瀟湘館112 / 數學及古代數學 / 《測圓海鏡》圓城圖之極弦及諸弦篇﹝2﹞說

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《測圓海鏡》圓城圖之極弦及諸弦篇﹝2﹞說

2021-01-12  瀟湘館112

測圓海鏡圓城圖之極弦諸弦篇﹝2

上傳書齋名:瀟湘館112  Xiāo Xiāng Guǎn 112

何世強 Ho Sai Keung

提要:《測圓海鏡》之“圓城圖式”含十四勾股形,連同原有之大勾股形共十五勾股形。此等勾股形三邊形成一系列之恆等式,本文主要談及各勾股形與極弦、虛弦明弦關之等式

關鍵詞:極弦、虛弦明弦

《測圓海鏡》乃金‧李冶所撰,書成於 1248 年,時為南宋淳祐八年。該書卷一“圓城圖式”主要討論與十五勾股形相關之等式,本文介紹其部分等式並作出証明。

本文所引用之勾股式源自“圓城圖式”之十五勾股形,a1b1c1 乃最大勾股形天地乾之勾、股及弦長。故 a1b1c1 又稱為大勾﹝地乾﹞、大股﹝天乾﹞及大弦﹝天地﹞。

《測圓海鏡》之〈弦〉篇涉及諸勾股形之斜邊,本文重點在於証明弦之等式,而弦之位置可參閱以下兩圖。筆者已有文談及此類等式名為〈測圓海鏡圓城圖之諸弦篇﹝1〉。

《測圓海鏡》涉及一系列之勾股恆等式,所有恆等式皆與十五勾股形有關。十五勾股形中最大者為天地乾,其三邊勾股弦分別以 a1b1c1 表之,其餘十四勾股形三邊勾股弦則分別以 aibici 表之,其中 1 < i 15。但 aibici 均可以 a1b1c1 表之,此乃《測圓海鏡》之精注意 i 為直角之點,亦可表示一勾股形。另外注意勾股定理成立,即 ai2 + bi2 = ci2

有關以 a1b1c1 aibici 之式可參閱筆者另文〈測圓海鏡》“圓城圖式”之十二勾股弦算法〉。

以下左為“圓城圖式”右為“圓城圖式十五句股形圖”

注意圓徑為 a1 + b1c1,見上圖之東南西北圓。至於各弦之名稱及位可參閱上兩圖。

以下為與諸有關之等式:

極弦乃髙股平勾共又為平弦明弦共又為髙弦弦共又為大差弦內減髙平二弦較又為小差弦內加髙平二弦較虛弦乃皇極黃方又為明勾股共又為髙弦內減明弦又為平弦內減明弦乃髙弦內減虛弦弦乃平弦內減虛黃黃廣弦黃長弦相並為大一虛弦共也以此數減於大和餘即虛和若以二弦相減餘即虛弦平弦共也﹝案:“虛弦平弦共此題數偶合云“二極差”﹞黃廣弦又為大差一虛弦共黃長弦又為小差一虛弦共以黃長弦減於大勾餘即虛勾以黃廣弦減於大股餘即虛股

以下為各條目之証明:

極弦乃髙股平勾共

已知股﹝在勾股形天日旦日山朱b6 = = (a1 + b1c1)

勾﹝在勾股形月川青川地夕a8 = = (a1 + b1c1)

髙股平勾共 = ( a1 +b1c1) + ( a1 + b1c1)

=(a1 + b1c1)[+]

=(a1 + b1c1)[b12 + a12]

=(a1 + b1c1)

已知日川皇極弦﹝簡皇極﹞:c12 =(a1 + b1c1)

所以極弦= 髙股 + 平勾

又為平弦明弦共

已知在勾股形月川青 8 川地夕 9=c8 = (a1 + b1c1)

在勾股形日月南 14= c14 =(c1a1)(b1 c1 + a1)

弦明弦共 = c8 + c14 =( a1 +b1c1) + (c1a1)(b1 c1 + a1)

