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[讨论] 四面体中费马点计算

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 楼主| 发表于 2015-9-5 11:39:20 | 显示全部楼层
更有趣的是:

对于已知空间的四条共点但不共面的四条线段,设\(AP=x,BP=y,CP=z,DP=w\),且\(P\)为四面体D-ABC内的一点,可设四面体的各棱长为\(a,b,c,a_1,b_1,c_1\)

\(S_{ABC}+S_{DBC}+S_{DAC}+S_{DAB}\)取极值条件?

由余弦公式及面积公式可以得到:

\(a^2 = y^2+z^2-2yz\cos(\alpha), b^2 = x^2+z^2-2xz\cos(\beta), c^2 = x^2+y^2-2yx\cos(\delta), a^2_1 = x^2+w^2-2xw\cos(\alpha_1), b^2_1 = y^2+w^2-2yw\cos(\beta_1), c^2_1 = z^2+w^2-2zw\cos(\delta_1)\)

\(4s_0 =4S_{ABC}= \sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}\)

\(4s_1 =4S_{DBC}= \sqrt{-a^4+2a^2b^2_1+2a^2c^2_1-b_1^4+2b_1^2c_1^2-c_1^4}\)

\(4s_2 =4S_{DAC}= \sqrt{-b^4+2b^2a^2_1+2b^2c^2_1-a_1^4+2a^2_1c^2_1-c_1^4}\)

\(4s_3 =4S_{DAB}=\sqrt{-c^4+2c^2a^2_1-a_1^4+2a^2_1b^2_1-b_1^4+2b^2_1c^2}\)

\(16s_0^2=4y^2x^2\cos(\delta)^2+4y^2z^2\cos(\alpha)^2+4x^2z^2\cos(\beta)^2-4y^2z^2-4y^2x^2-4z^2x^2+8y^2xz\cos(\beta)+8yz\cos(\alpha)x^2+8z^2yx\cos(\delta)-8yz^2\cos(\alpha)x\cos(\beta)-8y^2z\cos(\alpha)x\cos(\delta)-8x^2z\cos(\beta)y\cos(\delta)\)

\(16s_1^2=4y^2z^2\cos(\alpha)^2+4y^2w^2\cos(\beta_1)^2-4y^2z^2-4y^2w^2-4z^2w^2+4z^2w^2\cos(\delta_1)^2+8z^2yw\cos(\beta_1)+8yz\cos(\alpha)w^2+8y^2zw\cos(\delta_1)-8y^2z\cos(\alpha)w\cos(\beta_1)-8yz^2\cos(\alpha)w\cos(\delta_1)-8yw^2\cos(\beta_1)z\cos(\delta_1)\)

\(16s_2^2=4x^2z^2\cos(\beta)^2-4z^2x^2-4z^2w^2-4w^2x^2+4z^2w^2\cos(\delta_1)^2+4x^2w^2\cos(\alpha_1)^2+8w^2xz\cos(\beta)+8xw\cos(\alpha_1)z^2+8x^2zw\cos(\delta_1)-8x^2w\cos(\alpha_1)z\cos(\beta)-8xz^2\cos(\beta)w\cos(\delta_1)-8xw^2\cos(\alpha_1)z\cos(\delta_1)\)

\(16s_3^2=4y^2x^2\cos(\delta)^2+4y^2w^2\cos(\beta_1)^2-4y^2x^2-4y^2w^2-4w^2x^2+4x^2w^2\cos(\alpha_1)^2+8w^2yx\cos(\delta)+8xw\cos(\alpha_1)y^2+8x^2yw\cos(\beta_1)-8x^2w\cos(\alpha_1)y\cos(\delta)-8xw^2\cos(\alpha_1)y\cos(\beta_1)-8y^2x\cos(\delta)w\cos(\beta_1)\)

利用拉格朗日乘子法得到:

\(\frac{-(yz\cos(\alpha)-xz\cos(\beta)-yx\cos(\delta)+x^2)}{s_0}+\frac{(yw\cos(\beta_1)+zw\cos(\delta_1)-yz\cos(\alpha)-w^2)}{s_1}=0\) ................(1)

\(\frac{(yz\cos(\alpha)-xz\cos(\beta)+yx\cos(\delta)-y^2)}{s_0}+\frac{(zw\cos(\delta_1)-xz\cos(\beta)+xw\cos(\alpha_1)-w^2)}{s_2}=0\)  ................(2)

