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《測圓海鏡》圓城圖之黃廣弦及髙弦﹝4﹞說

 瀟湘館112 2021-01-16

測圓海鏡圓城圖之黃廣髙弦4

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黃廣弦、黃長弦及其關之等式

關鍵詞:黃廣弦、黃長弦髙弦平弦

《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》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

《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〉、〈測圓海鏡圓城圖之極弦諸弦篇﹝2〉及〈測圓海鏡圓城圖之邊及相關弦﹝3〉。

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。

以下為與諸有關之等式:

黃廣弦內減邊股即黃長弦內減底勾即明勾也

髙弦髙股共即邊股平弦平勾共即底勾髙弦髙勾共即底股平弦平股共即邊勾

上髙弦減於通股餘即邊股內減股也下平弦減通勾餘即邊勾內減明勾也髙弦平弦相併,即大弦內少個皇極弦也若以相數減於大和餘為皇極弦圓徑共也髙弦平弦相減餘即皇極差也又為皇極弦上減小差弦也若以相減數卻加於相即黃廣弦也

以下為各條目之証明:

黃廣弦內減邊股即

在勾股形天山金 4在勾股形天川西 2

已知天山弦﹝簡弦﹞= c4 = (a1 + b1c1)

天西股﹝簡股﹞:b2 = b1(a1 + b1c1) = (c1 + b1a1)

黃廣弦內減邊股,即:

 (a1 + b1c1) –(c1 + b1a1)

= [(2c1a1 + 2c1b1 – 2c12) – (a1c1 + a1b1a12)]

= (2c1a1 + 2c1b1 – 2c12a1c1a1b1 + a12)]

= (c1a1 + 2c1b1 – 2c12a1b1 + a12)

= (c1a1 + 2c1b1 – 2a12 – 2b12a1b1 + a12)

= (c1a1 + 2c1b1a12 – 2b12a1b1)

= (c1b1)(a1c1 + b1)

已知山東﹝又在勾股形山川東 15﹞:

b15 == (a1c1a12b1a1 – 2b12 + 2b1c1)

= (c1b1)(a1c1 + b1)

與前式比較,可知黃廣弦內減邊股 =

黃長弦內減底勾即明勾也

已知黃長在勾股形月地泉 5c5 = (a1 + b1c1)

北地勾﹝簡勾,在勾股形日地北 3﹞:
a3 = a1(a1 + b1c1) = (a1b1 + c1)

黃長弦內減底勾 = (a1 + b1c1) –(a1b1 + c1)

= (2a1c1 + 2b1c1 – 2c12b1a1 + b12b1c1)

=(2a1c1 + b1c1 – 2c12b1a1 + b12)

=(2a1c1 + b1c1c12a12b12b1a1 + b12)

=(2a1c1 + b1c1c12a12b1a1)

= (c1a1)(b1 c1 + a1)

已知南月勾﹝又勾在勾股形日月南 14﹞:a14 =(c1a1)(b1 c1 + a1)

比較兩式,可知黃長弦內減底勾 = 明勾

髙弦髙股共即邊股

“和”。

已知在勾股形天日旦 6 日山朱 7弦同

b6 = = ( a1 + b1c1)

天日或日山c6 = ( a1 +b1c1)

髙弦上股弦共=c6 + b6 = (a1 + b1c1) + ( a1 + b1c1)

= (a1 + b1c1)(c1 + b1)

= [a1 – (c1 b1)](c1 + b1)

= (a1c1 + a1b1c12 + b12)

= (a1c1 + a1b1a12)

= (c1 + b1a1)

已知b2 = b1(a1 + b1c1) = (c1 + b1a1)

比較兩式,所以髙弦上股弦共 = 股。

平弦平勾共即底勾

平弦平勾在勾股形月川青 8 川地夕 9

已知勾:a8 =  = ( a1 +b1c1)

弦:c8 = (a1 + b1c1)

