双心三角形顶点和对边切点连线的交点的轨迹
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 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。 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。 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
{Array{Cos, #]],Sin, #]]}&,3](*外接圆以原点为圆心，Subscript为半径*),Array[{d,0}+Subscript{Cos, #]],Sin, #]]}&,3,4](*内切圆以{d,0}为圆心，Subscript为半径*),{x,y}}//#/.{Cos, i_]]:>Subscript, i],Sin, i_]]:>Subscript, i]}&//Function[{ABCPoints,CDFPoints,g},{Subsets,CDFPoints}//(*每条边分配一个切点*)Transpose//Parallelize&//(*正弦余弦平方和为1*)Total[#^2]==1&,{Append,{p,Complement//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,Subscript, _]]]&}//#[]/.#[[-1]]&//Flatten//(Subtract@@#//Factor//Numerator)&/@#&//#/.(#~Function[{PolynomialList,\},Select]==1&][]//Solve[#==0,\]&]~Cases,Subscript, _]][])&]@@#&/@#]&]@@#&//#/.Solve(Subscript-2Subscript),Subscript]&//Flatten//Parallelize[(#/.{Subscript, i_]:>Cos, i]],Subscript, i_]:>Sin, i]]}/.{Subscript[\, i_]:>2ArcTan, i]]}//TrigExpand//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,(Select[#,Exponent[#,x]==1\Exponent[#,y]==1&][]//Solve[#=={0,0},{x,y}]&//Factor//Flatten)}&//{{#[]/.#[],Table[{x,y}/.#[]//(#-(#/.\))&,{\,Array, #]&,3]//Function[{\},Permutations[\]//({#,\}//Transpose//Rule@@#&/@#&)&/@#&]}]}//Flatten//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,Array[(SymmetricPolynomial[#,Array, #]&,3]]-Subscript)&,3]}&//Fold],Subscript, i]]//CoefficientList[#,Subscript, i]]&//Factor)&,XToSymmetricPolynomial,{2}]//(#//Flatten//DeleteCases[#,0]&)&/@#&],#,{1,2,3}]&//#[]&//{#,(Select[#,Exponent[#,Subscript]==1&]//First//Solve[#==0,Subscript]&//First)}&//{#[]/.#[]//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//PolynomialGCD@@#&//Solve[#==0,Subscript]&,#[]}&//{#[],#[]/.#[]}&//Flatten//(Subtract@@#//#/.Array[(Subscript->SymmetricPolynomial[#,Array, #]&,3]])&,3]&//Factor//Numerator)&/@#&,#[]}&//{#[],#[]}/.Solve[#[]==0,Subscript, 3]][]&//{#[]//Factor//Numerator,#[]//Parallelize[(#//Factor)&/@#]&}&//Function[{DividePolynomial,Rules},Parallelize[#[]->({Numerator[#[]],Denominator[#[]]}//PolynomialRemainder[#,DividePolynomial,Subscript, 2]]&/@#&//Divide@@#&//Factor//#/.{Subscript, 1]->Sin, 1]]/(1+Cos, 1]])}&//Factor//TrigReduce//FullSimplify//Factor//Which[#[]===Plus,Collect[#,{d,Subscript},FullSimplify],#[]===Power,Collect[#[],{d,Subscript},FullSimplify]^#[],True,#]&/@#&)&/@Rules]]@@#&
结果有d的7次和3倍\(\theta _1\)
\(\left\{x\to \frac{d \left(d^7 \cos \left(\theta _1\right)-11 d^5 \cos \left(\theta _1\right) r_o^2+4 d^4 \left(\cos \left(2 \theta _1\right)+2\right) r_o^3+d^3 \left(26 \cos \left(\theta _1\right)+\cos \left(3 \theta _1\right)\right) r_o^4-4 d^2 \left(5 \cos \left(2 \theta _1\right)+11\right) r_o^5+d \left(48 \cos \left(\theta _1\right)-\cos \left(3 \theta _1\right)\right) r_o^6-12 r_o^7\right)}{r_o \left(d-3 r_o\right) \left(d+3 r_o\right) \left(d^2-2 d \cos \left(\theta _1\right) r_o+r_o^2\right){}^2},y\to -\frac{d^2 \sin \left(\theta _1\right) \left(d-r_o\right) \left(d+r_o\right) \left(d^4-6 d^2 r_o^2+8 d \cos \left(\theta _1\right) r_o^3+\left(-2 \cos \left(2 \theta _1\right)-1\right) r_o^4\right)}{r_o \left(d-3 r_o\right) \left(d+3 r_o\right) \left(d^2-2 d \cos \left(\theta _1\right) r_o+r_o^2\right){}^2}\right\}\)
