证明正交四球间15个重心球及距离公式的算法——四维体积勾股定理的应用(公式二)

doi:10.12677/PM.2019.98115

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证明正交四球间15个重心球及距离公式的算法——四维体积勾股定理的应用(公式二)
The Proof of 15 Barycentric Spheres between Orthogonal Four Spheres and Algorithm of Distance Formula—Application of Pythagorean Theorem of Four Dimensional Volume (Formula 2)

DOI: 10.12677/PM.2019.98115, PDF, HTML, XML, 被引量下载: 738 浏览: 1,013
作者: 蔡国伟：上海汇美房产有限公司，上海
关键词: 体积勾股定理；应用；垂心四面体；重心球；间距公式；算法；Volume Pythagorean Theorem； Application； Vertical Tetrahedron； Barycentric Sphere；Spacing Formula； Algorithm

摘要: 1球至4球正交球心构成的垂心四面体，存在4点、6线、4面、1体15个重心球，这15个重心球间的105个间距，除了可用2点坐标的距离公式计算外；还可以摆脱坐标直接利用重心球间距公式计算。

Abstract: A vertical tetrahedron consisting of four orthogonal sphere centers from one sphere to four spheres has 15 barycentric spheres of four points, six lines, four planes and one body. The 105 spacing between these 15 barycentric spheres can be calculated by the distance formula of two coordinates. It can also get rid of coordinates and calculate directly by using the formula of center of gravity sphere spacing.

文章引用：蔡国伟. 证明正交四球间15个重心球及距离公式的算法——四维体积勾股定理的应用(公式二)[J]. 理论数学, 2019, 9(8): 880-889. https://doi.org/10.12677/PM.2019.98115

1. 引言

4球正交，球心间的垂心四面体构成的4个球心点(球)、6个两球心间连线(勾股定理)、4个3球心所围面(面积勾股定理 [1])、1个4球心所围体(四维体积勾股定理 [2] )均有各自的重心。那么这15个重心间的间距除了用2点坐标的距离公式计算外，是否能摆脱坐标，直接用重心球距离公式计算呢？

2. 重心球球心间存在间距公式的证明

4球正交，根据勾股4态 [3]，以及垂心四面体重心的性质 [4] 共有15点重心球球心(4点与球心共点、6点为2球心间连线中点、4点为3球心所围面、1点4球心所围体)，其间距数量根据等差求和公式为

$S_{n} = \frac{n (n + 1)}{2}, 1 + 2 + \dots + 13 + 14 = \frac{14 (14 + 1)}{2} = 105$

共有105个间距，这105个间距可摆脱坐标，直接用同构的重心球球心间距公式计算。

2.1. 任意2点重心的距离公式

定义：正交4球心间，存在15点重心，任意2重心间距的平方，等于2重心球半径的平方和减去2倍的2重心球维数积的倒数与2重心球交集的一维球半径的平方和之积。其公式为：

${(P_{1} P_{2})}^{2} = r_{p 1}^{2} + r_{p 2}^{2} - 2 {(n_{p 1} \times m_{p 2})}^{- 1} \sum_{r_{G} = p_{1} \cap p_{2}} r_{G}^{2}$ (1)

(这里： ${(P_{1} P_{2})}^{2}$ 为15点重心的任意2点重心距离的平方， $\sum_{r_{G} = p_{1} \cap p_{2}} r_{G}^{2}$ 为2重心球交集的一维球半径的平方和， ${(n_{p 1} \times m_{p 2})}^{- 1}$ 为2重心球维数积的倒数。)

重心球定义：一至四维重心球的平方等于维数平方分子并集一维球的平方和

$R_{G i}^{2} = \frac{1}{n^{2}} \sum r_{i}^{2} (n = 1, 2, 3, 4, \dots; i = a, b, c, d, \dots)$ (2)

例：4个一维重心球，因球心与重心共点，其重心球平方分别为： $a^{2}, b^{2}, c^{2}, d^{2}$ ；

6个二维重心球，其二维重心至两端相关球心距离的乘积为： $R_{A B}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2})$ ；

