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[原创] 双心三角形顶点和对边切点连线的交点的轨迹

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发表于 2019-3-3 21:49:05 | 显示全部楼层 |阅读模式

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bicentral-circles.png
如图双心三角形ABC三个顶点和对应边上切点D,E,F的连线AD、BE、CF三线共点,交于G点。
那么保持两个圆不动,在外圆上转动A点,由双心三角形的性质,无论A在何处,还是有新的三角形ABC同时内接外圆和外切内圆。
相对应的,G点也会相应的移动。请问,当A点移动时,G点的轨迹是什么?是一个圆吗?
由于G点轨迹直径极小,我一度以为G点是不动点,但是后来发现三角形ABC形状比较扁时,G点的变化范围会相对比较明显,比较容易观察出来
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 01:59:54 | 显示全部楼层
从图形上看是的,而且点A→B→C→A时,点G刚好各绕轨迹圆一周,总共饶了三周。

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参与人数 1威望 +3 金币 +3 贡献 +3 经验 +3 鲜花 +3 收起 理由
wayne + 3 + 3 + 3 + 3 + 3 观察细微,确实是绕了三圈

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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 02:06:39 | 显示全部楼层
易知如果轨迹是圆,那么圆心在外接圆和内切圆的连线上,只要取这条连线与外接圆的两个交点分别作为点A,得出的两个点G就可以得出轨迹圆的直径。
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 13:14:00 | 显示全部楼层
设置:外接圆以原点为圆心,\(r_o\)为半径;内切圆以\(\{d,0\} \)为圆心,\(r_i\)为半径,可通过\(d^2=r_o \left(r_o-2 r_i\right)\)消去\(r_i\)。
目的消元到以角A位置\(\theta _1\)作参数
  1. {Array[Subscript[r, o]{Cos[Subscript[\[Theta], #]],Sin[Subscript[\[Theta], #]]}&,3](*外接圆以原点为圆心,Subscript[r, o]为半径*),Array[{d,0}+Subscript[r, i]{Cos[Subscript[\[Theta], #]],Sin[Subscript[\[Theta], #]]}&,3,4](*内切圆以{d,0}为圆心,Subscript[r, i]为半径*),{x,y}}//#/.{Cos[Subscript[\[Theta], i_]]:>Subscript[Cos\[Theta], i],Sin[Subscript[\[Theta], i_]]:>Subscript[Sin\[Theta], i]}&//Function[{ABCPoints,CDFPoints,g},{Subsets[ABCPoints,{2}],CDFPoints}//(*每条边分配一个切点*)Transpose//Parallelize[Function[{l,p},{p-{d,0}//#/Subscript[r, i]&//(*正弦余弦平方和为1*)Total[#^2]==1&,{Append[l,p],{p,Complement[ABCPoints,l]//Flatten,g}}//(Subsets[#,{2}]//(Subtract@@#//Divide@@#&//Factor)&/@#&//Subsets[#,{2}]&//Equal@@#&/@#&(*共线条件*))&/@#&,{l,{p,{d,0}}}//(Subtract@@#//Divide@@#&//Factor)&/@#&//Times@@#&//(*边与相切半径垂直条件*)#==-1&//Solve[#,Cases[Variables[p],Subscript[Cos\[Theta], _]]]&}//#[[1;;-2]]/.#[[-1]]&//Flatten//(Subtract@@#//Factor//Numerator)&/@#&//#/.(#~Function[{PolynomialList,\[Zeta]},Select[PolynomialList,Exponent[#,\[Zeta]]==1&][[1]]//Solve[#==0,\[Zeta]]&]~Cases[Variables[p],Subscript[Sin\[Theta], _]][[1]])&]@@#&/@#]&]@@#&//#/.Solve[d^2==Subscript[r, o](Subscript[r, o]-2Subscript[r, i]),Subscript[r, i]]&//Flatten//Parallelize[(#/.{Subscript[Cos\[Theta], i_]:>Cos[Subscript[\[Theta], i]],Subscript[Sin\[Theta], i_]:>Sin[Subscript[\[Theta], i]]}/.{Subscript[\[Theta], i_]:>2ArcTan[Subscript[HalfTan\[Theta], i]]}//TrigExpand//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,(Select[#,Exponent[#,x]==1\[And]Exponent[#,y]==1&][[1;;2]]//Solve[#=={0,0},{x,y}]&//Factor//Flatten)}&//{{#[[1]]/.#[[2]],Table[{x,y}/.#[[2]]//(#-(#/.\[Rho]))&,{\[Rho],Array[Subscript[HalfTan\[Theta], #]&,3]//Function[{\[Xi]},Permutations[\[Xi]]//({#,\[Xi]}//Transpose//Rule@@#&/@#&)&/@#&]}]}//Flatten//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,Array[(SymmetricPolynomial[#,Array[Subscript[HalfTan\[Theta], #]&,3]]-Subscript[s, #])&,3]}&//Fold[Function[{XToSymmetricPolynomial,i},Map[(PolynomialRemainder[#,XToSymmetricPolynomial[[2,1]],Subscript[HalfTan\[Theta], i]]//CoefficientList[#,Subscript[HalfTan\[Theta], i]]&//Factor)&,XToSymmetricPolynomial,{2}]//(#//Flatten//DeleteCases[#,0]&)&/@#&],#,{1,2,3}]&//#[[1]]&//{#,(Select[#,Exponent[#,Subscript[s, 3]]==1&]//First//Solve[#==0,Subscript[s, 3]]&//First)}&//{#[[1]]/.#[[2]]//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//PolynomialGCD@@#&//Solve[#==0,Subscript[s, 2]]&,#[[2]]}&//{#[[1]],#[[2]]/.#[[1]]}&//Flatten//(Subtract@@#//#/.Array[(Subscript[s, #]->SymmetricPolynomial[#,Array[Subscript[HalfTan\[Theta], #]&,3]])&,3]&//Factor//Numerator)&/@#&,#[[2]]}&//{#[[1,2]],#[[2]]}/.Solve[#[[1,1]]==0,Subscript[HalfTan\[Theta], 3]][[1]]&//{#[[1]]//Factor//Numerator,#[[2]]//Parallelize[(#//Factor)&/@#]&}&//Function[{DividePolynomial,Rules},Parallelize[#[[1]]->({Numerator[#[[2]]],Denominator[#[[2]]]}//PolynomialRemainder[#,DividePolynomial,Subscript[HalfTan\[Theta], 2]]&/@#&//Divide@@#&//Factor//#/.{Subscript[HalfTan\[Theta], 1]->Sin[Subscript[\[Theta], 1]]/(1+Cos[Subscript[\[Theta], 1]])}&//Factor//TrigReduce//FullSimplify//Factor//Which[#[[0]]===Plus,Collect[#,{d,Subscript[r, o]},FullSimplify],#[[0]]===Power,Collect[#[[1]],{d,Subscript[r, o]},FullSimplify]^#[[2]],True,#]&/@#&)&/@Rules]]@@#&
复制代码


