求三角形内的相交三角形面积

数学星空

27#结果没问题，例如
a=b=c=1;k1=k2=k3=0.25,α=β=γ=π/6可以得到是s=3√3/64=0.0811898816047

现给一个数值解便于检验结果是否正确

\(a=8, b=5, c=6, k_1=\frac{1}{2}、  k_2=\frac{1}{3}、  k_3=\frac{1}{4}, \alpha=\arctan(4), \beta=\arctan(2) ,\gamma=\arctan(1)\)

\(s=\frac{98329671653329\sqrt{399}}{210472039308960}-\frac{20033376020153}{2192417076135}=0.19446552212026\)

TSC999

27#的公式中有个笔误，是在这里 N=4S(cos(α−γ)−cot(A)sin(α−γ))(cos(α−β)+cot(B)sin(α−β))(cot(C)sin(β−γ)+cos(β−γ))， 4S 后面的三个因式中，第一个因式中错了一个符号，应当是：
N=4S(cos(α−γ)+cot(A)sin(α−γ))(cos(α−β)+cot(B)sin(α−β))(cot(C)sin(β−γ)+cos(β−γ))。

数学星空

27#公式的下式是对的哈

\(N=4S(\cos(\alpha-\gamma)-\cot(A)\sin(\alpha-\gamma))(\cos(\alpha-\beta)+\cot(B)\sin(\alpha-\beta))(\cot(C)\sin(\beta-\gamma)+\cos(\beta-\gamma))\)

可以写成更对称的形式：

\(N=4S(\cos(\gamma-\alpha)+\cot(A)\sin(\gamma-\alpha))(\cos(\alpha-\beta)+\cot(B)\sin(\alpha-\beta))(\cot(C)\sin(\beta-\gamma)+\cos(\beta-\gamma))\)

注意：

\(\gamma-\alpha+\alpha-\beta+\beta-\gamma=0\)

creasson

记号$\lambda  = \frac{AF}{AB},\mu  = \frac{BD}{BC},v = \frac{CE}{CA}$,并记$R$为三角形$ABC$的外接圆半径，那么所求三角形面积有如下简化的表达式:
\[\frac{S}{{4{R^2}}} = \left( {1 - \lambda  - \mu  - v + \lambda \mu  + \mu v + v\lambda } \right)\sin A\sin B\sin C + \]
\[\frac{1}{{\sin \left( {A + \varphi  - \phi } \right)}}\left( {\lambda \sin C\sin \left( {A + \varphi } \right) - \left( {1 - v} \right)\sin B\sin \varphi } \right)\left( {\lambda \sin C\sin \phi  + \left( {1 - v} \right)\sin B\sin \left( {A - \phi } \right)} \right) + \]
\[\frac{1}{{\sin \left( {B + \phi  - \theta } \right)}}\left( {\mu \sin A\sin \left( {B + \phi } \right) - \left( {1 - \lambda } \right)\sin C\sin \phi } \right)\left( {\mu \sin A\sin \theta  + \left( {1 - \lambda } \right)\sin C\sin \left( {B - \theta } \right)} \right) + \]
\[\frac{1}{{\sin \left( {C + \theta  - \varphi } \right)}}\left( {v\sin B\sin \left( {C + \theta } \right) - \left( {1 - \mu } \right)\sin A\sin \theta } \right)\left( {v\sin B\sin \varphi  + \left( {1 - \mu } \right)\sin A\sin \left( {C - \varphi } \right)} \right)\]

creasson

上式利用$A + B + C = \pi $再化简可得
\[{S_{XYZ}} = M \times {N^2}\]
\[\begin{array}{l}
M = \frac{{{R^2}}}{{\sin \left( {A + \varphi  - \phi } \right)\sin \left( {B + \phi  - \theta } \right)\sin \left( {C + \theta  - \varphi } \right)}} \\
\end{array}\]
\[\begin{array}{l}
N = 2\lambda \sin \phi \sin C\sin \left( {C + \theta  - \varphi } \right) + 2\mu \sin \theta \sin A\sin \left( {A - \phi  + \varphi } \right) + 2\nu \sin \varphi \sin B\sin \left( {B - \theta  + \phi } \right) \\
+  \sin A\cos \left( {A + \theta  + \varphi  - \phi } \right) + \sin B\cos \left( {B + \varphi  + \phi  - \theta } \right) + \sin C\cos \left( {C + \phi  + \theta  - \varphi } \right) \\
\end{array}\]

数学星空

将楼上的结论替换为主题所设参数即下式：

\(s=\frac{R^2(2k_1\sin(\alpha)\sin(C)\sin(C+\beta-\gamma)+2k_2\sin(\beta)\sin(A)\sin(A+\gamma-\alpha)+2k_3\sin(\gamma)\sin(B)\sin(B+\alpha-\beta)+\sin(A)cos(A+\beta+\gamma-\alpha)+\sin(B)\cos(B+\gamma+\alpha-\beta)+\sin(C)\cos(C+\alpha+\beta-\gamma))^2}{2(\sin(A+\gamma-\alpha)\sin(B+\alpha-\beta)\sin(C+\beta-\gamma)}\)