4个三维重心球，其三维重心至2点3球面交点距离的乘积为： $R_{A B C}^{2} = \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2})$ ；

1个四维重心球，其四维重心任意二维重心距离平方为： $R_{G}^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2})$ ；

一至四维15个重心球及其重心坐标详见表1。

2.2. 在欧氏空间，建立正交4球间的15点重心球球心坐标

2.2.1. 设15点重心球球心坐标符号

建立坐标的目的，主要用于验证公式(1)，最终达到摆脱坐标，直接用重心球球心间距公式计算。除了体重心坐标外，设：点、线、面3态重心坐标符号展开图。见图1。

Figure 1. Symbolic expansion of point, line and surface 3-state barycentric coordinates

图1. 点、线、面3态重心坐标符号展开图

2.2.2. 建立正交4球心代数坐标

设：正交4球心为大写： $A, B, C, D$ ：及相对应的球半径为小写： $a, b, c, d$ ； $(A \in a; B \in b; C \in c; D \in d)$

设正交4球心坐标(与4球重心共点)为：

$A (a, 0, 0); B (0, b, 0); C (0, 0, c); D (b c t, a c t, a b t)$ (3)

(这里： $t = \frac{- d^{2}}{v + a b c}$ ， $v = \sqrt{a^{2} b^{2} c^{2} + a^{2} b^{2} d^{2} + a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2}} = 6 V_{A B C D}$ ；D坐标在第7象限。)

根据上述4点坐标，它们间6个间距，可用2点距离公式计算为：

$A B^{2} = {(a - 0)}^{2} + {(0 - b)}^{2} + {(0 - 0)}^{2} = a^{2} + b^{2}$

$A C^{2} = {(a - 0)}^{2} + {(0 - 0)}^{2} + {(0 - c)}^{2} = a^{2} + c^{2}$

$B C^{2} = {(0 - 0)}^{2} + {(b - 0)}^{2} + {(0 - c)}^{2} = b^{2} + c^{2}$

$\begin{matrix} A D^{2} = {(a - b c (\frac{- d^{2}}{v + a b c}))}^{2} + {(0 - a c (\frac{- d^{2}}{v + a b c}))}^{2} + {(0 - a b (\frac{- d^{2}}{v + a b c}))}^{2} \\ = a^{2} + \frac{d^{2} (2 a^{2} b^{2} c^{2} + a^{2} b^{2} d^{2} + a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = a^{2} + \frac{d^{2} (a^{2} b^{2} c^{2} + v^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = a^{2} + d^{2} \end{matrix}$ (4)

$\begin{matrix} B D^{2} = {(0 - b c (\frac{- d^{2}}{v + a b c}))}^{2} + {(b - a c (\frac{- d^{2}}{v + a b c}))}^{2} + {(0 - a b (\frac{- d^{2}}{v + a b c}))}^{2} \\ = b^{2} + \frac{d^{2} (2 a^{2} b^{2} c^{2} + a^{2} b^{2} d^{2} + a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = b^{2} + \frac{d^{2} (a^{2} b^{2} c^{2} + v^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = b^{2} + d^{2} \end{matrix}$ (5)

$\begin{matrix} C D^{2} = {(0 - b c (\frac{- d^{2}}{v + a b c}))}^{2} + {(0 - a c (\frac{- d^{2}}{v + a b c}))}^{2} + {(c - a b (\frac{- d^{2}}{v + a b c}))}^{2} \\ = c^{2} + \frac{d^{2} (2 a^{2} b^{2} c^{2} + a^{2} b^{2} d^{2} + a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = c^{2} + \frac{d^{2} (a^{2} b^{2} c^{2} + v^{2} + 2 a b c v)}{{(a b c + v)}^{2}} \\ = c^{2} + d^{2} \end{matrix}$ (6)

