mathe 发表于 2019-3-31 08:14
原来三坐标反演就是等角共轭
我们这里的定义是关于边的中点和无穷远点共轭调和，而等角共轭是关于角平分线 ...
等角共轭中所谓的三线坐标与重心坐标有所不同。
等角共轭的定义在仿射变换下无法保持，它的三线坐标在仿射变换下也无法保持。
但是等分共轭可以。
查看过定义了，三线坐标是点到三边距离，所以和面积坐标的区别是差了个边长做系数
为了将上面的三线坐标变换关系\(\frac{\lambda\lambda_1}{a^2}=\frac{\mu\mu_1}{b^2}=\frac{\nu\nu_1}{c^2}\)变成\(x_1x_2=y_1y_2=z_1z_2\)
我们将下面的重心坐标反转公式：
\(x_1 =\frac{ay}{2s}, y_1 = -\frac{a^2y+b^2y-c^2y-4as+4sx}{4as}, z_1 = -\frac{a^2y-b^2y+c^2y-4sx}{4as}\)
代入mathe 得到的下列关系中：
\(x_2 = \frac{a^2y_1z_1}{a^2y_1z_1+b^2x_1z_1+c^2x_1y_1}, y_2 =\frac{ b^2x_1z_1}{a^2y_1z_1+b^2x_1z_1+c^2x_1y_1}, z_2 =\frac{ c^2x_1y_1}{a^2y_1z_1+b^2x_1z_1+c^2x_1y_1}\)
我们便可以得到：
\(x_1 = \frac{(a^2y_0+b^2y_0-c^2y_0-4as+4sx_0)(a^2y_0-b^2y_0+c^2y_0-4sx_0)}{a^4y_0^2-2a^2b^2y_0^2-2a^2c^2y_0^2+b^4y_0^2-2b^2c^2y_0^2+c^4y_0^2-4a^3sy_0+4ab^2sy_0+4ac^2sy_0+16as^2x_0-16s^2x_0^2}, y_1 =\frac{ -2b^2y_0(a^2y_0-b^2y_0+c^2y_0-4sx_0)}{a^4y_0^2-2a^2b^2y_0^2-2a^2c^2y_0^2+b^4y_0^2-2b^2c^2y_0^2+c^4y_0^2-4a^3sy_0+4ab^2sy_0+4ac^2sy_0+16as^2x_0-16s^2x_0^2}, z_1 = \frac{-2c^2y_0(a^2y_0+b^2y_0-c^2y_0-4as+4sx_0)}{a^4y_0^2-2a^2b^2y_0^2-2a^2c^2y_0^2+b^4y_0^2-2b^2c^2y_0^2+c^4y_0^2-4a^3sy_0+4ab^2sy_0+4ac^2sy_0+16as^2x_0-16s^2x_0^2}\)
然后同上面51#的计算方案，这里由于当\(a=b=c=1\)时上面两种变换（交点变换和对称变换）将变成同一不变曲线，因此同样有\(a_1=a_2=a_3=a_4=a_5=a_6=\frac{2}{21}\),这样做就不必先求出6个不动点，可以得到如下不变曲线？
\(-2a^8c^2y-4a^7b^2xy+6a^6b^2c^2y+2a^6b^2x^2y+2a^6b^2y^3+6a^6c^4y+2a^6c^2x^2y+2a^6c^2y^3+12a^5b^4xy-46a^5b^2c^2xy-6a^4b^4c^2y-6a^4b^4x^2y-6a^4b^4y^3-4a^4b^2c^4y+46a^4b^2c^2x^2y+46a^4b^2c^2y^3-6a^4c^6y-6a^4c^4x^2y-6a^4c^4y^3-12a^3b^6xy+108a^3b^4c^2xy+104a^3b^2c^4xy+2a^2b^6c^2y+6a^2b^6x^2y+6a^2b^6y^3-2a^2b^4c^4y-106a^2b^4c^2x^2y+2a^2b^4c^2y^3-2a^2b^2c^6y-106a^2b^2c^4x^2y+2a^2b^2c^4y^3+2a^2c^8y+6a^2c^6x^2y+6a^2c^6y^3+4ab^8xy-62ab^6c^2xy+112ab^4c^4xy-54ab^2c^6xy-2b^8x^2y-2b^8y^3+58b^6c^2x^2y-50b^6c^2y^3-112b^4c^4x^2y+104b^4c^4y^3+58b