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[讨论] 四面体中费马点计算

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 楼主| 发表于 2015-8-29 08:27:08 | 显示全部楼层
最终的简化方程组:

1.\(-a^2wx+m^2yz-w^2yz+wxy^2+wxz^2-x^2yz=0\)

2.\(-b^2wy+n^2xz-w^2xz+wx^2y+wyz^2-xy^2z=0\)

3.\(-c^2wz+p^2xy-w^2xy+wx^2z+wy^2z-xyz^2=0\)

4.\(-a^2x-b^2y-c^2z+x^2y+x^2z+xy^2+2xyz+xz^2+y^2z+yz^2=0\)

5.\(-m^2yz-n^2xz-p^2xy+w^2xy+w^2xz+w^2yz+2wxyz+x^2yz+xy^2z+xyz^2=0\)

注: 其中4与5是等价的,因此谁有能力对{1,2,3,4}或者{1,2,3,5} 消元{y,z,w}得到x的表达式?

我计算了{1,2,3,4} 10小时也没有终止~
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 2015-8-30 19:55:16 | 显示全部楼层
经过几天的尝试,利用消元方法得到x的代数方程不现实..

不过我依然无法确定: hujunhua  在6#描述:
/forum.php ... 27&fromuid=1455
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?

有趣的定理:

若\(F\)是四面体\(A_0A_1A_2A_3\)的Fermat 点,射线\(A_iF\)交四面体各面\(S_i\)于\(B_i ,i=0...3\),则有:

\(|FA_0|+|FA_1|+|FA_2|+|FA_3| \geqslant 3(|FB_0|+|FB_1|+|FB_2|+|FB_3|)\)


则我们有:若F是Fermat点,又是外心,则F点必为重心;若F点既是外心,又是重心,则F为Fermat点
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 2015-8-31 21:23:19 | 显示全部楼层
根据15#的立体角定义,我们可以得到:

令\(2\alpha=\angle{BFC}=\angle{AFD},2\beta=\angle{AFC}=\angle{BFD},2\gamma=\angle{AFB}=\angle{CFD},2s = \alpha + \beta + \gamma\)

则有与三角形费马点类似的性质:

1.在三角形中\(\triangle ABC\) 及费马点F有:\(\angle AFC =\angle BFC=\angle AFB=\frac{2\pi}{3}\)

2. 在四面体中\(D-ABC\),及费马点\(F\)有\(\Omega_{FABC}=\Omega_{FBCD}=\Omega_{FACD}=\Omega_{FABD}=\Omega\)

    且\(\tan \frac{\Omega}{4} = \sqrt{ \tan (s) \tan(s - \alpha) \tan ( s - \beta) \tan(s - \gamma)}\)

3.\(wV_{F-ABC}=zV_{F-ABD}=yV_{F-ACD}=xV_{F-BCD}=\frac{xyzw}{6}\sqrt{1-\cos(2\alpha)^2-\cos(2\beta)^2-\cos(2\gamma)^2+2\cos(2\alpha)\cos(2\beta)\cos(2\gamma)}\)

   且\(\cos(2\alpha)+\cos(2\beta)+\cos(2\gamma)=-1\)

4. 设F点到四面体各面的距离为\(d_0,d_1,d_2,d_3\),各个项点到对面的距离为\(h_0,h_1,h_2,h_3\),四面体内切球半径为\(r\)

    \(\frac{1}{r}=\frac{1}{h_0}+\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}\)

   则有\(\frac{\mu_0}{d_0}+\frac{\mu_1}{d_1}+\frac{\mu_2}{d_2}+\frac{\mu_3}{d_3}=\frac{3}{r}\)

   其中:\(\mu_0=\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}},\mu_1=\frac{\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)

          \(\mu_2=\frac{\frac{1}{z}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}},\mu_3=\frac{\frac{1}{w}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)
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 2015-9-1 18:22:20 | 显示全部楼层
为了描述一般性的结论:

我们记:令

\(\alpha_{00}=\angle{BFC},\alpha_{01}=\angle{AFD},\beta_{00}=\angle{AFC},\beta_{01}=\angle{BFD},\delta_{00}=\angle{AFB},\delta_{01}=\angle{CFD}\)