=( a1 +b1c1) [1 + (c1a1)]

=(a1 + b1c1)(a1 + c1a1)

=(a1 + b1c1) × c1

= (a1 + b1c1)

已知 = c12 = (a1 + b1c1)

所以極弦= + 明弦

又為髙弦弦共

已知在勾股形天日旦 6 日山朱7= c6= ( a1 + b1c1)

*在勾股形山川東 15= c15 = (c1b1)(a1c1 + b1)

髙弦弦共 = c6 + c15

=( a1 + b1c1) + (c1b1)(a1c1 + b1)

=( a1 +b1c1)[1 + (c1b1)]

=( a1 +b1c1)[b1 + (c1b1)]

=(a1 + b1c1) × c1

= (a1 + b1c1)

所以極弦= 髙弦弦共

又為大差弦內減髙平二弦較

天月大差弦﹝簡大差弦,在勾股形天月坤 10= c10= (c1 a1)

已知在勾股形天日旦 6 日山朱7 = c6= ( a1 + b1c1)

在勾股形月川青 8 川地夕 9= c8 = (a1 + b1c1)

髙平二弦較 = c6c8

= ( a1 +b1c1) –(a1 + b1c1)

=( a1 + b1c1)[]

=(a1 + b1c1)(b1a1)

大差弦內減髙平二弦較 = (c1 a1) –(a1 + b1c1)(b1a1)

= [(c1 a1) –(a1 + b1c1)(b1a1)]

= [2a1c1 – 2a12 – (b12a12c1b1 + c1a1)]

= [2a1c1 – 2a12b12 + a12+ c1b1c1a1]

= [a1c1 a12b12 + c1b1]

= [a1c1 c12+ c1b1]

= (a1 + b1c1)

比較答案兩式,可知相等,所以極弦= 大差弦內減髙平二弦較

又為小差弦內加髙平二弦較

山地小差弦﹝簡小差在勾股形山地艮 11= c11 = (c1b1)

髙平二弦較 =(a1 + b1c1)(b1a1)﹝見前條﹞。

小差弦內加髙平二弦較 = (c1b1) + (a1 + b1c1)(b1a1)

= [(c1b1) + (a1 + b1c1)(b1a1)]

= (2b1c1 – 2b12 + b12a12c1b1 + a1c1)

= (b1c1b12a12 + a1c1)

= (b1c1c12+ a1c1)

= (a1 + b1c1)

所以極弦= 小差弦內加髙平二弦較

虛弦乃皇極黃方

已知弦﹝簡太虛弦,在勾股形月山泛 13= c13 = (c1b1)(c1a1)

皇極在勾股形日川心 12黃方 = 皇極弦三事

 = 弦和較 =b12 + a12c12

b12 + a12c12 = –(a1 + b1c1) + (a1 + b1c1) +(a1 + b1c1)

= (a1 + b1c1)[ –+ +]

= (a1 + b1c1)( – c1 + b1 + a1)

= (a1 + b1c1)2

= (a12 + b12+ c12 + 2a1b1 – 2a1c1 – 2b1c1)

= (2c12 + 2a1b1 – 2a1c1 – 2b1c1)

= (c12 + a1b1a1c1b1c1)

= (c1b1)(c1a1)

所以虛弦 = 皇極黃方面。

又為明勾股共

已知南月勾﹝又﹞:a14 = (c1a1)(b1 c1 + a1)

山東﹝又﹞:b15 = (c1b1)(a1c1 + b1)

明勾*股共 = a14 + b15

a14 + b15 = (c1a1)(b1 c1 + a1) + (c1b1)(a1c1 + b1)

= (b1 c1 + a1)[(c1a1) +(c1b1)]

= (a1 + b1c1)(a1c1a12 + b1c1b12)

= (a1 + b1c1)(a1c1c12 + b1c1)

=(a1 + b1c1)(a1c1 + b1)

=(a1 + b1c1)2

=(c1b1)(c1a1)