结果有d的7次和3倍\(\theta _1\)
\(\left\{x\to \frac{d \left(d^7 \cos \left(\theta _1\right)-11 d^5 \cos \left(\theta _1\right) r_o^2+4 d^4 \left(\cos \left(2 \theta _1\right)+2\right) r_o^3+d^3 \left(26 \cos \left(\theta _1\right)+\cos \left(3 \theta _1\right)\right) r_o^4-4 d^2 \left(5 \cos \left(2 \theta _1\right)+11\right) r_o^5+d \left(48 \cos \left(\theta _1\right)-\cos \left(3 \theta _1\right)\right) r_o^6-12 r_o^7\right)}{r_o \left(d-3 r_o\right) \left(d+3 r_o\right) \left(d^2-2 d \cos \left(\theta _1\right) r_o+r_o^2\right){}^2},y\to -\frac{d^2 \sin \left(\theta _1\right) \left(d-r_o\right) \left(d+r_o\right) \left(d^4-6 d^2 r_o^2+8 d \cos \left(\theta _1\right) r_o^3+\left(-2 \cos \left(2 \theta _1\right)-1\right) r_o^4\right)}{r_o \left(d-3 r_o\right) \left(d+3 r_o\right) \left(d^2-2 d \cos \left(\theta _1\right) r_o+r_o^2\right){}^2}\right\}\)

