四面体内心与外心的计算

数学星空

对于已知四面体D-ABC各边长,我们设各边长\(AB=c,AC=b,BC=a,AD=a_1,BD=b_1,CD=c_1\),

四个面\(\triangle{ABC},\triangle{BCD},\triangle{ACD},\triangle{ABD}\)的面积分别记为\(s_0,s_1,s_2,s_3\),四面体内一点\(P\),

\(PA=x,PB=y,PC=z,PD=w\),P点到四个面\(\triangle{ABC},\triangle{BCD},\triangle{ACD},\triangle{ABD}\)的垂足分别为\(H_0,H_1,H_2,H_3\),

现在我们主要讨论当\(P\)分别为外接球球心\(O\),及内切球球心\(I\) 时,\(x,y,z,w,h_0,h_1,h_2,h_3\)的计算公式.

  1.对于特殊点,外心\(O\)记\(OA=OB=OC=OD=R,PH_0=h_0,PH_1=h_1,PH_2=h_2,PH_3=h_3\),试给出\(h_0,h_1,h_2,h_3\)的计算公式?

  2.对于特殊点,内心\(I\)记\(IA=x,IB=y,IC=z,ID=w,IH_0=IH_1=IH_2=IH_3=r\)试给出\(x,y,z,w\)的计算公式?

  3.对于一般内点,我们记\(PA=x,PB=y,PC=z,PD=w,PH_0=h_0,PH_1=h_1,PH_2=h_2,PH_3=h_3\)

    3.1 若已知\(x,y,z,w\),试给出\(h_0,h_1,h_2,h_3\)的计算公式?

    3.2 若已知\(h_0,h_1,h_2,h_3\),试给出\(x,y,z,w\)的计算公式?

hejoseph

这个问题可以用重心坐标的结论去解决，不过结论很繁琐

数学星空

看来要想得到单个变量的方程不可能，只能通过求解代数方程方法来求解：

\(h_0^2(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4)+a^4x^2+a^2x^4-a^2x^2b^2-a^2x^2y^2-a^2x^2c^2-a^2x^2z^2-a^2b^2y^2+a^2b^2c^2+a^2y^2z^2-a^2c^2z^2-x^2b^2y^2+x^2b^2z^2+x^2y^2c^2-x^2c^2z^2+b^4y^2+b^2y^4-b^2y^2c^2-b^2y^2z^2-b^2c^2z^2-y^2c^2z^2+c^4z^2+c^2z^4=0\)

\(h_1^2(-a^4+2a^2b_1^2+2a^2c_1^2-b_1^4+2b_1^2c_1^2-c_1^4)+a^4w^2+a^2w^4-a^2w^2b_1^2-a^2w^2z^2-a^2w^2c_1^2-a^2w^2y^2-a^2b_1^2z^2+a^2b_1^2c_1^2+a^2y^2z^2-a^2c_1^2y^2-w^2b_1^2z^2+w^2b_1^2y^2+w^2z^2c_1^2-w^2c_1^2y^2+b_1^4z^2+b_1^2z^4-b_1^2z^2c_1^2-b_1^2z^2y^2-b_1^2c_1^2y^2-z^2c_1^2y^2+c_1^4y^2+c_1^2y^4=0\)

\(h_2^2(-a_1^4+2a_1^2b^2+2a_1^2c_1^2-b^4+2b^2c_1^2-c_1^4)+a_1^4z^2+a_1^2z^4-a_1^2z^2b^2-a_1^2z^2w^2-a_1^2z^2c_1^2-a_1^2z^2x^2-a_1^2b^2w^2+a_1^2b^2c_1^2+a_1^2w^2x^2-a_1^2c_1^2x^2-z^2b^2w^2+x^2b^2z^2+w^2z^2c_1^2-z^2c_1^2x^2+b^4w^2+b^2w^4-b^2w^2c_1^2-b^2w^2x^2-b^2c_1^2x^2-w^2c_1^2x^2+c_1^4x^2+c_1^2x^4=0\)

\(h_3^2(-a_1^4+2a_1^2b_1^2+2a_1^2c^2-b_1^4+2b_1^2c^2-c^4)+a_1^4y^2+a_1^2y^4-a_1^2y^2b_1^2-a_1^2y^2x^2-a_1^2y^2c^2-a_1^2y^2w^2-a_1^2b_1^2x^2+a_1^2b_1^2c^2+a_1^2w^2x^2-a_1^2c^2w^2-y^2b_1^2x^2+w^2b_1^2y^2+x^2y^2c^2-y^2c^2w^2+b_1^4x^2+b_1^2x^4-b_1^2x^2c^2-b_1^2x^2w^2-b_1^2c^2w^2-x^2c^2w^2+c^4w^2+c^2w^4=0\)

1.若已知

\(x=y=z=w=R\),则可以求得\(h_0,h_1,h_2,h_3\)

当然可以利用解析几何知识得到：

\(R^2=\frac{2(aa_1bb_1)^2+2(aa_1cc_1)^2+2(cc_1bb_1)^2-(aa_1)^4-(bb_1)^4-(cc_1)^4}{576V^2}\)

\(h_0^2=R^2-R_0^2,h_1^2=R^2-R_1^2,h_2^2=R^2-R_2^2,h_3^2=R^2-R_3^2\)

\(R_0^2=\frac{(abc)^2}{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}\)

\(R_1^2=\frac{(ab_1c_1)^2}{2a^2b_1^2+2a^2c_1^2+2b_1^2c_1^2-a^4-b_1^4-c_1^4}\)

\(R_2^2=\frac{(a_1bc_1)^2}{2a_1^2b^2+2a_1^2c_1^2+2b^2c_1^2-a_1^4-b^4-c_1^4}\)

\(R_3^2=\frac{(a_1b_1c)^2}{2a_1^2b_1^2+2a_1^2c^2+2b_1^2c^2-a_1^4-b_1^4-c^4}\)

\(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2-a^2b^2b_1^2+a^2b^2c^2+a^2b_1^2c_1^2-a^2c^2c_1^2-a_1^2b^2b_1^2+a_1^2b^2c_1^2+a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^4b_1^2+b^2b_1^4-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2-b_1^2c^2c_1^2+c^4c_1^2+c^2c_1^4+144V^2=0\)

