四面体中费马点计算

数学星空

更难以接受的是：

将\(\{a = 1, b = x, c = y, a1 = 1, b1 = x, c1 = y\}\) 代入上面两个约束式：

\(262144(x-1)^6(x+1)^6(3x^4-2x^2y^2+3y^4-6x^2-2y^2+3)＝0\)

\(262144(y-1)^6(y+1)^6(3x^4-2x^2y^2+3y^4-2x^2-6y^2+3)＝0\)

其中\(3x^4-2x^2y^2+3y^4-2x^2-6y^2+3=0\)的函数曲线为：



我们需要确认的是上面两个条件约束式是否正确？能否进一步简化？

数学星空

当然我们也可以利用下面的体积公式：

\[V=\frac{abc}{6}\sqrt{1-\cos(\alpha)^2-\cos(\beta)^2-\cos(\gamma)^2+2\cos(\alpha)\cos(\beta)\cos(\gamma)}\]

对四个四面体\(F-ABC,F-ABD,F-BCD,F-ACD\),使用上面的体积公式可以得到：

\[V=V_1+V_2+V_3+V_4=\frac{2xyzw}{9\sqrt{3}}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w})\]

\[12V=\sqrt
{-a^4a_1^2-a^2a_1^4+a^2a_1^2b^2+a^2a_1^2b_1^2+a^2a_1^2c^2+a^2a_1^2c_1^2+a^2b^2b_1^2-a^2b^2c^2-a^2b_1^2c_1^2+a^2c^2c_1^2+a_1^2b^2b_1^2-a_1^2b^2c_1^2-a_1^2b_1^2c^2+a_1^2c^2c_1^2-b^4b_1^2-b^2b_1^4+b^2b_1^2c^2+b^2b_1^2c_1^2+b^2c^2c_1^2+b_1^2c^2c_1^2-c^4c_1^2-c^2c_1^4}\]


参与上面的消元计算，或许得到的结论要简洁一些？

数学星空

对于楼上得到的两个约束条件：

1.对于正三棱锥，若令\(\{a = x, b = x, c = x, a_1 = y, b_1 = y, c_1 = y\}\),则恒为\(0\)


2.对于等四面体，若令\(\{a = x, b = y, c = z, a_1 = x, b_1 = y, c_1 = z\}\),则简化为：

    \(262144(x-y)^6(x+y)^6(3x^4-6x^2y^2-2x^2z^2+3y^4-2y^2z^2+3z^4)＝0\)

    \(262144(x-z)^6(x+z)^6(3x^4-2x^2y^2-6x^2z^2+3y^4-2y^2z^2+3z^4)＝0\)


3.对于垂心四面体，若令\(a = \sqrt{m^2+n^2}, a_1 =\sqrt{s^2+t^2}, b =\sqrt{m^2+s^2}, b_1 =\sqrt{n^2+t^2}, c = \sqrt{m^2+t^2}, c_1 =\sqrt{n^2+s^2}\),则简化为：

    \(768(n-s)^4(n+s)^4(m-t)^4(m+t)^4=0\)

    \(768(n-t)^4(n+t)^4(m-s)^4(m+s)^4=0\)

hujunhua

从17#的方程组中任选5个消去{x,y,z,w}, 都可以得到一个5棱长约束，共计6个五棱长约束。
这6个五棱长约束不是或的关系，而是与的关系。

hujunhua

计算了一下，不存在非正四面体的等面四面体具有等角中心，垂心四面体具有等角中心的只有正三棱锥。

倪举鹏

找两点，这两点所在的直线会通过四面体两个对棱，这线段与，这点与两个顶点连线段都成120度

数学星空

其实2#~3#已给出了求解方程:

对于四面体\(D-ABC\)的费马点\(F\),有\[\angle BFC=\angle AFD=\alpha,\angle AFC=\angle BFD=\beta,\angle AFB=\angle CFD=\gamma\]并且每对相等的角被两者所在平面的交线平分。即：
\[\left\{\begin{split}\frac{y^2+z^2-a^2}{2yz}=\frac{x^2+w^2-a_1^2}{2xw}=\cos(\alpha)\\ \frac{x^2+z^2-b^2}{2xz}=\frac{y^2+w^2-b_1^2}{2yw}=\cos(\beta)\\ \frac{x^2+y^2-c^2}{2xy}=\frac{z^2+w^2-c_1^2}{2zw}=\cos(\gamma)\end{split}\right.\]

\[\begin{vmatrix}
0 & c^2 & b^2 & a_1^2 & x^2 & 1\\
c^2 &  0 &  a^2 & b_1^2 & y^2 & 1\\
b^2 &  a^2 & 0 &  c_1^2 &  z^2 & 1\\
a_1^2 &  b_1^2 &  c_1^2 &  0 & w^2 & 1\\
x^2 &  y^2 &  z^2 &  w^2 &  0 & 1\\
1 & 1 &  1 & 1 & 1&  0
\end{vmatrix}=0\]

通过四个方程及\(a,b,c,a_1,b_1,c_1\) 可能求解四个未知数\(x,y,z,w\)

数学星空

对于特殊四面体:  \(a=a_1,b=b_1,c=c_1\) 费马点即为四面体的外心.

