过三点的椭圆参数方程

creasson

过三点$\left(0,0\right)$,$\left(1,0\right)$,$\left(x_0,y_0\right)$的二次曲线总可以表示为:$\left(\frac{pu^2+\left(x_0-p \right)u}{pu^2+qu+1-p-q},\frac{y_0u}{pu^2+qu+1-p-q} \right)$
在$q^2< 4p\left(1 - p - q\right)$时，属于椭圆。
椭圆的中心:$\left(\frac{-2p+2p^2+x_0q+pq}{-4p+4p^2+4pq+q^2},\frac{y_0q}{-4p+4p^2+4pq+q^2} \right)$
椭圆的半轴长$L$满足方程:
\[{\left( { - 4p + 4{p^2} + 4pq + {q^2}} \right)^3}{L^4} - 4{y_0}^2{p^2}{\left( { - 1 + p + q} \right)^2} + 4p\left( { - 1 + p + q} \right)\left( {{x_0}^2 + {y_0}^2 + p - 2{x_0}p - {x_0}q} \right)\left( { - 4p + 4{p^2} + 4pq + {q^2}} \right){L^2} = 0\]
由格林面积公式可以求得椭圆面积：$S = \frac{2p\left(-1+p+q \right)y_0\pi }{\left(-4p+4p^2+4pq+q^2\right)^{3/2}$
对$p,q$分别求导可知在$p=1,q=-1$时面积为最小。

creasson

与三点$\left(0,0\right)$,$\left(1,0\right)$,$\left(x_0,y_0\right)$的三条连线相切的二次曲线总可以表示为：$\left( \frac{\left(1 + p\right)u^2+ \left(p + q\right)x_0}{u^2 + 2pu + q},\frac{\left(p + q \right)y_0}{u^2 + 2pu + q}\right)$
它在$p^2 < q$时成为椭圆，其中心为$\left( \frac{\left( 1+p \right)q+\left( p+q \right)x_0}{2\left( q-p^2 \right)},\frac{\left( p+q \right)y_0}{2\left( q-p^2\right)} \right)$
半轴长$L$满足方程
\[4{{\left( q-{{p}^{2}} \right)}^{3}}{{L}^{4}}+y_{0}^{2}{{p}^{2}}{{\left( 1+p \right)}^{2}}{{\left( p+q \right)}^{2}}-\left( q-{{p}^{2}} \right)\left( {{\left( 1+p \right)}^{2}}{{q}^{2}}+2{{x}_{0}}\left( 1+p \right)\left( 2{{p}^{2}}-q \right)\left( p+q \right)+x_{0}^{2}{{\left( p+q \right)}^{2}}+y_{0}^{2}{{\left( p+q \right)}^{2}} \right){{L}^{2}}=0\]
三个切点分别为$\left(1 + p,0 \right)$,$\left( \frac{1 + p + \left(p + q\right)x_0}{1 + 2p + q},\frac{\left( p + q\right)y_0}{1 + 2p + q} \right)$,$\left(\frac{\left(p + q\right)x_0}{q},\frac{\left(p + q\right)y_0}q\right)$
椭圆面积为$\frac{p\left(1 + p\right)\left(p + q\right)y_0\pi}{\left(q - p^2\right)^{3/2}}$,并在$p=-\frac{1}{2},q = 1$时取得最大，此时中心为三点的重心.