2.对于\(h_0=h_1=h_2=h_3=r=\frac{3V}{s_0+s_1+s_2+s_3}\)没能得到相关的代数方程，只能通过解上面方程组求解

不知对于内心有没有好的几何性质，便于简单计算相关几何量，从而得到\(x,y,z,w\)？

即内心在各个面的投影点有什么几何性质？？

hejoseph

设 $P$ 是 $\triangle ABC$ 所在平面内一点，关于 $\triangle ABC$，点 $P$ 到直线 $BC$ 的有向距离如下确定：当点 $P$ 在直线 $BC$ 上有向距离是 0；当点 $P$ 不在直线 $B$C 上，且到直线 $BC$ 的距离是 $d$，则有向距离的绝对值等于 $d$，当点 $P$、$A$ 在直线 $BC$ 同侧时有向距离的符号是正，当点$P$、$A$ 在直线 $BC$ 异侧时有向距离的符号是负。关于 $\triangle ABC$，点 $P$ 到直线 $CA$、$AB$ 的有向距离类似进行定义。
设 $P$ 是 $\triangle ABC$ 所在平面内一点，关于 $\triangle ABC$，$\triangle PBC$ 的有向面积如下确定：当点 $P$ 在直线 $BC$ 上有向面积是 0；当点 $P$ 不在直线 $BC$ 上时，$\triangle PBC$ 的有向面积绝对值等于 $\triangle PBC$ 的面积，且符号与点 $P$ 到直线 $BC$ 的有向距离相同。关于 $\triangle ABC$，$\triangle PCA$、$\triangle PAB$ 的有向面积类似进行定义。
关于四面体 $ABCD$，点 $P$ 到平面 $BCD$ 的有向距离如下确定：当点 $P$ 在平面$BCD$ 内有向距离是 0；当点 $P$ 不在平面 $BCD$ 上，且到平面 $BCD$ 的距离是 $d$，则有向距离的绝对值等于 $d$，当点 $P$、$A$ 在平面 $BCD$ 同侧时有向距离的符号是正，当点 $P$、$A$ 在平面 $BCD$ 异侧时有向距离的符号是负。关于四面体 $ABCD$，点 $P$ 到平面 $ABC$、$ABD$、$ACD$ 的有向距离类似进行定义。
关于四面体 $ABCD$，四面体 $PBCD$ 的有向体积如下确定：当点 $P$ 在平面$BCD$ 内有向体积是 0；当点 $P$ 不在平面 $BCD$ 内时，四面体 $PBCD$ 的有向体积绝对值等于四面体 $PBCD$ 的体积，且符号与点 $P$ 到平面 $BCD$ 的有向距离相同。关于四面体 $ABCD$，四面体 $PABC$、$PABD$、$PACD$ 的有向体积类似进行定义。

hejoseph

设 $P$ 是空间一点，四面体 $ABCD$ 的体积是 $V$，$V_A$、$V_B$、$V_C$、$V_D$ 分别表示关于四面体 $ABCD$ 中四面体 $PBCD$、四面体 $PACD$、四面体 $PABD$、四面体 $PABC$ 的有向体积，则 $\left(V_A/V,V_B/V,V_C/V,V_D/V\right)$ 称为点 $P$ 关于四面体 $ABCD$ 的重心坐标。
若点 $P$ 关于四面体 $ABCD$ 的重心坐标是 $(\alpha,\beta,\gamma,\delta)$，则
\[
PQ^2=\alpha QA^2+\beta QB^2+\gamma QC^2+\delta QD^2-\alpha\beta AB^2-\alpha\gamma AC^2-\alpha\delta AD^2-\beta \gamma BC^2-\beta \delta BD^2-\gamma \delta CD^2
\]
取 $Q$ 分别是 $A$、$B$、$C$、$D$，利用上面的结论就可以得到点 $P$ 到各顶点的距离了，可以认为 3.2 就解决了。

hejoseph

设 $t=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha \delta AD^2+\beta \gamma BC^2+\beta \delta BD^2+\gamma \delta CD^2$，由上面的结论及重心坐标的定义得方程组
\[
\left\{
\begin{aligned}
&PA^2=\beta AB^2+\gamma AC^2+\delta AD^2-t\\
&PB^2=\alpha AB^2+\gamma BC^2+\delta BD^2-t\\
&PC^2=\alpha AC^2+\beta BC^2+\delta CD^2-t\\
&PD^2=\alpha AD^2+\beta BD^2+\gamma CD^2-t\\
&\alpha +\beta +\gamma +\delta=1
\end{aligned}
\right.
\]
解这个方程组，就可以得到 $\alpha$、$\beta$、$gamma$、$\delta$、$t$ 的值，此时就可以认为 3.1 就解决了。

数学星空

根据楼上的方法及之前creasson 在四面体的外心－内心公式

https://bbs.emath.ac.cn/forum.ph ... 03&fromuid=1455
(出处: 数学研发论坛)

给出的相关结论,我整一理了一下计算方法及结果:

1.对于\(\triangle BCD\)面上的高线AH,设\(H=\alpha B+\beta C+\gamma D\),

  则有\(AH^2=\alpha AB^2+\beta AC^2+\gamma AD^2-\alpha\beta BC^2-\alpha\gamma BD^2-\gamma\beta CD^2\)

  对上式分别对\(\alpha,\beta,\gamma\) 求导得到

  \(AB^2-\beta BC^2-\gamma BD^2=k\)

  \(AC^2-\alpha BC^2-\gamma CD^2=k\)

  \(AD^2-\alpha BD^2-\beta CD^2=k\)

  \(\alpha+\beta+\gamma=1\)


  对上面消元可以得到

\(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2+a^2b^2c^2-a^2b^2c_1^2-a^2b_1^2c^2+a^2b_1^2c_1^2+a_1^2b^2b_1^2-a_1^2b^2c_1^2-a_1^2b_1^2c^2+a_1^2c^2c_1^2+b^4c_1^2-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2+b^2c_1^4+b_1^4c^2+b_1^2c^4-b_1^2c^2c_1^2-(a+b_1+c_1)(a+b_1-c_1)(a-b_1-c_1)(a-b_1+c_1)h^2=0\)