^2c^6x^2y-50b^2c^6y^3-2c^8x^2y-2c^8y^3+8a^6c^2sx+8a^5b^2sx^2-8a^5b^2sy^2-16a^5c^2sx^2-16a^5c^2sy^2-16a^4b^2c^2sx-8a^4b^2sx^3-8a^4b^2sxy^2-16a^4c^4sx+8a^4c^2sx^3+8a^4c^2sxy^2-16a^3b^4sx^2+16a^3b^4sy^2+16a^3b^2c^2sx^2-184a^3b^2c^2sy^2+32a^3c^4sx^2+32a^3c^4sy^2+8a^2b^4c^2sx+16a^2b^4sx^3+16a^2b^4sxy^2-16a^2b^2c^4sx+8a^2c^6sx-16a^2c^4sx^3-16a^2c^4sxy^2+8ab^6sx^2-8ab^6sy^2-32ab^4c^2sx^2+200ab^4c^2sy^2+40ab^2c^4sx^2-208ab^2c^4sy^2-16ac^6sx^2-16ac^6sy^2-8b^6sx^3-8b^6sxy^2+24b^4c^2sx^3-408b^4c^2sxy^2-24b^2c^4sx^3+408b^2c^4sxy^2+8c^6sx^3+8c^6sxy^2=0\)
例如：取\(a=3,b=4,c=5,s=6\)可以得到如下曲线：
\(x^3+\frac{1286}{27}x^2y-\frac{73}{2}xy^2-\frac{64}{27}y^3-\frac{34}{3}x^2-\frac{1322}{9}xy+\frac{6137}{54}y^2+25x+\frac{100}{3}y=0\)
然而将下面对应变换关系代入上面不变曲线（对称变换不变曲线？）：
\(x = \frac{(x_0-3)(3x_0+4y_0)}{x_0^2+y_0^2-3x_0-4y_0}, y=\frac{(x_0-3)(4x_0-3y_0)}{x_0^2+y_0^2-3x_0-4y_0}\)
得到的曲线则和交点变换的不变曲线一样，有些奇怪！！！
\(x^2y-\frac{3}{4}xy^2-\frac{8}{27}x^2-\frac{25}{9}xy+\frac{25}{12}y^2+\frac{8}{9}x=0\)
下面是对称变换不变曲线（红色）与交点变换不变曲线图（绿色）：(下图中的点为不变点）
我们利用63#的计算方法,不变曲线方程选择57#的形式:
\(a_1\frac{x_1}{a}((\frac{y_1}{b})^2+(\frac{z_1}{c})^2)+a_2\frac{y_1}{b}((\frac{x_1}{a})^2+(\frac{z_1}{c})^2)+a_3\frac{z_1}{c}((\frac{x_1}{a})^2+(\frac{y_1}{b})^2)-a_4\frac{x_1}{a}\frac{y_1}{b}\frac{z_1}{c}=0\)
可以得到最终的不变曲线(已通过检验,正确)
\(-a^8cy-2a^7bxy+2a^7c^2y+3a^6b^2cy-21a^6bcxy+a^6bx^2y+a^6by^3+a^6c^3y-4a^6c^2xy+a^6cx^2y+a^6cy^3+6a^5b^3xy-4a^5b^2c^2y-4a^5b^2cxy+2a^5b^2x^2y-2a^5b^2y^3+2a^5bc^2xy+21a^5bcx^2y+21a^5bcy^3-4a^5c^4y+4a^5c^3xy+2a^5c^2x^2y-2a^5c^2y^3-3a^4b^4cy+42a^4b^3cxy-5a^4b^3x^2y-a^4b^3y^3+2a^4b^2c^3y+8a^4b^2c^2xy+a^4b^2cx^2y+5a^4b^2cy^3+42a^4bc^3xy+a^4bc^2x^2y+5a^4bc^2y^3+a^4c^5y+8a^4c^4xy-5a^4c^3x^2y-a^4c^3y^3-6a^3b^5xy+2a^3b^4c^2y+8a^3b^4cxy-4a^3b^4x^2y+4a^3b^4y^3+4a^3b^3c^2xy-42a^3b^3cx^2y-4a^3b^2c^4y-8a^3b^2c^2x^2y-8a^3b^2c^2y^3+2a^3bc^4xy-42a^3bc^3x^2