wayne

欢迎creasson 大神 归来

creasson

如果固定椭圆的两个焦点为0,1，则椭圆上任意一点可用参数表示为：$(\frac{- p^2u^2 + \left(1 + p\right)^2}{pu^2 + \left(p + 1\right)},\frac{2p\left( p + 1\right)u}{pu^2 + \left(p + 1\right)})$
由此可以推出，如果该椭圆内切三角形$ABC$,并使得切点$DEF$分线段$\frac{AD}{BD} = \lambda ,\frac{BE}{CE} = \mu ,\frac{CF}{AF} = \frac{1}{\lambda \mu}$，那么
\[\left\{ \begin{array}{l}
\mathop {{F_1}A}\limits^ \to   \times \mathop {{F_1}B}\limits^ \to   + \mathop {\lambda \mu {F_1}B}\limits^ \to   \times \mathop {{F_1}C}\limits^ \to   + \mu \mathop {{F_1}C}\limits^ \to   \times \mathop {{F_1}A}\limits^ \to   = 0\\
\mathop {{F_2}A}\limits^ \to   \times \mathop {{F_2}B}\limits^ \to   + \mathop {\lambda \mu {F_2}B}\limits^ \to   \times \mathop {{F_2}C}\limits^ \to   + \mu \mathop {{F_2}C}\limits^ \to   \times \mathop {{F_2}A}\limits^ \to   = 0
\end{array} \right.\]
由此可以看出，在复平面上，如果$z_1、z_2、z_3$是三角形的三个顶点，内切椭圆分三边比例为$\lambda 、\mu 、\frac{1}{\lambda \mu }$，则椭圆的焦点是方程
\[\left( {z - {z_1}} \right)\left( {z - {z_2}} \right) + \lambda \mu \left( {z - {z_2}} \right)\left( {z - {z_3}} \right) + \mu \left( {z - {z_3}} \right)\left( {z - {z_1}} \right) = 0\]
的两个根。当切点为三边中点时，即为marden定理。

creasson

定义$F\left(u,v\right): = a\left(uv\right)^2 + b\left( uv\right)\left(u + v\right) + c\left( u + v \right)^2 + d\left(uv\right) + e\left(u + v\right) + f$
彭赛列闭合定理的等价表述：
数列${\p_k\}$：满足$F\left(p_k,p_{k+1}\right) = 0$
如果存在初始值$p_0$,使得$p_{n + 1}= p_0$,那么对于任意的初始值，也总有$p_{n + 1}= p_0$.
注意到相邻两式相减可得到$p_{k + 2} = \frac{ - e - cp_k - 2cp_{k + 1} - dp_{k + 1} - bp_kp_{k + 1} - bp_{k + 1}^2 - ap_kp_{k + 1}^2}{c + bp_{k + 1} + ap_{k + 1}^2}$
我们可以就$n$的一些特殊值加以验证。