此即我们需要的最简最对称的表达式            
            

数学星空

对于同类问题：



我们可以得到

\(\frac{s}{S}=\frac{M}{N}\)

\(M=\lambda^2\lambda_1^2\mu^2\mu_1^2\nu^2\nu_1^2-2\lambda^2\lambda_1\mu^2\mu_1\nu^2\nu_1+2\lambda^2\lambda_1\mu^2\mu_1\nu\nu_1^2+2\lambda^2\lambda_1\mu\mu_1^2\nu^2\nu_1+2\lambda\lambda_1^2\mu^2\mu_1\nu^2\nu_1+\lambda^2\mu^2\nu^2-2\lambda^2\mu^2\nu\nu_1+\lambda^2\mu^2\nu_1^2-2\lambda^2\mu\mu_1\nu^2+2\lambda^2\mu\mu_1\nu\nu_1+\lambda^2\mu_1^2\nu^2-2\lambda\lambda_1\mu^2\nu^2+2\lambda\lambda_1\mu^2\nu\nu_1+2\lambda\lambda_1\mu\mu_1\nu^2+2\lambda\lambda_1\mu\mu_1\nu\nu_1+\lambda_1^2\mu^2\nu^2-2\lambda\mu\nu+2\lambda\mu\nu_1+2\lambda\mu_1\nu+2\lambda_1\mu\nu+1\)

\(N=(\lambda\lambda_1\mu\nu_1+\lambda\mu\nu_1+\lambda\mu-\lambda_1\mu+\lambda+1)(\lambda_1\mu\mu_1\nu+\lambda_1\mu\nu+\mu\nu-\mu_1\nu+\mu+1)(\lambda\mu_1\nu\nu_1+\lambda\mu_1\nu+\lambda\nu-\lambda\nu_1+\nu+1)\)

数学星空

当\(\lambda_1=\nu_1=\mu_1=0\)时

\(\frac{s}{S}=\frac{(\lambda^2\mu^2\nu^2-2\lambda\mu\nu+1)}{(\lambda\mu+\lambda+1)(\mu\nu+\mu+1)(\lambda\nu+\nu+1)}\)

数学星空

若设\(\lambda=\mu=\nu=k,\frac{s}{S}=\frac{1}{4}\)

我们可以得到：

\((k^2-3k+1)(k^2+k+1)^2=0\)

即\(k_1=\frac{3-\sqrt{5}}{2}=\phi^2,k_2=\frac{3+\sqrt{5}}{2}=\frac{1}{\phi^2}\)

数学星空

按照楼上38#的记号，我们可以算出：

我们先给出相关代号：

\(S_{D_1FZ}=S_{11},S_{E_1DX}=S_{12},S_{F_1EY}=S_{13},S_{D_1ZYEA}=T_{11},S_{E_1XZFB}=T_{12},S_{F_1YXDC}=T_{13}\)

\(S_{XYZ}=S_0,S_{ABC}=S,S_{AFZ}=S_{21},S_{BDX}=S_{22},S_{CEY}=S_{23},S_{AZYE}=T_{21},S_{BXZF}=T_{22},S_{CYXD}=T_{23}\)

经过复杂的计算，我们最终得到下列结论：

\(\frac{S_{11}}{S}=\frac{\lambda(\nu^2+\nu_1^2-2\nu_1\nu)}{(1+\nu)(1+\nu_1)(1+\nu-\lambda\nu_1+\lambda\mu_1\nu+\lambda\nu_1\mu_1\nu+\lambda\nu)}\)

\(\frac{S_{12}}{S}=\frac{\mu(\lambda-\lambda_1)^2}{(1+\lambda_1)(1+\lambda+\mu\lambda\nu_1+\mu\lambda-\mu\lambda_1+\mu\lambda_1\nu_1\lambda)(1+\lambda)}\)

\(\frac{S_{13}}{S}=\frac{\nu(\mu_1^2+\mu^2-2\mu\mu_1)}{(1+\mu)(1+\mu_1)(1+\mu\nu+\mu+\mu\lambda_1\nu\mu_1-\mu_1\nu+\lambda_1\nu\mu)}\)

\(\frac{S_0}{S}=\frac{(-2\mu^2\lambda\lambda_1\nu^2+\lambda^2\mu_1^2\nu^2+2\mu^2\nu_1\lambda\lambda_1\nu+2\mu_1\nu^2\nu_1\lambda\mu^2\lambda_1^2+\mu_1^2\nu^2\nu_1^2\lambda^2\mu^2\lambda_1^2+1-2\lambda^2\mu_1\nu^2\mu+2\mu^2\nu_1^2\lambda^2\lambda_1\mu_1\nu+2\nu_1\mu\lambda^2\mu_1\nu+2\mu\lambda_1\mu_1\nu\lambda\nu_1+2\mu\lambda_1\mu_1\nu^2\lambda+2\mu\lambda_1\mu_1^2\nu^2\lambda^2\nu_1+2\lambda_1\nu\mu+2\lambda\mu_1\nu+2\mu\lambda\nu_1-2\mu\lambda\nu-2\mu^2\nu_1\lambda^2\nu+\mu^2\nu_1^2\lambda^2-2\mu^2\lambda_1\mu_1\nu^2\nu_1\lambda^2+\nu^2\mu^2\lambda_1^2+\mu^2\lambda^2\nu^2)}{(1+\mu\nu+\mu+\mu\lambda_1\nu\mu_1-\mu_1\nu+\lambda_1\nu\mu)(1+\lambda+\mu\lambda\nu_1+\mu\lambda-\mu\lambda_1+\mu\lambda_1\nu_1\lambda)(1+\nu-\lambda\nu_1+\lambda\mu_1\nu+\lambda\nu_1\mu_1\nu+\lambda\nu)}\)

