+8$$
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$$v = 4(h^2+2t-7)h(t-1)$$
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\section{Triangle centers}
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We use homogenous barycentric coordinates to get parametrization of cartesian coordinates of a triangle center. Homogenous barycentric coordinates for $M = \alpha : \beta : \gamma$ are $(\frac{a \alpha}{n} , \frac{b \beta}{n} , \frac{c \gamma}{n})$ with $n = a \alpha + b \beta + c \gamma$.
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As an example we compute cartesian coordinates ($x_I,y_I$) of ABC incenter $I = X_1$ whose locus is drawn as the blue ellipse figure \ref{cartlay}.
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The subject was dealt by Olga Paris-Romaskevich in \cite{OR} teasing for use of \textit{complex reflection}. Ronaldo Garcia added explicit cartesian coordinates equation of the locus in \cite{RAG}.
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Incenter has constant coordinates : $\alpha = \beta = \gamma = 1$. From them we get barycentric coordinates with respect to A,B,C points :
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$$k_A = \frac{1+h^2}{3+h^2}$$
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$$k_B = \frac{3-h^2-h\sqrt{-h^2+2h^2+3}}{9-h^4}$$
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$$k_C = \frac{3-h^2+h\sqrt{-h^2+2h^2+3}}{9-h^4}$$
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and
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$$x_I = k_A x_A + k_B x_B + k_C x_C$$
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$$y_I = k_A y_A + k_B y_B + k_C y_C$$
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Usually these cartesian coordinates don't have simple closed form. Signs of A,B,C cartesian coordinates are troublesome too because depending on t interval values range.
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If we already know that geometric locus is a conic, it is easier. For an ellipse, we plug into cartesian coordinates formulas five values of t between $k-h$ and $k+h$ to get five points and then deduce ellipse equation.
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$$\frac{x^2}{l^2} + \frac{y^2}{m^2} = 1$$
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Cheating here a little because $I_1$ (for $t = k - h$) and $I_2$ (for $t = k + h$) are on minor and major axis we get easily :
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$$l^2 = \frac{16(1+h)h^2s^2}{(3+h)^2(3 - h)^3}$$
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$$m^2 = \frac{16(1-h)h^2s^2}{(3+h)^3(3 - h)^2}$$
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from plugging $t = k \pm h$ in $x^2_A(t)$, $y^2_A(t)$, $x^2_B(t)$, $y^2_B(t)$, $x^2_C(t)$, $y^2_C(t)$, square rooting and evaluating $x_I(t)$ and $y_I(t)$.
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From $l^2$ and $m^2$, we compute local distance : $4s(\sqrt{\frac{2h}{9-h^2}})^3$ which is one of result of \cite{RAG}.
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\section{Other formulas}
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From ellipse equation and squared distance formula
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$$ (X_9A)^2 = \frac{-(a^4 + b^4 + c^4 - 2(a^2 + 2bc)(b^2 + c^2) + 6b^2c^2 )bc}{(a^2 + b^2 + c^2 - 2(ab + bc + ca))^2} $$
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$$ = \frac{-2(h^4 + 4h^2t - 6h^2 - 12t - 3)s^2}{(h + 3)^2(h - 3)^2(1 + t)} = q $$
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we can deduce parameter t :
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$$t = - \frac{2(h^4 - 6h^2 - 3)s^2 + (h + 3)^2(h - 3)^2q}{8(h^2 - 3)s^2 + (h + 3)^2(h - 3)^2q} $$
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Squared distance from $X_9$ to $B$ and $C$ can be computed by permuting a,b,c cyclically in $(X_9A)^2$ formula or replacing $ t = \cos A$ with :
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$$\cos B = \frac{4(t + 1)(t - 1)^2 - (h^2 + 2t - 3)w}{2(t - 1)((h^2 + 2t - 3)(t + 1) - w)}$$
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or
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$$\cos C = \frac{4(t + 1)(t - 1)^2 + (h^2 + 2t - 3)w}{2(t - 1)((h^2 + 2t - 3)(t + 1) + w)}$$
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\section{Future work}
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We will compute radius of curvature, intersection point of A angle bissector with major axis, generalize the Ptolemy-Alhazen circle inner reflection problem to an ellipse and deal with n-orbits.
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\begin{thebibliography}{90}
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\bibitem{reznik} D. Reznik, R. Garcia, J. Koiller.
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{\it Can the elliptic billiard still surprise us?} arXiv:1911.01515.
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\bibitem{AST} A. Akopyan, R. Schwartz, S. Tabachnikov.
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{\it Billiards in ellipses revisited.} arXiv:2001.02934.
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\bibitem {Wolf} Wolfram Mathworld. - http://mathworld.wolfram.com/IsogonalConjugate.html {\it Isogonal conjugate.}
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\bibitem{OR} Olga Romaskevich
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{\it On the incenters of triangular orbits in elliptic billiard.} arXiv:1304.7588.
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\bibitem{RAG} Ronaldo A. Garcia.
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{\it Centers of inscribed circles in triangular orbits of an elliptic billiar.} arXiv:1607.00179.
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\end{thebibliography}
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\end{document}
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