\(\frac{yz\cos(\alpha)+xz\cos(\beta)-yx\cos(\delta)-z^2)}{s_0}+\frac{(yw\cos(\beta_1)-yx\cos(\delta)+xw\cos(\alpha_1)-w^2)}{s_3}=0\)  ................(3)

\(\frac{(zw\cos(\delta_1)+xz\cos(\beta)-xw\cos(\alpha_1)-z^2)}{s_2}+\frac{(yw\cos(\beta_1)+yx\cos(\delta)-xw\cos(\alpha_1)-y^2)}{s_3}=0\) ................(4)

\(\frac{-(yw\cos(\beta_1)-zw\cos(\delta_1)-yz\cos(\alpha)+z^2)}{s_1}-\frac{(yw\cos(\beta_1)-yx\cos(\delta)-xw\cos(\alpha_1)+x^2)}{s_3}=0\) ................(5)

\(\frac{(yw\cos(\beta_1)-zw\cos(\delta_1)+yz\cos(\alpha)-y^2)}{s_1}-\frac{(zw\cos(\delta_1)-xz\cos(\beta)-xw\cos(\alpha_1)+x^2)}{s_2}=0\) ................(6)

有谁能从(1)~(6)得到更简洁的约束条件?
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 楼主| 发表于 2015-9-5 13:59:55 | 显示全部楼层
41#更进一步简化为:

\(\frac{(a^2-c^2-b^2)}{s_0}-\frac{(-a^2+b_1^2+c_1^2)}{s_1}=0\)

\(\frac{-(-b^2+c^2+a^2)}{s_0}+\frac{(b^2-a_1^2-c_1^2)}{s_2}=0\)

\(\frac{(c^2-b^2-a^2)}{s_0}+\frac{(c^2-a_1^2-b_1^2)}{s_3}=0\)

\(\frac{-(-a_1^2+c_1^2+b^2)}{s_2}-\frac{(-a_1^2+b_1^2+c^2)}{s_3}=0\)

\(\frac{-(-b_1^2+c_1^2+a^2)}{s_1}-\frac{(-b_1^2+a_1^2+c^2)}{s_3}=0\)

\(\frac{(c_1^2-b_1^2-a^2)}{s_1}-\frac{(-c_1^2+a_1^2+b^2)}{s_2}=0\)


代入面积公式并分解化简可以得到下面两组解:

A组解 (此组解不满足上面简化方程)

\(b^2b_1c_1+bca^2-bcb_1^2-b_1c_1a^2+c^2b_1c_1-bcc_1^2=0\)

\(-aca_1^2+acb^2+a^2a_1c_1-b^2a_1c_1+c^2a_1c_1-acc_1^2=0\)

\(-aba_1^2+a^2a_1b_1+b^2a_1b_1-abb_1^2+abc^2-c^2a_1b_1=0\)

\(b^2b_1c-b_1ca_1^2+bc_1a_1^2-bc_1b_1^2-c^2bc_1+c_1^2b_1c=0\)

\(-ac_1a_1^2+a^2a_1c-b_1^2a_1c+ac_1b_1^2-c^2ac_1+c_1^2a_1c=0\)

\(-ab_1a_1^2+a^2a_1b-b^2ab_1+a_1bb_1^2-c_1^2a_1b+ab_1c_1^2=0\)




B组解:

\(-b^2b_1c_1+bca^2-bcb_1^2+b_1c_1a^2-c^2b_1c_1-bcc_1^2=0\)

\(aca_1^2-acb^2+a^2a_1c_1-b^2a_1c_1+c^2a_1c_1+acc_1^2=0\)

\(aba_1^2+a^2a_1b_1+b^2a_1b_1+abb_1^2-abc^2-c^2a_1b_1=0\)

\(-b^2b_1c+b_1ca_1^2+bc_1a_1^2-bc_1b_1^2-c^2bc_1-c_1^2b_1c=0\)

\(ac_1a_1^2+a^2a_1c-b_1^2a_1c-ac_1b_1^2+c^2ac_1+c_1^2a_1c=0\)

\(ab_1a_1^2+a^2a_1b+b^2ab_1+a_1bb_1^2-c_1^2a_1b-ab_1c_1^2=0\)

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 2015-9-7 18:00:01 | 显示全部楼层
关于A组解,我们可以得到,D-ABC为等面四面体

\(a_1=a,b_1=b,c_1=c,s_0=s_1=s_2=s_3\)

即问题转化为为等面四面体的最大表面积:

在下列方程的约束下:

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

求下面方程的最大值?

