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[讨论] 四面体中费马点计算

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发表于 2014-8-27 10:01:12 | 显示全部楼层
看到17#方程组,我猛然觉醒,发现自己又犯錯了。

1、不是任意的四面体都存在等角中心,即使四面体的四个立体角都小于 `\pi`。从12#的方程组中消去{x,y,z,w}, 应可得到存在等角中心的四面体的六棱长约束条件。
      不幸的是,我用Mathematica10试了一下,一直Runing, 不出结果。
2、对于存在等角中心的四面体,等角中心应该既是1#所要的费马点,即到四顶点距离之和最小的点,也是6#所说的6个三角形面积之和最小的点。
3、对于不存在等角中心的四面体,6楼所说的6个三角形面积之和最小的点的几何特征需要另行确定,但从该点所张的6个三角形之结构应该不是肥皂膜实验的结果(最小曲面)。
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 2014-8-27 19:52:36 | 显示全部楼层
关于\(w\),重新消元计算得到:


\(27(3a^2-3b_1^2+2b_1c_1-3c_1^2)^2(3a^2-3b_1^2-2b_1c_1-3c_1^2)^2+(12960a^6-40608a^4b_1^2-40608a^4c_1^2+42336a^2b_1^4+69696a^2b_1^2c_1^2+42336a^2c_1^4-14688b_1^6-31392b_1^4c_1^2-31392b_1^2c_1^4-14688c_1^6)w^2+(33408a^4-63744a^2b_1^2-63744a^2c_1^2+33408b_1^4+46848b_1^2c_1^2+33408c_1^4)w^4+(34816a^2-30720b_1^2-30720c_1^2)w^6+12288w^8=0\)

\(27(3a_1^2+2a_1c_1-3b^2+3c_1^2)^2(3a_1^2-2a_1c_1-3b^2+3c_1^2)^2+(-14688a_1^6+42336a_1^4b^2-31392a_1^4c_1^2-40608a_1^2b^4+69696a_1^2b^2c_1^2-31392a_1^2c_1^4+12960b^6-40608b^4c_1^2+42336b^2c_1^4-14688c_1^6)w^2+(33408a_1^4-63744a_1^2b^2+46848a_1^2c_1^2+33408b^4-63744b^2c_1^2+33408c_1^4)w^4+(-30720a_1^2+34816b^2-30720c_1^2)w^6+12288w^8=0\)

\(27(3a_1^2-2a_1b_1+3b_1^2-3c^2)^2(3a_1^2+2a_1b_1+3b_1^2-3c^2)^2+(-14688a_1^6-31392a_1^4b_1^2+42336a_1^4c^2-31392a_1^2b_1^4+69696a_1^2b_1^2c^2-40608a_1^2c^4-14688b_1^6+42336b_1^4c^2-40608b_1^2c^4+12960c^6)w^2+(33408a_1^4+46848a_1^2b_1^2-63744a_1^2c^2+33408b_1^4-63744b_1^2c^2+33408c^4)w^4+(-30720a_1^2-30720b_1^2+34816c^2)w^6+12288w^8=0\)



然后将上面两两消元,得到独立的约束条件(存在费马点的条件)

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复制代码
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 2014-8-27 20:01:47 | 显示全部楼层
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:

费马点.jpg

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?
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 2014-8-27 20:17:44 | 显示全部楼层
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?

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 2014-8-28 09:57:22 | 显示全部楼层
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:

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

    \(768(n-t)^4(n+t)^4(m-s)^4(m+s)^4=0\)
毋因群疑而阻独见  毋任己意而废人言
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发表于 2014-8-28 15:58:52 | 显示全部楼层
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。

点评

是啊,以上得到的两个约束条件需要同时满足哈。  发表于 2014-8-28 19:26
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 2014-8-29 02:18:56 | 显示全部楼层
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。

点评

23#的计算结果刚好能解释你的说法,说明四面体中的费马点并不常见……  发表于 2014-8-29 12:53
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 2014-8-29 20:02:50 | 显示全部楼层
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

点评

你理解错了,找两点 这两点与一个棱共面 同时跟对棱共面  发表于 2014-8-30 08:39
点错支持了。 某点与一三角形顶点连线段都成120度,就固定在平面内,不可能与平面外第4点都成120度。  发表于 2014-8-29 20:23
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 2015-8-10 19:43:50 | 显示全部楼层
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\(x,y,z,w\)
毋因群疑而阻独见  毋任己意而废人言
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 楼主| 发表于 2015-8-10 19:58:17 | 显示全部楼层
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:

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