Wang Xiaotong on right triangles:
Six problems from ‘Continuation of ancient mathematics’ (7th century AD)
[1]

Forthcoming in East Asian Science, Technology, and Medicine.

Tina Su-lyn Lim 林淑铃

4 May 2012

Click on any image to see it enlarged.

Tina Su-lyn Lim is a project manager with NNIT (Novo Nordisk Information Technology) in Copenhagen, Denmark. This article formed part of her M.Sc. thesis in Mathematics at the University of Copenhagen, 2006.

Donald B. Wagner has taught at the University of Copenhagen, the University of Victoria (British Columbia), and the Technical University of Berlin. He has written widely on the history of science and technology in China; his most recent book is the volume on Ferrous Metallurgy in Joseph Needham’s Science and Civilisation in China.

Abstract: Wang Xiaotong’s Jigu suanjing 緝古算經 is primarily concerned with problems in solid and plane geometry leading to polynomial equations which are to be solved numerically using a procedure similar to Horner’s Method. We translate and analyze here six problems in plane geometry. In each case the solution is derived using a dissection of a 3-dimensional object. We suggest an interpretation of one fragmentary comment which at first sight appears to refer to a dissection of a 4-dimensional object.

Wang Xiaotong (late 6th–7th century AD, exact dates unknown) served the Sui and Tang dynasties in posts concerned with calendrical calculations, and presented his book, Jigu suanshu 緝 古算, ‘Continuation of ancient mathematics’, to the Imperial court at some time after AD 626. In 656 it was made one of ten official ‘canons’ (jing ) for mathematical education, and was retitled Jigu suanjing.[2] The book contains 20 problems: one astronomical problem, then 13 on solid geometry, then six on right triangles. All but the first provide extensions of the methods in the mathematical classic Jiuzhang suanshu 九章算術 (‘Arithmetic in nine chapters’, perhaps 1st century AD): the solid-geometry methods in Chapter 5 and the right-triangle methods in Chapter 9,[3] thus ‘continuing’ ancient mathematics. All but the first require the extraction of a root of a cubic or (in two cases) quadratic equation. The present article is concerned with problems 15–20, on right triangles.

Each problem describes a geometric situation, states some given quantities, and asks for one or more other quantities. Then the answer is given, and finally an algorithm for computing the answer. Printed in smaller characters in the text are comments which generally give explanations of the algorithms of the main text. All commentators appear to agree that the comments are by Wang Xiaotong himself, but we have noticed some differences in terminology between the comments and the main text,[4] and feel therefore that the question of the authorship of the comments should be left open. Perhaps most but not all of the comments are by him.

The text

The history of the text is discussed by the modern editors, Qian Baocong (1963: 490–491) and Guo Shuchun and Liu Dun (1998: 1: 21–22); see also He Shaogeng (1989). All extant editions of Jigu suanjing go back to an edition printed in the early 13th century AD. One hand-copy of this edition survived to the 20th century; Qian Baocong appears to have seen it.[5] It is now lost, but in 1684 a copy was included in a collectaneum, Jiguge congshu 汲古閣叢書. This version, the oldest surviving text of the Jigu suanjing, is now available in facsimile on the World Wide Web.[6] The best Qing critical edition is that of Li Huang (1832);[7] others are by Dai Zhen (1777), Bao Tingbo (1780), Zhang Dunren (1803), and the Korean mathematician Nam Pyŏng-Gil (1820–1869).[8] All of these are available on the Web.

The standard modern edition has long been Qian Baocong’s (1963, 2: 487–527; important corrections, 1966), but that of Guo Shuchun and Liu Dun (1998) has much to recommend it. In the parts of the text treated in the present article there is no important difference between the two.

The 1684 edition is marred by numerous obvious scribal errors which must be corrected by reference to the mathematical context. Guo and Liu (1998, 1: 22) note that Li Huang introduced ca. 700 emendations to the text, and that Qian Baocong followed most of these but introduced 20 new emendations. The part of the text considered here gives special difficulties, for it appears that the last few pages of the Southern Song hand copy were damaged, and a great many characters are missing. The problems involved in reconstructing the text are considered further below.