\(\frac{T_{11}}{S}=\frac{(\lambda\nu_1\nu-\nu_1^2\lambda_1\nu\mu^2\lambda-2\nu_1^2\mu\lambda+2\nu_1\nu^2\mu\lambda_1\mu_1\lambda+\nu+2\mu\nu+2\nu_1\nu\mu^2\lambda-\mu_1\nu+\nu_1\nu+2\nu_1\mu\nu-\nu_1\mu_1\nu-\lambda\mu_1^2\nu^2-\lambda\nu_1^2\mu^2+\nu^2\lambda_1\mu+\nu^2\lambda_1\mu^2-\lambda\mu^2\nu^2-\nu^2\mu_1\nu_1+\mu\nu^2\nu_1+\lambda\nu_1\mu_1\nu+2\nu_1\lambda\mu_1\nu^2\mu-2\nu_1\lambda\mu_1^2\nu^2+2\nu_1\lambda\mu_1\nu^2\lambda_1\mu^2+2\lambda\mu_1\nu^2\mu+2\lambda\nu_1^2\mu_1\nu-\lambda\nu_1^2\lambda_1\nu\mu_1\mu^2+\mu\nu^2\lambda\nu_1+\nu_1^2\lambda\mu_1\nu^2\mu+\nu_1^2\lambda\mu_1\nu^2\lambda_1\mu+\nu_1^2\lambda\mu_1\nu^2\lambda_1\mu^2+\lambda\nu_1^2\mu_1\nu\mu+\nu^2\lambda_1\mu^2\nu_1-\lambda\nu_1^2\lambda_1\nu\mu_1\mu-\nu_1^2\lambda\mu_1^2\nu^2+\lambda\nu^2\lambda_1\mu\nu_1-\lambda\nu_1^2\lambda_1\nu\mu-\lambda\nu\nu_1^2\mu+\nu^2\lambda_1\mu\nu_1-\lambda\nu^2\mu_1\nu_1+2\lambda\nu\nu_1\mu+\nu^2\mu^2\lambda_1\nu_1\lambda-\nu^2\mu_1+\mu\nu^2-\lambda\nu_1^2)}{(1+\nu-\lambda\nu_1+\lambda\mu_1\nu+\lambda\nu_1\mu_1\nu+\lambda\nu)(1+\nu_1)(1+\mu\nu+\mu+\mu\lambda_1\nu\mu_1-\mu_1\nu+\lambda_1\nu\mu)(1+\mu)}\)

\(\frac{T_{12}}{S}=\frac{(\lambda-\lambda_1\nu_1\lambda+\nu^2\mu\lambda_1\mu_1\lambda^2+\lambda_1\lambda-2\lambda_1^2\mu\nu+2\lambda\nu-\lambda_1^2\lambda\mu\nu^2\mu_1+2\lambda^2\mu\lambda_1\nu_1\mu_1\nu-\lambda^2\nu_1-\mu\lambda^2\nu_1^2-\lambda^2\lambda_1\nu_1+\mu\lambda\lambda_1+\lambda^2\mu_1\nu+2\mu\lambda_1^2\nu_1\lambda+\lambda^2\nu^2\mu_1+2\lambda\lambda_1\nu-\mu\lambda^2\nu^2+2\mu\lambda\lambda_1\nu-\mu\lambda^2\lambda_1\nu_1+\mu\lambda^2\lambda_1\nu+\mu\lambda^2\lambda_1\mu_1\nu-\mu\lambda_1^2\lambda\nu_1\mu_1\nu-\mu\lambda_1^2\lambda\nu+\lambda^2\lambda_1\mu_1\nu-\mu\lambda_1^2\lambda\mu_1\nu+\lambda^2\lambda_1\nu-\mu\lambda_1^2\nu_1^2\lambda^2+2\mu\lambda_1\nu_1\lambda^2\nu-2\mu\lambda_1\nu_1^2\lambda^2-\mu\lambda_1^2\nu^2+\mu\lambda_1^2\nu_1\lambda^2\nu+\mu\lambda_1^2\nu_1\lambda\nu+\mu\lambda_1^2\nu_1\lambda^2\mu_1\nu+\mu\lambda_1\nu_1\lambda-\lambda\nu_1+2\mu\lambda^2\nu_1\nu+\lambda^2\lambda_1\nu^2\mu_1+2\mu\lambda^2\lambda_1\nu^2\nu_1\mu_1-\mu\lambda_1^2\nu^2\lambda\nu_1\mu_1+\mu\lambda^2\lambda_1^2\nu^2\nu_1\mu_1+2\lambda\nu^2\mu\lambda_1+\lambda^2\nu-\mu\lambda_1^2)}{((1+\nu)(1+\lambda+\mu\lambda\nu_1+\mu\lambda-\mu\lambda_1+\mu\lambda_1\nu_1\lambda)(1+\lambda_1)(1+\nu-\lambda\nu_1+\lambda\mu_1\nu+\lambda\nu_1\mu_1\nu+\lambda\nu))}\)

