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Ce Yuan Hai Jıng

Paul Yiu

Department of MathematicsFlorida Atlantic University

Summer 2005

Version 050825

Ce Yuan Hai Jıng

2

north

east

A

B C(1)

D(10)E

F (11)

G(15)

H(14)

I(2)

J(3)

K

L

M

NO(12)

P (6)

Q(9)

R(7)

S(8)

W (4)X(13)

Y (5)

The Circular City Diagram

Chapter 2

CYHJ Book I

2.1 The circular city diagram

Ceyuan Haıjıng begins with a section on terminology (Zong luu mıng hao), explaining the labels ofthe vertices in the circular city diagram and the names of various segments.

A = tian (heaven) B = dı (earth)C = qian D = kun E = xun F = genG = dong (east) H = nan (south) I = xı (west) J = bei (north)K = rı (sun) L = yue (moon) M = shan (mountain) N = chuan (river)O = xın (heart)P = dan (dawn) Q = xı (dusk)R = zhu (red) S = qıng (green)W = jın (gold) X = fan (flood) Y = quan (stream)

Remark. qian, kun, xun, gen are four of the eight trigrams which make up the 64 hexagrams inYıjıng. The eight trigrams (ba gua) with their names and what they symbolize are as follows.

qian kun zhen xun kan lı gen duı(heaven) (earth) (storm) (wind) (water) (fire) (mountain) (lake)

Each right triangle in the circular city diagram can be identified by its hypotenuse which is asegment of the main hypotenuse (t ong xian), the two legs being along the east-west and south-north directions respectively. These are namedt ong, bian, dı, ...,mıng, zhuan. We enumerate thesetriangles as∆n, n = 1, 2, . . . , 15, and attach these numbers to the corresponding right angle vertices.

4 Yiu: A commentary on Ce Yuan Haı Jıng

2.2 Definitions (Zong luu mıng hao)

Text

∆1 Tian zhı dı wei tong xian, ABmain tian zhı qian wei tong gu, AC

qian zhı dı wei tong gou. CB∆2 Tian zhı chuan wei bian xian, ANside tian zhı xı wei bian gu, AI

xı zhı chuan wei bian gou. IN∆3 Rı zhı bei wei dı xian, KJbase rı zhı dı wei dı gu, KB

dı zhı bei wei dı gou. BJ∆4 Tian zhı shan wei huang guang xian, AMyellow tian zhı jın wei gu, jı gu fang cha ye, AWwidth jın zhı shan wei gou. WM∆5 Yue zhı dı wei huang chang xian, LByellow yue zhı quan wei gu, LYlength quan zhı dı wei gou. Y B∆6 Tian zhı rı wei shang gao xian, AKupper tian zhı dan wei gu, APheight dan zhı rı wei gou. PK∆7 Rı zhı shan wei xia gao xian, KMlower rı zhı zhu wei gu, KRheight zhu zhı shan wei gou. RM∆8 Yue zhı chuan wei shang pıng xian, LN

yue zhı qıng wei gu, LSqıng zhı chuan wei gou. SN

∆9 Chuan zhı dı wei xia pıng xian, NBchuan zhı xı wei gu, NQxı zhı dı wei gou. QB

∆10 Tian zhı yue wei da cha xian, ALbig tian zhı kun wei gu, ADdifference kun zhı yue wei gou. DL∆11 Shan zhı dı wei xiao cha xian, MBsmall Shan zhı gen wei gu, MFdifference gen zhı dı wei gou. FB∆12 Rı zhı chuan wei huang jı xian, KNsupreme rı zhı xın wei gu, KOauthority xın zhı chuan wei gou. ON∆13 Yue zhı shan wei tai xu xian, LMgreat yue zhı fan wei gu, LXvoid fan zhı shan wei gou. XM∆14 Rı zhı yue wei mıng xian, KLbright rı zhı nan wei gu, KH

nan zhı yue wei gou. HL∆15 Shan zhı chuan wei zhuan xian, MNtiny shan zhı dong wei gu, MG

dong zhı chuan wei gou. GN

2.3 The data

Along with the sides of a right triangle are the five sums and five differences (wu huo wu jiao).