creasson

双心多边形
n边形外切于$x^2 + y^2= r^2$而内接于$(x-d)^2 + y^2= R^2$的条件可由如下数列给出：
$$a_k=-a_{k - 2}-\frac{2pqa_{k - 1}}{\left(p + 1\right)\left(q - 1\right)a_{k - 1}^2 - 1}$$
$a_0=\sqrt{\frac{q + 1}{q - 1}} , a_1=-a_0$，其中$p = \frac{R + d}{r},q = \frac{R - d}{r}$
利用对称性可得：如果$n=2m+1$,则闭合条件为 $a_{m+1}=0$, 如果$n=2m+2$,则闭合条件为 $a_{m+1}+a_{m+2}=0$
这里列举一些结论：$n=3$,由$a_2 = 0 \Rightarrow  p + q = pq$
$n=4$，由$a_2 + a_3 = 0 \Rightarrow p^2 + q^2 = p^2q^2$
$n=11$，由$a_6 = 0 \Rightarrow $
$$q^{15}p^{15}-3 q^{14} p^{15}-3 q^{13} p^{15}+17 q^{12} p^{15}-3 q^{11} p^{15}-39 q^{10} p^{15}+25 q^9 p^{15}+45 q^8 p^{15}-45 q^7 p^{15}-25 q^6 p^{15}+39 q^5 p^{15}+3 q^4 p^{15}-17 q^3 p^{15}+3 q^2 p^{15}+3 q p^{15}-p^{15}-3 q^{15} p^{14}-6 q^{14} p^{14}-q^{13} p^{14}+4 q^{12} p^{14}+21 q^{11} p^{14}+38 q^{10} p^{14}-17 q^9 p^{14}-72 q^8 p^{14}-17 q^7 p^{14}+38 q^6 p^{14}+21 q^5 p^{14}+4 q^4 p^{14}-q^3 p^{14}-6 q^2 p^{14}-3 q p^{14}-3 q^{15} p^{13}-q^{14} p^{13}+14 q^{13} p^{13}+2 q^{12} p^{13}-25 q^{11} p^{13}+5 q^{10} p^{13}+20 q^9 p^{13}-20 q^8 p^{13}-5 q^7 p^{13}+25 q^6 p^{13}-2 q^5 p^{13}-14 q^4 p^{13}+q^3 p^{13}+3 q^2 p^{13}+17 q^{15} p^{12}+4 q^{14} p^{12}+2 q^{13} p^{12}-76 q^{12} p^{12}-33 q^{11} p^{12}+72 q^{10} p^{12}+28 q^9 p^{12}+72 q^8 p^{12}-33 q^7 p^{12}-76 q^6 p^{12}+2 q^5 p^{12}+4 q^4 p^{12}+17 q^3 p^{12}-3 q^{15} p^{11}+21 q^{14} p^{11}-25 q^{13} p^{11}-33 q^{12} p^{11}+50 q^{11} p^{11}+34 q^{10} p^{11}-34 q^9 p^{11}-50 q^8 p^{11}+33 q^7 p^{11}+25 q^6 p^{11}-21 q^5 p^{11}+3 q^4 p^{11}-39 q^{15} p^{10}+38 q^{14} p^{10}+5 q^{13} p^{10}+72 q^{12} p^{10}+34 q^{11} p^{10}-220 q^{10} p^{10}+34 q^9 p^{10}+72 q^8 p^{10}+5 q^7 p^{10}+38 q^6 p^{10}-39 q^5 p^{10}+25 q^{15} p^9-17 q^{14} p^9+20 q^{13} p^9+28 q^{12} p^9-34 q^{11} p^9+34 q^{10} p^9-28 q^9 p^9-20 q^8 p^9+17 q^7 p^9-25 q^6 p^9+45 q^{15} p^8-72 q^{14} p^8-20 q^{13} p^8+72 q^{12} p^8-50 q^{11} p^8+72 q^{10} p^8-20 q^9 p^8-72 q^8 p^8+45 q^7 p^8-45 q^{15} p^7-17 q^{14} p^7-5 q^{13} p^7-33 q^{12} p^7+33 q^{11} p^7+5 q^{10} p^7+17 q^9 p^7+45 q^8 p^7-25 q^{15} p^6+38 q^{14} p^6+25 q^{13} p^6-76 q^{12} p^6+25 q^{11} p^6+38 q^{10} p^6-25 q^9 p^6+39 q^{15} p^5+21 q^{14} p^5-2 q^{13} p^5+2 q^{12} p^5-21 q^{11} p^5-39 q^{10} p^5+3 q^{15} p^4+4 q^{14} p^4-14 q^{13} p^4+4 q^{12} p^4+3 q^{11} p^4-17 q^{15} p^3-q^{14} p^3+q^{13} p^3+17 q^{12} p^3+3 q^{15} p^2-6 q^{14} p^2+3 q^{13} p^2+3 q^{15} p-3 q^{14} p-q^{15}=0$$

数学星空

根据楼上5#的结论可以得到：
\(n=3,pq - p - q=0\)

\(n=4,p^2q^2 - p^2 - q^2=0\)

\(n=5,p^3q^3 + p^3q^2 + p^2q^3 - p^3q - 2p^2q^2 - pq^3 - p^3 + p^2q + pq^2 - q^3=0\)

\(n=6,3p^4q^4 - 2p^4q^2 - 2p^2q^4 - p^4 + 2p^2q^2 - q^4=0\)

\(n=7,p^6q^6 - 2p^6q^5 - 2p^5q^6 - p^6q^4 - 2p^5q^5 - p^4q^6 + 4p^6q^3 + 4p^3q^6 - p^6q^2 + 2p^4q^4 - p^2q^6 - 2p^6q + 2p^5q^2 + 2p^2q^5 - 2pq^6 + p^6 + 2p^5q - p^4q^2 - 4p^3q^3 - p^2q^4 + 2pq^5 + q^6=0\)