\(\frac{T_{13}}{S}=\frac{(2\mu^2\mu_1\lambda_1\nu_1\lambda\nu+2\mu\lambda^2\mu_1\nu+\mu+\mu\lambda_1\nu\mu_1-2\mu_1^2\lambda\nu+2\mu^2\lambda\lambda_1\nu+2\mu^2\lambda^2\nu_1\lambda_1\nu\mu_1+2\mu^2\lambda_1\lambda\mu_1\nu-2\lambda_1^2\mu^2\nu\mu_1-\mu^2\lambda_1^2\nu-\mu\lambda_1-\mu_1^2\nu+\lambda^2\mu^2\nu_1\mu_1\nu-\mu_1^2\mu\lambda^2\nu_1\nu-\lambda_1\mu_1\mu^2\nu-\lambda_1^2\mu_1^2\mu^2\nu+2\mu\lambda_1\mu_1^2\nu+\lambda\mu_1\mu^2+\mu_1\mu\nu+\mu^2\lambda^2\nu_1+2\mu\lambda+\lambda\mu_1\mu^2\nu-\mu\lambda^2\lambda_1\nu_1\mu_1^2\nu+\mu^2\lambda^2\nu_1\lambda_1\nu\mu_1^2+\mu_1\mu^2\lambda\nu_1-\mu\lambda\mu_1^2\nu+\lambda\mu_1^2\mu^2\lambda_1\nu+\mu_1\mu^2\lambda\nu_1\nu+\mu^2\lambda\nu_1\lambda_1\nu\mu_1^2-\mu\lambda\nu_1\mu_1^2\nu-\mu\lambda_1\lambda\nu_1\mu_1^2\nu+\mu_1\mu^2\lambda^2\nu_1-\lambda^2\mu_1^2\nu+2\lambda\mu_1\mu-\lambda_1\mu_1\mu^2+\lambda\mu_1^2\mu\lambda_1\nu+2\mu\lambda\mu_1\nu+\mu^2\lambda-\lambda_1\mu_1\mu-\mu^2\lambda^2\nu+\mu^2\lambda\nu_1-\mu^2\lambda_1+\mu\mu_1)}{((1+\lambda+\mu\lambda\nu_1+\mu\lambda-\mu\lambda_1+\mu\lambda_1\nu_1\lambda)\)(1+\lambda)(1+\mu_1)(1+\mu\nu+\mu+\mu\lambda_1\nu\mu_1-\mu_1\nu+\lambda_1\nu\mu))}

进一步我们在上面令\(\lambda_1=0,\nu_1=0,\mu_1=0\)可以得到：

\(\frac{S_{21}}{S}＝\frac{S_{11}}{S}＝\frac{\nu^2\lambda}{(1+\nu)(1+\nu+\lambda\nu)}\)

\(\frac{S_{22}}{S}＝\frac{S_{12}}{S}＝\frac{\mu\lambda^2}{(1+\lambda)(1+\lambda+\mu\lambda)}\)

\(\frac{S_{23}}{S}＝\frac{S_{13}}{S}＝\frac{\mu^2\nu}{(1+\mu)(1+\mu\nu+\mu)}\)

\(\frac{S_0}{S}＝\frac{(1-2\mu\lambda\nu+\mu^2\lambda^2\nu^2)}{(1+\mu\nu+\mu)(1+\lambda+\mu\lambda)(1+\nu+\lambda\nu)}\)

\(\frac{T_{21}}{S}＝\frac{T_{11}}{S}=\frac{(\nu+2\mu\nu-\lambda\mu^2\nu^2+\mu\nu^2)}{(1+\nu+\lambda\nu)(1+\mu\nu+\mu)(1+\mu)}\)

\(\frac{T_{22}}{S}＝\frac{T_{12}}{S}=\frac{(\lambda+2\lambda\nu-\mu\lambda^2\nu^2+\lambda^2\nu)}{(1+\nu)(1+\lambda+\mu\lambda)(1+\nu+\lambda\nu)}\)

\(\frac{T_{23}}{S}＝\frac{T_{13}}{S}=\frac{(\mu+2\mu\lambda+\mu^2\lambda-\mu^2\lambda^2\nu)}{(1+\lambda+\mu\lambda)(1+\lambda)(1+\mu\nu+\mu))}\)

A.若进一步要求\(S_{11}=S_{12}=S_{13}=S_0\)