y+2a^3c^6y-8a^3c^5xy-4a^3c^4x^2y+4a^3c^4y^3+a^2b^6cy-21a^2b^5cxy+7a^2b^5x^2y-a^2b^5y^3-3a^2b^4c^3y-4a^2b^4c^2xy-5a^2b^4cx^2y-5a^2b^4cy^3+42a^2b^3c^3xy-2a^2b^3c^2x^2y+6a^2b^3c^2y^3+3a^2b^2c^5y+8a^2b^2c^4xy-2a^2b^2c^3x^2y+6a^2b^2c^3y^3-21a^2bc^5xy-5a^2bc^4x^2y-5a^2bc^4y^3-a^2c^7y-4a^2c^6xy+7a^2c^5x^2y-a^2c^5y^3+2ab^7xy-4ab^6cxy+2ab^6x^2y-2ab^6y^3-6ab^5c^2xy+21ab^5cx^2y-21ab^5cy^3+12ab^4c^3xy-2ab^4c^2x^2y+2ab^4c^2y^3+6ab^3c^4xy-42ab^3c^3x^2y+42ab^3c^3y^3-12ab^2c^5xy-2ab^2c^4x^2y+2ab^2c^4y^3-2abc^6xy+21abc^5x^2y-21abc^5y^3+4ac^7xy+2ac^6x^2y-2ac^6y^3-3b^7x^2y+b^7y^3+3b^6cx^2y-b^6cy^3+9b^5c^2x^2y-3b^5c^2y^3-9b^4c^3x^2y+3b^4c^3y^3-9b^3c^4x^2y+3b^3c^4y^3+9b^2c^5x^2y-3b^2c^5y^3+3bc^6x^2y-bc^6y^3-3c^7x^2y+c^7y^3+4a^6csx+4a^5bsx^2-4a^5bsy^2-8a^5csx^2-8a^5csy^2-8a^4b^2csx-84a^4bcsy^2-4a^4bsx^3-4a^4bsxy^2-8a^4c^3sx+16a^4c^2sy^2+4a^4csx^3+4a^4csxy^2-8a^3b^3sx^2+8a^3b^3sy^2+16a^3b^2csx^2+16a^3b^2sxy^2-8a^3bc^2sx^2-24a^3bc^2sy^2+16a^3c^3sx^2-16a^3c^2sxy^2+4a^2b^4csx+84a^2b^3csy^2+8a^2b^3sx^3-8a^2b^3sxy^2-8a^2b^2c^3sx+16a^2b^2c^2sy^2-8a^2b^2csx^3-24a^2b^2csxy^2-84a^2bc^3sy^2+8a^2bc^2sx^3+24a^2bc^2sxy^2+4a^2c^5sx-16a^2c^4sy^2-8a^2c^3sx^3+8a^2c^3sxy^2+4ab^5sx^2-4ab^5sy^2-8ab^4csx^2+8ab^4csy^2-16ab^4sxy^2-8ab^3c^2sx^2+8ab^3c^2sy^2-168ab^3csxy^2+16ab^2c^3sx^2-16ab^2c^3sy^2+4abc^4sx^2-4abc^4sy^2+168abc^3sxy^2-8ac^5sx^2+8ac^5sy^2+16ac^4sxy^2-4b^5sx^3+12b^5sxy^2+4b^4csx^3-12b^4csxy^2+8b^3c^2sx^3-24b^3c^2sxy^2-8b^2c^3sx^3+24b^2c^3sxy^2-4bc^4sx^3+12bc^4sxy^2+4c^5sx^3-12c^5sxy^2=0\)
例如:取\(a = 6, b = 5, c = 3, s = 2\sqrt{14}\)得到对称变换的不变曲线:
\(2372768y^3-594432\sqrt{14}xy^2-2231040x^2y+13429248xy-774144\sqrt{14}x+43008\sqrt{14}x^2-1981824\sqrt{14}y^2+14336\sqrt{14}x^3-1548288y=0\)
交点变换的不变曲线为:
\(168480y^3-41472\sqrt{14}xy^2-145152x^2y+899584xy-43008\sqrt{14}x+7168\sqrt{14}x^2-122368\sqrt{14}y^2-279552y=0\)