\(n=8,p^8q^8 - 4p^8q^6 - 4p^6q^8 + 6p^8q^4 - 4p^6q^6 + 6p^4q^8 - 4p^8q^2 + 4p^6q^4 + 4p^4q^6 - 4p^2q^8 + p^8 + 4p^6q^2 - 10p^4q^4 + 4p^2q^6 + q^8=0\)

\(n=9,p^9q^9 + 3p^9q^8 + 3p^8q^9 - 8p^9q^6 + 4p^8q^7 + 4p^7q^8 - 8p^6q^9 - 6p^9q^5 + 12p^7q^7 - 6p^5q^9 + 6p^9q^4 - 14p^8q^5 + 8p^7q^6 + 8p^6q^7 - 14p^5q^8 + 6p^4q^9 + 8p^9q^3 - 8p^7q^5 - 8p^5q^7 + 8p^3q^9 + 4p^8q^3 - 12p^7q^4 + 8p^6q^5 + 8p^5q^6 - 12p^4q^7 + 4p^3q^8 - 3p^9q - 4p^7q^3 + 14p^5q^5 - 4p^3q^7 - 3pq^9 - p^9 + 3p^8q - 8p^6q^3 + 6p^5q^4 + 6p^4q^5 - 8p^3q^6 + 3pq^8 - q^9=0\)

\(n=10,5p^{12}q^{12} - 10p^{12}q^{10} - 10p^{10}q^{12} - 9p^{12}q^8 + 34p^{10}q^{10} - 9p^8q^{12} + 36p^{12}q^6 - 36p^{10}q^8 - 36p^8q^{10} + 36p^6q^{12} - 29p^{12}q^4 + 4p^{10}q^6 + 50p^8q^8 + 4p^6q^{10} - 29p^4q^{12} + 6p^{12}q^2 + 14p^{10}q^4 - 20p^8q^6 - 20p^6q^8 + 14p^4q^{10} + 6p^2q^{12} + p^{12} - 6p^{10}q^2 + 15p^8q^4 - 20p^6q^6 + 15p^4q^8 - 6p^2q^{10} + q^{12}=0\)