也就是可以得到体积V及AH的表达式


2.对于四面体内任两点\(P,Q\)有

  \(PQ^2=\alpha QA^2+\beta QB^2+\gamma QC^2+\delta QD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

  a_1.若Q点分别位于四个顶点则有

  \(PA^2=\beta AB^2+\gamma AC^2+\delta AD^2-T\)

  \(PB^2=\alpha AB^2+\gamma BC^2+\delta BD^2-T\)

  \(PC^2=\alpha AC^2+\beta BC^2+\delta CD^2-T\)

  \(PD^2=\alpha AD^2+\beta BC^2+\gamma CD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

a_11.若P点位于外心O,则有:\(PA=OA=R,PB=OB=R,PC=OC=R,PD=OD=R\)

       可以得到:\(4a^4a_1^2+4a^2a_1^4-4a^2a_1^2b^2-4a^2a_1^2b_1^2-4a^2a_1^2c^2-4a^2a_1^2c_1^2-4a^2b^2b_1^2+4a^2b^2c^2+4a^2b_1^2c_1^2-4a^2c^2c_1^2-4a_1^2b^2b_1^2+4a_1^2b^2c_1^2+4a_1^2b_1^2c^2-4a_1^2c^2c_1^2+4b^4b_1^2+4b^2b_1^4-4b^2b_1^2c^2-4b^2b_1^2c_1^2-4b^2c^2c_1^2-4b_1^2c^2c_1^2+4c^4c_1^2+4c^2c_1^4)R^2-a^4a_1^4+2a^2a_1^2b^2b_1^2+2a^2a_1^2c^2c_1^2-b^4b_1^4+2b^2b_1^2c^2c_1^2-c^4c_1^4=0\)

       此时\(\alpha=\frac{h_1s_1}{3V},\beta=\frac{h_2s_2}{3V},\gamma=\frac{h_3s_3}{3V},\delta=\frac{h_0s_0}{3V}\)

       进一步可以得到:

       \(-3V(a^4a_1^2-a^2a_1^2b^2-a^2a_1^2c^2-a^2b^2b_1^2+2a^2b^2c^2-a^2c^2c_1^2+b^4b_1^2-b^2b_1^2c^2-b^2c^2c_1^2+c^4c_1^2)+2s_0(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2-a^2b^2b_1^2+a^2b^2c^2+a^2b_1^2c_1^2-a^2c^2c_1^2-a_1^2b^2b_1^2+a_1^2b^2c_1^2+a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^4b_1^2+b^2b_1^4-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2-b_1^2c^2c_1^2+c^4c_1^2+c^2c_1^4)h_0=0\)

       \(-3V(a^4a_1^2-a^2a_1^2b_1^2-a^2a_1^2c_1^2-a^2b^2b_1^2+2a^2b_1^2c_1^2-a^2c^2c_1^2+b^2b_1^4-b^2b_1^2c_1^2-b_1^2c^2c_1^2+c^2c_1^4)+2s_1(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2-a^2b^2b_1^2+a^2b^2c^2+a^2b_1^2c_1^2-a^2c^2c_1^2-a_1^2b^2b_1^2+a_1^2b^2c_1^2+a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^4b_1^2+b^2b_1^4-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2-b_1^2c^2c_1^2+c^4c_1^2+c^2c_1^4)h_1=0\)

       \(-3V(a^2a_1^4-a^2a_1^2b^2-a^2a_1^2c_1^2-a_1^2b^2b_1^2+2a_1^2b^2c_1^2-a_1^2c^2c_1^2+b^4b_1^2-b^2b_1^2c_1^2-b^2c^2c_1^2+c^2c_1^4)+2s_2(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2-a^2b^2b_1^2+a^2b^2c^2+a^2b_1^2c_1^2-a^2c^2c_1^2-a_1^2b^2b_1^2+a_1^2b^2c_1^2+a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^4b_1^2+b^2b_1^4-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2-b_1^2c^2c_1^2+c^4c_1^2+c^2c_1^4)h_2=0\)

       \(-3V(a^2a_1^4-a^2a_1^2b_1^2-a^2a_1^2c^2-a_1^2b^2b_1^2+2a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^2b_1^4-b^2b_1^2c^2-b_1^2c^2c_1^2+c^4c_1^2)+2s_3(a^4a_1^2+a^2a_1^4-a^2a_1^2b^2-a^2a_1^2b_1^2-a^2a_1^2c^2-a^2a_1^2c_1^2-a^2b^2b_1^2+a^2b^2c^2+a^2b_1^2c_1^2-a^2c^2c_1^2-a_1^2b^2b_1^2+a_1^2b^2c_1^2+a_1^2b_1^2c^2-a_1^2c^2c_1^2+b^4b_1^2+b^2b_1^4-b^2b_1^2c^2-b^2b_1^2c_1^2-b^2c^2c_1^2-b_1^2c^2c_1^2+c^4c_1^2+c^2c_1^4)h_3=0\)


a_12.若P点位于内心I,则有:\(PA=IA=x,PB=IB=y,PC=IC=z,PD=ID=w\),

      此时\(\alpha=\frac{s_1}{s_0+s_1+s_2+s_3},\beta=\frac{s_2}{s_0+s_1+s_2+s_3},\gamma=\frac{s_3}{s_0+s_1+s_2+s_3},\delta=\frac{s_0}{s_0+s_1+s_2+s_3}\)

      可以得到:

      \(IA^2s_0^2+2IA^2s_0s_1+2IA^2s_0s_2+2IA^2s_0s_3+IA^2s_1^2+2IA^2s_1s_2+2IA^2s_1s_3+IA^2s_2^2+2IA^2s_2s_3+IA^2s_3^2+a^2s_2s_3-a_1^2s_0^2-a_1^2s_0s_2-a_1^2s_0s_3-b^2s_0s_3-b^2s_2s_3-b^2s_3^2+b_1^2s_0s_2-c^2s_0s_2-c^2s_2^2-c^2s_2s_3+c_1^2s_0s_3=0\)

      \(IB^2s_0^2+2IB^2s_0s_1+2IB^2s_0s_2+2IB^2s_0s_3+IB^2s_1^2+2IB^2s_1s_2+2IB^2s_1s_3+IB^2s_2^2+2IB^2s_2s_3+IB^2s_3^2-a^2s_0s_3-a^2s_1s_3-a^2s_3^2+a_1^2s_0s_1+b^2s_1s_3-b_1^2s_0^2-b_1^2s_0s_1-b_1^2s_0s_3-c^2s_0s_1-c^2s_1^2-c^2s_1s_3+c_1^2s_0s_3=0\)