平弦上勾弦共 = c8 + a8= (a1 + b1c1) + (a1 + b1c1)

= (a1 + b1c1)(c1 + a1)

= [b1 – (c1a1)](c1 + a1)

= [b1c1 + b1a1 – (c12a12)]

= [b1c1 + b1a1b12]

= (a1b1 + c1)

已知底勾日地北之勾, = a3= a1(a1 + b1c1)

= (a1b1 + c1)

所以平弦上勾弦共 =

髙弦髙勾共即底股

髙弦髙勾在勾股形天日旦 6 日山朱 7

已知日旦或山朱a6 = (a1 + b1c1)

天日或日山c6 = ( a1 +b1c1)

髙弦上勾弦共=c6 + a6 = (a1 + b1c1) + (a1 + b1c1)

= (a1 + b1c1)( + 1)

= (a1 + b1c1)(c1 + a1)

= [b1 – (c1a1)](c1 + a1)

= [b1c1 + b1a1c12 + a12]

= [b1c1 + b1a1b12]

= (a1b1 + c1)

已知股:b3 = = (a1b1 + c1)

比較兩式,所以髙弦上勾弦共 = 股。

平弦平股共即邊勾

已知在勾股形月川青 8 川地夕 9弦同

b8 = (a1 + b1c1)弦:c8 = (a1 + b1c1)

平弦上弦共 = (a1 + b1c1) +(a1 + b1c1)

= (a1 + b1c1)(+ 1)

= (a1 + b1c1)(c1 + b1)

= (a1 + b1c1)(b1 + c1)

= (a1b1 + a1c1 + b12c12)

= (a1b1 + a1c1a12)

= (c1 + b1a1)

已知勾乃天川西之勾 = a2 = (c1 + b1a1)

以上兩式相同,所以平弦上弦共 = 勾。

上髙弦減於通股餘即邊股內減股也

已知通股在勾股形天地乾 1= b1

天日或日山= c6 = ( a1 +b1c1)

上髙弦減於通股=b1(a1 + b1c1)

= (2b1a1c1a1c1b1 + c12

= (2b1a1c1a1c1b1 + a12+ b12)  

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

= (b1 + a1)(b1 + a1c1)

已知天西股﹝簡股﹞:b2 = b1(a1 + b1c1) = (c1 + b1a1)

在勾股形日月南 14= b14 = (c1a1)(b1 c1 + a1)

  邊股內減= (c1 + b1a1) –(c1a1)(b1 c1 + a1)

= (a1c1 + a1b1a12) –(c1b1b1a1c12 + 2c1a1a12)

= (a1c1 + a1b1a12c1b1 + b1a1 + a12+ c12 – 2c1a1)

= (a1b1c1b1 + b1a1 + c12c1a1)

= (– c1a1c1b1 + c12 + 2a1b1)

= (b1 + a1)(b1 + a1c1)

比較兩式可知相同,所以上髙弦減於通股 = 邊股內減

下平弦減通勾餘即邊勾內減明勾也

已知下平在勾股形川地夕 9= c9 = (a1 + b1c1)

通勾在勾股形天地乾 1= a1

下平弦減通勾 = a1(a1 + b1c1)

= (2a1b1a1c1c1b1 + c12)

= (a12 + b12+ 2a1b1a1c1c1b1)

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

= (a1 + b1)(a1 + b1c1)

已知勾﹝川西﹞:a2 = (c1 + b1a1)

南月勾﹝又﹞:a14 = (c1a1)(b1 c1 + a1)

邊勾內減明勾,即:

(c1 + b1a1) –(c1a1)(b1 c1 + a1)

= [(a1c1 + a1b1a12) – (c1b1 c12+ c1a1a1b1 + c1a1a12)]

= (a1c1 + a1b1a12c1b1+ c12c1a1 + a1b1 c1a1 + a12)

= (2a1b1c1b1+ c12c1a1)

= (a1 + b1)(a1 + b1c1)