\(n=11,p^{15}q^{15} - 3p^{15}q^{14} - 3p^{14}q^{15} - 3p^{15}q^{13} - 6p^{14}q^{14} - 3p^{13}q^{15} + 17p^{15}q^{12} - p^{14}q^{13} - p^{13}q^{14} + 17p^{12}q^{15} - 3p^{15}q^{11} + 4p^{14}q^{12} + 14p^{13}q^{13} + 4p^{12}q^{14} - 3p^{11}q^{15} - 39p^{15}q^{10} + 21p^{14}q^{11} + 2p^{13}q^{12} + 2p^{12}q^{13} + 21p^{11}q^{14} - 39p^{10}q^{15} + 25p^{15}q^9 + 38p^{14}q^{10} - 25p^{13}q^{11} - 76p^{12}q^{12} - 25p^{11}q^{13} + 38p^{10}q^{14} + 25p^9q^{15} + 45p^{15}q^8 - 17p^{14}q^9 + 5p^{13}q^{10} - 33p^{12}q^{11} - 33p^{11}q^{12} + 5p^{10}q^{13} - 17p^9q^{14} + 45p^8q^{15} - 45p^{15}q^7 - 72p^{14}q^8 + 20p^{13}q^9 + 72p^{12}q^{10} + 50p^{11}q^{11} + 72p^{10}q^{12} + 20p^9q^{13} - 72p^8q^{14} - 45p^7q^{15} - 25p^{15}q^6 - 17p^{14}q^7 - 20p^{13}q^8 + 28p^{12}q^9 + 34p^{11}q^{10} + 34p^{10}q^{11} + 28p^9q^{12} - 20p^8q^{13} - 17p^7q^{14} - 25p^6q^{15} + 39p^{15}q^5 + 38p^{14}q^6 - 5p^{13}q^7 + 72p^{12}q^8 - 34p^{11}q^9 - 220p^{10}q^{10} - 34p^9q^{11} + 72p^8q^{12} - 5p^7q^{13} + 38p^6q^{14} + 39p^5q^{15} + 3p^{15}q^4 + 21p^{14}q^5 + 25p^{13}q^6 - 33p^{12}q^7 - 50p^{11}q^8 + 34p^{10}q^9 + 34p^9q^{10} - 50p^8q^{11} - 33p^7q^{12} + 25p^6q^{13} + 21p^5q^{14} + 3p^4q^{15} - 17p^{15}q^3 + 4p^{14}q^4 - 2p^{13}q^5 - 76p^{12}q^6 + 33p^{11}q^7 + 72p^{10}q^8 - 28p^9q^9 + 72p^8q^{10} + 33p^7q^{11} - 76p^6q^{12} - 2p^5q^{13} + 4p^4q^{14} - 17p^3q^{15} + 3p^{15}q^2 - p^{14}q^3 - 14p^{13}q^4 + 2p^{12}q^5 + 25p^{11}q^6 + 5p^{10}q^7 - 20p^9q^8 - 20p^8q^9 + 5p^7q^{10} + 25p^6q^{11} + 2p^5q^{12} - 14p^4q^{13} - p^3q^{14} + 3p^2q^{15} + 3p^{15}q - 6p^{14}q^2 + p^{13}q^3 + 4p^{12}q^4 - 21p^{11}q^5 + 38p^{10}q^6 + 17p^9q^7 - 72p^8q^8 + 17p^7q^9 + 38p^6q^{10} - 21p^5q^{11} + 4p^4q^{12} + p^3q^{13} - 6p^2q^{14} + 3pq^{15} - p^{15} - 3p^{14}q + 3p^{13}q^2 + 17p^{12}q^3 + 3p^{11}q^4 - 39p^{10}q^5 - 25p^9q^6 + 45p^8q^7 + 45p^7q^8 - 25p^6q^9 - 39p^5q^{10} + 3p^4q^{11} + 17p^3q^{12} + 3p^2q^{13} - 3pq^{14} - q^{15}=0\)

进一步代入得到：

\(n=3,R^2 - 2Rr - d^2=0\)

\(n=4,R^4 - 2R^2d^2 - 2R^2r^2 + d^4 - 2d^2r^2=0\)

\(n=5,R^6 + 2R^5r - 3R^4d^2 - 4R^4r^2 - 4R^3d^2r + 3R^2d^4 + 4R^2d^2r^2 + 2Rd^4r - 8Rd^2r^3 - d^6=0\)

\(n=6,3R^8 - 12R^6d^2 - 4R^6r^2 + 18R^4d^4 + 4R^4d^2r^2 - 12R^2d^6 + 4R^2d^4r^2 - 16R^2d^2r^4 + 3d^8 - 4d^6r^2=0\)

\(n=7,R^{12} - 4R^{11}r - 6R^{10}d^2 - 4R^{10}r^2 + 20R^9d^2r + 8R^9r^3 + 15R^8d^4 + 16R^8d^2r^2 - 40R^7d^4r - 20R^6d^6 - 24R^6d^4r^2 - 16R^6d^2r^4 + 40R^5d^6r - 48R^5d^4r^3 - 32R^5d^2r^5 + 15R^4d^8 + 16R^4d^6r^2 + 32R^4d^4r^4 + 64R^4d^2r^6 - 20R^3d^8r + 64R^3d^6r^3 - 6R^2d^{10} - 4R^2d^8r^2 - 16R^2d^6r^4 + 4Rd^{10}r - 24Rd^8r^3 + 32Rd^6r^5 + d^{12}=0\)