点评

上传不了图片,我怀疑是没有Flash的缘故。  发表于 2019-3-4 13:59
能做一个放大的图看看吗?  发表于 2019-3-4 13:28
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-3-4 13:18:41 | 显示全部楼层
那看来是不正确了,不知道为什么这个图看起来这么“圆”
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 13:19:31 | 显示全部楼层
本帖最后由 zeroieme 于 2019-3-4 13:20 编辑

完全消去参数\(\theta _1\)后,
  1. %//(Subtract@@#/.{Subscript[\[Theta], i_]:>2ArcTan[Subscript[HalfTan\[Theta], i]]}//TrigExpand//Factor//Numerator)&/@#&//FixedPoint[(#//SortBy[#,Exponent[#,Subscript[HalfTan\[Theta], 1]]&]&//Which[Exponent[#[[1]],Subscript[HalfTan\[Theta], 1]]==0,#,Exponent[#[[1]],Subscript[HalfTan\[Theta], 1]]==1,#[[2;;-1]]/.Solve[#[[1]]==0,Subscript[HalfTan\[Theta], 1]][[1]],True,Append[Table[PolynomialRemainder[\[Alpha],#[[1]],Subscript[HalfTan\[Theta], 1]],{\[Alpha],#[[2;;-1]]}],#[[1]]]]&//Parallelize[(#//Factor//Numerator//\[LightBulb] #&//Select[#,Variables[#]\[Intersection]{Subscript[HalfTan\[Theta], 1],x,y}!={}&]&//Collect[#,Subscript[HalfTan\[Theta], 1],Factor]&)&/@#]&)&,#]&
复制代码


\(-\left(d^6-9 r_o^2 d^4+8 x r_o^2 d^3+16 r_o^4 d^2-x^2 r_o^2 d^2-y^2 r_o^2 d^2-24 x r_o^4 d+9 x^2 r_o^4+9 y^2 r_o^4\right) \left(d^9-3 r_o d^8-10 r_o^2 d^7+30 r_o^3 d^6+8 x r_o^2 d^6+57 r_o^4 d^5-8 x r_o^3 d^5-x^2 r_o^2 d^5-y^2 r_o^2 d^5-171 r_o^5 d^4-80 x r_o^4 d^4+3 x^2 r_o^3 d^4+3 y^2 r_o^3 d^4-80 r_o^6 d^3+144 x r_o^5 d^3+26 x^2 r_o^4 d^3+18 y^2 r_o^4 d^3+304 r_o^7 d^2+136 x r_o^6 d^2-14 x^2 r_o^5 d^2-54 y^2 r_o^5 d^2-456 x r_o^7 d-57 x^2 r_o^6 d-81 y^2 r_o^6 d+171 x^2 r_o^7+243 y^2 r_o^7\right){}^2 \left(x^2 d^9+y^2 d^9+60 r_o^2 d^9+16 x r_o d^9-180 r_o^3 d^8-48 x r_o^2 d^8-3 x^2 r_o d^8-3 y^2 r_o d^8-360 r_o^4 d^7-48 x r_o^3 d^7-12 y^2 r_o^2 d^7+1144 r_o^5 d^6+632 x r_o^4 d^6+64 x^2 r_o^3 d^6+36 y^2 r_o^3 d^6+92 r_o^6 d^5-1032 x r_o^5 d^5-122 x^2 r_o^4 d^5-18 y^2 r_o^4 d^5-660 r_o^7 d^4-1248 x r_o^6 d^4-18 x^2 r_o^5 d^4+54 y^2 r_o^5 d^4+1488 r_o^8 d^3+2016 x r_o^7 d^3+816 x^2 r_o^6 d^3+324 y^2 r_o^6 d^3-3120 r_o^9 d^2-1896 x r_o^8 d^2-1104 x^2 r_o^7 d^2-972 y^2 r_o^7 d^2+4680 x r_o^9 d+585 x^2 r_o^8 d+729 y^2 r_o^8 d-1755 x^2 r_o^9-2187 y^2 r_o^9\right){}^2\)