      \(IC^2s_0^2+2IC^2s_0s_1+2IC^2s_0s_2+2IC^2s_0s_3+IC^2s_1^2+2IC^2s_1s_2+2IC^2s_1s_3+IC^2s_2^2+2IC^2s_2s_3+IC^2s_3^2-a^2s_0s_2-a^2s_1s_2-a^2s_2^2+a_1^2s_0s_1-b^2s_0s_1-b^2s_1^2-b^2s_1s_2+b_1^2s_0s_2+c^2s_1s_2-c_1^2s_0^2-c_1^2s_0s_1-c_1^2s_0s_2=0\)

      \(ID^2s_0^2+2ID^2s_0s_1+2ID^2s_0s_2+2ID^2s_0s_3+ID^2s_1^2+2ID^2s_1s_2+2ID^2s_1s_3+ID^2s_2^2+2ID^2s_2s_3+ID^2s_3^2+a^2s_2s_3-a_1^2s_1^2-a_1^2s_1s_2-a_1^2s_1s_3+b^2s_1s_3-b_1^2s_1s_2-b_1^2s_2^2-b_1^2s_2s_3+c^2s_1s_2-c_1^2s_1s_3-c_1^2s_2s_3-c_1^2s_3^2=0\)


a_13.若P点位于重心G,则有:\(PA=GA=x,PB=GB=y,PC=GC=z,PD=GD=w\),

      此时\(\alpha=\frac{1}{4},\beta=\frac{1}{4},\gamma=\frac{1}{4},\delta=\frac{1}{4}\)

      可以得到:

      \(16x^2=-a^2+3a_1^2+3b^2-b_1^2+3c^2-c_1^2\)

      \(16y^2=3a^2-a_1^2-b^2+3b_1^2+3c^2-c_1^2\)

      \(16z^2=3a^2-a_1^2+3b^2-b_1^2-c^2+3c_1^2\)

      \(16w^2=-a^2+3a_1^2-b^2+3b_1^2-c^2+3c_1^2\)

数学星空

对于四面体内任两点\(P,Q\)有

  \(PQ^2=\alpha QA^2+\beta QB^2+\gamma QC^2+\delta QD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)


1.若Q点分别位于重心G，P点位于外心O,

  \(OG^2=\alpha OA^2+\beta OB^2+\gamma OC^2+\delta OD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

  \(\alpha=\frac{1}{4},\beta=\frac{1}{4},\gamma=\frac{1}{4},\delta=\frac{1}{4},OA=R,OB=R,OC=R,OD=R\)

  则可以消元得到：

   \(OG^2=R^2-\frac{a^2+a_1^2+b^2+b_1^2+c^2+c_1^2}{16}\)


2.若Q点分别位于重心G，P点位于内心I,

  \(IG^2=\alpha GA^2+\beta GB^2+\gamma GC^2+\delta GD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

  \(\alpha=\frac{s_1}{s_0+s_1+s_2+s_3},\beta=\frac{s_2}{s_0+s_1+s_2+s_3},\gamma=\frac{s_3}{s_0+s_1+s_2+s_3},\delta=\frac{s_0}{s_0+s_1+s_2+s_3}\)

  \(GA,GB,GC,GD\)见楼上计算式

  则可以消元得到：

  \(IG^2=\frac{(-a^2+3a_1^2-b^2+3b_1^2-c^2+3c_1^2)s_0^2+((-2a^2-10a_1^2+2b^2+2b_1^2+2c^2+2c_1^2)s_1+(2a^2+2a_1^2-2b^2-10b_1^2+2c^2+2c_1^2)s_2+2s_3(a^2+a_1^2+b^2+b_1^2-c^2-5c_1^2))s_0+(-a^2+3a_1^2+3b^2-b_1^2+3c^2-c_1^2)s_1^2+((2a^2+2a_1^2+2b^2+2b_1^2-10c^2-2c_1^2)s_2+2s_3(a^2+a_1^2-5b^2-b_1^2+c^2+c_1^2))s_1+(3a^2-a_1^2-b^2+3b_1^2+3c^2-c_1^2)s_2^2-s_3(10a^2-2b^2-2c^2+2a_1^2-2b_1^2-2c_1^2)s_2+3s_3^2(3a^2+3b^2-c^2-a_1^2-b_1^2+3c_1^2)}{16(s_0+s_1+s_2+s_3)^2}\)

3.若Q点分别位于外心O，P点位于内心I,

  \(IO^2=\alpha OA^2+\beta OB^2+\gamma OC^2+\delta OD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

  \(\alpha=\frac{s_1}{s_0+s_1+s_2+s_3},\beta=\frac{s_2}{s_0+s_1+s_2+s_3},\gamma=\frac{s_3}{s_0+s_1+s_2+s_3},\delta=\frac{s_0}{s_0+s_1+s_2+s_3}\)

  则可以消元得到：

  \(OI^2=R^2+\frac{(-a_1^2s_1-b_1^2s_2-c_1^2s_3)s_0+(-b^2s_3-c^2s_2)s_1-a^2s_2s_3}{(s_0+s_1+s_2+s_3)^2}\)

数学星空

对于四面体内任一点\(P\)有

  \(PA^2=\beta AB^2+\gamma AC^2+\delta AD^2-T\)

  \(PB^2=\alpha AB^2+\gamma BC^2+\delta BD^2-T\)

  \(PC^2=\alpha AC^2+\beta BC^2+\delta CD^2-T\)

  \(PD^2=\alpha AD^2+\beta BC^2+\gamma CD^2-T\)

  \(T=\alpha\beta AB^2+\alpha\gamma AC^2+\alpha\delta AD^2+\beta\gamma BC^2+\beta\delta BD^2+\gamma\delta CD^2\)

  \(\alpha+\beta+\gamma+\delta=1\)

  \(\alpha=\frac{h_1s_1}{3V},\beta=\frac{h_2s_2}{3V},\gamma=\frac{h_3s_3}{3V},\delta=\frac{h_0s_0}{3V}\)