\(-16s^2=a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2\)

若我们进一步设

\(\cos(\alpha)=\cos(\alpha_1)=t\)

\(\cos(\beta)=\cos(\beta_1)=t\)

\(\cos(\delta)=\cos(\delta_1)=t\)

可以得到:

\((-x^2+w^2-y^2-z^2)(-x^4z^2-y^2x^4-x^2z^4+y^2x^2w^2+x^2w^2z^2-y^4x^2+5x^2z^2y^2-y^4z^2-y^2z^4+y^2z^2w^2)+(-x^2+w^2-y^2-z^2)(5x^2z^2y^2-x^2z^4-y^2x^4-x^4z^2-y^4x^2-y^2z^4-y^4z^2+4y^3z^3+2yzx^4+x^2w^2z^2+4y^3x^3+4x^3z^3+2y^4zx-2yz^2x^3-2y^2z^3x+2yxz^4-2y^3zx^2+y^2z^2w^2-2z^2yxw^2-2yzx^2w^2-2y^2xzw^2-2y^3z^2x-2yx^2z^3+y^2x^2w^2-2y^2zx^3)t+(-4y^3xz^4+4y^5x^2z-4x^5z^3-4yx^3z^4+4y^3z^3w^2-4y^3z^5+4x^4z^4-4y^5z^3+4y^4z^4+4y^4x^4-4y^3x^5-4y^5x^3-4x^3z^5+12y^3z^3x^2-4y^4zx^3-4y^4z^3x+4y^5xz^2+4y^3x^3w^2+12y^3z^2x^3+4y^2xz^5-4yz^3x^4+4x^3z^3w^2-4y^2zx^3w^2+12y^2z^3x^3+4yz^5x^2+4yz^2x^5+8x^2w^2z^2y^2-4y^3xz^2w^2-4y^3x^2zw^2-4y^2xz^3w^2-4yx^2z^3w^2-4yz^2x^3w^2-4y^3x^4z+4y^2x^5z)t^2+(4(-yz+yx-xz))(yz+yx-xz)(yz+yx+xz)(-yz+yx+xz)t^3=0\)

\(a^2 = y^2+z^2-2yzt\)

\(b^2 = x^2+z^2-2xzt\)

\(c^2 = y^2+x^2-2yxt\)

\(16s^2 = -y^2z^2-x^2z^2-y^2x^2+2xyz(y+z+x)t+(x^2z^2+y^2x^2+y^2z^2-2y^2xz-2yzx^2-2z^2yx)t^2\)

注:由于A组解得到的是

\(\frac{(a^2-c^2-b^2)}{s_0}+\frac{(-a^2+b_1^2+c_1^2)}{s_1}=0\)

\(\frac{(-b^2+c^2+a^2)}{s_0}+\frac{(b^2-a_1^2-c_1^2)}{s_2}=0\)

\(\frac{(c^2-b^2-a^2)}{s_0}-\frac{(c^2-a_1^2-b_1^2)}{s_3}=0\)

\(\frac{-(-a_1^2+c_1^2+b^2)}{s_2}+\frac{(-a_1^2+b_1^2+c^2)}{s_3}=0\)

\(\frac{(-b_1^2+c_1^2+a^2)}{s_1}-\frac{(-b_1^2+a_1^2+c^2)}{s_3}=0\)

\(\frac{(c_1^2-b_1^2-a^2)}{s_1}+\frac{(-c_1^2+a_1^2+b^2)}{s_2}=0\)

与约束条件符号不一样,故可舍掉

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 2015-9-7 18:16:29 | 显示全部楼层
关于B组解:我们还没有找到合适的方法来求解.

其实关于42#的约束方程,我们还可以转化为四个顶点角(12个)的角度关系:

\(\beta+\beta_0-\pi=0\)

\(\delta+\delta_0-\pi=0\)

\(\alpha_0+\alpha-\pi=0\)

\(\alpha_1+\delta_1-\pi=0\)

\(\alpha_2+\beta_2-\pi=0\)

\(\beta_1+\delta_2-\pi=0\)

\(\alpha+\beta+\delta-\pi=0\)

\(\alpha_1+\beta_1+\delta_0-\pi=0\)

\(\beta_0+\alpha_2+\delta_2-\pi=0\)

\(\delta_1+\alpha_0+\beta_2-\pi=0\)

其中

\(\angle BAC=\alpha,\angle BCA=\delta,\angle ABC=\beta\)

\(\angle ADB=\delta_0,\angle ADC=\beta_0,\angle BDC=\alpha_0\)

\(\angle DAB=\alpha_1,\angle DBA=\beta_1,\angle DCB=\delta_1\)

\(\angle DAC=\alpha_2,\angle DBC=\beta_2,\angle DCA=\delta_2\)
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 2015-9-8 18:16:38 | 显示全部楼层
对于40#问题更详细的说明:

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?