\(\alpha_0=\cos(\alpha{00}),\alpha_1=\cos(\alpha_{01}),\beta_0=\cos(\beta_{00}),\beta_1=\cos(\beta_{01}),\delta_0=\cos(\delta_{00}),\delta_1=\cos(\delta_{01})\)

\(\sin(\Omega_1)^2 = 2\alpha_1\beta_0\gamma_0-\alpha_1^2-\beta_0^2-\gamma_0^2+1, \sin(\Omega_2)^2 = 2\alpha_0\beta_1\gamma_0-\alpha_0^2-\beta_1^2-\gamma_0^2+1\)

\(\sin(\Omega_3)^2 = 2\alpha_0\beta_0\gamma_1-\alpha_0^2-\beta_0^2-\gamma_1^2+1, \sin(\Omega_0)^2 = 2\alpha_1\beta_1\gamma_1-\alpha_1^2-\beta_1^2-\gamma_1^2+1\)

注:对四面体内点与棱的张角有下列条件:

\(\alpha_0^2 \alpha_1^2-2 \alpha_0 \alpha_1 \beta_0 \beta_1-2 \alpha_0 \alpha_1 \delta_0 \delta_1+\beta_0^2 \beta_1^2-2 \beta_0 \beta_1 \delta_0 \delta_1+\delta_0^2 \delta_1^2+2 \alpha_0 \beta_0 \delta_0+2 \alpha_0 \beta_1 \delta_1+2 \alpha_1 \beta_0 \delta_1+2 \alpha_1 \beta_1 \delta_0-\alpha_0^2-\alpha_1^2-\beta_0^2-\beta_1^2-\delta_0^2-\delta_1^2+1=0\)
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 2015-9-1 19:28:34 | 显示全部楼层
对于已知的四面体D-ABC,及内点P,设\(AB=c,AC=b,BC=a,AD=a_1,BD=b_1,CD=c_1,AP=x,BP=y,CP=z,DP=w\)有:

\(x^n+y^n+z^n+w^n\)取极值条件?

\(\frac{x^{n-1}}{\sin(\Omega_1)}=\frac{y^{n-1}}{\sin(\Omega_2)}=\frac{z^{n-1}}{\sin(\Omega_3)}=\frac{w^{n-1}}{\sin(\Omega_0)}\)

对于\(n=1\)

我们易得:\(\sin(\Omega_0)=\sin(\Omega_1)=\sin(\Omega_2)=\sin(\Omega_3)\)

可化简为\(\alpha_0=\alpha_1,\beta_0=\beta_1,\delta_0=\delta_1\)

其它相关的计算可见:33#

对于\(n=2\),我们可以得到:

\(\frac{xyzw}{yzw\sin(\Omega_1)}=\frac{xyzw}{xzw\sin(\Omega_2)}=\frac{xyzw}{xyw\sin(\Omega_3)}=\frac{xyzw}{xyz\sin(\Omega_0)}\)

即\(V_{P-BCD}=V_{P-ACD}=V_{P-ABD}=V_{P-ABC}\)

即仅当\(P\)为四面体重心时,\(x^2+y^2+z^2+w^2\)取最小值.
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 2015-9-1 19:39:35 | 显示全部楼层
对于37#的逆问题:

即空间共点但不共面的四条线段AP,BP,CP,DP可绕P点旋转,且\(AP=x,BP=y,CP=z,DP=w\),

则\(a^n+b^n+c^n+a_1^n+b_1^n+c_1^n\)取极值的条件为:

\(yz\sin(\alpha_{00})a^{n-1}=xw\sin(\alpha_{01})a_1^{n-1}=xz\sin(\beta_{00})b^{n-1}=yw\sin(\beta_{01})b_1^{n-1}=xy\sin(\delta_{00})c^{n-1}=zw\sin(\delta_{01})c_1^{n-1}\)