(注：上述6间距，可直接用重心球距离公式计算。因均无交集球，重心球距离公式中的第3项均为零，中间的坐标计算均可省略。)

通过上述计算，4坐标满足4球正交，球心间连线构成垂心四面体，即对棱的平方和相等为4球半径的平方和为：

$A B^{2} + C D^{2} = A C^{2} + B D^{2} = B C^{2} + A D^{2} = a^{2} + b^{2} + c^{2} + d^{2}$

2.2.3. 根据正交4球心勾股4态 [3] (点、线、面、体)的重心代数坐标及对应的重心球半径的平方列表

根据正交4球心构成的(点、线、面、体)的15点重心代数坐标，均为点态坐标的算术平均数。因此将15个重心坐标及对映的重心球半径平方列表1。

3. 重心球球心间距公式的验证

3.1. 一维至四维重心自身距离为零

$A A^{2} = {(a - a)}^{2} + {(0 - 0)}^{2} + {(0 - 0)}^{2} = 0$ (2点坐标距离公式)

$A A^{2} = a^{2} + a^{2} - 2 {(1 \times 1)}^{- 1} a^{2} = 0$ (重心球距离公式)

Table 1. 15-point barycentric coordinates and the square of radius of barycentric sphere

表1. 15点重心坐标和对映的重心球半径的平方

表内： $t = \frac{- d^{2}}{v + a b c}$ 。D坐标在第7象限， $v = \sqrt{a^{2} b^{2} c^{2} + a^{2} b^{2} d^{2} + a^{2} c^{2} d^{2} + b^{2} c^{2} d^{2}} = 6 V_{A B C D}$ 。

$G_{A B} G_{A B}^{2} = {(\frac{a}{2} - \frac{a}{2})}^{2} + {(\frac{b}{2} - \frac{b}{2})}^{2} + {(0 - 0)}^{2} = 0$ (2点坐标距离公式)

$G_{A B} G_{A B}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (a^{2} + b^{2}) - 2 {(2 \times 2)}^{- 1} (a^{2} + b^{2}) = 0$ (重心球距离公式)

$G_{A B C} G_{A B C}^{2} = {(\frac{a}{3} - \frac{a}{3})}^{2} + {(\frac{b}{3} - \frac{b}{3})}^{2} + {(\frac{c}{3} - \frac{c}{3})}^{2} = 0$ (2点坐标距离公式)

$G_{A B C} G_{A B C}^{2} = \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2}) + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2}) - 2 {(3 \times 3)}^{- 1} (a^{2} + b^{2} + c^{2}) = 0$ (重心球距离公式)

$G G^{2} = {(\frac{a + b c t}{4} - \frac{a + b c t}{4})}^{2} + {(\frac{b + a c t}{4} - \frac{b + a c t}{4})}^{2} + {(\frac{c + a b t}{4} - \frac{c + a b t}{4})}^{2} = 0$ (2点坐标距离公式)

$G G^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) + \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) - 2 {(4 \times 4)}^{- 1} (a^{2} + b^{2} + c^{2} + d^{2}) = 0$ (重心球距离公式)

3.2. 任意一维与对平面7点因无交集球，其距离的平方为2重心球半径的平方和

以D点与对平面 $A, B, C, G_{A B}, G_{A C}, G_{B C}, G_{A B C}$ 为例：

3.2.1. D点与对平面3个一维重心间距平方

D点与对平面一维重心球间距 $A, B, C$ 已在公式(4)，公式(5)，公式(6)已证，不再累述

3.2.2. D点与对平面3个二维重心间距平方

$D G_{A B}^{2} = {(b c t - \frac{a}{2})}^{2} + {(a c t - \frac{b}{2})}^{2} + {(a b t - 0)}^{2} = d^{2} + \frac{1}{2^{2}} (a^{2} + b^{2})$ (2点坐标距离公式)