我们得到：

\(1-\lambda\nu_1\nu^2\lambda_1^2\mu^2+\nu_1-\mu^2\lambda\lambda_1\nu^2-\nu^2\lambda\mu+\lambda\mu_1\nu^2+2\nu_1^2\lambda_1\nu\mu^2\lambda+2\mu\lambda_1\mu_1\nu\lambda\nu_1+\nu_1^2\mu\lambda+2\nu_1\nu\mu\lambda_1+2\nu_1\nu^2\mu\lambda_1\mu_1\lambda+\lambda\mu_1\nu+\nu_1\nu^2\lambda_1^2\mu^2+\nu_1^2\nu^2\mu^2\lambda_1^2\mu_1\lambda+2\lambda_1\nu\mu-2\mu\lambda\nu+\lambda\nu_1^2\mu\lambda_1-\lambda\nu+\nu^2\mu^2\lambda_1^2+2\mu\lambda\nu_1+\lambda\nu_1^2\mu^2\lambda_1^2\nu\mu_1+\lambda\nu_1-2\nu^2\mu^2\lambda_1\nu_1\lambda+\lambda\nu_1^2\lambda_1^2\nu\mu^2+\lambda\nu_1^2\mu^2\lambda_1-\lambda\nu^2\mu\lambda_1=0\)

\(1+\lambda_1-\mu\lambda^2\mu_1\nu+\lambda^2\mu_1^2\nu^2-\lambda^2\mu_1\nu^2\mu+\nu_1\mu\lambda^2+2\mu\lambda_1\mu_1\nu\lambda\nu_1-2\nu^2\mu\lambda_1\mu_1\lambda^2-\mu\lambda_1\mu_1^2\nu^2\lambda^2+2\lambda\mu_1\nu+2\lambda_1\nu\mu-2\mu\lambda\nu+\lambda_1^2\mu\nu+2\lambda_1\lambda\mu_1\nu+\lambda_1^2\lambda^2\mu\nu^2\mu_1^2\nu_1+2\lambda_1^2\lambda\mu\nu^2\mu_1+\lambda_1\lambda^2\mu_1^2\nu^2+2\lambda^2\mu\lambda_1\nu_1\mu_1\nu+\mu\lambda_1^2\nu^2\mu_1^2\lambda\nu_1+\mu\lambda_1^2\nu^2\mu_1^2\lambda+\mu\nu^2\lambda_1^2\mu_1+\mu\lambda_1^2\nu\mu_1+\mu\lambda_1+\mu\lambda\nu_1-\mu\lambda-\lambda^2\mu\nu=0\)

\(1+\nu\mu\lambda_1\mu_1^2\lambda^2\nu_1^2+\mu^2\mu_1^2\lambda_1\nu_1^2\lambda^2\nu+2\mu^2\mu_1\lambda_1\nu_1\lambda\nu+\nu\mu_1^2\lambda\nu_1+\nu\mu_1^2\mu\lambda^2\nu_1^2+\nu\lambda^2\mu_1^2\nu_1+\lambda_1\nu\mu^2-\mu^2\nu_1^2\lambda^2\mu_1\nu+\mu^2\nu_1^2\lambda^2+2\mu\lambda_1\mu_1\nu\lambda\nu_1-\mu\nu+2\lambda\mu_1\nu-\nu_1\nu\mu^2\lambda+\mu_1+\lambda_1\nu\mu+\mu_1\nu-2\mu\lambda\nu-\mu^2\nu_1\lambda^2\nu+\mu_1^2\lambda\nu+2\mu\lambda\nu_1-2\lambda^2\mu^2\nu_1\mu_1\nu+2\mu_1^2\mu\lambda^2\nu_1\nu-\mu^2\nu\lambda+2\mu\mu_1\lambda\nu_1+\mu^2\mu_1\lambda^2\nu_1^2=0\)

B.若进一步要求\(T_{11}=T_{12}=T_{13}\)