对于一般的四面体,我可以求解楼上的方程算出费马点

例如: 取\(a=3,b=4,c=5,a_1=6,b_1=6,c_1=6\)

为了方便输入:记\(\cos(\alpha)=k_1,\cos(\beta)=k_2,\cos(\gamma)=k_3\)

\(-a^4a_1^4+2a^4a_1^2w^2+2a^4a_1^2x^2-a^4w^4+2a^4w^2x^2-a^4x^4+2a^2a_1^4y^2+2a^2a_1^4z^2+2a^2a_1^2b^2b_1^2-2a^2a_1^2b^2w^2-2a^2a_1^2b^2y^2-2a^2a_1^2b_1^2x^2-2a^2a_1^2b_1^2z^2+2a^2a_1^2c^2c_1^2-2a^2a_1^2c^2w^2-2a^2a_1^2c^2z^2-2a^2a_1^2c_1^2x^2-2a^2a_1^2c_1^2y^2+4a^2a_1^2w^2x^2-2a^2a_1^2w^2y^2-2a^2a_1^2w^2z^2-2a^2a_1^2x^2y^2-2a^2a_1^2x^2z^2+4a^2a_1^2y^2z^2-2a^2b^2b_1^2w^2-2a^2b^2b_1^2x^2+4a^2b^2c^2w^2+2a^2b^2w^4-2a^2b^2w^2x^2-2a^2b^2w^2y^2+2a^2b^2x^2y^2+4a^2b_1^2c_1^2x^2-2a^2b_1^2w^2x^2+2a^2b_1^2w^2z^2+2a^2b_1^2x^4-2a^2b_1^2x^2z^2-2a^2c^2c_1^2w^2-2a^2c^2c_1^2x^2+2a^2c^2w^4-2a^2c^2w^2x^2-2a^2c^2w^2z^2+2a^2c^2x^2z^2-2a^2c_1^2w^2x^2+2a^2c_1^2w^2y^2+2a^2c_1^2x^4-2a^2c_1^2x^2y^2-a_1^4y^4+2a_1^4y^2z^2-a_1^4z^4-2a_1^2b^2b_1^2y^2-2a_1^2b^2b_1^2z^2+4a_1^2b^2c_1^2y^2-2a_1^2b^2w^2y^2+2a_1^2b^2w^2z^2+2a_1^2b^2y^4-2a_1^2b^2y^2z^2+4a_1^2b_1^2c^2z^2+2a_1^2b_1^2x^2y^2-2a_1^2b_1^2x^2z^2-2a_1^2b_1^2y^2z^2+2a_1^2b_1^2z^4-2a_1^2c^2c_1^2y^2-2a_1^2c^2c_1^2z^2+2a_1^2c^2w^2y^2-2a_1^2c^2w^2z^2-2a_1^2c^2y^2z^2+2a_1^2c^2z^4-2a_1^2c_1^2x^2y^2+2a_1^2c_1^2x^2z^2+2a_1^2c_1^2y^4-2a_1^2c_1^2y^2z^2-b^4b_1^4+2b^4b_1^2w^2+2b^4b_1^2y^2-b^4w^4+2b^4w^2y^2-b^4y^4+2b^2b_1^4x^2+2b^2b_1^4z^2+2b^2b_1^2c^2c_1^2-2b^2b_1^2c^2w^2-2b^2b_1^2c^2z^2-2b^2b_1^2c_1^2x^2-2b^2b_1^2c_1^2y^2-2b^2b_1^2w^2x^2+4b^2b_1^2w^2y^2-2b^2b_1^2w^2z^2-2b^2b_1^2x^2y^2+4b^2b_1^2x^2z^2-2b^2b_1^2y^2z^2-2b^2c^2c_1^2w^2-2b^2c^2c_1^2y^2+2b^2c^2w^4-2b^2c^2w^2y^2-2b^2c^2w^2z^2+2b^2c^2y^2z^2+2b^2c_1^2w^2x^2-2b^2c_1^2w^2y^2-2b^2c_1^2x^2y^2+2b^2c_1^2y^4-b_1^4x^4+2b_1^4x^2z^2-b_1^4z^4-2b_1^2c^2c_1^2x^2-2b_1^2c^2c_1^2z^2+2b_1^2c^2w^2x^2-2b_1^2c^2w^2z^2-2b_1^2c^2x^2z^2+2b_1^2c^2z^4+2b_1^2c_1^2x^4-2b_1^2c_1^2x^2y^2-2b_1^2c_1^2x^2z^2+2b_1^2c_1^2y^2z^2-c^4c_1^4+2c^4c_1^2w^2+2c^4c_1^2z^2-c^4w^4+2c^4w^2z^2-c^4z^4+2c^2c_1^4x^2+2c^2c_1^4y^2-2c^2c_1^2w^2x^2-2c^2c_1^2w^2y^2+4c^2c_1^2w^2z^2+4c^2c_1^2x^2y^2-2c^2c_1^2x^2z^2-2c^2c_1^2y^2z^2-c_1^4x^4+2c_1^4x^2y^2-c_1^4y^4=0\)