$D G_{A B}^{2} = d^{2} + \frac{1}{2^{2}} (a^{2} + b^{2}) - 2 {(1 \times 2)}^{- 1} (0) = d^{2} + \frac{1}{2^{2}} (a^{2} + b^{2})$ (重心球距离公式)

$D G_{A C}^{2} = {(b c t - \frac{a}{2})}^{2} + {(a c t - 0)}^{2} + {(a b t - \frac{c}{2})}^{2} = d^{2} + \frac{1}{2^{2}} (a^{2} + c^{2})$ (2点坐标距离公式)

$D G_{A C}^{2} = d^{2} + \frac{1}{2^{2}} (a^{2} + c^{2}) - 2 {(1 \times 2)}^{- 1} (0) = d^{2} + \frac{1}{2^{2}} (a^{2} + c^{2})$ (重心球距离公式)

$D G_{B C}^{2} = {(b c t - 0)}^{2} + {(a c t - \frac{b}{2})}^{2} + {(a b t - \frac{c}{2})}^{2} = d^{2} + \frac{1}{2^{2}} (b^{2} + c^{2})$ (2点坐标距离公式)

$D G_{B C}^{2} = d^{2} + \frac{1}{2^{2}} (b^{2} + c^{2}) - 2 {(1 \times 2)}^{- 1} (0) = d^{2} + \frac{1}{2^{2}} (b^{2} + c^{2})$ (重心球距离公式)

3.2.3. D点与对平面1个三维重心间距平方

$D G_{A B C}^{2} = {(b c t - \frac{a}{3})}^{2} + {(a c t - \frac{b}{3})}^{2} + {(a b t - \frac{c}{3})}^{2} = d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2})$ (2点坐标距离公式)

$D G_{A B C}^{2} = d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2}) - 2 {(1 \times 3)}^{- 1} (0) = d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2})$ (重心球距离公式)

3.3. 任意一维与除了对平面7点外其它7点重心距离的平方

以D点与除对平面其它7点 $G_{A D}, G_{B D}, G_{C D}, G_{A B D}, G_{A C D}, G_{B C D}, G$ 间距平方为：

3.3.1. D点与除对平面3个二维重心间距平方

$D G_{A D}^{2} = {(b c t - \frac{a + b c t}{2})}^{2} + {(a c t - \frac{a c t}{2})}^{2} + {(a b t - \frac{a b t}{2})}^{2} = \frac{1}{2^{2}} (a^{2} + d^{2})$ (2点坐标距离公式)

$D G_{A D}^{2} = d^{2} + \frac{1}{2^{2}} (a^{2} + d^{2}) - 2 {(1 \times 2)}^{- 1} (d^{2}) = \frac{1}{2^{2}} (a^{2} + d^{2})$ (重心球距离公式)

$D G_{B D}^{2} = {(b c t - \frac{b c t}{2})}^{2} + {(a c t - \frac{b + a c t}{2})}^{2} + {(a b t - \frac{a b t}{2})}^{2} = \frac{1}{2^{2}} (b^{2} + d^{2})$ (2点坐标距离公式)

$D G_{B D}^{2} = d^{2} + \frac{1}{2^{2}} (b^{2} + d^{2}) - 2 {(1 \times 2)}^{- 1} (d^{2}) = \frac{1}{2^{2}} (b^{2} + d^{2})$ (重心球距离公式)

$D G_{C D}^{2} = {(b c t - \frac{b c t}{2})}^{2} + {(a c t - \frac{a c t}{2})}^{2} + {(a b t - \frac{c + a b t}{2})}^{2} = \frac{1}{2^{2}} (c^{2} + d^{2})$ (2点坐标距离公式)

$D G_{C D}^{2} = d^{2} + \frac{1}{2^{2}} (c^{2} + d^{2}) - 2 {(1 \times 2)}^{- 1} (d^{2}) = \frac{1}{2^{2}} (c^{2} + d^{2})$ (重心球距离公式)

