四面体中费马点计算

数学星空

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~

数学星空

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:
http://bbs.emath.ac.cn/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

数学星空

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

          \(\mu_2=\frac{\frac{1}{z}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}},\mu_3=\frac{\frac{1}{w}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)

数学星空

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:

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

数学星空

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.

数学星空

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:\(a^2=y^2+z^2-2yz\alpha_0,a_1^2=x^2+w^2-2xw\alpha_1,b^2=x^2+z^2-2xz\beta_0,b_1^2=y^2+w^2-2yw\beta_1,c^2=x^2+y^2-2xy\delta_0,c_1^2=z^2+w^2-2zw\delta_1\)

数学星空

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:

http://bbs.emath.ac.cn/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:

\(a = 7.520014150, b = 6.089230677, c = 4.892424418, a_1 = 7.207657106, b_1= 8.638440579, c_1= 9.835246838\)

数学星空

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.

数学星空

\(S_{PAB}+S_{PAC}+S_{PBC}+S_{PAD}+S_{PBD}+S_{PCD}\)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.

http://bbs.emath.ac.cn/forum.php ... 75&fromuid=1455

数学星空

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