且:\(a^2=y^2+z^2-2yz\alpha_0,a_1^2=x^2+w^2-2xw\alpha_1,b^2=x^2+z^2-2xz\beta_0,b_1^2=y^2+w^2-2yw\beta_1,c^2=x^2+y^2-2xy\delta_0,c_1^2=z^2+w^2-2zw\delta_1\)
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 2015-9-1 19:52:16 | 显示全部楼层
对于38#取极值条件可化为:

\[\frac{y^2z^2(1-\alpha_0^2)}{(y^2+z^2-2yz\alpha_0)^{2-n}}=\frac{x^2w^2(1-\alpha_1^2)}{(x^2+w^2-2xw\alpha_1)^{2-n}}=\frac{x^2z^2(1-\beta_0^2)}{(x^2+z^2-2xz\beta_0)^{2-n}}=\frac{y^2w^2(1-\beta_1^2)}{(y^2+w^2-2yw\beta_1)^{2-n}}=\frac{x^2y^2(1-\delta_0^2)}{(x^2+y^2-2yz\delta_0)^{2-n}}=\frac{z^2w^2(1-\delta_1^2)}{(z^2+w^2-2zw\delta_1)^{2-n}}\]

当\(n=1\)时,即下面问题:

/forum.php ... 76&fromuid=1455

\(-y^2z^2\alpha_0^2+2kyz\alpha_0+y^2z^2-ky^2-kz^2=0\)

\(-w^2x^2\alpha_1^2+2kwx\alpha_1+w^2x^2-kw^2-kx^2=0\)

\(-x^2z^2\beta_0^2+2kxz\beta_0+x^2z^2-kx^2-kz^2=0\)

\(-w^2y^2\beta_1^2+2kwy\beta_1+w^2y^2-kw^2-ky^2=0\)

\(-x^2y^2\delta_0^2+2kxy\delta_0+x^2y^2-kx^2-ky^2=0\)

\(-w^2z^2\delta_1^2+2kwz\delta_1+w^2z^2-kw^2-kz^2=0\)

\(\alpha_0^2 \alpha_1^2-2 \alpha_0 \alpha_1 \beta_0 \beta_1-2 \alpha_0 \alpha_1 \delta_0 \delta_1+\beta_0^2 \beta_1^2-2 \beta_0 \beta_1 \delta_0 \delta_1+\delta_0^2 \delta_1^2+2 \alpha_0 \beta_0 \delta_0+2 \alpha_0 \beta_1 \delta_1+2 \alpha_1 \beta_0 \delta_1+2 \alpha_1 \beta_1 \delta_0-\alpha_0^2-\alpha_1^2-\beta_0^2-\beta_1^2-\delta_0^2-\delta_1^2+1=0\)

例如:取\(x = 3, y = 4, z = 5, w = 6\)

易得:

\(k = 6.004260491,\alpha_0= -0.3887653204, \alpha_1 = -0.1930644712, \beta_0 = -0.1026243414, \beta_1 = -0.4713053258, \delta_0 = 0.4434097118, \delta_1= -0.5955346727\)

进一步解得:

\(a = 7.520014150, b = 6.089230677, c = 4.892424418, a_1 = 7.207657106, b_1= 8.638440579, c_1= 9.835246838\)
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 2015-9-1 20:12:10 | 显示全部楼层
对于\(n=2\)可以得到\(a^2+b^2+c^2+a_1^2+b_1^2+c_1^2\)的极值条件:

\(x^2 w^2 (1-\alpha_1^2)-k^2=0\)

\(x^2 z^2 (1-\beta_0^2)-k^2=0\)

\(y^2 w^2 (1-\beta_1^2)-k^2=0\)

\(y^2 x^2 (1-\delta_0^2)-k^2=0\)

\(y^2 z^2 (1-\alpha_0^2)-k^2=0\)

\(z^2 w^2 (1-\delta_1^2)-k^2=0\)

\(\alpha_0^2 \alpha_1^2-2 \alpha_0 \alpha_1 \beta_0 \beta_1-2 \alpha_0 \alpha_1 \delta_0 \delta_1+\beta_0^2 \beta_1^2-2 \beta_0 \beta_1 \delta_0 \delta_1+\delta_0^2 \delta_1^2+2 \alpha_0 \beta_0 \delta_0+2 \alpha_0 \beta_1 \delta_1+2 \alpha_1 \beta_0 \delta_1+2 \alpha_1 \beta_1 \delta_0-\alpha_0^2-\alpha_1^2-\beta_0^2-\beta_1^2-\delta_0^2-\delta_1^2+1=0\)

例如:取\(w = 6, x = 3, y = 4, z = 5\)

得:\(k = 10.59206549, \alpha_0 = 0.8482454665, \alpha_1 =0 .8085350045,\beta_0=0 .7080745364, \beta_1 =0 .8973418970, \delta_0 = 0.4699892770, \delta_1 =0 .9355973425\)