注意等式 (c1b1)(c1a1) = (a1 + b1c1)2

所以虛弦 = 明勾*股共

又為髙弦內減明弦

已知在勾股形天日旦 6 日山朱 7= c6 = ( a1 +b1c1)

日月為明弦﹝簡弦,在勾股形日月南 14=
c14 =(c1a1)(b1 c1 + a1)

髙弦內減明弦 = c6 c14 = ( a1 +b1c1) –(c1a1)(b1 c1 + a1)

= ( a1 +b1c1)[1 –(c1a1)]

= ( a1 +b1c1)(b1 c1 + a1)

= ( a1 +b1c1)2

=(c1b1)(c1a1)

所以虛弦 = 髙弦內減明弦

又為平弦內減

已知平弦在勾股形月川青 8 川地夕 9 = c8 = (a1 + b1c1)

* 在勾股形山川東 15= c15 = (c1b1)(a1c1 + b1)

平弦內減*= c8c15 = (a1 + b1c1) –(c1b1)(a1c1 + b1)

= (a1 + b1c1)[1 – (c1b1)]

=(a1 + b1c1)(a1c1 + b1)

=(a1 + b1c1)2

= (c1b1)(c1a1)

所以虛弦 = 平弦內減*

明弦乃髙弦內減虛弦

已知日月為明弦﹝簡﹞,在勾股形日月南 14

= c14= (c1a1)(b1 c1 + a1)

在勾股形天日旦 6 日山朱 7= c6 = ( a1 +b1c1)

在勾股形月山泛 13= c13 = (c1b1)(c1a1)

髙弦內減虛弦 = c6c13= ( a1 + b1c1) –(c1b1)(c1a1)

= [( a1 + b1c1) –(a1 + b1c1)2]

= ( a1 + b1c1)[1–(a1 + b1c1)]

= ( a1 + b1c1)(b1a1b1 + c1)

= (c1a1)(b1 c1 + a1)

所以明弦 = 髙弦內減虛弦

*弦乃平弦內減虛弦

已知* 在勾股形山川東 15= c15 = (c1b1)(a1c1 + b1)

平弦在勾股形月川青 8 川地夕 9=c8 = (a1 + b1c1)

在勾股形月山泛 13= c13 = (c1b1)(c1a1)

平弦內減虛弦 = c8c13

c8c13= (a1 + b1c1) –(c1b1)(c1a1)

=(a1 + b1c1) –(a1 + b1c1)2

=(a1 + b1c1)[1 –(a1 + b1c1)]

=(a1 + b1c1)(a1a1b1 + c1)

=(c1b1)(a1c1 + b1)

所以*= 平弦內減虛弦

黃廣弦黃長弦相為大弦虛弦共也

已知天山弦﹝簡在勾股形天山金 4﹞:c4 = (a1 + b1c1)

月地黃長弦﹝簡黃長在勾股形月地泉 5﹞:c5 = (a1 + b1c1)

黃廣弦黃長弦相 = c4 + c5

= (a1 + b1c1) + (a1 + b1c1)

= c1(a1 + b1c1)[+]

= (a1 + b1c1)(a1 + b1)

已知大弦= c1虛弦 = c13= (c1b1)(c1a1)

大弦虛弦共 = c1 + c13= c1 +(c1b1)(c1a1)

=c1 [1 + (c1b1)(c1a1)]

=(a1b1 + c12c1a1b1c1 + b1a1)

=(2a1b1 + c12c1a1b1c1)

=(a12 + b12+ 2a1b1c1a1c1b1)

=[(a1 + b1)2c1(a1 + b1)]

=(a1 + b1)(a1 + b1c1)

所以黃廣弦黃長弦相= 大弦虛弦共

以此數減於大和餘即虛和

大和”即通弦上勾股和 = a1 + b1

此數減於大和(a1 + b1) –(a1 + b1)(a1 + b1c1)

= (a1 + b1)[1 –(a1 + b1c1)]

=(a1 + b1)(a1b1c1a1c1b1 + c12)