我们得到：

\(-\mu^2\mu_1\lambda_1\nu_1\lambda\nu-\mu^2\lambda^3\nu_1^3\lambda_1\nu\mu_1+2\nu\mu_1^2\mu\lambda^2\nu_1^2-3\lambda^2\nu_1\nu+4\mu\lambda^2\mu_1\nu-\lambda^2\nu_1^3\mu_1\nu\mu-\lambda^2\nu_1^3\mu_1\nu\mu\lambda_1-\mu\lambda^3\nu_1^2\lambda_1\nu-\lambda^2\nu_1^2\mu^2\lambda_1-3\lambda\nu_1\nu-2\lambda^3\nu_1\mu_1\nu-2\mu\lambda^3\nu_1\nu-2\lambda^2\nu_1^2\mu_1\nu+2\mu^2\nu_1^2\lambda^2\mu_1\nu+\mu-\mu^2\nu_1^2\lambda^2\lambda_1\mu_1\nu-2\nu_1\mu\lambda^2-\nu_1^2\lambda_1\nu\mu^2\lambda+\nu_1^3\mu^2\lambda^3\lambda_1\mu_1-\mu\lambda_1\mu_1\nu\lambda\nu_1+\lambda^2\nu_1^3\mu\mu_1+2\mu\lambda^3\nu_1^2\mu_1+\lambda^3\nu_1^3\mu\mu_1+\mu\nu_1-\nu-2\mu^2\lambda^3\nu_1^2\lambda_1\nu\mu_1+2\lambda^2\nu_1^2\mu_1+\lambda^3\nu_1^2\mu_1+\lambda\nu_1^2\mu_1-\mu\lambda_1\nu\mu_1-\lambda^2\nu_1^2\mu^2\lambda_1\mu_1-2\mu\lambda_1\lambda\mu_1\nu+\mu_1\mu^2\lambda\nu_1^2\nu-\mu\lambda_1\lambda\nu_1^2\mu_1^2\nu-\lambda_1\mu_1\mu^2\nu_1+\lambda^2\mu_1\mu\nu_1^2-\lambda_1\mu_1\mu\nu_1-\nu_1\nu-2\lambda\nu+2\lambda^2\mu\lambda_1\nu_1\mu_1\nu-2\lambda^2\nu_1^2\nu\mu+\nu_1^3\mu^2\lambda^2\lambda_1+\lambda^2\mu_1^2\mu^2\nu+\mu_1^2\mu^2\lambda^2\nu_1^3\nu-2\mu^2\lambda^2\nu_1^2\lambda_1\nu+\lambda\nu_1^2\mu^2+2\lambda^2\nu_1^2+2\mu^2\lambda^2\nu_1\lambda_1\nu\mu_1-2\mu^2\lambda_1\lambda\mu_1\nu-\mu\lambda_1-2\mu\lambda_1\nu_1^2\lambda^3\mu_1\nu+\mu_1^2\mu^2\lambda^3\nu_1^3\nu-4\mu\lambda^3\mu_1\nu\nu_1^2+\mu\lambda^2\nu_1^2+\mu\lambda\nu_1+4\lambda^2\mu^2\nu_1\mu_1\nu+\lambda^2\nu_1^3\mu\lambda_1+4\mu_1^2\mu\lambda^2\nu_1\nu-\lambda_1\mu_1\mu^2\nu+\lambda\mu_1\mu^2+\mu_1\mu\nu-\lambda^3\nu_1^3\mu_1\nu\mu-\lambda^3\nu_1^3\mu_1\nu\mu\lambda_1+\nu_1^3\mu^2\lambda^2\lambda_1\mu_1+2\mu\lambda+2\lambda^2\mu^2\mu_1\nu+\lambda\mu_1\mu^2\nu+2\mu_1\mu^2\lambda\nu_1+\mu\lambda\mu_1^2\nu-\lambda\mu_1^2\mu^2\lambda_1\nu-\mu\lambda_1\nu_1^2\lambda^2+2\mu_1\mu^2\lambda\nu_1\nu-2\mu^2\lambda\nu_1\lambda_1\nu\mu_1^2+2\mu\lambda\nu_1\mu_1^2\nu-2\mu\lambda_1\lambda\nu_1\mu_1^2\nu+2\lambda\mu_1\mu-\lambda_1\mu_1\mu^2+2\mu_1^2\mu\lambda^2\nu+\mu\lambda_1\nu_1\lambda-\lambda\mu_1^2\mu\lambda_1\nu+4\mu\lambda\mu_1\nu+3\mu\lambda\nu_1\mu_1\nu+3\mu_1^2\mu^2\lambda^2\nu_1^2\nu-\mu^2\nu_1^3\lambda^2\lambda_1\mu_1\nu+\mu_1\mu\nu\nu_1+\mu_1\mu^2\lambda\nu_1^2-\lambda^3\nu_1^2\mu_1\nu+\nu_1^3\mu^2\lambda^3\lambda_1-2\lambda^2\nu_1^2\nu\mu\lambda_1+\lambda^3\nu_1^3\mu\lambda_1\mu_1+3\lambda^2\mu_1^2\mu^2\nu_1\nu-2\lambda^2\mu_1\mu\nu_1+\nu\mu_1^2\lambda\nu_1^2\mu-\mu^2\lambda^3\nu_1^2\lambda_1\nu-4\lambda^2\nu_1\mu_1\nu-4\mu\lambda^2\nu_1\nu-\mu\lambda^3\nu_1^2\nu-2\lambda\nu_1\mu_1\nu-3\lambda^2\nu_1^2\mu_1\nu\mu-\lambda^2\nu_1^2\mu_1\nu\mu\lambda_1+\mu^2\lambda-\lambda_1\mu_1\mu+2\mu^2\lambda\nu_1-\mu^2\lambda_1-\mu\lambda_1\nu_1+\mu\mu_1\nu_1-\mu^2\lambda_1\nu_1-\mu^2\lambda_1\lambda\nu_1^2\mu_1^2\nu+\mu\mu_1\lambda\nu_1+\lambda^2\nu_1^3\mu-\lambda^3\nu_1\nu-\lambda\nu_1^2\mu_1\nu-\lambda\nu_1^2\lambda_1\nu\mu-\lambda\nu\nu_1^2\mu-2\lambda\nu\nu_1\mu-\nu_1\mu_1\nu\mu\lambda_1+2\mu\lambda^3\nu_1^2+\lambda^3\nu_1^3\mu+\mu\mu_1+\lambda_1\mu_1\mu\lambda\nu_1+\mu^2\lambda_1\lambda\nu_1+\lambda^3\nu_1^2+\lambda\nu_1^2+\mu^2\mu_1\lambda_1\nu_1\lambda+\lambda^3\nu_1^3\mu\lambda_1-\mu^2\mu_1\lambda_1\nu_1\nu+\lambda^2\nu_1^3\mu\lambda_1\mu_1-\mu\lambda_1\nu_1^2\lambda^2\mu_1-\lambda^2\nu+\mu_1^2\mu^2\lambda^3\nu_1\nu+2\mu_1^2\mu^2\lambda^3\nu_1^2\nu-4\mu\lambda^3\mu_1\nu\nu_1＝0\)