注:\(s_1=S_{PBC},s_2=S_{PAC},s_3=S_{PAB},s_4=S_{PAD},s_5=S_{PBD},s_6=S_{PCD}\)

我们由余弦公式有:

\(a^2 = y^2+z^2-2yz\cos(\alpha)\)

\(b^2 = x^2+z^2-2xz\cos(\beta)\)

\(c^2 = y^2+x^2-2yx\cos(\delta)\)

\(a_1^2 = x^2+w^2-2xw\cos(\alpha_1)\)

\(b_1^2 = y^2+w^2-2yw\cos(\beta_1)\)

\(c_1^2 = z^2+w^2-2zw\cos(\delta_1)\)

由面积公式有:

\(16s_1^2 = -(y^4+z^4+a^4)+2y^2z^2+2y^2a^2+2z^2a^2\)

\(16s_2^2 = -(x^4+z^4+b^4)+2x^2z^2+2x^2b^2+2z^2b^2\)

\(16s_3^2 = -(x^4+y^4+c^4)+2x^2y^2+2x^2c^2+2y^2c^2\)

\(16s_4^2 = -(x^4+w^4+a_1^4)+2x^2a_1^2+2w^2a_1^2+2x^2w^2\)

\(16s_5^2 = -(y^4+w^4+b_1^4)+2y^2b_1^2+2w^2b_1^2+2y^2w^2\)

\(16s_6^2 = -(z^4+w^4+c_1^4)+2z^2c_1^2+2w^2c_1^2+2z^2w^2\)

利用拉格朗日乘子法,求导得:

\(\frac{(-x^2+z^2+b^2)}{s_2}+\frac{(-x^2+y^2+c^2)}{s_3}+\frac{(-x^2+a_1^2+w^2)}{s_4}=0\)

\(\frac{(-y^2+z^2+a^2)}{s_1}+\frac{(-y^2+x^2+c^2)}{s_3}+\frac{(-y^2+b_1^2+w^2)}{s_5}=0\)

\(\frac{(-z^2+y^2+a^2)}{s_1}+\frac{(-z^2+x^2+b^2)}{s_2}+\frac{(-z^2+c_1^2+w^2)}{s_6}=0\)

\(\frac{(-w^2+a_1^2+x^2)}{s_4}+\frac{(-w^2+b_1^2+y^2)}{s_5}+\frac{(-w^2+c_1^2+z^2)}{s_6}=0\)

即:

\(\frac{(-x^2+z^2+b^2)}{\sqrt{-(x^4+z^4+b^4-2x^2z^2-2x^2b^2-2z^2b^2)}}+\frac{(-x^2+y^2+c^2)}{\sqrt{-(x^4+y^4+c^4-2x^2y^2-2x^2c^2-2y^2c^2)}}+\frac{(-x^2+a_1^2+w^2)}{\sqrt{-(x^4+w^4+a_1^4-2x^2a_1^2-2w^2a_1^2-2x^2w^2)}}=0\)

\(\frac{(-y^2+z^2+a^2)}{\sqrt{-(y^4+z^4+a^4-2y^2z^2-2y^2a^2-2z^2a^2)}}+\frac{(-y^2+x^2+c^2)}{\sqrt{-(x^4+y^4+c^4-2x^2y^2-2x^2c^2-2y^2c^2)}}+\frac{(-y^2+b_1^2+w^2)}{\sqrt{-(y^4+w^4+b_1^4-2y^2b_1^2-2w^2b_1^2-2y^2w^2)}}=0\)

\(\frac{(-z^2+y^2+a^2)}{\sqrt{-(y^4+z^4+a^4-2y^2z^2-2y^2a^2-2z^2a^2)}}+\frac{(-z^2+x^2+b^2)}{\sqrt{-(x^4+z^4+b^4-2x^2z^2-2x^2b^2-2z^2b^2)}}+\frac{(-z^2+c_1^2+w^2)}{\sqrt{-(z^4+w^4+c_1^4-2z^2c_1^2-2w^2c_1^2-2z^2w^2)}}=0\)