3.3.2. D点与除对平面3个三维重心间距平方

$\begin{matrix} D G_{A B D}^{2} = {(b c t - \frac{a + b c t}{3})}^{2} + {(a c t - \frac{b + a c t}{3})}^{2} + {(a b t - \frac{a b t}{3})}^{2} \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} D G_{A B D}^{2} = d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + d^{2}) - 2 {(1 \times 3)}^{- 1} (d^{2}) \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (a^{2} + b^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} D G_{A C D}^{2} = {(b c t - \frac{a + b c t}{3})}^{2} + {(a c t - \frac{a c t}{3})}^{2} + {(a b t - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (a^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} D G_{A C D}^{2} = d^{2} + \frac{1}{3^{2}} (a^{2} + c^{2} + d^{2}) - 2 {(1 \times 3)}^{- 1} (d^{2}) \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (a^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} D G_{B C D}^{2} = {(b c t - \frac{b c t}{3})}^{2} + {(a c t - \frac{b + a c t}{3})}^{2} + {(a b t - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (b^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} D G_{B C D}^{2} = d^{2} + \frac{1}{3^{2}} (b^{2} + c^{2} + d^{2}) - 2 {(1 \times 3)}^{- 1} (d^{2}) \\ = \frac{1}{3} d^{2} + \frac{1}{3^{2}} (b^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

3.3.3. D点与除对平面1个四维重心间距平方

$\begin{matrix} D G^{2} = {(b c t - \frac{a + b c t}{4})}^{2} + {(a c t - \frac{b + a c t}{4})}^{2} + {(a b t - \frac{c + a b t}{4})}^{2} \\ = \frac{1}{2} d^{2} + \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} D G^{2} = d^{2} + \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) - 2 {(1 \times 4)}^{- 1} (d^{2}) \\ = \frac{1}{2} d^{2} + \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

其它A点、B点、C点与其它14点同理(略)。

3.4. 验证任意二维与除了4点一维外的10点距离的平方

以 $G_{A B}$ 对 $G_{C D}, G_{A C}, G_{A D}, G_{B C}, G_{B D}, G_{A B C}, G_{A B D}, G_{A C D}, G_{B C D}, G$ 。

3.4.1. G_AB点与对棱1个二维重心G_CD间距平方

G_AB与G_CD因对棱无交集球，其距离的平方为2重心球半径的平方和

$G_{A B} G_{C D}^{2} = {(\frac{a}{2} - \frac{b c t}{2})}^{2} + {(\frac{b}{2} - \frac{a c t}{2})}^{2} + {(0 - \frac{c + a b t}{2})}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2} + c^{2} + d^{2})$ (2点坐标距离公式)

$G_{A B} G_{C D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (c^{2} + d^{2}) - 2 {(2 \times 2)}^{- 1} (0) = \frac{1}{2^{2}} (a^{2} + b^{2} + c^{2} + d^{2})$ (重心球距离公式)

3.4.2. G_AB点与其它4个旁棱二维重心间距平方

$G_{A B} G_{A C}^{2} = {(\frac{a}{2} - \frac{a}{2})}^{2} + {(\frac{b}{2} - 0)}^{2} + {(0 - \frac{c}{2})}^{2} = \frac{1}{4} (b^{2} + c^{2})$ (2点坐标距离公式)

$G_{A B} G_{A C}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (a^{2} + c^{2}) - 2 {(2 \times 2)}^{- 1} (a^{2}) = \frac{1}{4} (b^{2} + c^{2})$ (重心球距离公式)

$G_{A B} G_{A D}^{2} = {(\frac{a}{2} - \frac{a + b c t}{2})}^{2} + {(\frac{b}{2} - \frac{a c t}{2})}^{2} + {(0 - \frac{a b t}{2})}^{2} = \frac{1}{4} (b^{2} + d^{2})$ (2点坐标距离公式)

$G_{A B} G_{A D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (a^{2} + d^{2}) - 2 {(2 \times 2)}^{- 1} (a^{2}) = \frac{1}{4} (b^{2} + d^{2})$ (重心球距离公式)

