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## Triangular orbits in elliptic billiards - Triangle geometry

### Purpose

I made quicky current public worsheet to share about some of my computations.

### Parametrization in triangle geometry

I got following parametrization of a 3-orbit triangle ABC in the circumbilliard :

$w = \sqrt{ (1 - t^2)( 1 - 2k - (t - k)^2) }$ $a = s \frac{ 2(t - 1) }{ 2t - k - 2 }$ $b = s \frac{ t^2 - (t + 1)k - 1 + w }{ 2 t^2 - (t + 1)k - 2 }$ $c = s \frac{ t^2 - (t + 1)k - 1 - w }{ 2 t^2 - (t + 1)k - 2 }$

with parameters $k$ as ratio of inradius and circumradius ($k = \frac{r}{R}$), $t$ as cosine of internal angle at A vertex, and $s$ as semiperimeter. It comes from a Ravi substitution ($a = x + y$, $b = y + z$, $c = z + x$) and three equations in x,y,z unknowns given parameters t,s,k :

$x + y + z = s$ $\frac{4 x y z}{(x + y)(x + z)(y + z)} = k$ $\frac{z^2 + x z + y z - x y }{(x + z) (y + z)} = t$

Solving is best (simplest) for the couple of first and third equations, expressing y,z from x.

Putting back expression into second equation we have the binomial equation :

$x^2 - m x + n = 0$

where sum of roots is $m = \frac{2 (1 - t) s}{2 (1 - t) + k}$ and product is $n = \frac{(1 - t) k s^2}{2 (1 - t^2) + (t + 1) k}$

Choosing one root or the other is only changing sign of w (or swapping b and c).

### Cosine circle

Circle centered at Mittenpunk $M_p$ and with radius $\frac{2s}{k + 4}$ is the cosine circle for the excentral triangle T.

The excentral triangle T has for vertices, $X_A$ $X_B$ $X_C$, the excenters of the reference ABC triangle.

Its symmetrical image T' wrt Mittenpunkt $M_$ has vertices $Y_A$, $Y_B$ and $Y_C$.

Mittenpunkt has trilinear coordinates : $M_p = z : x : y$. Triangle T and its symmetrical image T' intersect at six points $M_1,...,M_6$ which are on the cosine circle.

Trilinear coordinates of these points are :

$M_1 = a - b - 3 c : -a + b - c : a - b + c$

$M_2 = -a - b + c : 3 a - b + c : a + b - c$

$M_3 = -a + b + c : a - b - c : a + 3b - c$

and

$M_4 = -a + b - c : a - b + c : 3 a + b - c$

$M_5 = a - 3 b - c : a + b - c : -a - b + c$

$M_6 = -a + b + c : a - b + 3 c : a - b - c$

Computing square of quadrance (=squared distance) between $M_p$ and $M_1$ we get squared radius of cosine circle :

$rM_p^2 = (\frac{-2 a b c}{a^2 - 2 ab + b^2 - 2 ac - 2 bc + c^2})^2 = (\frac{(x + y)(x + z)(y + z)}{2(xy + xz + yz)})^2$

and using equations for $k$ and $s$ : $rM_p = \frac{2s}{k + 4}$

Parallels with respect to ABC edges through $M_p$, cut disk in six sectors, and cutting points on cosine circle are intersection of T and T' edges.

Squared cosines of internal angles of external triangle T are :

$cos(X_A)^2 = \frac{x y}{(x + z)(y + z)}$ $cos(X_B)^2 = \frac{y z}{(y + x)(z + x)}$ $cos(X_C)^2 = \frac{z x}{(z + y)(x + y)}$

Product : $cos(X_A)^2 cos(X_B)^2 cos(X_C)^2 = (\frac{x y z }{(x + y)(y + z)(z + x)})^2 = (\frac{k}{4})^2$.

Sum of squared cosines is : $cos(X_A)^2 + cos(X_B)^2 + cos(X_C)^2 = 1 - \frac{k}{2}$.

Sum of product 2 by 2 of squared cosines is : $\frac{1}{16}(k^2 - 8 k + \frac{s^2 k^2}{r^2})$ where $r^2$ is the changing squared inradius of ABC : $r^2 = \frac{x y z}{x + y + z}$.

$r_a = \frac{r s}{s - a} = r \frac{x + y + z}{x}$ $r_b = \frac{r s}{s - b} = r \frac{x + y + z}{y}$ $r_c = \frac{r s}{s - c} = r \frac{x + y + z}{z}$