有“三组”二次因子,哪个才是真方程?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 13:41:08 | 显示全部楼层
  1. %//List@@#&//Parallelize[Which[#[[0]]===Plus,(MonomialList[#,{x,y}]//Simplify//Total),#[[0]]===Power,(MonomialList[#[[1]],{x,y}]//Simplify//Total)^#[[2]],True,#]&/@#]&//Times@@#&
复制代码


\(\left(d^6-9 d^4 r_o^2+x^2 \left(9 r_o^4-d^2 r_o^2\right)+8 d x r_o^2 \left(d^2-3 r_o^2\right)+y^2 \left(9 r_o^4-d^2 r_o^2\right)+16 d^2 r_o^4\right) \left(x^2 r_o^2 \left(d+3 r_o\right){}^2 \left(-d^3+9 d^2 r_o-19 d r_o^2+19 r_o^3\right)+8 d x r_o^2 \left(d^5-d^4 r_o-10 d^3 r_o^2+18 d^2 r_o^3+17 d r_o^4-57 r_o^5\right)+d^2 \left(d^7-3 d^6 r_o-10 d^5 r_o^2+30 d^4 r_o^3+57 d^3 r_o^4-171 d^2 r_o^5-80 d r_o^6+304 r_o^7\right)-y^2 r_o^2 \left(d+3 r_o\right){}^2 \left(d-3 r_o\right){}^3\right){}^2 \left(y^2 \left(d^3+3 d^2 r_o+3 d r_o^2+9 r_o^3\right){}^2 \left(d-3 r_o\right){}^3+x^2 \left(d+3 r_o\right){}^2 \left(d^7-9 d^6 r_o+45 d^5 r_o^2-125 d^4 r_o^3+223 d^3 r_o^4-231 d^2 r_o^5+195 d r_o^6-195 r_o^7\right)+4 d^2 r_o^2 \left(15 d^7-45 d^6 r_o-90 d^5 r_o^2+286 d^4 r_o^3+23 d^3 r_o^4-165 d^2 r_o^5+372 d r_o^6-780 r_o^7\right)+8 d x r_o \left(2 d^8-6 d^7 r_o-6 d^6 r_o^2+79 d^5 r_o^3-129 d^4 r_o^4-156 d^3 r_o^5+252 d^2 r_o^6-237 d r_o^7+585 r_o^8\right)\right){}^2\)

这里面真的有个圆

点评

三个因子,第一个\(\left(d^6-9 d^4 r_o^2+x^2 \left(9 r_o^4-d^2 r_o^2\right)+8 d x r_o^2 \left(d^2-3 r_o^2\right)+y^2 \left(9 r_o^4-d^2 r_o^2\right)+16 d^2 r_o^4\right)\),\(x^2\)与\(y^2\)系数相等。  发表于 2019-3-4 16:02
形式太复杂了,怎么确定有圆的?  发表于 2019-3-4 15:45
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 17:12:37 | 显示全部楼层
猜测对于偶数边的双心多边形,点G是固定的,对于奇数边,轨迹是一个圆。

补充内容 (2019-3-5 12:11):
五边形顶点和对应切点的连线已经不交于一点了,也不存在轨迹问题了
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-3-4 17:31:50 来自手机 | 显示全部楼层
偶数边就不能顶点和切点相连接了。如果是顶点和相对的顶点连接,或切点和切点连接,的确交于一点
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-3-4 17:36:15 | 显示全部楼层
本帖最后由 zeroieme 于 2019-3-4 17:42 编辑