\(a = 2.658981467, b = 3.571799644, c = 3.704086576, a_1= 3.986569934, b_1= 2.987908563, c_1= 2.205483947\)

显然取极值条件为P点对各个棱张成的三角形面积相等.
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 2015-9-1 20:27:26 | 显示全部楼层
\(S_{PAB}+S_{PAC}+S_{PBC}+S_{PAD}+S_{PBD}+S_{PCD}\)取极值条件:

\(yz\alpha_0=xw\alpha_1=xz\beta_0=yw\beta_1=xy\delta_0=zw\delta_1\)

得到:

\(a =\sqrt{y^2+z^2-2k}, b =\sqrt{x^2+z^2-2k}, c =\sqrt{x^2+y^2-2k}, a_1 = \sqrt{w^2+x^2-2k}, b_1 = \sqrt{w^2+y^2-2k}, c_1= \sqrt{w^2+z^2-2k}\)

\(-3k^4+(2w^2+2x^2+2y^2+2z^2)k^3+(-w^2x^2-w^2y^2-w^2z^2-x^2y^2-x^2z^2-y^2z^2)k^2+x^2z^2y^2w^2=0\)


消元结果为:

\(-3a^8+(-4w^2-4x^2+8y^2+8z^2)a^6+(-4w^2x^2+8w^2y^2+8w^2z^2+8x^2y^2+8x^2z^2-6y^4-16y^2z^2-6z^4)a^4+(4(y^2+z^2))(2w^2x^2-w^2y^2-w^2z^2-x^2y^2-x^2z^2+2y^2z^2)a^2-(y-z)^2(z+y)^2(2wx-y^2-z^2)(2wx+y^2+z^2)=0\)

\(-3b^8-4b^6w^2+8b^6x^2-4b^6y^2+8b^6z^2+8b^4w^2x^2-4b^4w^2y^2+8b^4w^2z^2-6b^4x^4+8b^4x^2y^2-16b^4x^2z^2+8b^4y^2z^2-6b^4z^4-4b^2w^2x^4+8b^2w^2x^2y^2-8b^2w^2x^2z^2+8b^2w^2y^2z^2-4b^2w^2z^4-4b^2x^4y^2+8b^2x^4z^2-8b^2x^2y^2z^2+8b^2x^2z^4-4b^2y^2z^4-4w^2x^4y^2+8w^2x^2y^2z^2-4w^2y^2z^4+x^8-2x^4z^4+z^8=0\)

\(-3c^8-4c^6w^2+8c^6x^2+8c^6y^2-4c^6z^2+8c^4w^2x^2+8c^4w^2y^2-4c^4w^2z^2-6c^4x^4-16c^4x^2y^2+8c^4x^2z^2-6c^4y^4+8c^4y^2z^2-4c^2w^2x^4-8c^2w^2x^2y^2+8c^2w^2x^2z^2-4c^2w^2y^4+8c^2w^2y^2z^2+8c^2x^4y^2-4c^2x^4z^2+8c^2x^2y^4-8c^2x^2y^2z^2-4c^2y^4z^2-4w^2x^4z^2+8w^2x^2y^2z^2-4w^2y^4z^2+x^8-2x^4y^4+y^8=0\)

\(w^8-2w^4x^4+8w^4x^2a_1^2-4w^4y^2z^2-4w^4y^2a_1^2-4w^4z^2a_1^2-6w^4a_1^4+8w^2x^4a_1^2+8w^2x^2y^2z^2-8w^2x^2y^2a_1^2-8w^2x^2z^2a_1^2-16w^2x^2a_1^4+8w^2y^2z^2a_1^2+8w^2y^2a_1^4+8w^2z^2a_1^4+8w^2a_1^6+x^8-4x^4y^2z^2-4x^4y^2a_1^2-4x^4z^2a_1^2-6x^4a_1^4+8x^2y^2z^2a_1^2+8x^2y^2a_1^4+8x^2z^2a_1^4+8x^2a_1^6-4y^2z^2a_1^4-4y^2a_1^6-4z^2a_1^6-3a_1^8=0\)