  则\(PA=x,PB=y,PC=z,PD=w,PH_0=h_0,PH_1=h_1,PH_2=h_2,PH_3=h_3\) 需满足下列方程

\(-2a^6a_1^2h_2s_2-2a^4a_1^4h_2s_2-2a^4a_1^2b^2h_1s_1+2a^4a_1^2b^2h_2s_2+2a^4a_1^2b_1^2h_2s_2+2a^4a_1^2c^2h_2s_2-2a^4a_1^2c_1^2h_0s_0+2a^4a_1^2c_1^2h_2s_2+2a^4b^2b_1^2h_2s_2-2a^4b^2c^2h_2s_2-2a^4b_1^2c_1^2h_2s_2+2a^4c^2c_1^2h_2s_2-2a^2a_1^4b^2h_1s_1-2a^2a_1^4c_1^2h_0s_0+2a^2a_1^2b^4h_1s_1+2a^2a_1^2b^2b_1^2h_1s_1+2a^2a_1^2b^2b_1^2h_2s_2+2a^2a_1^2b^2c^2h_1s_1+2a^2a_1^2b^2c_1^2h_0s_0+2a^2a_1^2b^2c_1^2h_1s_1-2a^2a_1^2b^2c_1^2h_2s_2-2a^2a_1^2b_1^2c^2h_2s_2+2a^2a_1^2b_1^2c_1^2h_0s_0+2a^2a_1^2c^2c_1^2h_0s_0+2a^2a_1^2c^2c_1^2h_2s_2+2a^2a_1^2c_1^4h_0s_0+2a^2b^4b_1^2h_1s_1-2a^2b^4b_1^2h_2s_2-2a^2b^4c^2h_1s_1-2a^2b^2b_1^4h_2s_2+2a^2b^2b_1^2c^2h_2s_2+2a^2b^2b_1^2c_1^2h_0s_0-2a^2b^2b_1^2c_1^2h_1s_1+2a^2b^2b_1^2c_1^2h_2s_2-2a^2b^2c^2c_1^2h_0s_0+2a^2b^2c^2c_1^2h_1s_1+2a^2b^2c^2c_1^2h_2s_2+2a^2b_1^2c^2c_1^2h_2s_2-2a^2b_1^2c_1^4h_0s_0-2a^2c^4c_1^2h_2s_2+2a^2c^2c_1^4h_0s_0-2a^2c^2c_1^4h_2s_2+2a_1^2b^4b_1^2h_1s_1-2a_1^2b^4c_1^2h_1s_1-2a_1^2b^2b_1^2c^2h_1s_1+2a_1^2b^2b_1^2c_1^2h_0s_0+2a_1^2b^2c^2c_1^2h_1s_1-2a_1^2b^2c_1^4h_0s_0-2a_1^2b_1^2c^2c_1^2h_0s_0+2a_1^2c^2c_1^4h_0s_0-2b^6b_1^2h_1s_1-2b^4b_1^4h_1s_1+2b^4b_1^2c^2h_1s_1-2b^4b_1^2c_1^2h_0s_0+2b^4b_1^2c_1^2h_1s_1+2b^4c^2c_1^2h_1s_1-2b^2b_1^4c_1^2h_0s_0+2b^2b_1^2c^2c_1^2h_0s_0+2b^2b_1^2c^2c_1^2h_1s_1+2b^2b_1^2c_1^4h_0s_0-2b^2c^4c_1^2h_1s_1+2b^2c^2c_1^4h_0s_0-2b^2c^2c_1^4h_1s_1+2b_1^2c^2c_1^4h_0s_0-2c^4c_1^4h_0s_0-2c^2c_1^6h_0s_0+3Va^4a_1^4-3Va^4a_1^2w^2-3Va^4a_1^2x^2+6Va^4a_1^2z^2-3Va^2a_1^4y^2+3Va^2a_1^4z^2-6Va^2a_1^2b^2b_1^2+3Va^2a_1^2b^2w^2+3Va^2a_1^2b^2y^2-6Va^2a_1^2b^2z^2+3Va^2a_1^2b_1^2x^2-3Va^2a_1^2b_1^2z^2-6Va^2a_1^2c^2c_1^2+3Va^2a_1^2c^2w^2-3Va^2a_1^2c^2z^2+3Va^2a_1^2c_1^2x^2+3Va^2a_1^2c_1^2y^2-6Va^2a_1^2c_1^2z^2+3Va^2b^2b_1^2w^2+3Va^2b^2b_1^2x^2-6Va^2b^2b_1^2z^2-6Va^2b^2c^2w^2+6Va^2b^2c^2z^2-6Va^2b_1^2c_1^2x^2+6Va^2b_1^2c_1^2z^2+3Va^2c^2c_1^2w^2+3Va^2c^2c_1^2x^2-6Va^2c^2c_1^2z^2+3Va_1^2b^2b_1^2y^2-3Va_1^2b^2b_1^2z^2-6Va_1^2b^2c_1^2y^2+6Va_1^2b^2c_1^2z^2+3Va_1^2c^2c_1^2y^2-3Va_1^2c^2c_1^2z^2+3Vb^4b_1^4-3Vb^4b_1^2w^2-3Vb^4b_1^2y^2+6Vb^4b_1^2z^2-3Vb^2b_1^4x^2+3Vb^2b_1^4z^2-6Vb^2b_1^2c^2c_1^2+3Vb^2b_1^2c^2w^2-3Vb^2b_1^2c^2z^2+3Vb^2b_1^2c_1^2x^2+3Vb^2b_1^2c_1^2y^2-6Vb^2b_1^2c_1^2z^2+3Vb^2c^2c_1^2w^2+3Vb^2c^2c_1^2y^2-6Vb^2c^2c_1^2z^2+3Vb_1^2c^2c_1^2x^2-3Vb_1^2c^2c_1^2z^2+3Vc^4c_1^4-3Vc^4c_1^2w^2+3Vc^4c_1^2z^2-3Vc^2c_1^4x^2-3Vc^2c_1^4y^2+6Vc^2c_1^4z^2=0\)