2.3 The data 5

gou a

gu b

xian c

gou gu huo or (bıng) a + b huo or bınggou gu jiao or (cha) b − a jiao or chagou xian huo or (bıng) a + c

gou xian jiao or (cha) c − a da chagu xian huo or (bıng) a + c

gu xian jiao or (cha) c − a xiao chaxian jiao huo c + (b − a) xjhxian jiao jiao c − (b − a) xjjxian huo huo or san shı huo (a + b) + c xhhxian huo jiao (a + b) − c xhj

Remarks. (1) Huo andbıng both mean “sum”. They are used interchangeably.(2) Cha means difference andjiao means comparison.(3) San shı means “three things”. They refer to thegou, gu andxian (hypotenuse).

2.3.1 The standard set (Jın wen zheng shu)

The main right triangle in the circular city diagram hasg ou 320, gu 600, hypotenuse680, and incirclediameter240. The various lengths associated with the different right triangles in the circular citydiagrams are listed (jın wen zheng shu).

c a b a + b b − a a + c c − a b + c c − b xjh xjj xhh xhj

∆1 680 320 600 920 280 1000 360 1280 80 960 400 1600 240∆2 544 256 480 736 224 800 288 1024 64 768 320 1280 192∆3 425 200 375 575 175 625 225 800 50 600 250 1000 150∆4 510 240 450 690 210 750 270 960 60 720 300 1200 180∆5 272 128 240 368 112 400 144 512 32 384 160 640 96∆6 255 120 225 345 105 375 135 480 30 360 150 600 90∆7 255 120 225 345 105 375 135 480 30 360 150 600 90∆8 136 64 120 184 56 200 72 256 16 192 80 320 48∆9 136 64 120 184 56 200 72 256 16 192 80 320 48∆10 408 192 360 552 168 600 216 768 48 576 240 960 144∆11 170 80 150 230 70 250 90 320 20 240 100 400 60∆12 289 136 255 391 119 425 153 544 34 408 170 680 102∆13 102 48 90 138 42 150 54 192 12 144 60 240 36∆14 153 72 135 207 63 225 81 288 18 216 90 360 54∆15 34 16 30 46 14 50 18 64 4 48 20 80 12


2.3.2 Four new sets (Xın she sı luu)

The Qing commentators appended at the very end of Book I, afterSBZJ, the data for four moreright triangles. These are the right triangles(6, 8, 10), (10, 24, 26), (14, 48, 50), and(18, 80, 82)magnified60, 30, 28, and45 times respectively. It was remarked that the diameter of the circularcity is the same as the standard one (jın wen zheng sh`) for the first two and different for the thirdand the fourth.

The first new set


∆1 600 360 480 840 120 960 240 1080 120 720 480 1440 240∆2 450 270 360 630 90 720 180 810 90 540 360 1080 180∆3 400 240 320 560 80 640 160 720 80 480 320 960 160∆4 400 240 320 560 80 640 160 720 80 480 320 960 160∆5 300 180 240 420 60 480 120 540 60 360 240 720 120∆6 200 120 160 280 40 320 80 360 40 240 160 480 80∆7 200 120 160 280 40 320 80 360 40 240 160 480 80∆8 150 90 120 210 30 240 60 270 30 180 120 360 60∆9 150 90 120 210 30 240 60 270 30 180 120 360 60∆10 300 180 240 420 60 480 120 540 60 360 240 720 120∆11 200 120 160 280 40 320 80 360 40 240 160 480 80∆12 250 150 200 350 50 400 100 450 50 300 200 600 100∆13 100 60 80 140 20 160 40 180 20 120 80 240 40∆14 100 60 80 140 20 160 40 180 20 120 80 240 40∆15 50 30 40 70 10 80 20 90 10 60 40 120 20