=(a1 + b1)[–a1(c1b1) + c1(c1b1)]

=(a1 + b1)(c1b1)(c1a1)

虛和”即太虛勾股和 = b13 + a13

太虛勾股和=(c1b1)(c1a1) +(c1b1)(c1a1)

= (c1b1)(c1a1)[+]

=(c1b1)(c1a1)(b1 + a1)

比較答案兩式,可知大弦虛弦共= 太虛勾股和

若以二弦相減餘即虛弦平弦共也﹝案:“虛弦平弦共此題數偶合云“二極差”﹞

二弦相減”即黃廣弦黃長弦之差。

二弦相減 = c4c5

= (a1 + b1c1) – (a1 + b1c1)

=c1(a1 + b1c1)[]

=(a1 + b1c1)(b1a1)

已知 = c13= (c1b1)(c1a1)

= c8 = (a1 + b1c1)

虛弦平弦共 = c13 +c8 =(c1b1)(c1a1) + (a1 + b1c1)

=(a1 + b1c1)2 + (a1 + b1c1)

=(a1 + b1c1)[(a1 + b1c1) + 1]

=(a1 + b1c1)(a1 + b1c1 + a1)

=(a1 + b1c1)(2a1 + b1c1)

以上不能得二弦相減之值。

極差”即皇極勾股較

皇極勾股較= b12a12 = (a1 + b1c1) – (a1 + b1c1)

= (a1 + b1c1)[ – ]

= (a1 + b1c1)(b1a1)

二極差”即乘以 2 即:

 2 × (a1 + b1c1)(b1a1) = (a1 + b1c1)(b1a1)

所以黃廣弦黃長弦之差 = 二極差。本條以“案語”為是。

黃廣弦又為大差弦虛弦共

已知在勾股形天山金 4 = c4 = (a1 + b1c1)

大差在勾股形天月坤 10=c10 = (c1 a1)

在勾股形月山泛 13= c13 = (c1b1)(c1a1)

大差弦虛弦共 = c10+ c13 = (c1 a1) + (c1b1)(c1a1)

= (c1 a1)[ 1 + (c1b1)]

= (c1 a1)( a1 + c1b1)

= (c12 a12c1b1 + a1b1)

= (b12 c1b1 + a1b1)

= (a1 + b1c1)

所以黃廣弦 = 大差弦虛弦共

黃長弦又為小差弦虛弦共

已知黃長在勾股形月地泉 5c5 = (a1 + b1c1)

又已知小差在勾股形山地艮 11= c11= (c1b1)

在勾股形月山泛 13= c13 = (c1b1)(c1a1)

小差弦虛弦共 = c11 + c13 = (c1b1) + (c1b1)(c1a1)

= (c1b1)[1 + (c1a1)]

= (c1b1)[b1 + (c1a1)]

= (c1b1)[c1 + b1a1]

= (c12b12+ a1b1c1a1)

= (a12 + a1b1c1a1)

= (a1 + b1c1)

所以黃長弦 = 小差弦虛弦共

以黃長弦減大勾餘即虛勾

已知虛勾在勾股形月山泛 13= a13 = (c1b1)(c1a1) 大勾 = a1

黃長弦減大勾 = a1c5

= a1(a1 + b1c1)

= (a1b1c1a1c1b1 + c12)

= [– a1(c1b1) + c1(c1b1)]

= (c1b1)(c1a1)

所以黃長弦減大勾 = 虛勾

以黃廣弦減於大股餘即虛股

已知月泛﹝又太虛在勾股形月山泛 13= b13

b13 = = (c1b1)(c1a1)

在勾股形天山金 4= c4 = (a1 + b1c1) 大股 = b1

黃廣弦減於大股 = b1c4

= b1(a1 + b1c1)

= (a1b1c1a1c1b1 + c12)

= [– a1(c1b1) + c1(c1b1)]

= (c1b1)(c1a1)

所以黃廣弦減於大股 = 虛股

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