\(n=8,R^{16} - 8R^{14}d^2 - 8R^{14}r^2 + 28R^{12}d^4 + 40R^{12}d^2r^2 + 8R^{12}r^4 - 56R^{10}d^6 - 72R^{10}d^4r^2 + 48R^{10}d^2r^4 + 70R^8d^8 + 40R^8d^6r^2 - 264R^8d^4r^4 - 128R^8d^2r^6 - 56R^6d^{10} + 40R^6d^8r^2 + 416R^6d^6r^4 + 128R^6d^4r^6 + 128R^6d^2r^8 + 28R^4d^{12} - 72R^4d^{10}r^2 - 264R^4d^8r^4 + 128R^4d^6r^6 - 8R^2d^{14} + 40R^2d^{12}r^2 + 48R^2d^{10}r^4 - 128R^2d^8r^6 + 128R^2d^6r^8 + d^{16} - 8d^{14}r^2 + 8d^{12}r^4=0\)

\(n=9,R^{18} + 6R^{17}r - 9R^{16}d^2 - 48R^{15}d^2r - 8R^{15}r^3 + 36R^{14}d^4 + 168R^{13}d^4r - 8R^{13}d^2r^3 - 84R^{12}d^6 - 96R^{12}d^2r^4 - 336R^{11}d^6r + 216R^{11}d^4r^3 + 32R^{11}d^2r^5 + 126R^{10}d^8 + 480R^{10}d^4r^4 + 256R^{10}d^2r^6 + 420R^9d^8r - 680R^9d^6r^3 + 32R^9d^4r^5 - 126R^8d^{10} - 960R^8d^6r^4 - 512R^8d^4r^6 - 256R^8d^2r^8 - 336R^7d^{10}r + 1000R^7d^8r^3 - 448R^7d^6r^5 + 128R^7d^4r^7 + 84R^6d^{12} + 960R^6d^8r^4 + 168R^5d^{12}r - 792R^5d^{10}r^3 + 832R^5d^8r^5 - 384R^5d^6r^7 - 36R^4d^{14} - 480R^4d^{10}r^4 + 512R^4d^8r^6 - 48R^3d^{14}r + 328R^3d^{12}r^3 - 608R^3d^{10}r^5 + 384R^3d^8r^7 - 512R^3d^6r^9 + 9R^2d^{16} + 96R^2d^{12}r^4 - 256R^2d^{10}r^6 + 256R^2d^8r^8 + 6Rd^{16}r - 56Rd^{14}r^3 + 160Rd^{12}r^5 - 128Rd^{10}r^7 - d^{18}=0\)

\(n=10,5R^{24} - 60R^{22}d^2 - 20R^{22}r^2 + 330R^{20}d^4 + 180R^{20}d^2r^2 + 16R^{20}r^4 - 1100R^{18}d^6 - 700R^{18}d^4r^2 - 304R^{18}d^2r^4 + 2475R^{16}d^8 + 1500R^{16}d^6r^2 + 1872R^{16}d^4r^4 + 1152R^{16}d^2r^6 - 3960R^{14}d^{10} - 1800R^{14}d^8r^2 - 5952R^{14}d^6r^4 - 5760R^{14}d^4r^6 - 1792R^{14}d^2r^8 + 4620R^{12}d^{12} + 840R^{12}d^{10}r^2 + 11424R^{12}d^8r^4 + 10368R^{12}d^6r^6 + 3328R^{12}d^4r^8 + 1024R^{12}d^2r^{10} - 3960R^{10}d^{14} + 840R^{10}d^{12}r^2 - 14112R^{10}d^{10}r^4 - 5760R^{10}d^8r^6 + 2816R^{10}d^6r^8 + 2475R^8d^{16} - 1800R^8d^{14}r^2 + 11424R^8d^{12}r^4 - 5760R^8d^{10}r^6 - 8704R^8d^8r^8 - 1024R^8d^6r^{10} - 1100R^6d^{18} + 1500R^6d^{16}r^2 - 5952R^6d^{14}r^4 + 10368R^6d^{12}r^6 + 2816R^6d^{10}r^8 - 1024R^6d^8r^{10} + 4096R^6d^6r^{12} + 330R^4d^{20} - 700R^4d^{18}r^2 + 1872R^4d^{16}r^4 - 5760R^4d^{14}r^6 + 3328R^4d^{12}r^8 - 60R^2d^{22} + 180R^2d^{20}r^2 - 304R^2d^{18}r^4 + 1152R^2d^{16}r^6 - 1792R^2d^{14}r^8 + 1024R^2d^{12}r^{10} + 5d^{24} - 20d^{22}r^2 + 16d^{20}r^4=0\)