$G_{A B} G_{B C}^{2} = {(\frac{a}{2} - 0)}^{2} + {(\frac{b}{2} - \frac{b}{2})}^{2} + {(0 - \frac{c}{2})}^{2} = \frac{1}{4} (a^{2} + c^{2})$ (2点坐标距离公式)

$G_{A B} G_{B C}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (b^{2} + c^{2}) - 2 {(2 \times 2)}^{- 1} (b^{2}) = \frac{1}{4} (a^{2} + c^{2})$ (重心球距离公式)

$G_{A B} G_{B D}^{2} = {(\frac{a}{2} - \frac{b c t}{2})}^{2} + {(\frac{b}{2} - \frac{b + a c t}{2})}^{2} + {(0 - \frac{a b t}{2})}^{2} = \frac{1}{4} (a^{2} + d^{2})$ (2点坐标距离公式)

$G_{A B} G_{B D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{2^{2}} (b^{2} + d^{2}) - 2 {(2 \times 2)}^{- 1} (b^{2}) = \frac{1}{4} (a^{2} + d^{2})$ (重心球距离公式)

3.4.3. G_AB点与4个三维重心间距平方

$G_{A B} G_{A B C}^{2} = {(\frac{a}{2} - \frac{a}{3})}^{2} + {(\frac{b}{2} - \frac{b}{3})}^{2} + {(0 - \frac{c}{3})}^{2} = \frac{1}{36} (a^{2} + b^{2} + 4 c^{2})$ (2点坐标距离公式)

$\begin{matrix} G_{A B} G_{A B C}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2}) - 2 {(2 \times 3)}^{- 1} (a^{2} + b^{2}) \\ = \frac{1}{36} (a^{2} + b^{2} + 4 c^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G_{A B} G_{A B D}^{2} = {(\frac{a}{2} - \frac{a + b c t}{3})}^{2} + {(\frac{b}{2} - \frac{b + a c t}{3})}^{2} + {(0 - \frac{a b t}{3})}^{2} \\ = \frac{1}{36} (a^{2} + b^{2} + 4 d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G_{A B} G_{A B D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{3^{2}} (a^{2} + b^{2} + d^{2}) - 2 {(2 \times 3)}^{- 1} (a^{2} + b^{2}) \\ = \frac{1}{36} (a^{2} + b^{2} + 4 d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G_{A B} G_{A C D}^{2} = {(\frac{a}{2} - \frac{a + b c t}{3})}^{2} + {(\frac{b}{2} - \frac{a c t}{3})}^{2} + {(0 - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{36} (a^{2} + 9 b^{2} + 4 c^{2} + 4 d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G_{A B} G_{A C D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{3^{2}} (a^{2} + c^{2} + d^{2}) - 2 {(2 \times 3)}^{- 1} (a^{2}) \\ = \frac{1}{36} (a^{2} + 9 b^{2} + 4 c^{2} + 4 d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G_{A B} G_{B C D}^{2} = {(\frac{a}{2} - \frac{b c t}{3})}^{2} + {(\frac{b}{2} - \frac{b + a c t}{3})}^{2} + {(0 - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{36} (9 a^{2} + b^{2} + 4 c^{2} + 4 d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G_{A B} G_{B C D}^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{3^{2}} (b^{2} + c^{2} + d^{2}) - 2 {(2 \times 3)}^{- 1} (b^{2}) \\ = \frac{1}{36} (9 a^{2} + b^{2} + 4 c^{2} + 4 d^{2}) \end{matrix}$ (重心球距离公式)

3.4.4. G_AB点与1个四维重心间距平方(任意二维与四维重心间距均相同)

$\begin{matrix} G_{A B} G^{2} = {(\frac{a}{2} - \frac{a + b c t}{4})}^{2} + {(\frac{b}{2} - \frac{b + a c t}{4})}^{2} + {(0 - \frac{c + a b t}{4})}^{2} \\ = \frac{1}{16} (a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G_{A B} G^{2} = \frac{1}{2^{2}} (a^{2} + b^{2}) + \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) - 2 {(2 \times 4)}^{- 1} (a^{2} + b^{2}) \\ = \frac{1}{16} (a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

