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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=(1−t2)(1−2k−(t−k)2)w = \sqrt{ (1 - t^2)( 1 - 2k - (t - k)^2) }a=s2(t−1)2t−k−2a = s \frac{ 2(t - 1) }{ 2t - k - 2 }b=st2−(t+1)k−1+w2t2−(t+1)k−2b = s \frac{ t^2 - (t + 1)k - 1 + w }{ 2 t^2 - (t + 1)k - 2 }c=st2−(t+1)k−1−w2t2−(t+1)k−2c = s \frac{ t^2 - (t + 1)k - 1 - w }{ 2 t^2 - (t + 1)k - 2 }

with parameters kk as ratio of inradius and circumradius (k=rRk = \frac{r}{R}), tt as cosine of internal angle at A vertex, and ss as semiperimeter.

Elliptic billiard

It comes from a Ravi substitution (a=x+ya = x + y, b=y+zb = y + z, c=z+xc = z + x) and three equations in x,y,z unknowns given parameters t,s,k :

x+y+z=sx + y + z = s4xyz(x+y)(x+z)(y+z)=k\frac{4 x y z}{(x + y)(x + z)(y + z)} = kz2+xz+yz−xy(x+z)(y+z)=t\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 :

x2−mx+n=0x^2 - m x + n = 0

where sum of roots is m=2(1−t)s2(1−t)+k m = \frac{2 (1 - t) s}{2 (1 - t) + k} and product is n=(1−t)ks22(1−t2)+(t+1)k 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 MpM_p and with radius 2sk+4\frac{2s}{k + 4} is the cosine circle for the excentral triangle T.

The excentral triangle T has for vertices, XAX_A XBX_B XCX_C, the excenters of the reference ABC triangle.

Its symmetrical image T' wrt Mittenpunkt ParseError: KaTeX parse error: Expected group after '_' at position 2: M_̲ has vertices YAY_A, YB Y_B and YCY_C.

Mittenpunkt has trilinear coordinates : Mp=z:x:yM_p = z : x : y.

Cosine Circle

Triangle T and its symmetrical image T' intersect at six points M1,...,M6M_1,...,M_6 which are on the cosine circle.

Trilinear coordinates of these points are :

M1=a−b−3c:−a+b−c:a−b+cM_1 = a - b - 3 c : -a + b - c : a - b + cM2=−a−b+c:3a−b+c:a+b−cM_2 = -a - b + c : 3 a - b + c : a + b - cM3=−a+b+c:a−b−c:a+3b−cM_3 = -a + b + c : a - b - c : a + 3b - c

and

M4=−a+b−c:a−b+c:3a+b−cM_4 = -a + b - c : a - b + c : 3 a + b - cM5=a−3b−c:a+b−c:−a−b+cM_5 = a - 3 b - c : a + b - c : -a - b + cM6=−a+b+c:a−b+3c:a−b−cM_6 = -a + b + c : a - b + 3 c : a - b - c

Computing square of quadrance (=squared distance) between MpM_p and M1M_1 we get squared radius of cosine circle :

rMp2=(−2abca2−2ab+b2−2ac−2bc+c2)2=((x+y)(x+z)(y+z)2(xy+xz+yz))2rM_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 kk and ss : rMp=2sk+4rM_p = \frac{2s}{k + 4}

Parallels with respect to ABC edges through MpM_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(XA)2=xy(x+z)(y+z)cos(X_A)^2 = \frac{x y}{(x + z)(y + z)}cos(XB)2=yz(y+x)(z+x)cos(X_B)^2 = \frac{y z}{(y + x)(z + x)}cos(XC)2=zx(z+y)(x+y)cos(X_C)^2 = \frac{z x}{(z + y)(x + y)}

Product : cos(XA)2cos(XB)2cos(XC)2=(xyz(x+y)(y+z)(z+x))2=(k4)2 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(XA)2+cos(XB)2+cos(XC)2=1−k2cos(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 : 116(k2−8k+s2k2r2) \frac{1}{16}(k^2 - 8 k + \frac{s^2 k^2}{r^2}) where r2r^2 is the changing squared inradius of ABC : r2=xyzx+y+zr^2 = \frac{x y z}{x + y + z}.

Exradii are :

ra=rss−a=rx+y+zxr_a = \frac{r s}{s - a} = r \frac{x + y + z}{x}rb=rss−b=rx+y+zyr_b = \frac{r s}{s - b} = r \frac{x + y + z}{y}rc=rss−c=rx+y+zzr_c = \frac{r s}{s - c} = r \frac{x + y + z}{z}