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\(\alpha,\beta,\gamma,y,z,w\) ?
令\
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\在上述条件下的极值，余下略.
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~
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?mod=redirect&goto=findpost&ptid=5795&pid=55427&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
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:\(\mu_0=\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}},\mu_1=\frac{\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)
\(\mu_2=\frac{\frac{1}{z}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}},\mu_3=\frac{\frac{1}{w}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)
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:
\(\alpha_0^2 \alpha_1^2-2 \alpha_0 \alpha_1 \beta_0 \beta_1-2 \alpha_0 \alpha_1 \delta_0 \delta_1+\beta_0^2 \beta_1^2-2 \beta_0 \beta_1 \delta_0 \delta_1+\delta_0^2 \delta_1^2+2 \alpha_0 \beta_0 \delta_0+2 \alpha_0 \beta_1 \delta_1+2 \alpha_1 \beta_0 \delta_1+2 \alpha_1 \beta_1 \delta_0-\alpha_0^2-\alpha_1^2-\beta_0^2-\beta_1^2-\delta_0^2-\delta_1^2+1=0\)
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.
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:\(a^2=y^2+z^2-2yz\alpha_0,a_1^2=x^2+w^2-2xw\alpha_1,b^2=x^2+z^2-2xz\beta_0,b_1^2=y^2+w^2-2yw\beta_1,c^2=x^2+y^2-2xy\delta_0,c_1^2=z^2+w^2-2zw\delta_1\)
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?mod=redirect&goto=findpost&ptid=5190&pid=51176&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:
\(a = 7.520014150, b = 6.089230677, c = 4.892424418, a_1 = 7.207657106, b_1= 8.638440579, c_1= 9.835246838\)
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.