画图得到( 红色为交点变换的不变曲线,绿色为对称变换的不变曲线，孤立的点为不动点)
我们还可以有对偶三坐标反演
如图三角形ABC中，红色直线和三边分别交于H,I,J三点。分别做三点关于三个中点的对称点N,O,P,那么NOP三点共紫线，这条紫线就是红色的对偶三坐标反演。
当红线绕着固定点转动时，紫线会和紫色椭圆相切，也就是说，会将线束映射成和三边相切的椭圆，同样对偶三坐标反演会将和三边相切的圆锥曲线的二次线束映射为一个线束。
对偶三坐标反演示范图：
如图橙色双曲线有黑色外切三角形。
双曲线上有红绿蓝三动点，它们各自有同色切线（虚线）交三角形三边，这些交点关于三边中点形成对称点。连接这些对称点各自得到同色实线。
所有这些同色实线必然经过紫色大点。验证了和三角形三边相切的圆锥曲线的二次线束的对偶三坐标反演为过定点的一次线束。
这个紫色定点可以称为圆锥曲线关于这个外切三角形的对偶三坐标反演中心
设\(A,B,C\),定点\(G\),过\(G\)点直线斜率为\(k\),则我们可以算得外切三角形\(ABC\)的椭圆方程：
\(a^{10}-2a^9x-2a^9x_0-4a^8b^2-4a^8c^2+a^8x^2+2a^8xx_0+a^8x_0^2-a^8y^2+6a^8yy_0-a^8y_0^2+8a^7b^2x+8a^7b^2x_0+8a^7c^2x+8a^7c^2x_0-4a^7xyy_0+4a^7xy_0^2+4a^7x_0y^2-4a^7x_0yy_0+6a^6b^4+4a^6b^2c^2-4a^6b^2x^2-8a^6b^2xx_0-4a^6b^2x_0^2-24a^6b^2yy_0+6a^6c^4-4a^6c^2x^2-8a^6c^2xx_0-4a^6c^2x_0^2+4a^6c^2y^2+4a^6c^2y_0^2-4a^6x^2y_0^2+8a^6xx_0yy_0-4a^6x_0^2y^2-12a^5b^4x-12a^5b^4x_0-8a^5b^2c^2x-8a^5b^2c^2x_0+20a^5b^2xyy_0-4a^5b^2xy_0^2-4a^5b^2x_0y^2+20a^5b^2x_0yy_0-12a^5c^4x-12a^5c^4x_0-4a^5c^2xyy_0-12a^5c^2xy_0^2-12a^5c^2x_0y^2-4a^5c^2x_0yy_0-4a^4b^6+4a^4b^4c^2+6a^4b^4x^2+12a^4b^4xx_0+6a^4b^4x_0^2+2a^4b^4y^2+28a^4b^4yy_0+2a^4b^4y_0^2+4a^4b^2c^4+4a^4b^2c^2x^2+8a^4b^2c^2xx_0+4a^4b^2c^2x_0^2+4a^4b^2c^2y^2-8a^4b^2c^2yy_0+4a^4b^2c^2y_0^2+8a^4b^2x^2y_0^2-16a^4b^2xx_0yy_0+8a^4b^2x_0^2y^2-4a^4c^6+6a^4c^4x^2+12a^4c^4xx_0+6a^4c^4x_0^2-6a^4c^4y^2-20a^4c^4yy_0-6a^4c^4y_0^2+8a^4c^2x^2y_0^2-16a^4c^2xx_0yy_0+8a^4c^2x_0^2y^2+8a^3b^6x+8a^3b^6x_0-8a^3b^4c^2x-8a^3b^4c^2x_0-28a^3b^4xyy_0-4a^3b^4xy_0^2-4a^3b^4x_0y^2-28a^3b^4x_0yy_0-8a^3b^2c^4x-8a^3b^2c^4x_0+8a^3b^2c^2xyy_0-8a^3b^2c^2xy_0^2-8a^3b^2c^2x_0y^2+8a^3b^2c^2x_0yy_0+8a^3c^6x+8a^3c^6x_0+20a^3c^4xyy_0+12a^3c^4xy_0^2+12a^3c^4x_0y^2+20a^3c^4x_0yy_0+a^2b^8-4a^2b^6c^2-4a^2b^6x^2-8a^2b^6xx_0-4a^2b^6x_0^2-8a^2b^6yy_0+6a^2b^4c^4+4a^2b^4c^2x^2+8a^2b^4c^2xx_0+4a^2b^4c^2x_0^2+4a^2b^4c^2y^2+32a^2b^4c^2yy_0+4a^2b^4c^2y_0^2-4a^2b^4x^2y_0^2+8a^2b