比較答案兩式,可知相等,所以下平弦減通勾 = 邊勾內減明勾

髙弦平弦相併,即大弦內少個皇極弦也

併”即“加”。

已知天日或日山c6 = ( a1 +b1c1)

在勾股形月川青 8 川地夕 9c8 = (a1 + b1c1)

髙弦平弦相 = (a1 + b1c1) + (a1 + b1c1)

= (a1 + b1c1)[+]

= (a1 + b1c1)(a1 + b1)

已知大弦 = c1

皇極弦在勾股形日川心 12c12 = (a1 + b1c1)

大弦內少個皇極弦=c1(a1 + b1c1)

= c1[1 –(a1 + b1c1)]

= [2a1b1c1(a1 + b1c1)]

= (2a1b1c1a1c1b1 + c12)

= (2a1b1c1a1c1b1 + a12+ b12)

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

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

= (a1 + b1)(a1 + b1c1)

比較兩式可知相同,所以髙弦平弦相= 大弦內少個皇極弦

若以相數減於大和餘為皇極弦圓徑共也

已知“大和”即通勾通股併 = a1 + b1

= (a1 + b1c1)(a1 + b1)﹝見前﹞。

數減於大和

= (a1 + b1) – (a1 + b1c1)(a1 + b1)

= (a1 + b1)[1 –(a1 + b1c1)]

=(a1 + b1)(2a1b1c1a1c1b1 + c12)

=(a1 + b1)[2a1b1 + a12 + b12c1(a1 + b1)]

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

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

=(a1 + b1c1)(a1 + b1)2

已知皇極在勾股形日川心 12c12 = (a1 + b1c1)

已知圓徑 = (a1 + b1c1)

皇極弦圓徑共=(a1 + b1c1) + (a1 + b1c1)

= (a1 + b1c1)(1 + )

=(a1 + b1c1)(c12+ 2a1b1)

=(a1 + b1c1)(a12+ b12 + 2a1b1)

=(a1 + b1c1)(a1 + b1)2

所以數減於大和 = 皇極弦 + 圓徑

髙弦平弦相減餘即皇極差也

已知天日或日山c6 = ( a1 +b1c1)

弦:c8 = (a1 + b1c1)

髙弦平弦相減 = (a1 + b1c1) –(a1 + b1c1)

=(a1 + b1c1) []

=(a1 + b1c1)(b1a1)

皇極差”即皇極勾股較在勾股形日川心 12

= b12a12 = (a1 + b1c1) – (a1 + b1c1)

= (a1 + b1c1)[]

= (a1 + b1c1)(b1a1)

比較兩式,可知髙弦平弦相減 = 皇極差

又為皇極弦上減小差弦也

已知皇極弦在勾股形日川心 12c12 = (a1 + b1c1)

山地小差弦﹝簡小差在勾股形山地艮 11﹞:c11 = (c1b1)

皇極弦上減小差弦=c12c11 =(a1 + b1c1) –(c1b1)

= [(a1 + b1c1) – (c1b1)]

= [(c1a1 + c1b1c12) – 2b1(c1b1)]

= [c1a1 + c1b1c12 – 2b1c1 + 2b12]

= [c1a1a12b1c1 + b12]

= [(b1 + a1)(b1a1) – c1(b1a1)]

= (b1a1)(b1 + a1c1)

所以髙弦平弦相減 = 皇極弦上減小差弦

若以相減數加於相即黃廣弦也

已知 = (a1 + b1c1)(a1 + b1)﹝見前﹞。

相減 = (b1a1)(b1 + a1c1)﹝亦見前﹞。

相減數 +

=(a1 + b1c1)(a1 + b1) +(b1a1)(b1 + a1c1)

=(a1 + b1c1)(a1 + b1 + b1a1)

=(a1 + b1c1)(2b1)

=(a1 + b1c1)

已知天山弦﹝簡弦﹞= c4= (a1 + b1c1)

比較兩式,可知相減數加於相 = 黃廣弦

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