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、k1=1/2、k2=1/3、k3=1/4、 α=arctan(4)、β=arctan(2),、 γ=arctan(1)
s=98329671653329√399/210472039308960-20033376020153/2192417076135=0.19446552212026
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
\(a=8, b=5, c=6, k_1=\frac{1}{2}、k_2=\frac{1}{3}、k_3=\frac{1}{4}, \alpha=\arctan(4), \beta=\arctan(2) ,\gamma=\arctan(1)\)
\(s=\frac{98329671653329\sqrt{399}}{210472039308960}-\frac{20033376020153}{2192417076135}=0.19446552212026\)
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))。
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：
\(\gamma-\alpha+\alpha-\beta+\beta-\gamma=0\)
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:
\[\frac{S}{{4{R^2}}} = \left( {1 - \lambda- \mu- v + \lambda \mu+ \mu v + v\lambda } \right)\sin A\sin B\sin C + \]
\[\frac{1}{{\sin \left( {A + \varphi- \phi } \right)}}\left( {\lambda \sin C\sin \left( {A + \varphi } \right) - \left( {1 - v} \right)\sin B\sin \varphi } \right)\left( {\lambda \sin C\sin \phi+ \left( {1 - v} \right)\sin B\sin \left( {A - \phi } \right)} \right) + \]
\[\frac{1}{{\sin \left( {B + \phi- \theta } \right)}}\left( {\mu \sin A\sin \left( {B + \phi } \right) - \left( {1 - \lambda } \right)\sin C\sin \phi } \right)\left( {\mu \sin A\sin \theta+ \left( {1 - \lambda } \right)\sin C\sin \left( {B - \theta } \right)} \right) + \]
\[\frac{1}{{\sin \left( {C + \theta- \varphi } \right)}}\left( {v\sin B\sin \left( {C + \theta } \right) - \left( {1 - \mu } \right)\sin A\sin \theta } \right)\left( {v\sin B\sin \varphi+ \left( {1 - \mu } \right)\sin A\sin \left( {C - \varphi } \right)} \right)\]
本帖最后由 creasson 于 2015-12-27 00:04 编辑
上式利用$A + B + C = \pi $再化简可得
\[{S_{XYZ}} = M \times {N^2}\]
\[\begin{array}{l}
M = \frac{{{R^2}}}{{\sin \left( {A + \varphi- \phi } \right)\sin \left( {B + \phi- \theta } \right)\sin \left( {C + \theta- \varphi } \right)}} \\
\end{array}\]
\[\begin{array}{l}
N = 2\lambda \sin \phi \sin C\sin \left( {C + \theta- \varphi } \right) + 2\mu \sin \theta \sin A\sin \left( {A - \phi+ \varphi } \right) + 2\nu \sin \varphi \sin B\sin \left( {B - \theta+ \phi } \right) \\
+\sin A\cos \left( {A + \theta+ \varphi- \phi } \right) + \sin B\cos \left( {B + \varphi+ \phi- \theta } \right) + \sin C\cos \left( {C + \phi+ \theta- \varphi } \right) \\
\end{array}\]
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
对于同类问题：
我们可以得到
\(\frac{s}{S}=\frac{M}{N}\)
\(M=\lambda^2\lambda_1^2\mu^2\mu_1^2\nu^2\nu_1^2-2\lambda^2\lambda_1\mu^2\mu_1\nu^2\nu_1+2\lambda^2\lambda_1\mu^2\mu_1\nu\nu_1^2+2\lambda^2\lambda_1\mu\mu_1^2\nu^2\nu_1+2\lambda\lambda_1^2\mu^2\mu_1\nu^2\nu_1+\lambda^2\mu^2\nu^2-2\lambda^2\mu^2\nu\nu_1+\lambda^2\mu^2\nu_1^2-2\lambda^2\mu\mu_1\nu^2+2\lambda^2\mu\mu_1\nu\nu_1+\lambda^2\mu_1^2\nu^2-2\lambda\lambda_1\mu^2\nu^2+2\lambda\lambda_1\mu^2\nu\nu_1+2\lambda\lambda_1\mu\mu_1\nu^2+2\lambda\lambda_1\mu\mu_1\nu\nu_1+\lambda_1^2\mu^2\nu^2-2\lambda\mu\nu+2\lambda\mu\nu_1+2\lambda\mu_1\nu+2\lambda_1\mu\nu+1\)
\(N=(\lambda\lambda_1\mu\nu_1+\lambda\mu\nu_1+\lambda\mu-\lambda_1\mu+\lambda+1)(\lambda_1\mu\mu_1\nu+\lambda_1\mu\nu+\mu\nu-\mu_1\nu+\mu+1)(\lambda\mu_1\nu\nu_1+\lambda\mu_1\nu+\lambda\nu-\lambda\nu_1+\nu+1)\)
当\(\lambda_1=\nu_1=\mu_1=0\)时
\(\frac{s}{S}=\frac{(\lambda^2\mu^2\nu^2-2\lambda\mu\nu+1)}{(\lambda\mu+\lambda+1)(\mu\nu+\mu+1)(\lambda\nu+\nu+1)}\)
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\(k_1=\frac{3-\sqrt{5}}{2}=\phi^2,k_2=\frac{3+\sqrt{5}}{2}=\frac{1}{\phi^2}\)