^4xx_0yy_0-4a^2b^4x_0^2y^2-4a^2b^2c^6+4a^2b^2c^4x^2+8a^2b^2c^4xx_0+4a^2b^2c^4x_0^2-8a^2b^2c^4y^2-40a^2b^2c^4yy_0-8a^2b^2c^4y_0^2+8a^2b^2c^2x^2y_0^2-16a^2b^2c^2xx_0yy_0+8a^2b^2c^2x_0^2y^2+a^2c^8-4a^2c^6x^2-8a^2c^6xx_0-4a^2c^6x_0^2+4a^2c^6y^2+16a^2c^6yy_0+4a^2c^6y_0^2-4a^2c^4x^2y_0^2+8a^2c^4xx_0yy_0-4a^2c^4x_0^2y^2-2ab^8x-2ab^8x_0+8ab^6c^2x+8ab^6c^2x_0+12ab^6xyy_0+4ab^6xy_0^2+4ab^6x_0y^2+12ab^6x_0yy_0-12ab^4c^4x-12ab^4c^4x_0-36ab^4c^2xyy_0-12ab^4c^2xy_0^2-12ab^4c^2x_0y^2-36ab^4c^2x_0yy_0+8ab^2c^6x+8ab^2c^6x_0+36ab^2c^4xyy_0+12ab^2c^4xy_0^2+12ab^2c^4x_0y^2+36ab^2c^4x_0yy_0-2ac^8x-2ac^8x_0-12ac^6xyy_0-4ac^6xy_0^2-4ac^6x_0y^2-12ac^6x_0yy_0+b^8x^2+2b^8xx_0+b^8x_0^2-b^8y^2-2b^8yy_0-b^8y_0^2-4b^6c^2x^2-8b^6c^2xx_0-4b^6c^2x_0^2+4b^6c^2y^2+8b^6c^2yy_0+4b^6c^2y_0^2+6b^4c^4x^2+12b^4c^4xx_0+6b^4c^4x_0^2-6b^4c^4y^2-12b^4c^4yy_0-6b^4c^4y_0^2-4b^2c^6x^2-8b^2c^6xx_0-4b^2c^6x_0^2+4b^2c^6y^2+8b^2c^6yy_0+4b^2c^6y_0^2+c^8x^2+2c^8xx_0+c^8x_0^2-c^8y^2-2c^8yy_0-c^8y_0^2+8a^7sy+8a^7sy_0-8a^6sxy-24a^6sxy_0-24a^6sx_0y-8a^6sx_0y_0-8a^5b^2sy-8a^5b^2sy_0-24a^5c^2sy-24a^5c^2sy_0+16a^5sx^2y_0+16a^5sxx_0y+16a^5sxx_0y_0+16a^5sx_0^2y+32a^5sy^2y_0+32a^5syy_0^2+8a^4b^2sxy+40a^4b^2sxy_0+40a^4b^2sx_0y+8a^4b^2sx_0y_0+24a^4c^2sxy+56a^4c^2sxy_0+56a^4c^2sx_0y+24a^4c^2sx_0y_0-8a^3b^4sy-8a^3b^4sy_0-16a^3b^2c^2sy-16a^3b^2c^2sy_0-32a^3b^2sx^2y_0-32a^3b^2sxx_0y-32a^3b^2sxx_0y_0-32a^3b^2sx_0^2y+24a^3c^4sy+24a^3c^4sy_0-32a^3c^2sx^2y_0-32a^3c^2sxx_0y-32a^3c^2sxx_0y_0-32a^3c^2sx_0^2y+8a^2b^4sxy-8a^2b^4sxy_0-8a^2b^4sx_0y+8a^2b^4sx_0y_0+16a^2b^2c^2sxy+48a^2b^2c^2sxy_0+48a^2b^2c^2sx_0y+16a^2b^2c^2sx_0y_0-24a^2c^4sxy-40a^2c^4sxy_0-40a^2c^4sx_0y-24a^2c^4sx_0y_0+8ab^6sy+8ab^6sy_0-24ab^4c^2sy-24ab^4c^2sy_0+16ab^4sx^2y_0+16ab^4sxx_0y+16ab^4sxx_0y_0+16ab^4sx_0^2y-32ab^4sy^2y_0-32ab^4syy_0^2+24ab^2c^4sy+24ab^2c^4sy_0-32ab^2c^2sx^2y_0-32ab^2c^2sxx_0y-32