\(\frac{(-w^2+a_1^2+x^2)}{\sqrt{-(x^4+w^4+a_1^4-2x^2a_1^2-2w^2a_1^2-2x^2w^2)}}+\frac{(-w^2+b_1^2+y^2)}{\sqrt{-(y^4+w^4+b_1^4-2y^2b_1^2-2w^2b_1^2-2y^2w^2)}}+\frac{(-w^2+c_1^2+z^2)}{\sqrt{-(z^4+w^4+c_1^4-2z^2c_1^2-2w^2c_1^2-2z^2w^2)}}=0\)

将上面六个方程相加得到:

\(\frac{a^2}{s_1}+\frac{b^2}{s_2}+\frac{c^2}{s_3}+\frac{a_1^2}{s_4}+\frac{b_1^2}{s_5}+\frac{c_1^2}{s_6}=0\)

若转化为角度关系即为:

\(\cot(\angle PCA)+\cot(\angle PBA )+\cot(\angle PDA) = 0\)

\(\cot(\angle PCB)+\cot(\angle PAB)+\cot(\angle PDB) = 0\)

\(\cot(\angle PAC)+\cot(\angle PBC)+\cot(\angle PDC) = 0\)

\(\cot(\angle PAD)+\cot(\angle PBD)+\cot(\angle PCD) = 0\)
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 2019-1-26 17:48:31 | 显示全部楼层
经过长时间的探索及计算,终于得到了有关\(x,y,z,w\)单变量的代数方程:

次数为16,保存在TXT文件中有15M,现在计算过程及结果总结如下:

结论1:在四面体D-ABC中有一点P设\(PA=x,PB=y,PC=z,PD=w\),则

      \(x^n+y^n+z^n+w^n\)取极值的条件为:

      \(\frac{x^{n-1}}{\sin{A_0}}=\frac{y^{n-1}}{\sin{B_0}}=\frac{z^{n-1}}{\sin{C_0}}=\frac{w^{n-1}}{\sin{D_0}}\)

      其中\(A_0,B_0,C_0,D_0\)分别为四面体\(P-BCD,P-ACD,P-ABD,P-ABC\)的顶点空间角

      \(\sin{A_0}=\sqrt{1-\cos{\alpha_0}^2-\cos{\beta_1}^2-\cos{\gamma_1}^2+2\cos{\alpha_0}{\beta_1}{\gamma_1}}\)

      \(\sin{B_0}=\sqrt{1-\cos{\alpha_1}^2-\cos{\beta_0}^2-\cos{\gamma_1}^2+2\cos{\alpha_1}{\beta_0}{\gamma_1}}\)

      \(\sin{C_0}=\sqrt{1-\cos{\alpha_1}^2-\cos{\beta_1}^2-\cos{\gamma_0}^2+2\cos{\alpha_1}{\beta_1}{\gamma_0}}\)
     
      \(\sin{D_0}=\sqrt{1-\cos{\alpha_0}^2-\cos{\beta_0}^2-\cos{\gamma_0}^2+2\cos{\alpha_0}{\beta_0}{\gamma_0}}\)

      其中

      \(\alpha_0=\angle BPC,\alpha_1=\angle APD,\beta_0=\angle APC,\beta_1=\angle BPD,\gamma_0=\angle APB,\gamma_1=\angle CPD\)


为了方便我们约定:\(\cos{\alpha_0}=m,\cos{\beta_0}=n,\cos{\gamma_0}=p\)


结论2:关于n=1时四面体D-ABC中费马点的问题即

     \(x+y+z+w\)取极值的条件为:

     \(\sin{A_0}=\sin{B_0}=\sin{C_0}=\sin{D_0}\)

     即\(\alpha_0=\alpha_1,\beta_0=\beta_1,\gamma_0=\gamma_1\)

     此时内点P\(x,y,z,w\)满足的约束方程: \(m+n+p+1=0\)


    即:

    \(y^2+z^2-a^2=2yzm\)

    \(x^2+w^2-a1^2=2xwm\)

    \(x^2+z^2-b^2=2xzn\)

    \(y^2+w^2-b1^2=2ywn\)

    \(x^2+y^2-c^2=2xyp\)

    \(z^2+w^2-c1^2=2zwp\)