\(-\lambda-\mu_1^3\nu^3\lambda+2\mu\lambda_1\mu_1^2\nu^3\lambda^2\nu_1-\mu_1^3\nu^3\lambda\nu_1+2\lambda\mu_1\nu^2+\mu+4\mu\lambda_1\mu_1\nu^3\lambda+\mu\lambda_1\mu_1^2\nu^2\lambda^2\nu_1+\lambda_1\nu_1\lambda-\mu\lambda_1^2\nu^3\mu_1^3\lambda^2-2\mu\lambda_1^2\nu^3\mu_1^2\lambda^2+\mu\lambda_1\mu_1\nu\lambda\nu_1-2\mu_1^2\nu^2-2\nu_1\nu^2\mu\lambda_1\mu_1\lambda-\lambda^2\lambda_1^2\nu^3\mu_1\mu-4\nu^2\mu\lambda_1\mu_1\lambda^2-2\mu\lambda_1\mu_1^2\nu^2\lambda^2-2\mu_1^2\nu^3\lambda-\mu_1^3\nu^2\lambda\nu_1+2\mu\nu-\lambda\mu_1\nu-2\lambda_1^2\lambda\mu\nu^2-2\nu^2\mu\lambda_1\lambda^2+2\mu\lambda_1\nu\mu_1-3\mu\lambda_1\lambda\mu_1\nu+\lambda^2\mu\lambda_1\nu_1\mu_1+\mu\mu_1^2\nu^3\lambda-\lambda^2\mu_1^3\nu^3\lambda_1\nu_1+\mu\lambda^2\lambda_1^2\nu_1\mu_1^2\nu+\mu\lambda_1^2\lambda\nu_1\mu_1^2\nu-\lambda\mu_1\mu\lambda_1+\mu\lambda^2\mu_1^2\nu^3\nu_1-\lambda_1\lambda+\mu\nu_1\mu_1^2\nu^3\lambda+2\mu\nu_1\mu_1^2\nu^3\lambda_1\lambda-\lambda_1\lambda\mu_1\nu-\lambda\mu_1^2\mu\lambda_1^2\nu-2\lambda\nu-4\lambda_1^2\lambda\mu\nu^2\mu_1+\lambda^2\mu\lambda_1\nu_1\mu_1\nu+2\mu_1^2\mu\lambda^2\nu_1\nu^2+2\mu\lambda_1\mu_1^2\nu^2+2\mu_1^2\mu\nu^2\lambda+2\mu\lambda_1\nu^3\mu_1-\lambda_1\mu_1^3\nu^2\lambda+\lambda^2\nu_1-\lambda^2\lambda_1^2\nu^2\mu-\lambda\mu_1^2\nu^2-2\mu\lambda_1^2\nu^2\mu_1^2\lambda-\lambda^2\lambda_1\nu\mu_1^2+2\mu\lambda_1\lambda\nu_1\nu+\lambda^2\lambda_1\nu_1-\mu\lambda\lambda_1-2\lambda^2\mu_1\nu-\mu_1^2\nu+\mu_1^2\mu\lambda^2\nu_1\nu-2\lambda\lambda_1\nu+\mu\lambda_1\mu_1^3\nu^2\lambda+\mu\lambda_1\mu_1^3\nu^2\lambda\nu_1+\mu\lambda_1\mu_1^2\nu+3\mu_1\mu\nu-4\mu\lambda\lambda_1\nu+\mu\lambda^2\lambda_1\nu_1-\mu\lambda^2\lambda_1\nu+\mu_1\mu\nu^3-\mu_1^2\nu^3\lambda_1-\lambda_1\lambda\mu_1-2\mu\lambda^2\lambda_1\mu_1\nu+2\mu\lambda_1^2\lambda\nu_1\mu_1\nu-\mu\lambda_1^2\lambda\nu-2\lambda^2\lambda_1\mu_1\nu-2\mu\lambda_1^2\lambda\mu_1\nu-\lambda^2\lambda_1\nu+\mu\lambda\mu_1^2\nu+2\mu\lambda_1\nu_1\lambda^2\nu+\mu\lambda\nu_1\mu_1^2\nu-\lambda^2\mu_1^2\nu+\mu\lambda_1^2\nu_1\lambda^2\nu+\mu\lambda_1^2\nu_1\lambda\nu+2\mu\lambda_1^2\nu_1\lambda^2\mu_1\nu+\mu_1\lambda\nu_1-\mu\lambda^2\lambda_1\mu_1^2\nu+\mu\lambda_1\nu_1\lambda-\lambda^2\mu_1^3\lambda_1\nu^2\nu_1-\lambda\mu_1+\mu_1\lambda^2\nu_1+2\mu\lambda\mu_1\nu-\mu_1^2\nu^3+2\mu_1\mu\nu^3\lambda-\mu_1^3\nu^3\lambda_1\lambda-2\mu_1^2\nu^3\lambda_1\lambda+\lambda\nu_1+2\mu_1^2\mu\nu^2\lambda\nu_1-\mu_1\lambda_1\nu_1\lambda\nu-\lambda^2\nu_1\mu_1\nu+\mu\lambda_1\mu_1^2\nu^3-\lambda_1\mu_1^3\nu^2\lambda\nu_1-2\mu\lambda^2\lambda_1\nu^2\nu_1\mu_1-\lambda\nu_1\mu_1\nu-\mu_1^3\nu^2\lambda+\mu\lambda_1\nu^3\mu_1^3\lambda^2\nu_1-\lambda^2\mu_1^3\mu\lambda_1^2\nu^2-3\lambda^2\mu_1^2\mu\lambda_1^2\nu^2+3\mu_1\mu\nu^2+\mu_1\lambda_1\nu_1\lambda+\mu_1^2\lambda_1\nu_1\lambda^2\nu^2+\mu\lambda_1\mu_1^3\nu^3\lambda+\mu\lambda_1\mu_1^3\nu^3\lambda\nu_1+4\mu\lambda_1\mu_1^2\nu^3\lambda-3\lambda_1^2\lambda^2\mu\nu^2\mu_1+\nu_1\lambda\mu_1^2\nu^2+4\lambda\mu_1\nu^2\mu+2\lambda_1\lambda\mu_1\nu^2-2\mu_1^2\nu^2\lambda_1-4\lambda\nu^2\mu\lambda_1+\mu\mu_1-\lambda^2\mu_1^3\nu^3\nu_1+\lambda_1\mu_1\mu\lambda\nu_1+\lambda^2\mu_1^2\nu^2\nu_1+\mu\nu^2-\lambda\mu_1^2\nu^2\lambda_1+4\mu\lambda_1\nu^2\mu_1-\mu_1^3\lambda^2\nu^2\nu_1-\lambda^2\nu+\lambda^2\lambda_1\nu_1\mu_1-\lambda_1\mu_1^2\nu+\mu_1^2\nu^2\lambda_1\lambda\nu_1-\mu_1^3\nu^3\lambda_1\lambda\nu_1+\mu\lambda_1\lambda^2\nu_1\mu_1^3\nu^2-\lambda^2\lambda_1\nu_1\mu_1\nu+3\mu\lambda_1\nu^2\mu_1^2\lambda+\mu\lambda_1\nu^2\mu_1^2\lambda\nu_1=0\)