ab^2c^2sxx_0y_0-32ab^2c^2sx_0^2y+64ab^2c^2sy^2y_0+64ab^2c^2syy_0^2-8ac^6sy-8ac^6sy_0+16ac^4sx^2y_0+16ac^4sxx_0y+16ac^4sxx_0y_0+16ac^4sx_0^2y-32ac^4sy^2y_0-32ac^4syy_0^2-8b^6sxy-8b^6sxy_0-8b^6sx_0y-8b^6sx_0y_0+24b^4c^2sxy+24b^4c^2sxy_0+24b^4c^2sx_0y+24b^4c^2sx_0y_0-24b^2c^4sxy-24b^2c^4sxy_0-24b^2c^4sx_0y-24b^2c^4sx_0y_0+8c^6sxy+8c^6sxy_0+8c^6sx_0y+8c^6sx_0y_0=0\)
取\(a=5,b=4,c=3,s=6,x_0=-2,y_0=-2,k=\frac{1}{2}\)
可得到椭圆方程：
\(1115136x^2+884736xy+451584y^2-11759616x-5879808y+31002624=0\)
画图得到：
上面椭圆方程简化为重心坐标方程：
\(-4a^4x_1^2y_0^2+(a^6-2a^5x_0-2a^4b^2-2a^4c^2+a^4x_0^2-a^4y_0^2+4a^3b^2x_0+4a^3c^2x_0+a^2b^4-2a^2b^2c^2-2a^2b^2x_0^2-2a^2b^2y_0^2+a^2c^4-2a^2c^2x_0^2+2a^2c^2y_0^2-2ab^4x_0+4ab^2c^2x_0-2ac^4x_0+b^4x_0^2-b^4y_0^2-2b^2c^2x_0^2+2b^2c^2y_0^2+c^4x_0^2-c^4y_0^2+8a^3sy_0-8a^2sx_0y_0+8ab^2sy_0-8ac^2sy_0-8b^2sx_0y_0+8c^2sx_0y_0)y_1^2+(a^4x_0^2-a^4y_0^2-2a^2b^2x_0^2+2a^2b^2y_0^2-2a^2c^2x_0^2-2a^2c^2y_0^2+b^4x_0^2-b^4y_0^2-2b^2c^2x_0^2+2b^2c^2y_0^2+c^4x_0^2-c^4y_0^2+8a^2sx_0y_0-8b^2sx_0y_0+8c^2sx_0y_0)z_1^2+(-4a^4y_0^2-4a^2b^2y_0^2+4a^2c^2y_0^2+16a^3sy_0-16a^2sx_0y_0)x_1y_1+(-4a^4y_0^2+4a^2b^2y_0^2-4a^2c^2y_0^2+16a^2sx_0y_0)x_1z_1+(-2a^5x_0+2a^4x_0^2+2a^4y_0^2+4a^3b^2x_0+4a^3c^2x_0-4a^2b^2x_0^2-4a^2c^2x_0^2-2ab^4x_0+4ab^2c^2x_0-2ac^4x_0+2b^4x_0^2-2b^4y_0^2-4b^2c^2x_0^2+4b^2c^2y_0^2+2c^4x_0^2-2c^4y_0^2-8a^3sy_0+8ab^2sy_0-8ac^2sy_0-16b^2sx_0y_0+16c^2sx_0y_0)y_1z_1=0\)
对偶三坐标反演和三坐标反演同样采用面积坐标，其不动线无穷远直线对偶于重心，三条中位线满足方程x+y=z等，所以对偶于另外三个不动点。它们构成了四个不动线。可以看出，对偶者的坐标形式一模一样。所以如果我们把圆锥曲线用其切线满足的方程表示，得出对偶三坐标形式各种性质会同原先三坐标反演下的一模一样
为了计算方便，我们先设三角形\(ABC,A,B,C\),及定点\(G\)反向解析外切的圆锥曲线：
\(a_1x^2+2a_2xy+a_3y^2+2a_4x+2a_5y+a_6=0\)
由于圆锥曲线的切线为AB,AC,BC,及过G点的任何三线\(L_1: y-y_0=k_1(x-x_0) ,L_2: y-y_0=k_2(x-x_0) ,L_3: y-y_0=k_3(x-x_0)\) 的关于各边中点的对称线，主要是针对任意的\(k_1,k_2,k_3\)成立；我们求出满足条件的圆锥曲线：