\(-2k_1yz-a^2+y^2+z^2=0\)
\(-2k_1wx-a_1^2+w^2+x^2=0\)
\(-2k_2xz-b^2+x^2+z^2=0\)
\(-2k_2wy-b_1^2+w^2+y^2=0\)
\(-2k_3xy-c_1^2+x^2+y^2=0\)
\(-2k_3wz-c_1^2+w^2+z^2=0\)

代入得到:

\(576w^4-576w^2x^2-576w^2y^2+1215x^4+288x^2y^2-2142x^2z^2+2048y^4-3808y^2z^2+2975z^4-27072w^2-20736x^2-20736y^2+746496=0\)

\(-2k_1yz+y^2+z^2-9=0\)
\(-2k_1wx+w^2+x^2-36=0\)
\(-2k_2xz+x^2+z^2-16=0\)
\(-2k_2wy+w^2+y^2-36=0\)
\(-2k_3xy+x^2+y^2-25=0\)
\(-2k_3wz+w^2+z^2-36=0\)

求解得到:

\(k_1 = -0.08125763422, k_2 = -0.3028690928, k_3 = -0.6158732730, w = 4.838906457, x = 3.176057999, y = 2.372782500, z = 1.653020392,\alpha = 94.660858238723315798^{\circ}, \beta = 107.63000925585309530^{\circ},\delta = 128.01540128977782951^{\circ}\)

数学星空

其实楼上的方程可约化为下面方程:

\(-\alpha yz-a^2+y^2+z^2=0\)

\(-\alpha wx-a_1^2+w^2+x^2=0\)

\(-\beta xz-b^2+x^2+z^2=0\)

\(-\beta wy-b_1^2+w^2+y^2=0\)

\(-\gamma xy-c^2+x^2+y^2=0\)

\(-\gamma wz-c_1^2+w^2+z^2=0\)

\(\alpha+\beta+\gamma+2=0\)

谁有兴趣消元\(\alpha,\beta,\gamma,y,z,w\) ?

creasson

令\[P = \alpha A + \beta B + \gamma C + \delta D\]
于是有五式\[\left\{ \begin{array}{l}
\alpha  + \beta  + \gamma  + \delta  = 1  \\
P{A^2} = \beta A{B^2} + \gamma A{C^2} + \delta A{D^2} - \left\{ {\alpha \beta A{B^2} + \alpha \gamma A{C^2} + \alpha \delta A{D^2} + \beta \gamma B{C^2} + \beta \delta B{D^2} + \gamma \delta C{D^2}} \right\} \\
P{B^2} = \alpha A{B^2} + \gamma B{C^2} + \delta B{D^2} - \left\{ {\alpha \beta A{B^2} + \alpha \gamma A{C^2} + \alpha \delta A{D^2} + \beta \gamma B{C^2} + \beta \delta B{D^2} + \gamma \delta C{D^2}} \right\} \\
P{C^2} = \alpha A{C^2} + \beta B{C^2} + \delta C{D^2} - \left\{ {\alpha \beta A{B^2} + \alpha \gamma A{C^2} + \alpha \delta A{D^2} + \beta \gamma B{C^2} + \beta \delta B{D^2} + \gamma \delta C{D^2}} \right\} \\
P{D^2} = \alpha A{D^2} + \beta B{D^2} + \gamma C{D^2} - \left\{ {\alpha \beta A{B^2} + \alpha \gamma A{C^2} + \alpha \delta A{D^2} + \beta \gamma B{C^2} + \beta \delta B{D^2} + \gamma \delta C{D^2}} \right\} \\
\end{array} \right.\]
即为求\[L =  PA + PB + PC + PD\]在上述条件下的极值，余下略.