3.5. 验证任意三维与四维重心距离的平方

$\begin{matrix} G G_{A B C}^{2} = {(\frac{a + b c t}{4} - \frac{a}{3})}^{2} + {(\frac{b + a c t}{4} - \frac{b}{3})}^{2} + {(\frac{c + a b t}{4} - \frac{c}{3})}^{2} \\ = \frac{1}{144} (a^{2} + b^{2} + c^{2} + 9 d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G G_{A B C}^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) + \frac{1}{3^{2}} (a^{2} + b^{2} + c^{2}) - 2 {(4 \times 3)}^{- 1} (a^{2} + b^{2} + c^{2}) \\ = \frac{1}{144} (a^{2} + b^{2} + c^{2} + 9 d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G G_{A B D}^{2} = {(\frac{a + b c t}{4} - \frac{a + b c t}{3})}^{2} + {(\frac{b + a c t}{4} - \frac{b + a c t}{3})}^{2} + {(\frac{c + a b t}{4} - \frac{a b t}{3})}^{2} \\ = \frac{1}{144} (a^{2} + b^{2} + 9 c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G G_{A B D}^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) + \frac{1}{3^{2}} (a^{2} + b^{2} + d^{2}) - 2 {(4 \times 3)}^{- 1} (a^{2} + b^{2} + d^{2}) \\ = \frac{1}{144} (a^{2} + b^{2} + 9 c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G G_{A C D}^{2} = {(\frac{a + b c t}{4} - \frac{a + b c t}{3})}^{2} + {(\frac{b + a c t}{4} - \frac{a c t}{3})}^{2} + {(\frac{c + a b t}{4} - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{144} (a^{2} + 9 b^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G G_{A C D}^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) + \frac{1}{3^{2}} (a^{2} + c^{2} + d^{2}) - 2 {(4 \times 3)}^{- 1} (a^{2} + c^{2} + d^{2}) \\ = \frac{1}{144} (a^{2} + 9 b^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

$\begin{matrix} G G_{B C D}^{2} = {(\frac{a + b c t}{4} - \frac{b c t}{3})}^{2} + {(\frac{b + a c t}{4} - \frac{b + a c t}{3})}^{2} + {(\frac{c + a b t}{4} - \frac{c + a b t}{3})}^{2} \\ = \frac{1}{144} (9 a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (2点坐标距离公式)

$\begin{matrix} G G_{B C D}^{2} = \frac{1}{4^{2}} (a^{2} + b^{2} + c^{2} + d^{2}) + \frac{1}{3^{2}} (b^{2} + c^{2} + d^{2}) - 2 {(4 \times 3)}^{- 1} (b^{2} + c^{2} + d^{2}) \\ = \frac{1}{144} (9 a^{2} + b^{2} + c^{2} + d^{2}) \end{matrix}$ (重心球距离公式)

4. 小结

通过上述验证，证明了4球正交存在勾股四态点、线、面、体15个重心及对映的重心球，这些重心球间105个间距除了利用重心坐标计算外，更可以摆脱坐标，利用重心球距离公式直接计算。而且该公式(1)中的维数乘积的倒数，揭示了4球正交即四维空间。而重心球半径平方公式(2)也可向更高维拓展。

参考文献

[1]	陶杰. 勾股定理的新探索——把勾股定理推广到三维空间[J]. 中等数学, 1983(2): 44.
[2]	蔡国伟. 体积勾股定理的证明[J]. 理论数学, 2019, 9(6): 723-729.
[3]	蔡国伟. 论勾股四态、以及正交球心间同构的场方程[J]. 理论数学, 2019, 9(7): 763-770.
[4]	耿恒考. 四面体的重心与垂心的性质[J]. 数学通报, 2010, 49(10): 55-57.

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