The Jigu suanjing has not been much studied in modern times. The two critical editions, already mentioned, do not explicitly comment on the mathematical content. Lin Yanquan (2001) translates the text into modern Chinese, expresses the calculations in modern notation, and gives derivations of some of the formulas. Deeper studies of individual parts of the text are by Shen Kangshen (1964), Qian Baocong (1966), He Shaogeng (1989), Wang Rongbin (1990), Guo Shiying (1994), and Andrea Bréard (1999: 95–99, 333–336, 353–356). The derivations of Problems 15 and 17 (Figures 2 and 3–4 below) have been reconstructed by Lin Yanquan, by He Shaogeng, and by Wang Rongbin; our reconstruction of the derivation of Problem 19 (Figure 5) is new.

The comments in the text give special difficulties, for two reasons. Being written in smaller characters, they are more subject to banal scribal errors, and their content is more abstract and complex than the main text. They have hardly been studied at all by modern scholars: the most recent serious study we have found is that of Luo Tengfeng (1770–1841) [1993].[9]

Solving cubic equations by ‘Horner’s method’

In the problem solutions given by Wang Xiaotong the coefficients of a cubic equation are calculated, after which the reader is instructed to ‘extract the cube root’. Wang Xiaotong gives no indication of how this was done, but it was obviously a well-known procedure, for algorithms are described in detail in the Jiuzhang suanshu and the Zhang Qiujian suanjing 張 邱建算經 (5th century AD) for the special case of extracting a cube root (the cubic equation x3 = A), and extensions of the latter algorithm to general polynomials of all orders are given in several Chinese books from after Wang Xiaotong’s time.[10]

So much has been written about Chinese methods for extracting the roots of polynomial equations that it is not necessary to go into detail here. Let it suffice here to say that the algorithm is equivalent to ‘Horner’s method’ as sometimes taught in modern schools and colleges.[11] The first digit of the root is determined, the roots of the equation are reduced by the value of that digit (an operation which amounts to a change of variable), the next digit is determined, and this step is repeated until the desired precision is reached. The operation is carried out on paper in the Western version, in the Chinese version with ‘calculating rods’ laid out on a table.

Wang Xiaotong’s text describes how each of the elements of the cubic equation is calculated, using the following terminology:

shi , the constant term

fangfa 方 法, the linear coefficient

lianfa 廉 法, the quadratic coefficient

The cubic coefficient is always 1, and is never mentioned.[12]

The triangle problems

The Chinese text of problems 15–20 and our translation are given in the final section of this article. Here we give an overview using modern terminology.

In pre-modern Chinese mathematics the sides of a right triangle are referred to as gou , ‘shorter leg’, gu , ‘longer leg’, and xian , ‘hypotenuse’. We translate these terms as ‘base’, ‘leg’, and ‘hypotenuse’, and in equations denote them as a, b, and c respectively; see Figure 1.

Figure 1

A peculiarity of the six triangle problems is that no units are given for the quantities involved. This is virtually unique in all of pre-modern Chinese mathematics (including Wang Xiaotong’s Problems 1–14), in which problems are normally stated with reference to practical situations.

In Problem 15 the given quantities are:

ab = 7061/50

ca = 369/10

and the values of a, b, and c are required. In the solution the coefficients of a cubic equation are calculated:

(1)

This has one real root,

a = 147/20

whereafter it is straightforward to calculate the remaining quantities:

b =  = 491/5

c = a + (ca) = 511/4

A comment, printed in smaller characters, explains (1) in a mixture of algebraic and geometric reasoning. It first notes that

(ab)2 = a2b2

The comment refers to a fang (rectangular parallelipiped) and to quantities ‘lined up’; this suggests a geometric rather than algebraic derivation of the method. We reconstruct this derivation as in Figure 2. We note that

(which Wang Xiaotong does not state explicitly). Then using Figure 2,

It is not known whether Wang Xiaotong’s book originally included illustrations like Figure 2. We think it probably did not; he seems simply to give verbal descriptions of the geometric constructions which he uses.