合并4#,6#,7#的代码,同时把外接圆半径定为1。
  1. {Array[Subscript[r,o]{Cos[Subscript[\[Theta],#]],Sin[Subscript[\[Theta],#]]}&,3](*外接圆以原点为圆心,Subscript[r,o]为半径*),Array[{d,0}+Subscript[r,i]{Cos[Subscript[\[Theta],#]],Sin[Subscript[\[Theta],#]]}&,3,4](*内切圆以{d,0}为圆心,Subscript[r,i]为半径*),{x,y}}//#/.{Cos[Subscript[\[Theta],i_]]:>Subscript[Cos\[Theta],i],Sin[Subscript[\[Theta],i_]]:>Subscript[Sin\[Theta],i]}&//Function[{ABCPoints,CDFPoints,g},{Subsets[ABCPoints,{2}],CDFPoints}//(*每条边分配一个切点*)Transpose//Parallelize[Function[{l,p},{p-{d,0}//#/Subscript[r,i]&//(*正弦余弦平方和为1*)Total[#^2]==1&,{Append[l,p],{p,Complement[ABCPoints,l]//Flatten,g}}//(Subsets[#,{2}]//(Subtract@@#//Divide@@#&//Factor)&/@#&//Subsets[#,{2}]&//Equal@@#&/@#&(*共线条件*))&/@#&,{l,{p,{d,0}}}//(Subtract@@#//Divide@@#&//Factor)&/@#&//Times@@#&//(*边与相切半径垂直条件*)#==-1&//Solve[#,Cases[Variables[p],Subscript[Cos\[Theta],_]]]&}//#[[1;;-2]]/.#[[-1]]&//Flatten//(Subtract@@#//Factor//Numerator)&/@#&//#/.(#~Function[{PolynomialList,\[Zeta]},Select[PolynomialList,Exponent[#,\[Zeta]]==1&][[1]]//Solve[#==0,\[Zeta]]&]~Cases[Variables[p],Subscript[Sin\[Theta],_]][[1]])&]@@#&/@#]&]@@#&//#/.Solve[d^2==Subscript[r,o](Subscript[r,o]-2Subscript[r,i]),Subscript[r,i]]&//Flatten//Parallelize[(#/.{Subscript[Cos\[Theta],i_]:>Cos[Subscript[\[Theta],i]],Subscript[Sin\[Theta],i_]:>Sin[Subscript[\[Theta],i]]}/.{Subscript[\[Theta],i_]:>2ArcTan[Subscript[HalfTan\[Theta],i]]}//TrigExpand//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,(Select[#,Exponent[#,x]==1\[And]Exponent[#,y]==1&][[1;;2]]//Solve[#=={0,0},{x,y}]&//Factor//Flatten)}&//{{#[[1]]/.#[[2]],Table[{x,y}/.#[[2]]//(#-(#/.\[Rho]))&,{\[Rho],Array[Subscript[HalfTan\[Theta],#]&,3]//Function[{\[Xi]},Permutations[\[Xi]]//({#,\[Xi]}//Transpose//Rule@@#&/@#&)&/@#&]}]}//Flatten//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,Array[(SymmetricPolynomial[#,Array[Subscript[HalfTan\[Theta],#]&,3]]-Subscript[s,#])&,3]}&//Fold[Function[{XToSymmetricPolynomial,i},Map[(PolynomialRemainder[#,XToSymmetricPolynomial[[2,1]],Subscript[HalfTan\[Theta],i]]//CoefficientList[#,Subscript[HalfTan\[Theta],i]]&//Factor)&,XToSymmetricPolynomial,{2}]//(#//Flatten//DeleteCases[#,0]&)&/@#&],#,{1,2,3}]&//#[[1]]&//{#,(Select[#,Exponent[#,Subscript[s,3]]==1&]//First//Solve[#==0,Subscript[s,3]]&//First)}&//{#[[1]]/.#[[2]]//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//PolynomialGCD@@#&//Solve[#==0,Subscript[s,2]]&,#[[2]]}&//{#[[1]],#[[2]]/.#[[1]]}&//Flatten//(Subtract@@#//#/.Array[(Subscript[s,#]->SymmetricPolynomial[#,Array[Subscript[HalfTan\[Theta],#]&,3]])&,3]&//Factor//Numerator)&/@#&,#[[2]]}&//{#[[1,2]],#[[2]]}/.Solve[#[[1,1]]==0,Subscript[HalfTan\[Theta],3]][[1]]&//{#[[1]]//Factor//Numerator,#[[2]]//Parallelize[(#//Factor)&/@#]&}&//Function[{DividePolynomial,Rules},Parallelize[#[[1]]->({Numerator[#[[2]]],Denominator[#[[2]]]}//PolynomialRemainder[#,DividePolynomial,Subscript[HalfTan\[Theta],2]]&/@#&//Divide@@#&//Factor//#/.{Subscript[HalfTan\[Theta],1]->Sin[Subscript[\[Theta],1]]/(1+Cos[Subscript[\[Theta],1]])}&//Factor//TrigReduce//FullSimplify//Factor//Which[#[[0]]===Plus,Collect[#,{d,Subscript[r,o]},FullSimplify],#[[0]]===Power,Collect[#[[1]],{d,Subscript[r,o]},FullSimplify]^#[[2]],True,#]&/@#&)&/@Rules]]@@#&//#/.{Subscript[r,o]->1}&//{#//Parallelize[Simplify/@#]&,#//(Subtract@@#/.{Subscript[\[Theta],i_]:>2ArcTan[Subscript[HalfTan\[Theta],i]]}//TrigExpand//Factor//Numerator)&/@#&//FixedPoint[(#//SortBy[#,Exponent[#,Subscript[HalfTan\[Theta],1]]&]&//Which[Exponent[#[[1]],Subscript[HalfTan\[Theta],1]]==0,#,Exponent[#[[1]],Subscript[HalfTan\[Theta],1]]==1,#[[2;;-1]]/.Solve[#[[1]]==0,Subscript[HalfTan\[Theta],1]][[1]],True,Append[Table[PolynomialRemainder[\[Alpha],#[[1]],Subscript[HalfTan\[Theta],1]],{\[Alpha],#[[2;;-1]]}],#[[1]]]]&//Parallelize[(#//Factor//Numerator//\[LightBulb] #&//Select[#,Variables[#]\[Intersection]{Subscript[HalfTan\[Theta],1],x,y}!={}&]&//Collect[#,Subscript[HalfTan\[Theta],1],Factor]&)&/@#]&)&,#]&//MapAll[If[#[[0]]===Plus,(MonomialList[#,{x,y}]//Collect[#,{x,y},Simplify]&/@#&//Total),#]&,#]&//First}&
复制代码