那看来是不正确了，不知道为什么这个图看起来这么“圆” 本帖最后由 zeroieme 于 2019-3-4 13:20 编辑
完全消去参数\(\theta _1\)后，%//(Subtract@@#/.{Subscript[\, i_]:>2ArcTan, i]]}//TrigExpand//Factor//Numerator)&/@#&//FixedPoint[(#//SortBy[#,Exponent[#,Subscript, 1]]&]&//Which],Subscript, 1]]==0,#,Exponent[#[],Subscript, 1]]==1,#[]/.Solve[#[]==0,Subscript, 1]][],True,Append,#[],Subscript, 1]],{\,#[]}],#[]]]&//Parallelize[(#//Factor//Numerator//\ #&//Select[#,Variables[#]\{Subscript, 1],x,y}!={}&]&//Collect[#,Subscript, 1],Factor]&)&/@#]&)&,#]&
\(-\left(d^6-9 r_o^2 d^4+8 x r_o^2 d^3+16 r_o^4 d^2-x^2 r_o^2 d^2-y^2 r_o^2 d^2-24 x r_o^4 d+9 x^2 r_o^4+9 y^2 r_o^4\right) \left(d^9-3 r_o d^8-10 r_o^2 d^7+30 r_o^3 d^6+8 x r_o^2 d^6+57 r_o^4 d^5-8 x r_o^3 d^5-x^2 r_o^2 d^5-y^2 r_o^2 d^5-171 r_o^5 d^4-80 x r_o^4 d^4+3 x^2 r_o^3 d^4+3 y^2 r_o^3 d^4-80 r_o^6 d^3+144 x r_o^5 d^3+26 x^2 r_o^4 d^3+18 y^2 r_o^4 d^3+304 r_o^7 d^2+136 x r_o^6 d^2-14 x^2 r_o^5 d^2-54 y^2 r_o^5 d^2-456 x r_o^7 d-57 x^2 r_o^6 d-81 y^2 r_o^6 d+171 x^2 r_o^7+243 y^2 r_o^7\right){}^2 \left(x^2 d^9+y^2 d^9+60 r_o^2 d^9+16 x r_o d^9-180 r_o^3 d^8-48 x r_o^2 d^8-3 x^2 r_o d^8-3 y^2 r_o d^8-360 r_o^4 d^7-48 x r_o^3 d^7-12 y^2 r_o^2 d^7+1144 r_o^5 d^6+632 x r_o^4 d^6+64 x^2 r_o^3 d^6+36 y^2 r_o^3 d^6+92 r_o^6 d^5-1032 x r_o^5 d^5-122 x^2 r_o^4 d^5-18 y^2 r_o^4 d^5-660 r_o^7 d^4-1248 x r_o^6 d^4-18 x^2 r_o^5 d^4+54 y^2 r_o^5 d^4+1488 r_o^8 d^3+2016 x r_o^7 d^3+816 x^2 r_o^6 d^3+324 y^2 r_o^6 d^3-3120 r_o^9 d^2-1896 x r_o^8 d^2-1104 x^2 r_o^7 d^2-972 y^2 r_o^7 d^2+4680 x r_o^9 d+585 x^2 r_o^8 d+729 y^2 r_o^8 d-1755 x^2 r_o^9-2187 y^2 r_o^9\right){}^2\)
有“三组”二次因子，哪个才是真方程？ %//List@@#&//Parallelize]===Plus,(MonomialList[#,{x,y}]//Simplify//Total),#[]===Power,(MonomialList[#[],{x,y}]//Simplify//Total)^#[],True,#]&/@#]&//Times@@#&
\(\left(d^6-9 d^4 r_o^2+x^2 \left(9 r_o^4-d^2 r_o^2\right)+8 d x r_o^2 \left(d^2-3 r_o^2\right)+y^2 \left(9 r_o^4-d^2 r_o^2\right)+16 d^2 r_o^4\right) \left(x^2 r_o^2 \left(d+3 r_o\right){}^2 \left(-d^3+9 d^2 r_o-19 d r_o^2+19 r_o^3\right)+8 d x r_o^2 \left(d^5-d^4 r_o-10 d^3 r_o^2+18 d^2 r_o^3+17 d r_o^4-57 r_o^5\right)+d^2 \left(d^7-3 d^6 r_o-10 d^5 r_o^2+30 d^4 r_o^3+57 d^3 r_o^4-171 d^2 r_o^5-80 d r_o^6+304 r_o^7\right)-y^2 r_o^2 \left(d+3 r_o\right){}^2 \left(d-3 r_o\right){}^3\right){}^2 \left(y^2 \left(d^3+3 d^2 r_o+3 d r_o^2+9 r_o^3\right){}^2 \left(d-3 r_o\right){}^3+x^2 \left(d+3 r_o\right){}^2 \left(d^7-9 d^6 r_o+45 d^5 r_o^2-125 d^4 r_o^3+223 d^3 r_o^4-231 d^2 r_o^5+195 d r_o^6-195 r_o^7\right)+4 d^2 r_o^2 \left(15 d^7-45 d^6 r_o-90 d^5 r_o^2+286 d^4 r_o^3+23 d^3 r_o^4-165 d^2 r_o^5+372 d r_o^6-780 r_o^7\right)+8 d x r_o \left(2 d^8-6 d^7 r_o-6 d^6 r_o^2+79 d^5 r_o^3-129 d^4 r_o^4-156 d^3 r_o^5+252 d^2 r_o^6-237 d r_o^7+585 r_o^8\right)\right){}^2\)