\(-2a^5y^2y_0-2a^5yy_0^2+2ab^4y^2y_0+2ab^4yy_0^2-4ab^2c^2y^2y_0-4ab^2c^2yy_0^2+2ac^4y^2y_0+2ac^4yy_0^2-a^4sy^2+6a^4syy_0-a^4sy_0^2-4a^3sxyy_0+4a^3sxy_0^2+4a^3sx_0y^2-4a^3sx_0yy_0-2a^2b^2sy^2-12a^2b^2syy_0-2a^2b^2sy_0^2+2a^2c^2sy^2+12a^2c^2syy_0+2a^2c^2sy_0^2-4a^2sx^2y_0^2+8a^2sxx_0yy_0-4a^2sx_0^2y^2+12ab^2sxyy_0+4ab^2sxy_0^2+4ab^2sx_0y^2+12ab^2sx_0yy_0-12ac^2sxyy_0-4ac^2sxy_0^2-4ac^2sx_0y^2-12ac^2sx_0yy_0-b^4sy^2-2b^4syy_0-b^4sy_0^2+2b^2c^2sy^2+4b^2c^2syy_0+2b^2c^2sy_0^2-c^4sy^2-2c^4syy_0-c^4sy_0^2+8a^3s^2y+8a^3s^2y_0-8a^2s^2xy-24a^2s^2xy_0-24a^2s^2x_0y-8a^2s^2x_0y_0+8ab^2s^2y+8ab^2s^2y_0-8ac^2s^2y-8ac^2s^2y_0+16as^2x^2y_0+16as^2xx_0y+16as^2xx_0y_0+16as^2x_0^2y-8b^2s^2xy-8b^2s^2xy_0-8b^2s^2x_0y-8b^2s^2x_0y_0+8c^2s^2xy+8c^2s^2xy_0+8c^2s^2x_0y+8c^2s^2x_0y_0-16a^2s^3+32as^3x+32as^3x_0-16s^3x^2-32s^3xx_0-16s^3x_0^2=0\)
我们分别取
\(a=5,b=4,c=3,s=6,x_0=2,y_0=2,k_1=\frac{1}{5},k_2=\frac{1}{3},k_3=-\frac{1}{4}\)
\(-96x^2+384xy-12384y^2+384x+4032y-384=0\)
画图得到：
\(a=5,b=4,c=3,s=6,x_0=-2,y_0=2,k_1=\frac{1}{5},k_2=\frac{1}{3},k_3=-\frac{1}{4}\)
\(-96x^2-20736xy-27744y^2+4992x+102336y-64896=0\)
画图得到：
我们利用重心坐标反演公式：
\(x=-\frac{a^2x_1+2a^2y_1+b^2x_1-c^2x_1-2a^2}{2a},x_0=-\frac{a^2x_2+2a^2y_2+b^2x_2-c^2x_2-2a^2}{2a}, y=\frac{2x_1s}{a}, y_0=\frac{2x_2s}{a},x_1+y_1+z_1=1,x_2+y_2+z_2=1\)
代入上面圆锥曲线方程得到：
\(x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2-2x_1^2y_2-2x_1x_2y_1-2x_1x_2y_2-2x_1y_1y_2-2x_1y_2^2-2x_2^2y_1-2x_2y_1^2-2x_2y_1y_2+x_1^2+2x_1x_2+2x_1y_1+4x_1y_2+x_2^2+4x_2y_1+2x_2y_2+y_1^2+2y_1y_2+y_2^2-2x_1-2x_2-2y_1-2y_2+1=0\)
进一步简化得到：
\(x_1^2x_2^2+y_1^2y_2^2+z_1^2z_2^2-2x_1x_2y_1y_2-2x_1x_2z_1z_2-2y_1y_2z_1z_2=0\)
我们下一步要搞清楚：
\(x_1^2x_2^2+y_1^2y_2^2+z^2z_2^2-2x_1x_2y_1y_2-2x_1x_2z_1z_2-2y_1y_2z_1z_2=0\)
1. 如何变换成对合变换的基本关系：
\(xx_1=yy_1=zz_1\)
2.上面圆锥曲线的不动直线为：
\(y=\frac{a}{s}\)
\(y=-\frac{2s(a-2x)}{a^2-b^2+c^2}\)
\(y=\frac{2s(a-2x)}{a^2+b^2-c^2}\)
是否也存在不变3次曲线？