结果
\(\left\{\left\{x\to \frac{d \left(8 d^4+d^3 \cos \left(3 \theta _1\right)+4 \left(d^2-5\right) d^2 \cos \left(2 \theta _1\right)-44 d^2+\left(d^6-11 d^4+26 d^2+48\right) d \cos \left(\theta _1\right)-d \cos \left(3 \theta _1\right)-12\right)}{\left(d^2-9\right) \left(d^2-2 d \cos \left(\theta _1\right)+1\right){}^2},y\to -\frac{(d-1) d^2 (d+1) \sin \left(\theta _1\right) \left(d^4-6 d^2+8 d \cos \left(\theta _1\right)-2 \cos \left(2 \theta _1\right)-1\right)}{\left(d^2-9\right) \left(d^2-2 d \cos \left(\theta _1\right)+1\right){}^2}\right\},\left(-(d+3)^2 \left(d^3-9 d^2+19 d-19\right) x^2+8 d \left(d^5-d^4-10 d^3+18 d^2+17 d-57\right) x+d^2 \left(d^7-3 d^6-10 d^5+30 d^4+57 d^3-171 d^2-80 d+304\right)-(d+3)^2 (d-3)^3 y^2\right)^2 \color{red}{\left(\left(9-d^2\right) x^2+8 d\left(d^2-3\right)  x+\left(9-d^2\right) y^2+\left(d^4-9 d^2+16\right) d^2\right)} \left(\left(d^3+3 d^2+3 d+9\right)^2 (d-3)^3 y^2+(d+3)^2 \left(d^7-9 d^6+45 d^5-125 d^4+223 d^3-231 d^2+195 d-195\right) x^2+4 d^2 \left(15 d^7-45 d^6-90 d^5+286 d^4+23 d^3-165 d^2+372 d-780\right)+8 d \left(2 d^8-6 d^7-6 d^6+79 d^5-129 d^4-156 d^3+252 d^2-237 d+585\right) x\right)^2\right\}\)

外接圆半径定为1后,圆因子跑到第二位了。其他因子是增根吗?
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