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[几何] 四面体托勒密不等式 chordal metric

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hbghlyj 发表于 2022-10-23 01:09 |阅读模式
Ptolemy's Inequality and the Chordal Metric
Author(s): Tom M. Apostol
JSTOR
4. Ptolemy's inequality and the chordal metric. The chordal distance $\chi(a, b)$ between two complex numbers $a$ and $b$ (see [1], p. 81, or [5], p. 42) is given by the equation
\[\tag6
\chi(a, b)=\frac{|a-b|}{\sqrt{1+|a|^2} \sqrt{1+|b|^2}} .
\]
In this section we prove that when $a, b, c$ are three noncollinear points in the complex plane, the triangle inequality for the chordal metric,
\[\tag7
\chi(a, b)<\chi(a, c)+\chi(c, b),
\]
is equivalent to the tetrahedral theorem discussed in the foregoing section. Using (6) we see that (7) is equivalent to the inequality
$$\tag8|a-b| \sqrt{1+|c|^2}<|a-c| \sqrt{1+|b|^2}+|c-b| \sqrt{1+|a|^2}$$
Construct a tetrahedron using as vertices the three points $a, b, c$ in the complex plane and a fourth point $d$ located at a distance 1 above the origin of the complex plane. The edges of this tetrahedron in the complex plane have lengths $|a-b|,|a-c|$, and $|c-b|$. The other three edges meeting at $d$ have lengths $\sqrt{1+|c|^2}, \sqrt{1+|b|^2}$, and $\sqrt{1+|a|^2}$. Therefore we see at once that Ptolemy's tetrahedral inequality implies (8). Conversely, inequality (8) implies Ptolemy's tetrahedral inequality for a tetrahedron with altitude 1. This, in turn, implies Ptolemy's inequality for a general tetrahedron.

4.托勒密不等式和弦度量. 两个复数 $a$ 和 $b$ 之间的弦距离 $\chi(a, b)$(参见 [1], p. 81 或 [5], p . 42) 由如下等式给出\[\tag6
\chi(a, b)=\frac{|a-b|}{\sqrt{1+|a|^2} \sqrt{1+|b|^2}} .\]在本节中,我们证明当$a,b,c$是复平面上的三个非共线点时,弦度量的三角不等式\[\tag7
\chi(a, b)<\chi(a, c)+\chi(c, b),
\]等价于上一节讨论的四面体的托勒密不等式。使用 (6) 我们看到 (7) 等价于不等式$$\tag8|a-b| \sqrt{1+|c|^2}<|a-c| \sqrt{1+|b|^2}+|c-b| \sqrt{1+|a|^2}$$使用复平面中的三个点 $a、b、c$ 和位于复平面原点上方距离 1 处的第四点 $d$ 作为顶点构造一个四面体。这个四面体“在复平面中的棱” 长为 $|a-b|、|a-c|$ 和 $|c-b|$, “在 $d$ 处相交的其他三条棱” 长为 $\sqrt{1+|c|^2}$、$\sqrt{1+|b|^2}$ 和 $\sqrt{1+|a|^2}$. 因此,我们立即看到四面体的托勒密不等式蕴含(8)。相反,不等式 (8) 蕴含对高度为 1 的四面体的托勒密四面体不等式。这反过来又蕴含对一般四面体的托勒密不等式。

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 楼主| hbghlyj 发表于 2022-10-23 01:15
本层隐藏作者栏

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 楼主| hbghlyj 发表于 2022-10-23 01:22
chordal distance公式(6)的证明,见这帖中,
LEMMA For any $z, w \in \mathbb{C}$ we have$$d(z, w)=\frac{2|z-w|}{\sqrt{1+|z|^{2}} \sqrt{1+|w|^{2}}}$$

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 楼主| hbghlyj 发表于 2022-10-23 01:24
加“chordal distance”标签被空格分成两个了

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