由三角形引导的一个几何变换——三坐标反演

hujunhua

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。

mathe

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

数学星空

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）

数学星空

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)

mathe

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。

mathe

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

数学星空

设\(A[x_1,y_1],B[0,0],C[a,0]\),定点\(G[x_0,y_0]\),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：

\(-4a^4x_1^2y_0^2+(a^6-2a^5x_0-2a^4b^2-2a^4c^2+a^4x_0^2-a^4y_0^2+4a^3b^2x_0+4a^3c^2x_0+a^2b^4-2a^2b^2c^2-2a^2b^2x_0^2-2a^2b^2y_0^2+a^2c^4-2a^2c^2x_0^2+2a^2c^2y_0^2-2ab^4x_0+4ab^2c^2x_0-2ac^4x_0+b^4x_0^2-b^4y_0^2-2b^2c^2x_0^2+2b^2c^2y_0^2+c^4x_0^2-c^4y_0^2+8a^3sy_0-8a^2sx_0y_0+8ab^2sy_0-8ac^2sy_0-8b^2sx_0y_0+8c^2sx_0y_0)y_1^2+(a^4x_0^2-a^4y_0^2-2a^2b^2x_0^2+2a^2b^2y_0^2-2a^2c^2x_0^2-2a^2c^2y_0^2+b^4x_0^2-b^4y_0^2-2b^2c^2x_0^2+2b^2c^2y_0^2+c^4x_0^2-c^4y_0^2+8a^2sx_0y_0-8b^2sx_0y_0+8c^2sx_0y_0)z_1^2+(-4a^4y_0^2-4a^2b^2y_0^2+4a^2c^2y_0^2+16a^3sy_0-16a^2sx_0y_0)x_1y_1+(-4a^4y_0^2+4a^2b^2y_0^2-4a^2c^2y_0^2+16a^2sx_0y_0)x_1z_1+(-2a^5x_0+2a^4x_0^2+2a^4y_0^2+4a^3b^2x_0+4a^3c^2x_0-4a^2b^2x_0^2-4a^2c^2x_0^2-2ab^4x_0+4ab^2c^2x_0-2ac^4x_0+2b^4x_0^2-2b^4y_0^2-4b^2c^2x_0^2+4b^2c^2y_0^2+2c^4x_0^2-2c^4y_0^2-8a^3sy_0+8ab^2sy_0-8ac^2sy_0-16b^2sx_0y_0+16c^2sx_0y_0)y_1z_1=0\)

mathe

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

数学星空

为了计算方便，我们先设三角形\(ABC,A[x_1,y_1],B[0,0],C[a,0]\),及定点\(G[x_0,y_0]\)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：

\(x_1^2x_2^2+y_1^2y_2^2+z_1^2z_2^2-2x_1x_2y_1y_2-2x_1x_2z_1z_2-2y_1y_2z_1z_2=0\)

数学星空

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？