    \(m+n+p+1=0\)


    也可以化成下列方程组:

    \((-a^2+y^2+z^2)xw-(-a1^2+w^2+x^2)yz=0\)

    \((-b^2+x^2+z^2)yw-(-b1^2+w^2+y^2)xz=0\)

    \((-c^2+x^2+y^2)zw-(-c1^2+w^2+z^2)xy=0\)

    \((-a^2+y^2+z^2)x+(-b^2+x^2+z^2)y+(-c^2+x^2+y^2)z+2xyz=0\)

    \((-a1^2+w^2+x^2)yz+(-b1^2+w^2+y^2)xz+(-c1^2+w^2+z^2)xy+2xyzw=0\)

   以上方程组因不对称,经过近1个星期日夜不间断消元计算最终却因内存不足中断!!


   通过反复思索,考虑空间角的定义及对称性,我们可以得到


  \((-a^2+y^2+z^2)x+(-b^2+x^2+z^2)y+(-c^2+x^2+y^2)z+2xyz=0\)

  \((-c1^2+w^2+z^2)y+(-b1^2+w^2+y^2)z+(-a^2+y^2+z^2)w+2yzw=0\)

  \((-b1^2+w^2+y^2)x+(-a1^2+w^2+x^2)y+(-c^2+x^2+y^2)w+2xyw=0\)

  \((-c1^2+w^2+z^2)x+(-a1^2+w^2+x^2)z+(-b^2+x^2+z^2)w+2xzw=0\)

  以上方程组消元过程很快得到了结果!!

  我给几个计算得到的结果:

\({a = 6, a1 = 8, b = 7, b1 = 10, c = 9, c1 = 11, m = 0.1405248549, n = -0.3114323364, p = -0.8290925185, w = 7.292002433, x = 4.470970219, y = 4.939026612, z = 4.170797268}\)

\({a = 6, a1 = 9, b = 7, b1 = 10, c = 10, c1 = 8, m = -0.05282713200, n = -0.3834540341, p = -0.5637188345, w = 6.540726076, x = 5.846264500, y = 5.461204845, z = 2.213194391}\)

\({a = 6, a1 = 9, b = 7, b1 = 10, c = 11, c1 = 5, m = -0.8632740175, n = -0.1689698674, p = 0.03224388450, w = 2.829162049, x = 6.443651215, y = 9.125308744, z = -4.032387350}\) 此解无效!

\({a = 6, a1 = 7, b = 8, b1 = 10, c = 9, c1 = 11, m = 0.2621092795, n = -0.4573814604, p = -0.8047278190, w = 6.678496951, x = 4.482146588, y = 4.990774641, z = 4.886307806}\)

\({a = 6, a1 = 9, b = 8, b1 = 7, c = 10, c1 = 11, m = -0.01408510133, n = 0.02273924854, p = -1.008654146, w = 6.146588896, x = 6.488144455, y = 3.492227650, z = 4.830029337}\)

\({a = 6, a1 = 9, b = 8, b1 = 10, c = 11, c1 = 7, m = 0.1027117933, n = -0.4953240922, p = -0.5149470870, w = 5.846451281, x = 6.902755076, y = 5.718354435, z = 1.876389951}\)

\({a = 6, a1 = 7, b = 9, b1 = 8, c = 10, c1 = 11, m =0 .2533576736, n = -0.2839459657, p = -0.9694117070, w = 5.655430780, x = 5.799670870, y = 4.275898905, z = 5.429628652}\)

\({a = 6, a1 = 7, b = 9, b1 = 8, c = 11, c1 = 10, m = 0.2522660493, n = -0.2845263382, p = -0.9677397110, w = 5.125016959, x = 6.233081163, y = 4.855319304, z = 4.956599685}\)

\({a = 8, a1 = 6, b = 9, b1 = 7, c = 10, c1 = 11, m = 0.1321925780, n = -0.1395888214, p = -0.9926037570, w = 4.141243296, x = 4.923489650, y = 5.095046950, z = 6.877891640}\)

\({a = 8, a1 = 6, b = 9, b1 = 7, c = 11, c1 = 10, m = 0.1365103938, n = -0.1362491840, p = -1.000261210, w = 3.483595101, x = 5.383775750, y = 5.615506312, z = 6.515811976}\)

\({a = 9, a1 = 6, b = 10, b1 = 7, c = 11, c1 = 8, m = 0.1194918729, n = -.3122928028, p = -0.6996563845, w = 1.846681557, x = 5.730853794, y = 6.199898236, z = 6.598403653}\)