\(n=11,R^{30} - 6R^{29}r - 15R^{28}d^2 - 12R^{28}r^2 + 84R^{27}d^2r + 32R^{27}r^3 + 105R^{26}d^4 + 156R^{26}d^2r^2 + 16R^{26}r^4 - 546R^{25}d^4r - 280R^{25}d^2r^3 - 32R^{25}r^5 - 455R^{24}d^6 - 936R^{24}d^4r^2 - 240R^{24}d^2r^4 + 2184R^{23}d^6r + 864R^{23}d^4r^3 - 384R^{23}d^2r^5 + 1365R^{22}d^8 + 3432R^{22}d^6r^2 + 1584R^{22}d^4r^4 + 1408R^{22}d^2r^6 - 6006R^{21}d^8r - 176R^{21}d^6r^3 + 5088R^{21}d^4r^5 + 1792R^{21}d^2r^7 - 3003R^{20}d^{10} - 8580R^{20}d^8r^2 - 6160R^{20}d^6r^4 - 12480R^{20}d^4r^6 - 4864R^{20}d^2r^8 + 12012R^{19}d^{10}r - 7040R^{19}d^8r^3 - 22720R^{19}d^6r^5 - 11264R^{19}d^4r^7 - 3072R^{19}d^2r^9 + 5005R^{18}d^{12} + 15444R^{18}d^{10}r^2 + 15840R^{18}d^8r^4 + 48960R^{18}d^6r^6 + 27648R^{18}d^4r^8 + 7168R^{18}d^2r^{10} - 18018R^{17}d^{12}r + 26136R^{17}d^{10}r^3 + 54720R^{17}d^8r^5 + 26496R^{17}d^6r^7 + 12288R^{17}d^4r^9 + 2048R^{17}d^2r^{11} - 6435R^{16}d^{14} - 20592R^{16}d^{12}r^2 - 28512R^{16}d^{10}r^4 - 111360R^{16}d^8r^6 - 57600R^{16}d^6r^8 - 20480R^{16}d^4r^{10} - 4096R^{16}d^2r^{12} + 20592R^{15}d^{14}r - 52800R^{15}d^{12}r^3 - 78336R^{15}d^{10}r^5 - 21504R^{15}d^8r^7 - 12288R^{15}d^6r^9 + 6435R^{14}d^{16} + 20592R^{14}d^{14}r^2 + 36960R^{14}d^{12}r^4 + 161280R^{14}d^{10}r^6 + 37632R^{14}d^8r^8 + 12288R^{14}d^6r^{10} - 18018R^{13}d^{16}r + 70752R^{13}d^{14}r^3 + 63168R^{13}d^{12}r^5 - 21504R^{13}d^{10}r^7 - 12800R^{13}d^8r^9 - 24576R^{13}d^6r^{11} - 5005R^{12}d^{18} - 15444R^{12}d^{16}r^2 - 34848R^{12}d^{14}r^4 - 153216R^{12}d^{12}r^6 + 48384R^{12}d^{10}r^8 - 5120R^{12}d^8r^{10} - 4096R^{12}d^6r^{12} + 12012R^{11}d^{18}r - 66528R^{11}d^{16}r^3 - 14976R^{11}d^{14}r^5 + 64512R^{11}d^{12}r^7 + 33792R^{11}d^{10}r^9 + 57344R^{11}d^8r^{11} + 16384R^{11}d^6r^{13} + 3003R^{10}d^{20} + 8580R^{10}d^{18}r^2 + 23760R^{10}d^{16}r^4 + 94080R^{10}d^{14}r^6 - 112896R^{10}d^{12}r^8 + 56320R^{10}d^{10}r^{10} + 28672R^{10}d^8r^{12} + 16384R^{10}d^6r^{14} - 6006R^9d^{20}r + 44440R^9d^{18}r^3 - 24480R^9d^{16}r^5 - 59136R^9d^{14}r^7 - 19968R^9d^{12}r^9 - 55296R^9d^{10}r^{11} - 24576R^9d^8r^{13} - 32768R^9d^6r^{15} - 1365R^8d^{22} - 3432R^8d^{20}r^2 - 11440R^8d^{18}r^4 - 34560R^8d^{16}r^6 + 91392R^8d^{14}r^8 - 104448R^8d^{12}r^{10} + 2184R^7d^{22}r - 20768R^7d^{20}r^3 + 30080R^7d^{18}r^5 + 21504R^7d^{16}r^7 - 2048R^7d^{14}r^9 + 24576R^7d^{12}r^{11} + 455R^6d^{24} + 936R^6d^{22}r^2 + 3696R^6d^{20}r^4 + 5760R^6d^{18}r^6 - 34560R^6d^{16}r^8 + 71680R^6d^{14}r^{10} - 45056R^6d^{12}r^{12} - 546R^5d^{24}r + 6480R^5d^{22}r^3 - 16032R^5d^{20}r^5 + 2304R^5d^{18}r^7 + 4608R^5d^{16}r^9 - 4096R^5d^{14}r^{11} + 8192R^5d^{12}r^{13} - 105R^4d^{26} - 156R^4d^{24}r^2 - 720R^4d^{22}r^4 + 320R^4d^{20}r^6 + 4608R^4d^{18}r^8 - 17408R^4d^{16}r^{10} + 24576R^4d^{14}r^{12} - 16384R^4d^{12}r^{14} + 84R^3d^{26}r - 1216R^3d^{24}r^3 + 4416R^3d^{22}r^5 - 4096R^3d^{20}r^7 + 15R^2d^{28} + 12R^2d^{26}r^2 + 64R^2d^{24}r^4 - 192R^2d^{22}r^6 + 256R^2d^{20}r^8 - 6Rd^{28}r + 104Rd^{26}r^3 - 512Rd^{24}r^5 + 896Rd^{22}r^7 - 512Rd^{20}r^9 - d^{30}=0\)