Figure 2

Problem 16 is equivalent to Problem 15. The given quantities are

ab = 40361/5

cb = 61/5

and c is required. The cubic equation here is

(2)

This has one real root,

b = 1081/2

whence

c = b + (cb) = 1147/10

Problem 17 gives the quantities

ac = 13371/20

cb = 11/10

and b is required. The cubic equation arrived at is

(3)

This has one real root,

b = 922/5

A comment explains (3) geometrically, but here our textual problems start, for a large number of characters are missing in the extant editions. In spite of the lacunae in the comment text it is clear that the geometrical construction is similar to that shown in Figures 3 and 4. The solid in Figure 3 has volume

which is the right side of (3). The sum of the volumes of the blocks into which the solid is divided (Figure 4) is

which is the left side of (3).

Note that the dimensions of the triangles in Problems 15–17 are based on to Pythagorean triples:

15.        [147/20, 491/5, 511/4] = [7, 24, 25] × 41/20

16.        [371/5, 1081/2, 1147/10] = [12, 35, 37] × 31/10                                     (4)

17.        [143/10, 922/5, 931/2] = [13, 84, 85] × 11/10

Figure 3

Figure 4

In Problems 18–20 so much is missing from the text that any reconstruction will to some extent be speculative. The Chinese mathematician Zhang Dunren 張敦 仁 (1754-1834) and the Korean mathematician Nam Pyŏng-Gil 南秉吉 (1820–1869) have given two reconstructions which agree in principle but differ in detail. Neither explains how he arrived at his reconstruction, but some matters seem clear.

Any reconstruction should satisfy the approximate number of characters seen to be missing from the text.[13] In addition, certain assumptions can safely be made: (1) The problems all concern right triangles; (2) the dimensions of the triangles are derived from Pythagorean triples, as in Problems 15–17; (3) the problem statements all follow the same rigid pattern as in Problems 15–17; and (4) the problems come in pairs, so that, just as 15 and 16 are equivalent, so are the pairs 17–18 and 19–20.

Assumption (4) implies that the fragmentary problem statements are

17. Given ac and cb, determine b.

18. Given bc and ca, [determine either a or b].

19. Given bc [and a], determine b.

20. Given [ac and] b, [determine a].

What remains of the ‘method’ for Problem 18 indicates that the quantity to be determined is b.

Zhang Dunren and Nam Pyŏng-Gil agree on the form of the problem statements as given here, but they differ on the numerical values of the given quantities in Problems 18 and 19. Their reconstructions are as follows:[14]

 Nam Pyŏng-Gil Zhang Dunren 18 Given bc = 44283/5 47393/5 c–a = 55 542/5 Result b = 66 68 Other quantities a = 121/10 153/10 c = 671/10 697/10 19 Given bc = 3/50 726 a = 7/100 77/10 Result b = 6/25 262/5 Other quantity c = 1/4 271/2 20 Given ac = 16414/25 b = 161/2 Result a = 84/5 Other quantity c = 187/10

The Pythagorean triples on which these reconstructions are based are:

18.        [121/10, 66, 671/10] = [11, 60, 61] × 11/10

[153/10, 68, 697/10] = [9, 40, 41] × 17/10

19          [7/100, 6/25, 1/4] = [7, 24, 25] × 1/100

[77/10, 262/5, 271/2] = [7, 24, 25] × 11/10

20.         [84/5, 161/2, 187/10] = [8, 15, 17] × 11/10

In the following we have arbitrarily chosen to follow Zhang Dunren’s reconstructions of the given quantities. Zhang Dunren also attempts – with much less success – to reconstruct large parts of the ‘method’ sections, and we have not followed him in the translation. Nam Pyŏng-Gil’s edition does not include Wang Xiaotong’s methods, but instead gives his own, which appear to follow the methods of Li Ye 李冶 (1192–1279) in Ceyuan haijing 測圓海 鏡.[15]

Problem 18 is, by assumption (4) above, equivalent to Problem 17. It results in the cubic equation

(cf. (3) above). In Zhang Dunren’s reconstruction,

This has one real root,

a = 153/10

And

= 68

Problems 19 and 20 are extremely fragmentary, but the assumption that they are equivalent makes Zhang Dunren’s and Nam Pyŏng-Gil’s reconstructions of the problems and their methods plausible. The method for Problem 19 leads in each case to the quadratic equation