\(-2a^6a_1^2h_3s_3-2a^4a_1^4h_3s_3+2a^4a_1^2b^2h_3s_3-2a^4a_1^2b_1^2h_0s_0+2a^4a_1^2b_1^2h_3s_3-2a^4a_1^2c^2h_1s_1+2a^4a_1^2c^2h_3s_3+2a^4a_1^2c_1^2h_3s_3+2a^4b^2b_1^2h_3s_3-2a^4b^2c^2h_3s_3-2a^4b_1^2c_1^2h_3s_3+2a^4c^2c_1^2h_3s_3-2a^2a_1^4b_1^2h_0s_0-2a^2a_1^4c^2h_1s_1+2a^2a_1^2b^2b_1^2h_0s_0+2a^2a_1^2b^2b_1^2h_3s_3+2a^2a_1^2b^2c^2h_1s_1-2a^2a_1^2b^2c_1^2h_3s_3+2a^2a_1^2b_1^4h_0s_0+2a^2a_1^2b_1^2c^2h_0s_0+2a^2a_1^2b_1^2c^2h_1s_1-2a^2a_1^2b_1^2c^2h_3s_3+2a^2a_1^2b_1^2c_1^2h_0s_0+2a^2a_1^2c^4h_1s_1+2a^2a_1^2c^2c_1^2h_1s_1+2a^2a_1^2c^2c_1^2h_3s_3-2a^2b^4b_1^2h_3s_3+2a^2b^2b_1^4h_0s_0-2a^2b^2b_1^4h_3s_3-2a^2b^2b_1^2c^2h_0s_0+2a^2b^2b_1^2c^2h_1s_1+2a^2b^2b_1^2c^2h_3s_3+2a^2b^2b_1^2c_1^2h_3s_3-2a^2b^2c^4h_1s_1+2a^2b^2c^2c_1^2h_3s_3-2a^2b_1^4c_1^2h_0s_0+2a^2b_1^2c^2c_1^2h_0s_0-2a^2b_1^2c^2c_1^2h_1s_1+2a^2b_1^2c^2c_1^2h_3s_3+2a^2c^4c_1^2h_1s_1-2a^2c^4c_1^2h_3s_3-2a^2c^2c_1^4h_3s_3+2a_1^2b^2b_1^4h_0s_0+2a_1^2b^2b_1^2c^2h_1s_1-2a_1^2b^2b_1^2c_1^2h_0s_0-2a_1^2b^2c^2c_1^2h_1s_1-2a_1^2b_1^4c^2h_0s_0-2a_1^2b_1^2c^4h_1s_1+2a_1^2b_1^2c^2c_1^2h_0s_0+2a_1^2c^4c_1^2h_1s_1-2b^4b_1^4h_0s_0-2b^4b_1^2c^2h_1s_1-2b^2b_1^6h_0s_0+2b^2b_1^4c^2h_0s_0-2b^2b_1^4c^2h_1s_1+2b^2b_1^4c_1^2h_0s_0+2b^2b_1^2c^4h_1s_1+2b^2b_1^2c^2c_1^2h_0s_0+2b^2b_1^2c^2c_1^2h_1s_1+2b^2c^4c_1^2h_1s_1+2b_1^4c^2c_1^2h_0s_0-2b_1^2c^4c_1^2h_0s_0+2b_1^2c^4c_1^2h_1s_1-2b_1^2c^2c_1^4h_0s_0-2c^6c_1^2h_1s_1-2c^4c_1^4h_1s_1+3Va^4a_1^4-3Va^4a_1^2w^2-3Va^4a_1^2x^2+6Va^4a_1^2y^2+3Va^2a_1^4y^2-3Va^2a_1^4z^2-6Va^2a_1^2b^2b_1^2+3Va^2a_1^2b^2w^2-3Va^2a_1^2b^2y^2+3Va^2a_1^2b_1^2x^2-6Va^2a_1^2b_1^2y^2+3Va^2a_1^2b_1^2z^2-6Va^2a_1^2c^2c_1^2+3Va^2a_1^2c^2w^2-6Va^2a_1^2c^2y^2+3Va^2a_1^2c^2z^2+3Va^2a_1^2c_1^2x^2-3Va^2a_1^2c_1^2y^2+3Va^2b^2b_1^2w^2+3Va^2b^2b_1^2x^2-6Va^2b^2b_1^2y^2-6Va^2b^2c^2w^2+6Va^2b^2c^2y^2-6Va^2b_1^2c_1^2x^2+6Va^2b_1^2c_1^2y^2+3Va^2c^2c_1^2w^2+3Va^2c^2c_1^2x^2-6Va^2c^2c_1^2y^2-3Va_1^2b^2b_1^2y^2+3Va_1^2b^2b_1^2z^2+6Va_1^2b_1^2c^2y^2-6Va_1^2b_1^2c^2z^2-3Va_1^2c^2c_1^2y^2+3Va_1^2c^2c_1^2z^2+3Vb^4b_1^4-3Vb^4b_1^2w^2+3Vb^4b_1^2y^2-3Vb^2b_1^4x^2+6Vb^2b_1^4y^2-3Vb^2b_1^4z^2-6Vb^2b_1^2c^2c_1^2+3Vb^2b_1^2c^2w^2-6Vb^2b_1^2c^2y^2+3Vb^2b_1^2c^2z^2+3Vb^2b_1^2c_1^2x^2-3Vb^2b_1^2c_1^2y^2+3Vb^2c^2c_1^2w^2-3Vb^2c^2c_1^2y^2+3Vb_1^2c^2c_1^2x^2-6Vb_1^2c^2c_1^2y^2+3Vb_1^2c^2c_1^2z^2+3Vc^4c_1^4-3Vc^4c_1^2w^2+6Vc^4c_1^2y^2-3Vc^4c_1^2z^2-3Vc^2c_1^4x^2+3Vc^2c_1^4y^2=0\)