C.若进一步要求\(S_{21}=S_{22}=S_{23}=S_0\)

我们得到：

\(1-\nu^2\lambda\mu-2\mu\lambda\nu-\lambda\nu＝0\)

\(1-2\mu\lambda\nu-\mu\lambda-\lambda^2\mu\nu=0\)

\(1-\mu\nu-2\mu\lambda\nu-\mu^2\nu\lambda=0\)

D.若进一步要求\(T_{21}=T_{22}=T_{23}\)

我们得到：

\(\mu-\nu-2\lambda\nu+2\mu\lambda+\mu^2\lambda-\lambda^2\nu=0\)

\(-\lambda+\mu+2\mu\nu-2\lambda\nu+\mu\nu^2-\lambda^2\nu=0\)

E.若进一步要求\(\lambda=p, \mu=p, \nu=p, \lambda_1=q, \mu_1=q, \nu_1=q\)

我们得到：

\(T_{11}=T_{12}=T_{13}=\frac{(q^2p^2+p^2q-p^2+3pq+p-2q^2+1)p}{(1+p)(1+q)(1+p^2+p+q^2p^2-pq+p^2q)}\)

\(S_{11}=S_{12}=S_{13}=\frac{p(p-q)^2}{(1+p)(1+q)(1+p^2+p+q^2p^2-pq+p^2q)}\)

\(S_0=\frac{(pq-p+1)^2}{1+p^2+p+q^2p^2-pq+p^2q}\)

F.若进一步要求

1.\(\lambda=p, \mu=p, \nu=p, \lambda_1=q, \mu_1=q, \nu_1=q\)

2.\(S_{21}=S_{22}=S_{23}=S_0\)

3.\(T_{21}=T_{22}=T_{23}\)

我们得到：

\(p^3q^3-p^3q^2-p^3q+p^2q^3+q^2p^2+p^2q-p^2+q^2p+pq-p+1+q=0\)