(5)

which in Zhang Dunren’s version gives

x2 + 5929/100 x = 52706

where x = b2. This has one positive root.

b = 262/5

It is easy to derive (5) algebraically, for

but what little remains of the comment appears to imply a geometric explanation. Commentators have had difficulty here, for a geometric explanation would seem to involve four-dimensional quantities. The lack of units in the quantities involved, however, makes the construction shown in Figure 5 plausible, in which both b and b2 are used as linear measures. The volume of the solid is

b2c2 = (bc)2

which is the right side of (5), and the sum of the two blocks into which the solid is divided is

which is the left side of (5). If our reconstruction is correct, the ‘length’ mentioned in one of the remaining fragments of the comment is probably x, the length of the solid.

Figure 5

Problem 20, assumed to be equivalent to Problem 19, results in the quadratic equation

where x = a2. In Zhang Dunren’s reconstrruction,

x2 + 2721/4 x = 27079621/625

This has one positive root,

a = 84/5

Concluding remarks

How did Wang Xiaotong arrive at the results in his book? The question is probably unanswerable, but some guesses may be permissible.

We presume that he started with one or more general methods and derived a number of concrete geometric constructions which could solved by their use. Our study of several of his problems in solid geometry[16] suggests that his most important method was dissections of 3-dimensional objects leading to (in modern terminology) cubic equations. (We hope that further research will lead to a more specific description.)

Our guess is then that Wang Xiaotong’s study of the possibilities of this method led, in addition to the problems in solid geometry, to the three dissections related to right triangles reconstructed here.

Having found a relation between the dimensions of a right triangle and a cubic equation, Wang Xiaotong then needed only to fit a specific right triangle to it. In each case he used a Pythagorean triple, making the student’s work more difficult by multiplying by a fraction. A general method for generating Pythagorean triples seems to have been known to Chinese mathematicians since the Jiuzhang suanshu,[17] but the three triples known to have been used by Wang Xiaotong (see (4) above) all have cb equal to 1 or 2. They could therefore be generated using the simpler special cases ascribed by Proclus to Pythagoras and Plato respectively:[18]

We confidently expect that careful study of the solid-geometry problems in the book will reveal further examples of disguised Pythagorean triples, and that these will provide useful clues to Wang Xiaotong’s general methods.

A further question is the function of this book in mathematical education. Though we know that it was adopted as one of the textbooks of Imperial mathematical education, we do not know whether it was originally written for this purpose, or perhaps was intended as a tour de force in pure mathematics. The main governmental uses of advanced mathematics – beyond accounting and the like – were in astronomy and in public works. It is difficult to see any use for these geometrical methods in astronomy, but in public works a thorough understanding of dissection methods (not necessarily the specific problems given in the book) might have been very useful to an official calculating the volumes of earthworks and their labour requirements.

Chinese text and translation

In the Chinese text we indicate text variants with the notation {a/b/c}, where a is the 1684 version, b is Qian Baocong’s version, which is usually the same as Zhang Dunren’s, and c, if present, is from some other version, indicated in a footnote. We use Qian Baocong’s punctuation.

Our translations of the smaller-character comments in the Chinese text are printed in a sans-serif font and with a sideline. Our mathematical comments are given indented in the text, while a few philological comments are given in footnotes.

Problem 15

[In a right triangle] the area obtained by multiplying the base [a] by the leg [b] is [ab =] 7061/50 and the hypotenuse [c] is greater than the base [a] by [ca =] 369/10. How large are the three quantities?

base, [a =] 147/20

leg, [b =] 491/5

hypotenuse, [c =] 511/4

Method: Multiply the area [ab] by itself and divide by twice the difference [2(ca)] to make the shi [the constant term]. Halve the difference [ca] to make the lianfa [the quadratic coefficient]. Extract the cube root; this is the base [a].

Add the hypotenuse difference [ca]; this is the hypotenuse [c]. Divide the product [ab] by the base [a]; this is the leg [b].

c = a + (ca)

[Comment:] The product of the base and the leg, multiplied by itself [(ab)2], is the ‘volume’ [ji ] obtained by multiplying the area of [the square on] the base by the area of [the square on] the leg [a2b2].