\(-2a^4a_1^4h_0s_0-2a^4a_1^2b^2h_3s_3-2a^4a_1^2c^2h_2s_2-2a^2a_1^6h_0s_0+2a^2a_1^4b^2h_0s_0-2a^2a_1^4b^2h_3s_3+2a^2a_1^4b_1^2h_0s_0+2a^2a_1^4c^2h_0s_0-2a^2a_1^4c^2h_2s_2+2a^2a_1^4c_1^2h_0s_0+2a^2a_1^2b^4h_3s_3+2a^2a_1^2b^2b_1^2h_0s_0+2a^2a_1^2b^2b_1^2h_3s_3-2a^2a_1^2b^2c^2h_0s_0+2a^2a_1^2b^2c^2h_2s_2+2a^2a_1^2b^2c^2h_3s_3+2a^2a_1^2b^2c_1^2h_3s_3+2a^2a_1^2b_1^2c^2h_2s_2-2a^2a_1^2b_1^2c_1^2h_0s_0+2a^2a_1^2c^4h_2s_2+2a^2a_1^2c^2c_1^2h_0s_0+2a^2a_1^2c^2c_1^2h_2s_2+2a^2b^4b_1^2h_3s_3-2a^2b^4c^2h_3s_3+2a^2b^2b_1^2c^2h_2s_2-2a^2b^2b_1^2c_1^2h_3s_3-2a^2b^2c^4h_2s_2+2a^2b^2c^2c_1^2h_3s_3-2a^2b_1^2c^2c_1^2h_2s_2+2a^2c^4c_1^2h_2s_2+2a_1^4b^2b_1^2h_0s_0-2a_1^4b^2c_1^2h_0s_0-2a_1^4b_1^2c^2h_0s_0+2a_1^4c^2c_1^2h_0s_0-2a_1^2b^4b_1^2h_0s_0+2a_1^2b^4b_1^2h_3s_3-2a_1^2b^4c_1^2h_3s_3-2a_1^2b^2b_1^4h_0s_0+2a_1^2b^2b_1^2c^2h_0s_0+2a_1^2b^2b_1^2c^2h_2s_2-2a_1^2b^2b_1^2c^2h_3s_3+2a_1^2b^2b_1^2c_1^2h_0s_0+2a_1^2b^2c^2c_1^2h_0s_0-2a_1^2b^2c^2c_1^2h_2s_2+2a_1^2b^2c^2c_1^2h_3s_3-2a_1^2b_1^2c^4h_2s_2+2a_1^2b_1^2c^2c_1^2h_0s_0-2a_1^2c^4c_1^2h_0s_0+2a_1^2c^4c_1^2h_2s_2-2a_1^2c^2c_1^4h_0s_0-2b^6b_1^2h_3s_3-2b^4b_1^4h_3s_3-2b^4b_1^2c^2h_2s_2+2b^4b_1^2c^2h_3s_3+2b^4b_1^2c_1^2h_3s_3+2b^4c^2c_1^2h_3s_3-2b^2b_1^4c^2h_2s_2+2b^2b_1^2c^4h_2s_2+2b^2b_1^2c^2c_1^2h_2s_2+2b^2b_1^2c^2c_1^2h_3s_3+2b^2c^4c_1^2h_2s_2-2b^2c^4c_1^2h_3s_3-2b^2c^2c_1^4h_3s_3+2b_1^2c^4c_1^2h_2s_2-2c^6c_1^2h_2s_2-2c^4c_1^4h_2s_2+3Va^4a_1^4-3Va^4a_1^2w^2+3Va^4a_1^2x^2+6Va^2a_1^4x^2-3Va^2a_1^4y^2-3Va^2a_1^4z^2-6Va^2a_1^2b^2b_1^2+3Va^2a_1^2b^2w^2-6Va^2a_1^2b^2x^2+3Va^2a_1^2b^2y^2-3Va^2a_1^2b_1^2x^2+3Va^2a_1^2b_1^2z^2-6Va^2a_1^2c^2c_1^2+3Va^2a_1^2c^2w^2-6Va^2a_1^2c^2x^2+3Va^2a_1^2c^2z^2-3Va^2a_1^2c_1^2x^2+3Va^2a_1^2c_1^2y^2+3Va^2b^2b_1^2w^2-3Va^2b^2b_1^2x^2-6Va^2b^2c^2w^2+6Va^2b^2c^2x^2+3Va^2c^2c_1^2w^2-3Va^2c^2c_1^2x^2-6Va_1^2b^2b_1^2x^2+3Va_1^2b^2b_1^2y^2+3Va_1^2b^2b_1^2z^2+6Va_1^2b^2c_1^2x^2-6Va_1^2b^2c_1^2y^2+6Va_1^2b_1^2c^2x^2-6Va_1^2b_1^2c^2z^2-6Va_1^2c^2c_1^2x^2+3Va_1^2c^2c_1^2y^2+3Va_1^2c^2c_1^2z^2+3Vb^4b_1^4-3Vb^4b_1^2w^2+6Vb^4b_1^2x^2-3Vb^4b_1^2y^2+3Vb^2b_1^4x^2-3Vb^2b_1^4z^2-6Vb^2b_1^2c^2c_1^2+3Vb^2b_1^2c^2w^2-6Vb^2b_1^2c^2x^2+3Vb^2b_1^2c^2z^2-3Vb^2b_1^2c_1^2x^2+3Vb^2b_1^2c_1^2y^2+3Vb^2c^2c_1^2w^2-6Vb^2c^2c_1^2x^2+3Vb^2c^2c_1^2y^2-3Vb_1^2c^2c_1^2x^2+3Vb_1^2c^2c_1^2z^2+3Vc^4c_1^4-3Vc^4c_1^2w^2+6Vc^4c_1^2x^2-3Vc^4c_1^2z^2+3Vc^2c_1^4x^2-3Vc^2c_1^4y^2=0\)