\(w^8-4w^4x^2z^2-4w^4x^2b_1^2-2w^4y^4+8w^4y^2b_1^2-4w^4z^2b_1^2-6w^4b_1^4+8w^2x^2y^2z^2-8w^2x^2y^2b_1^2+8w^2x^2z^2b_1^2+8w^2x^2b_1^4+8w^2y^4b_1^2-8w^2y^2z^2b_1^2-16w^2y^2b_1^4+8w^2z^2b_1^4+8w^2b_1^6-4x^2y^4z^2-4x^2y^4b_1^2+8x^2y^2z^2b_1^2+8x^2y^2b_1^4-4x^2z^2b_1^4-4x^2b_1^6+y^8-4y^4z^2b_1^2-6y^4b_1^4+8y^2z^2b_1^4+8y^2b_1^6-4z^2b_1^6-3b_1^8=0\)

\(-3c_1^8+(8w^2-4x^2-4y^2+8z^2)c_1^6+(-6w^4+8w^2x^2+8w^2y^2-16w^2z^2-4x^2y^2+8x^2z^2+8y^2z^2-6z^4)c_1^4-(4(w^2+z^2))(w^2x^2+w^2y^2-2w^2z^2-2x^2y^2+x^2z^2+y^2z^2)c_1^2+(w-z)^2(w+z)^2(w^2-2xy+z^2)(w^2+2xy+z^2)=0\)

显然此时为垂心四面体,且\(a^2+a_1^2=b^2+b_1^2=c^2+c_1^2=x^2+y^2+z^2+w^2-4k\)

由下面知:此时D-ABC所形成的体积也是最大的.

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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
 楼主| 发表于 2015-9-5 09:58:39 | 显示全部楼层
对于已知四面体D-ABC \(a,b,c,a_1,b_1,c_1\) 内的一点\(P\):,设\(AP=x,BP=y,CP=z,DP=w\)

\(S_{PAC}+S_{PAB}+S_{PBC}+S_{PAD}+S_{PBD}+S_{PCD}\)取极值条件?


\(\frac{1}{4}(\sqrt{-(a^4+y^4+z^4-2a^2y^2-2a^2z^2-2y^2z^2)}+\sqrt{-(x^4+b^4+z^4-2x^2b^2-2b^2z^2-2x^2z^2}+\sqrt{x^4+y^4+c^4-2c^2x^2-2c^2y^2-2x^2y^2)}+

\sqrt{-(a_1^4+x^4+w^4-2a_1^2x^2-2a_1^2w^2-2x^2w^2)}+\sqrt{-(b_1^4+y^4+w^4-2b_1^2y^2-2b_1^2w^2-2y^2w^2)}+\sqrt{-(c_1^4+z^4+w^4-2c_1^2z^2-2c_1^2w^2-2x^2w^2)})\)

利用拉格朗日乘子法可以得到:

设\(S_{PBC}=S_1,S_{PAC}=S_2,S_{PBC}=S_3,S_{PAD}=S_{11},S_{PBD}=S_{12},S_{PCD}=S_{13}\)

\(\frac{x^2-y^2-c^2}{S_3}+\frac{x^2-z^2-b^2}{S_2}+\frac{x^2-a_1^2-w^2}{S_{11}}=0\)       ................(1)

\(\frac{y^2-z^2-a^2}{S_1}+\frac{y^2-x^2-c^2}{S_3}+\frac{y^2-b_1^2-w^2}{S_{12}}=0\)       ................(2)

\(\frac{z^2-y^2-a^2}{S_1}+\frac{z^2-x^2-b^2}{S_2}+\frac{z^2-c_1^2-w^2}{S_{13}}=0\)       ................(4)

\(\frac{w^2-x^2-a_1^2}{S_{11}}+\frac{w^2-y^2-b_1^2}{S_{12}}+\frac{w^2-z^2-c_1^2}{S_{13}}=0\)       ................(5)

(1)+(2)+(3)+(4)+(5)得:

\(-2(\frac{a^2}{S_1}+\frac{b^2}{S_2}+\frac{c^2}{S_3}+\frac{a_1^2}{S_{11}}+\frac{b_1^2}{S_{12}}+\frac{c_1^2}{S_{13}})=0\) ...............(6)

对于已知四面体显然上式(6)不可能成立.是否不存在\(S_{PAC}+S_{PAB}+S_{PBC}+S_{PAD}+S_{PBD}+S_{PCD}\)的极值
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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