(ab)2 = a2b2

In the following see Figure 2.

Therefore dividing by the doubled difference between the base and the hypotenuse [2(ca)] gives the base and the halved difference lined up [a + (ca)/2] multiplied by the area of [the square on] the base [a2] to make a box [fang , rectangular parallelipiped].

This is why the difference is halved to make the lianfa, and the extraction of the cube root is carried out.

Problem 16

[In a right triangle], the area obtained by multiplying the base [a] by the leg [b] is [ab=] 40361/5. The leg is less than the hypotenuse [c] by [cb =] 61/5. How large is the hypotenuse?

Answer: The hypotenuse is [c=] 1147/10.

This problem is equivalent to Problem 15.

Method: Multiply the area [ab] by itself and divide by twice the difference [cb] to make the shi [the constant term]. Halve the difference [cb] to make the lianfa [the quadratic coefficient]. Extract the cube root; this is the leg [b].

Add the difference [cb] to obtain the hypotenuse [c].

c = b + (cb) = 1147/10

Problem 17

[In a right triangle] the area obtained by multiplying the base [a] by the hypotenuse [c] is [ac =] 13371/20, and the hypotenuse is greater than the leg by [cb =] 11/10. How large is the leg [b]?

Method: Multiply the area [ac] by itself and divide by twice the difference [cb] to make a volume . Further multiply the difference [cb] twice by itself, halve this, and subtract from the volume [V]. The difference is the shi [the constant term].

Further multiply the difference [cb] by itself and double it to make the fangfa [the linear coefficient].

Further lay out the difference [cb], multiply by 5, and divide by 2 to make the lianfa [the quadratic coefficient].

Extract the cube root. This is the leg [b].

[Comment:] Multiplying the area [ac] obtained by multiplying the base by the hypotenuse by itself gives the ‘volume’ [ji ] obtained by multiplying the area of [the square on] the base [a2] by the area of [the square on] the hypotenuse [c2].

(ac)2  =  a2c2

From here on the text is very fragmentary.

Therefore dividing by twice the difference [c-b] gives one leg [b] and . . .
[5 characters missing] . . . to make a box [fang
]. The difference [cb], multiplied twice by itself and halved, is the corner-piece [yu ] . . .
[5 characters missing] . . . two standing side-pieces [li lian

[11 characters missing] . . . doubled to make the longitudinal corner-piece/s [zong yu

[11 characters missing] . . . the difference [cb] makes the upper side-piece/s [shang lian

[9 characters missing] . . . fa
[coefficient?]. Therefore it is multiplied by 5 and divided by 2.  . . .
[9 characters missing].

Problem 18

[In a right triangle] the area obtained by multiplying the leg [b] by the hypotenuse [c] is [bc =] . . .
[9 characters missing] . . . 3. The base [a] is less than the hypotenuse [c] by [ca =] 50 . . .
[characters missing] . . . Answer: 6 . . .
[characters missing] . . . Method: Multiply the area [bc] by itself . . .
[11 characters missing] . . . multiply twice by itself and halve it . . .
[10 characters missing] . . . multiply and double it to make the fangfa [the linear coefficient].  . . .
[10 characters missing] . . . lianfa [the quadratic coefficient]. Carry out the extraction of the cube root . . .
[10 characters missing] . . . the area gives the leg.

Problem 19

[In a right triangle] the area obtained by multiplying the leg [b] by the hypotenuse [c] is [bc =] . . .
[10 characters missing] . . . 7. How large is the leg [b]?
Answer: The leg is 20 . . .
[characters missing] . . . Method:
The area . . . by itself . . .
[12 characters missing] . . . divide it by . . . ; the result . . .
[characters missing] . . .

[Comment:] [characters missing] . . . the number is also the leg [b] . . .
[characters missing] . . . to make the length [chang
], . . . by the leg [b] . . .

[characters missing] . . . to obtain the area of [the square on] the leg [b]. Extract the . . . root . . .

[characters missing] . . . gu bei fen mu chang 股 北分母常 [?]

