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楼主: 数学星空

[讨论] 四面体中费马点计算

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发表于 2014-8-26 15:17:55 | 显示全部楼层
四面体的等角中心与四面体的六条棱所张的六个三角形的面积之和最小?
前提,要等角中心在四面体内。

另 六个三角形的面积之和最小保证能等角中心是费马点?

点评

前提,要等角中心在四面体内. 谢谢补充。  发表于 2014-8-26 16:44
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 2014-8-26 20:54:52 | 显示全部楼层
http://zh.wikipedia.org/wiki/%E7%AB%8B%E4%BD%93%E8%A7%92 中提到了立体角的概念及四面体立体角计算公式:

        立体角,常用字母\(\Omega\)表示,是一个物体对特定点的三维空间的角度,是平面角在三维空间中的类比。它描述的是站在某一点的观察者测量到的物体大小的尺度。例如,对于一个特定的观察点,一个在该观察点附近的小物体有可能和一个远处的大物体有着相同的立体角。立体角的定义是,物体在一个以观测点为球心的球的投影面积与球半径平方值的比: \[\Omega=\frac{S}{r^{2}} = \iint_S \frac { \vec{r} \cdot \textrm{d}\vec{S}}{r^3}\]这和“平面角是圆的弧长与半径的比”类似。

       立体角的国际制单位是球面度(steradian,sr),一个完整的球面对于球内任意一点的立体角为\(4\pi  sr\)(对于球外任意一点的立体角为\(0  sr\))。立体角有一个非国际制单位平方度,\(1 sr =(\frac{180}{2\pi})^2\)  square degree。

在球坐标系中 \( \dif{\Omega}= \sin(\theta)\dif{\theta}\dif{\varphi}\)

     任意四面体的立体角公式

      对于任意一个四面体\(OABC\),其中\(O,A,B,C\)分别为四面体的四个顶点。令\(2\alpha=\angle{BOC},2\beta=\angle{AOC},2\gamma=\angle{AOB}\), \( 2s = \alpha + \beta + \gamma\),
那么从\(O\)点观察三角形ABC的立体角 \(\Omega\) 的公式如下:\[\tan \frac{\Omega}{4} = \sqrt{ \tan (s) \tan(s - \alpha) \tan ( s - \beta) \tan(s - \gamma)}\]
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 2014-8-26 21:17:32 | 显示全部楼层
若按照12# hujunhua 大哥的描述应该有:

\(\Omega=\pi\)
\(\D\alpha=\beta=\gamma=\frac12\arccos(-\frac13)\)
\(\D s=\frac32\alpha=\frac34\arccos(-\frac13)\), \(\tan(s)=\sqrt{26+15\sqrt3}=\left(\sqrt{2+\sqrt3}\right)^3\)   (*直接用Mathematica10算得*)
\(\D s-\alpha=s-\beta=s-\gamma=\frac14\arccos(-\frac13)\), \(\tan(s-\alpha)=\sqrt{2-\sqrt3}\)   (*直接用Mathematica10算得*)
代入楼上的立体角计算公式有:
左边=\(\D\tan(\frac{\Omega}{4})=\tan(\frac{\pi}{4})=1\)
右边=\(\D\sqrt{\tan(s)\tan^3(s-\alpha)}=\left(\left(2+\sqrt3\right)\left(2-\sqrt3\right)\right)^{3/2}=1\)
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 2014-8-26 21:42:08 | 显示全部楼层
根据12#的结论:

1、每个三角形的顶角(以`P`为顶点)都等于`\arccos(-1/3)`, 约`109\degree28'16''`, 由此可以得到6个方程:
\begin{equation*}\left\{\begin{split}x^2+y^2+\frac{2xy}{3}&=c^2\\x^2+z^2+\frac{2xz}{3}&=b^2\\y^2+z^2+\frac{2yz}{3}&=a^2\\x^2+w^2+\frac{2xw}{3}&=a_1^2\\y^2+w^2+\frac{2yw}{3}&=b_1^2\\z^2+w^2+\frac{2zw}{3}&=c_1^2\end{split}\right.\end{equation*}
现在问题是6个方程,4个参数\(\{x,y,z,w\}\),说明还有两个独立的约束恒等式?

2、每2个不共边的三角形(称为相对三角形,共3对)互相垂直平分

等面四面体的三双对棱中点的连线两两垂直平分,还存在哪些四面体满足同样条件?
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 2014-8-26 22:30:00 | 显示全部楼层
对于楼上消元的结果:

\(12288x^8+(34816a^2-30720b^2-30720c^2)x^6+(33408a^4-63744a^2b^2-63744a^2c^2+33408b^4+46848b^2c^2+33408c^4)x^4+(12960a^6-40608a^4b^2-40608a^4c^2+42336a^2b^4+69696a^2b^2c^2+42336a^2c^4-14688b^6-31392b^4c^2-31392b^2c^4-14688c^6)x^2+2187a^8-8748a^6b^2-8748a^6c^2+13122a^4b^4+24300a^4b^2c^2+13122a^4c^4-8748a^2b^6-22356a^2b^4c^2-22356a^2b^2c^4-8748a^2c^6+2187b^8+6804b^6c^2+9666b^4c^4+6804b^2c^6+2187c^8=0\)