这里面真的有个圆 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 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 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。
{Array{Cos,#]],Sin,#]]}&,3](*外接圆以原点为圆心，Subscript为半径*),Array[{d,0}+Subscript{Cos,#]],Sin,#]]}&,3,4](*内切圆以{d,0}为圆心，Subscript为半径*),{x,y}}//#/.{Cos,i_]]:>Subscript,i],Sin,i_]]:>Subscript,i]}&//Function[{ABCPoints,CDFPoints,g},{Subsets,CDFPoints}//(*每条边分配一个切点*)Transpose//Parallelize&//(*正弦余弦平方和为1*)Total[#^2]==1&,{Append,{p,Complement//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,Subscript,_]]]&}//#[]/.#[[-1]]&//Flatten//(Subtract@@#//Factor//Numerator)&/@#&//#/.(#~Function[{PolynomialList,\},Select]==1&][]//Solve[#==0,\]&]~Cases,Subscript,_]][])&]@@#&/@#]&]@@#&//#/.Solve(Subscript-2Subscript),Subscript]&//Flatten//Parallelize[(#/.{Subscript,i_]:>Cos,i]],Subscript,i_]:>Sin,i]]}/.{Subscript[\,i_]:>2ArcTan,i]]}//TrigExpand//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,(Select[#,Exponent[#,x]==1\Exponent[#,y]==1&][]//Solve[#=={0,0},{x,y}]&//Factor//Flatten)}&//{{#[]/.#[],Table[{x,y}/.#[]//(#-(#/.\))&,{\,Array,#]&,3]//Function[{\},Permutations[\]//({#,\}//Transpose//Rule@@#&/@#&)&/@#&]}]}//Flatten//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//{#,Array[(SymmetricPolynomial[#,Array,#]&,3]]-Subscript)&,3]}&//Fold],Subscript,i]]//CoefficientList[#,Subscript,i]]&//Factor)&,XToSymmetricPolynomial,{2}]//(#//Flatten//DeleteCases[#,0]&)&/@#&],#,{1,2,3}]&//#[]&//{#,(Select[#,Exponent[#,Subscript]==1&]//First//Solve[#==0,Subscript]&//First)}&//{#[]/.#[]//Parallelize[(#//Factor//Numerator)&/@#]&//Complement[#,{0}]&//PolynomialGCD@@#&//Solve[#==0,Subscript]&,#[]}&//{#[],#[]/.#[]}&//Flatten//(Subtract@@#//#/.Array[(Subscript->SymmetricPolynomial[#,Array,#]&,3]])&,3]&//Factor//Numerator)&/@#&,#[]}&//{#[],#[]}/.Solve[#[]==0,Subscript,3]][]&//{#[]//Factor//Numerator,#[]//Parallelize[(#//Factor)&/@#]&}&//Function[{DividePolynomial,Rules},Parallelize[#[]->({Numerator[#[]],Denominator[#[]]}//PolynomialRemainder[#,DividePolynomial,Subscript,2]]&/@#&//Divide@@#&//Factor//#/.{Subscript,1]->Sin,1]]/(1+Cos,1]])}&//Factor//TrigReduce//FullSimplify//Factor//Which[#[]===Plus,Collect[#,{d,Subscript},FullSimplify],#[]===Power,Collect[#[],{d,Subscript},FullSimplify]^#[],True,#]&/@#&)&/@Rules]]@@#&//#/.{Subscript->1}&//{#//Parallelize&,#//(Subtract@@#/.{Subscript[\,i_]:>2ArcTan,i]]}//TrigExpand//Factor//Numerator)&/@#&//FixedPoint[(#//SortBy[#,Exponent[#,Subscript,1]]&]&//Which],Subscript,1]]==0,#,Exponent[#[],Subscript,1]]==1,#[]/.Solve[#[]==0,Subscript,1]][],True,Append,#[],Subscript,1]],{\,#[]}],#[]]]&//Parallelize[(#//Factor//Numerator//\ #&//Select[#,Variables[#]\{Subscript,1],x,y}!={}&]&//Collect[#,Subscript,1],Factor]&)&/@#]&)&,#]&//MapAll]===Plus,(MonomialList[#,{x,y}]//Collect[#,{x,y},Simplify]&/@#&//Total),#]&,#]&//First}&
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？