Problem 20

[In a right triangle] the leg [b] is 16 and [one] half . . .
[10 characters missing] . . . 4 [and] . . . twenty-fifths . . .
[characters missing] . . .
[characters missing] . . .
Method:
The area multiplied by itself . . .
[11 characters missing] . . . extract the . . . root and extract the square root of the result . . .
[characters missing]

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[1] We are grateful to Karine Chemla for drawing our attention to Wang Xiaotong’s book, and for detailed comments on this article; to Jesper Lützen and Ivan Tafteberg for comments on an earlier version; and to Mark Elvin, Jia-ming Ying 英家 銘, and four anonymous reviewers for additional comments.

[2] Jiu Tang shu 1975, 32: 1168, 47: 2039, 4: 76, 44: 1892; Qian Baocong 1963: 487.

[3] Chemla and Guo 2004: 387–457, 661–745; see also Cullen 1993; Shen Kangshen et al. 1999: 250–306, 439–517.

[4] E.g. in the commentary to problem 3, Qian Baocong 1963: 506, lines 5–8. Here bienao 鼈 臑 is written in full, while it is always abbreviated as bie in the main text; yanchu 羡除 is once written in full and twice abbreviated as chu, while in the main text it is always abbreviated as yan.

[5] Qian Baocong describes the physical condition of the manuscript, and refers to it throughout his critical edition. He Shaogeng (1989: 37) states that the manuscript is ‘now’ in the Palace Museum in Beijing, but he does not appear to have seen it himself. Guo Shuchun and Liu Dun state that it is now lost.

[6] This is the Tianlu Linlang congshu 天祿琳瑯叢書 edition, a facsimile printed in 1931, available at http://www.scribd.com/doc/78515037.

[7] On Li Huang’s edition see Huang Juncai 2008.

[8] On Nam Pyŏng-Gil see Jia-ming Ying 2011; Horng Wann-sheng 2002; Zhang Fukai 2005.

[9] The only exceptions to this statement appear to be the highly speculative reconstructions of the fragmentary commentary to Problem 17 by He Shaogeng (1989) and Wang Rongbin (1990).

[10] Chemla & Guo 2004: 322–335, 363–379; Shen et al. 1999: 175–195, 204–226; Wang & Needham 1955; Lam Lay Yong 1970; 1977: 195–196, 251–285; 1986; Libbrecht 1973: 175–191; Chemla 1994; Martzloff 1997: 221–249.

[11] See e.g. Rees & Sparks 1967: 294–297 as well as numerous pages on the World Wide Web. Horner (1819) presented a procedure for approximating roots of any infinitely differentiable function, but modern descriptions of ‘Horner’s method’ consider only the special case of polynomial functions.

[12] Commentators are divided on the function of the word cong / zong in the phrase 從開立方除. Qian Baocong (1966: 46–47) reviews the various attempts to explain it, and comes to the conclusion that in each case it belongs at the end of the previous sentence, and means ‘follow, accompany’ (gensui 跟随). This is difficult for us to understand, and we omit the word in our translation. Another interpretation is given by Chemla and Guo (2004: 65, 912).

[13] Assuming that the Tianlu Linlang congshu edition used the same page and line lengths as the Southern Song edition. This assumption is not certain, but the counts of missing characters may be taken as approximations.

[14] In all but one case the two editors insert the same number of characters counted by the Zhibuzuzhai edition. In giving the value of bc in Problem 18, both insert 10 characters where the Zhibuzhai edition indicates only 9 missing.

[15] See e.g. Mei Rongzhao 1966; Chemla 1982; Martzloff 1997: 143–149; Zhang Fukai 2005.

[16] These are Problems 2, 3, 4, 5, and 7, each of which consists of several related but distinct problems. Our interpretation of the four problems of Problem 2 is given in a forthcoming article, ‘The Grand Astrologer’s platform and ramp’, at present under review by another journal.

[17] Chemla and Guo (2004: 727), under the interpretation of van der Waerden (1983: 5–7).

[18] Heath 1921, 1: 80–81.

[19] Zhang Dunren 1803.

[20] Guo Shuchun & Liu Dun 1998, 2: 18.