(怎么都一个月没更新了)
今天是兀节
觉得该发点什么
但是想不到
不如就发兀
三点一四一五九
二六五三五
还是来点正经的吧. 提到
这个定理最早由希腊-罗马数学家, 天文学家, 占星学家, 地理学家和音乐理论家 Claudius Ptolemy (c. 100-160s/170s AD) 给出. 他最著名的作品是《天文学大成》 (Almagest, 或《至大论》), 全书十三卷, 详细介绍了球面三角学, 一般运动的描述, 太阳运动与春分和岁差, 月球运动与远地点, 日食月食, 星图, 五大行星的三维运动等理论, 并附有详尽的数据和表格, 是古代天文学的巅峰之作. Ptolemy 也将地心说的思想完善成了理论, 统治天文学理论一千多年, 后世也用他的名字来指代地心说 (Ptolemaic model, Ptolemaic system, etc.)
第一卷中, 为了后面的计算, 他推导了一套三角函数表, 其使用的工具正是 Ptolemy 定理. 我们来看一看 (经过修订后的) 原文是怎么说的:
ὃν δὲ τρόπον ἀπὸ τούτων καὶ αἱ λοιπαὶ τῶν κατὰ μέρος δοθήσονται, δείξομεν ἐϕεξῆς προεκθέμενοι λημμάτιον εὔχρηστον πάνυ πρὸς τὴν παροῦσαν πραγματείαν.
ἔστω γὰρ κύκλος ἐγγεγραμμένον ἔχων τετράπλευρον τυχὸν τὸ ΑΒΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΓ καὶ ΒΔ. δεικτέον, ὅτι τὸ ὑπὸ τῶν ΑΓ καὶ ΒΔ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ συναμϕοτέροις τῷ τε ὑπὸ τῶν ΑΒ, ΔΓ καὶ τῷ ὑπὸ τῶν ΑΔ, ΒΓ. κείσθω γὰρ τῇ ὑπὸ τῶν ΔΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΑΒΕ. ἐὰν οὖν κοινήν προσθῶμεν τὴν ὑπὸ ΕΒΔ, ἔσται καὶ ἡ ὑπὸ ΑΒΔ γωνία ἴση τῇ ὑπὸ ΕΒΓ. ἔστιν δὲ καὶ ἡ ὑπὸ ΒΔΑ τῇ ὑπὸ ΒΓΕ ἴση [Eucl. III, 21]· τὸ γὰρ αὐτὸ τμῆμα ὑποτείνουσιν· ἰσογώνιον ἄρα ἐστὶν τὸ ΑΒΔ τρίγωνιον τῷ ΒΓΕ τριγώνῳ. ὥστε καὶ ἀνάλογόν ἐστιν, ὡς ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ [Eucl. VI, 4]· τὸ ἄρα ὑπὸ ΒΓ, ΑΔ ἴσον ἐστὶν τῷ ὑπὸ ΒΔ, ΓΕ [Eucl. VI, 16]. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΔΒΓ γωνίᾳ, ἔστιν δὲ καὶ ἡ ὑπὸ ΒΑΕ ἴση τῇ ὑπὸ ΒΔΓ, ἰσογώνιον ἄρα ἐστὶν τὸ ΑΒΕ τρίγωνον τῷ ΒΓΔ τριγώνῳ· ἀνάλογον ἄρα ἐστίν, ὡς ἡ ΒΑ πρὸς ΑΕ, ἡ ΒΔ πρὸς ΔΓ· τὸ ἄρα ὑπὸ ΒΑ, ΔΓ ἴσον ἐστὶν τῷ ὑπὸ ΒΔ, ΑΕ. ἐδείχθη δὲ καὶ τὸ ὑπὸ ΒΓ, ΑΔ ἴσον τῷ ὑπὸ ΒΔ, ΓΕ· καὶ ὅλον [Eucl. II, 1] ἄρα τὸ ὑπὸ ΑΓ, ΒΔ ἴσον ἐστὶν συναμϕοτέροις τῷ τε ὑπὸ ΑΒ, ΔΓ καὶ τῷ ὑπὸ ΑΔ, ΒΓ· ὅπερ ἔδει δεῖξαι.
相信大家都看不懂古希腊语. 为此我在这里提供一份英文翻译:
And in what way the remaining [properties] of the parts will also be given starting from these [principles], we shall show in sequence, by first setting out a very useful lemma for the present subject matter.
For let there be a circle with an inscribed quadrilateral, let’s say
, and let the diagonals and be drawn. We must show that the rectangle contained by and is equal to the sum of the rectangles contained by , and , . For let angle be constructed equal to angle . If, therefore, we add the common angle to both, then angle will also be equal to angle . And also angle is equal to angle [Eucl. III, 21], because they subtend the same arc. Therefore, triangle is similar to triangle . Therefore, they are proportional: as is to , so is to [Eucl. VI, 4]. Consequently, the product of and is equal to the product of and [Eucl. VI, 16]. Again, since angle is equal to angle , and also angle is equal to angle , therefore triangle is similar to triangle . Therefore, they are proportional: as is to , so is to . Consequently, the product of and is equal to the product of and . And it was also shown that the product of and is equal to the product of and . And so, the whole rectangle contained by and is equal to the sum of the rectangle contained by and and the rectangle contained by and ; which was to be demonstrated.
和一份中文翻译:
我们将通过依次阐述, 展示如何从这些基本原理出发推导出其余部分性质. 首先, 我们提出一个对当前主题极为有用的引理.
设有一个圆内接四边形
, 连接对角线 与 . 我们需证明: 由 与 构成的矩形, 等于 与 构成的矩形加上 与 构成的矩形之和. 为此, 作角 使其等于角 . 若在两边同时添加公共角 , 则角 也将等于角 . 此外, 角 与角 相等 [欧几里得 III.21], 因为它们对同一段圆弧. 因此, 三角形 与三角形 相似. 由此可得比例关系: 与 之比等于 与 之比 [欧几里得 VI.4]. 故 与 的乘积等于 与 的乘积 [欧几里得 VI.16]. 再观之, 因角 等于角 , 且角 等于角 , 故三角形 与三角形 相似. 由此可得比例关系: 与 之比等于 与 之比. 故 与 的乘积等于 与 的乘积. 前已证得 与 的乘积等于 与 的乘积. 综上, 由 与 构成的整体矩形, 等于 与 构成的矩形与 与 构成的矩形之和. 此即所求证之结论.
这下大家都能看出来这就是 Ptolemy 定理了. 顺便一提这也是我数学史 III 论文的出发点.
附上个人找到的一些《天文学大成》的版本.
Almagest (1-6) 1898 greek (pdf)
Almagest 1515 latin (pdf)
Almagest 1515 latin (djvu)
Almagest 1528 latin (pdf)
Almagest 1528 latin (another version of the same book) (pdf)
Almagest 1528 latin (yet another version) (pdf)