creasson

$n$边形外切于椭圆$\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$而内接于椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$的条件可由如下数列给出：
$$u_k=-u_{k-2}-\frac{2\left(a^2 q^2-a^2 b^2-b^2 p^2\right)u_{k-1}}{a^2 q^2-a^2 b^2 u_{k-1}^2+b^2 p^2  u_{k-1}^2}$$
$u_0=\frac{\sqrt{a+p}}{\sqrt{a-p}}$, $u_1=-u_0$
如果n=2m+1,则闭合条件为 $u_{m+1}=0$
如果n=2m+2,则闭合条件为 $u_{m+1}+u_{m+2}=0$
例举一些结果：
$n=3$，$a b - b p - a q=0$
$n=4$，$a^2 b^2 - b^2 p^2 - a^2 q^2=0$
$n=5$,  $a^3 b^3 + a^2 b^3 p - a b^3 p^2 - b^3 p^3 + a^3 b^2 q - 2 a^2 b^2 p q + a b^2 p^2 q - a^3 b q^2 + a^2 b p q^2 - a^3 q^3=0$

creasson

闭合定理的一般情形：
给定函数$F\left(u+v,uv\right)=0$, 如果存在函数$g$，使得$\frac{d s}{d u}=g(u)$，$\frac{d s}{d v}=wg(v)$, 其中$w$是某个$m$次单位根，则
对于这样定义的数列${a_k}$：$F\left(a_{k}+a_{k+1},a_k a_{k+1}\right)=0$，如果对于某个初值，$a_{k}$有周期$T（T\equiv 0(mod m)）$，则对其他任意初值，也总有周期$T$。
$F$是二次时对应彭赛列闭合定理。

数学星空

@mathe
或许你能给出楼上新方法与椭圆曲线论中除子的有趣联系～
https://bbs.emath.ac.cn/thread-5490-2-1.html