\(-2a^4a_1^4h_1s_1-2a^4a_1^2b_1^2h_2s_2-2a^4a_1^2c_1^2h_3s_3-2a^2a_1^6h_1s_1+2a^2a_1^4b^2h_1s_1+2a^2a_1^4b_1^2h_1s_1-2a^2a_1^4b_1^2h_2s_2+2a^2a_1^4c^2h_1s_1+2a^2a_1^4c_1^2h_1s_1-2a^2a_1^4c_1^2h_3s_3+2a^2a_1^2b^2b_1^2h_1s_1+2a^2a_1^2b^2b_1^2h_2s_2-2a^2a_1^2b^2c^2h_1s_1+2a^2a_1^2b^2c_1^2h_3s_3+2a^2a_1^2b_1^4h_2s_2+2a^2a_1^2b_1^2c^2h_2s_2-2a^2a_1^2b_1^2c_1^2h_1s_1+2a^2a_1^2b_1^2c_1^2h_2s_2+2a^2a_1^2b_1^2c_1^2h_3s_3+2a^2a_1^2c^2c_1^2h_1s_1+2a^2a_1^2c^2c_1^2h_3s_3+2a^2a_1^2c_1^4h_3s_3+2a^2b^2b_1^4h_2s_2-2a^2b^2b_1^2c^2h_2s_2+2a^2b^2b_1^2c_1^2h_3s_3-2a^2b^2c^2c_1^2h_3s_3-2a^2b_1^4c_1^2h_2s_2+2a^2b_1^2c^2c_1^2h_2s_2-2a^2b_1^2c_1^4h_3s_3+2a^2c^2c_1^4h_3s_3+2a_1^4b^2b_1^2h_1s_1-2a_1^4b^2c_1^2h_1s_1-2a_1^4b_1^2c^2h_1s_1+2a_1^4c^2c_1^2h_1s_1-2a_1^2b^4b_1^2h_1s_1-2a_1^2b^2b_1^4h_1s_1+2a_1^2b^2b_1^4h_2s_2+2a_1^2b^2b_1^2c^2h_1s_1+2a_1^2b^2b_1^2c_1^2h_1s_1-2a_1^2b^2b_1^2c_1^2h_2s_2+2a_1^2b^2b_1^2c_1^2h_3s_3+2a_1^2b^2c^2c_1^2h_1s_1-2a_1^2b^2c_1^4h_3s_3-2a_1^2b_1^4c^2h_2s_2+2a_1^2b_1^2c^2c_1^2h_1s_1+2a_1^2b_1^2c^2c_1^2h_2s_2-2a_1^2b_1^2c^2c_1^2h_3s_3-2a_1^2c^4c_1^2h_1s_1-2a_1^2c^2c_1^4h_1s_1+2a_1^2c^2c_1^4h_3s_3-2b^4b_1^4h_2s_2-2b^4b_1^2c_1^2h_3s_3-2b^2b_1^6h_2s_2+2b^2b_1^4c^2h_2s_2+2b^2b_1^4c_1^2h_2s_2-2b^2b_1^4c_1^2h_3s_3+2b^2b_1^2c^2c_1^2h_2s_2+2b^2b_1^2c^2c_1^2h_3s_3+2b^2b_1^2c_1^4h_3s_3+2b^2c^2c_1^4h_3s_3+2b_1^4c^2c_1^2h_2s_2-2b_1^2c^4c_1^2h_2s_2-2b_1^2c^2c_1^4h_2s_2+2b_1^2c^2c_1^4h_3s_3-2c^4c_1^4h_3s_3-2c^2c_1^6h_3s_3+3Va^4a_1^4+3Va^4a_1^2w^2-3Va^4a_1^2x^2+6Va^2a_1^4w^2-3Va^2a_1^4y^2-3Va^2a_1^4z^2-6Va^2a_1^2b^2b_1^2-3Va^2a_1^2b^2w^2+3Va^2a_1^2b^2y^2-6Va^2a_1^2b_1^2w^2+3Va^2a_1^2b_1^2x^2+3Va^2a_1^2b_1^2z^2-6Va^2a_1^2c^2c_1^2-3Va^2a_1^2c^2w^2+3Va^2a_1^2c^2z^2-6Va^2a_1^2c_1^2w^2+3Va^2a_1^2c_1^2x^2+3Va^2a_1^2c_1^2y^2-3Va^2b^2b_1^2w^2+3Va^2b^2b_1^2x^2+6Va^2b_1^2c_1^2w^2-6Va^2b_1^2c_1^2x^2-3Va^2c^2c_1^2w^2+3Va^2c^2c_1^2x^2-6Va_1^2b^2b_1^2w^2+3Va_1^2b^2b_1^2y^2+3Va_1^2b^2b_1^2z^2+6Va_1^2b^2c_1^2w^2-6Va_1^2b^2c_1^2y^2+6Va_1^2b_1^2c^2w^2-6Va_1^2b_1^2c^2z^2-6Va_1^2c^2c_1^2w^2+3Va_1^2c^2c_1^2y^2+3Va_1^2c^2c_1^2z^2+3Vb^4b_1^4+3Vb^4b_1^2w^2-3Vb^4b_1^2y^2+6Vb^2b_1^4w^2-3Vb^2b_1^4x^2-3Vb^2b_1^4z^2-6Vb^2b_1^2c^2c_1^2-3Vb^2b_1^2c^2w^2+3Vb^2b_1^2c^2z^2-6Vb^2b_1^2c_1^2w^2+3Vb^2b_1^2c_1^2x^2+3Vb^2b_1^2c_1^2y^2-3Vb^2c^2c_1^2w^2+3Vb^2c^2c_1^2y^2-6Vb_1^2c^2c_1^2w^2+3Vb_1^2c^2c_1^2x^2+3Vb_1^2c^2c_1^2z^2+3Vc^4c_1^4+3Vc^4c_1^2w^2-3Vc^4c_1^2z^2+6Vc^2c_1^4w^2-3Vc^2c_1^4x^2-3Vc^2c_1^4y^2=0\)

  

hejoseph

若点$P$关于四面体 $ABCD$ 的重心坐标是 $\left(\alpha_P,\beta_P,\gamma_P,\delta_P\right)$，点 $Q$ 关于四面体 $ABCD$ 的重心坐标是 $\left(\alpha_Q,\beta_Q,\gamma_Q,\delta_Q\right)$，则
\begin{align*}
PQ^2={}&\left(\alpha_P\beta_Q+\alpha_Q\beta_P-\alpha_P\beta_P-\alpha_Q\beta_Q\right)AB^2+\left(\alpha_P\gamma_Q+\alpha_Q\gamma_P-\alpha_P\gamma_P-\alpha_Q\gamma_Q\right)AC^2\\
&{}+\left(\alpha_P\delta_Q+\alpha_Q\delta_P-\alpha_P\delta_P-\alpha_Q\delta_Q\right)AD^2+\left(\beta_P\gamma_Q+\beta_Q\gamma_P-\beta_P\gamma_P-\beta_Q\gamma_Q\right)BC^2\\
&{}+\left(\beta_P\delta_Q+\beta_Q\delta_P-\beta_P\delta_P-\beta_Q\delta_Q\right)BD^2+\left(\gamma_P\delta_Q+\gamma_Q\delta_P-\gamma_P\delta_P-\gamma_Q\delta_Q\right)CD^2
\end{align*}
如果确定了两点的重心坐标，用这个结论就很方便了。