\(12288y^8+(-30720a^2+34816b^2-30720c^2)y^6+(33408a^4-63744a^2b^2+46848a^2c^2+33408b^4-63744b^2c^2+33408c^4)y^4+(-14688a^6+42336a^4b^2-31392a^4c^2-40608a^2b^4+69696a^2b^2c^2-31392a^2c^4+12960b^6-40608b^4c^2+42336b^2c^4-14688c^6)y^2+2187a^8-8748a^6b^2+6804a^6c^2+13122a^4b^4-22356a^4b^2c^2+9666a^4c^4-8748a^2b^6+24300a^2b^4c^2-22356a^2b^2c^4+6804a^2c^6+2187b^8-8748b^6c^2+13122b^4c^4-8748b^2c^6+2187c^8=0\)



\(12288z^8+(-30720a^2-30720b^2+34816c^2)z^6+(33408a^4+46848a^2b^2-63744a^2c^2+33408b^4-63744b^2c^2+33408c^4)z^4+(-14688a^6-31392a^4b^2+42336a^4c^2-31392a^2b^4+69696a^2b^2c^2-40608a^2c^4-14688b^6+42336b^4c^2-40608b^2c^4+12960c^6)z^2+2187a^8+6804a^6b^2-8748a^6c^2+9666a^4b^4-22356a^4b^2c^2+13122a^4c^4+6804a^2b^6-22356a^2b^4c^2+24300a^2b^2c^4-8748a^2c^6+2187b^8-8748b^6c^2+13122b^4c^4-8748b^2c^6+2187c^8=0\)



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  2. 42336a^2a1^2c^4-8748a^2b^6-22356a^2b^4c^2-22356a^2b^2c^4-8748a^2c^6+12288a1^8-30720a1^6b^2-30720a1^6c^2+33408a1^4b^4+46848a1^4b^2c^2+33408a1^4c^4-14688a1^2b^6-31392a1^2b^4c^2-31392a1^2b^2c^4-14688a1^2c^6+2187b^8+
  3. 6804b^6c^2+9666b^4c^4+6804b^2c^6+2187c^8)^2+(-3571283520a^14-40361172480a^12a1^2+25475155776a^12b^2+25475155776a^12c^2-207170555904a^10a1^4+248213652480a^10a1^2b^2+248213652480a^10a1^2c^2-77853980736a^10b^4-
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  15. 826326282240a1^4b^8c^2+1436507136000a1^4b^6c^4+1436507136000a1^4b^4c^6+826326282240a1^4b^2c^8+231697428480a1^4c^10-48100843008a1^2b^12-208749394944a1^2b^10c^2-437688893952a1^2b^8c^4-552527972352a1^2b^6c^6-
  16. 437688893952a1^2b^4c^8-208749394944a1^2b^2c^10-48100843008a1^2c^12+4047454656b^14+21242523456b^12c^2+53451679680b^10c^4+81784842048b^8c^6+81784842048b^6c^8+53451679680b^4c^10+
  17. 21242523456b^2c^12+4047454656c^14)w^2+(16089041664a^12+174223208448a^10a1^2-99933792768a^10b^2-99933792768a^10c^2+781580648448a^8a1^4-879124451328a^8a1^2b^2-879124451328a^8a1^2c^2+258803631360a^8b^4+
  18. 482471935488a^8b^2c^2+258803631360a^8c^4+1783670243328a^6a1^6-3030502146048a^6a1^4b^2-3030502146048a^6a1^4c^2+1788550668288a^6a1^2b^4+3267299524608a^6a1^2b^2c^2+1788550668288a^6a1^2c^4-357657431040a^6b^6-
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  20. 1835335065600a^4a1^2b^6-4612702298112a^4a1^2b^4c^2-4612702298112a^4a1^2b^2c^4-1835335065600a^4a1^2c^6+278152807680a^4b^8+904785251328a^4b^6c^2+1273242115584a^4b^4c^4+904785251328a^4b^2c^6+278152807680a^4c^8+
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  36. 1287118651392a^2a1^2c^2-355255713792a^2b^4-636369764352a^2b^2c^2-355255713792a^2c^4-684913065984a1^6+1080368824320a1^4b^2+1080368824320a1^4c^2-588370673664a1^2b^4-1099205443584a1^2b^2c^2-588370673664a1^2c^4+
  37. 116117471232b^6+282446266368b^4c^2+282446266368b^2c^4+116117471232c^6)w^10+(112142057472a^4+392737849344a^2a1^2-199898431488a^2b^2-199898431488a^2c^2+342456532992a1^4-346533396480a1^2b^2-
  38. 346533396480a1^2c^2+90398785536b^4+172455100416b^2c^2+90398785536c^4)w^12+(-53905195008a^2-97844723712a1^2+47563407360b^2+47563407360c^2)w^14+12230590464w^16=0
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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 2014-8-27 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:

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

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