另外给一个代数方程例子:

\(a = 4, a1 = 7, b = 5, b1 = 7, c = 6, c1 = 8, w = 5.80900791844013, x = 3.63561649601680, y = 2.81284104014185, z = 2.71146353336776\)

\(7342246912x^{12}-5549176949632x^{10}+1480426084849020x^8-165525804129759300x^6+7362076621084959375x^4-133362251200973025000x^2+815782152217032000000=0\)

\(10037248y^{12}-7910548480y^{10}+1808092648680y^8-113720578637715y^6+1834323625397520y^4-8298482327771328y^2+311146027315200=0\)

\(140521472z^{12}-97385655552z^{10}+20553096720720z^8-1265327133835675z^6+21392443001490000z^4-101787678636600000z^2+36888760832000000=0\)

\(35130368w^{12}-47717590912w^{10}+22589949024380w^8-4486158475703300w^6+382150984577361055w^4-14195249488593355952w^2+188983767435968913408=0\)

若我们利用重心坐标的方法计算,可以得到:

  \(x^2=\beta c^2+\gamma b^2+\delta a1^2-T\)

  \(y^2=\alpha c^2+\gamma a^2+\delta b1^2-T\)

  \(z^2=\alpha b^2+\beta a^2+\delta c1^2-T\)

  \(w^2=\alpha a1^2+\beta b1^2+\gamma c1^2-T\)

  \(T=\alpha\beta c^2+\alpha\gamma b^2+\alpha\delta a1^2+\beta\gamma a^2+\beta\delta b1^2+\gamma\delta c1^2\)

  \(\alpha=\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)

  \(\beta=\frac{\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)

  \(\gamma=\frac{\frac{1}{z}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)

  \(\delta=\frac{\frac{1}{w}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}\)


显然得到的式子并不简单!


我们可以推广到N维空间(单纯形几何)的费马点极值问题

其实极值条件主要取决于空间角的:

对于N=3(三维)

\[\begin{vmatrix}
1&\cos(\alpha)&\cos(\beta)\\
\cos(\alpha)&1&\cos(\gamma)\\
\cos(\beta)&\cos(\gamma)&1\\
\end{vmatrix}=\sin(A_0)^2\]

在正四面体中:

\(\cos(\alpha)=\cos(\beta)=\cos(\gamma)=-\frac{1}{3},\sin(A_0)^2=\frac{16}{27}\)


对于N=4(四维)

\[\begin{vmatrix}
1&\cos(\alpha)&\cos(\beta)&\cos(\gamma)\\
\cos(\alpha)&1&\cos(\gamma)&\cos(\delta)\\
\cos(\beta)&\cos(\gamma)&1&\cos(\alpha)\\
\cos(\gamma)&\cos(\delta)&\cos(\alpha)&1
\end{vmatrix}=\sin(A_0)^2\]

在四维超正四面体中:

\(\cos(\alpha)=\cos(\beta)=\cos(\gamma)=\cos(\delta)=-\frac{1}{4},\sin(A_0)^2=\frac{125}{256}\)

对于N=5(五维)

\[\begin{vmatrix}
1&\cos(\alpha)&\cos(\beta)&\cos(\gamma)&\cos(\delta)\\
\cos(\alpha)&1&\cos(\gamma)&\cos(\delta)&\cos(\mu)\\
\cos(\beta)&\cos(\gamma)&1&\cos(\mu)&\cos(\alpha)\\
\cos(\gamma)&\cos(\delta)&\cos(\mu)&1&\cos(\beta)\\
\cos(\delta)&\cos(\mu)&\cos(\alpha)&\cos(\beta)&1\\
\end{vmatrix}=\sin(A_0)^2\]

在五维超正四面体中:

\(\cos(\alpha)=\cos(\beta)=\cos(\gamma)=\cos(\delta)=\cos(\mu)=-\frac{1}{5},\sin(A_0)^2=\frac{1296}{3125}\)



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 2019-1-28 10:19:38 | 显示全部楼层
用距离表示距离导致这个问题很复杂,如果用坐标表示坐标几何意义就很明显了:以费马点为球心做一个小球,费马点与四面体四个顶点的连线与小球有四个交点,以费马点为起点,四个交点为终点的四个向量之和为0.

这个结论可以推广到任意多个点上。

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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
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