The second new set


∆1 780 300 720 1020 420 1080 480 1500 60 1200 360 1800 240∆2 650 250 600 850 350 900 400 1250 50 1000 300 1500 200∆3 468 180 432 612 252 648 288 900 36 720 216 1080 144∆4 624 240 576 816 336 864 384 1200 48 960 288 1440 192∆5 260 100 240 340 140 360 160 500 20 400 120 600 80∆6 312 120 288 408 168 432 192 600 24 480 144 720 96∆7 312 120 288 408 168 432 192 600 24 480 144 720 96∆8 130 50 120 170 70 180 80 250 10 200 60 300 40∆9 130 50 120 170 70 180 80 250 10 200 60 300 40∆10 520 200 480 680 280 720 320 1000 40 800 240 1200 160∆11 156 60 144 204 84 216 96 300 12 240 72 360 48∆12 338 130 312 442 182 468 208 650 26 520 156 780 104∆13 104 40 96 136 56 144 64 200 8 160 48 240 32∆14 208 80 192 272 112 288 128 400 16 320 96 480 64∆15 26 10 24 34 14 36 16 50 2 40 12 60 8

2.3 The data 7

The third new set


∆1 1400 392 1344 1736 952 1792 1008 2744 56 2352 448 3136 336∆2 1225 343 1176 1519 833 1568 882 2401 49 2058 392 2744 294∆3 800 224 768 992 544 1024 576 1568 32 1344 256 1792 192∆4 1200 336 1152 1488 816 1536 864 2352 48 2016 384 2688 288∆5 350 98 336 434 238 448 252 686 14 588 112 784 84∆6 600 168 576 744 408 768 432 1176 24 1008 192 1344 144∆7 600 168 576 744 408 768 432 1176 24 1008 192 1344 144∆8 175 49 168 217 119 224 126 343 7 294 56 392 42∆9 175 49 168 217 119 224 126 343 7 294 56 392 42∆10 1050 294 1008 1302 714 1344 756 2058 42 1764 336 2352 252∆11 200 56 192 248 136 256 144 392 8 336 64 448 48∆12 625 175 600 775 425 800 450 1225 25 1050 200 1400 150∆13 150 42 144 186 102 192 108 294 6 252 48 336 36∆14 450 126 432 558 306 576 324 882 18 756 144 1008 108∆15 25 7 24 31 17 32 18 49 1 42 8 56 6

The fourth new set


∆1 3690 810 3600 4410 2790 4500 2880 7290 90 6480 900 8100 720∆2 3321 729 3240 3969 2511 4050 2592 6561 81 5832 810 7290 648∆3 2050 450 2000 2450 1550 2500 1600 4050 50 3600 500 4500 400∆4 3280 720 3200 3920 2480 4000 2560 6480 80 5760 800 7200 640∆5 738 162 720 882 558 900 576 1458 18 1296 180 1620 144∆6 1640 360 1600 1960 1240 2000 1280 3240 40 2880 400 3600 320∆7 1640 360 1600 1960 1240 2000 1280 3240 40 2880 400 3600 320∆8 369 81 360 441 279 450 288 729 9 648 90 810 72∆9 369 81 360 441 279 450 288 729 9 648 90 810 72∆10 2952 648 2880 3528 2232 3600 2304 5832 72 5184 720 6480 576∆11 410 90 400 490 310 500 320 810 10 720 100 900 80∆12 1681 369 1640 2009 1271 2050 1312 3321 41 2952 410 3690 328∆13 328 72 320 392 248 400 256 648 8 576 80 720 64∆14 1312 288 1280 1568 992 1600 1024 2592 32 2304 320 2880 256∆15 41 9 40 49 31 50 32 81 1 72 10 90 8


2.4

We label these right triangles∆n for n = 1, 2, . . . , 15. Thegou, gu and hypotenuses are labelledan, bn andcn respectively. The triangles∆6 and∆7 are congruent. Their side lengths will simplybe writtenag, bg andcg. 1 Likewise, ∆8 and∆9 are congruent. Their side lengths will simplybe writtenap, bp and cp. 2 It is also clear that∆4 and ∆g have similarlity ratio 2, so do∆6

and∆p. Therefore, concerning lengths, we shall only consider the eleven triangles∆ n for n =1, 2, 3, g, p, 10, 11, 12, 13, 14 and15.

It is well known that an integer right triangle can be constructed from two positive integersnumbersp > q by takinga = 2pq, b = p2 − q2 andc = p2 + q2. 3 The main triangle inCYHJcorresponds to(p, q) = (4, 1), magnified40 times so that all the segments associated with the∆n

have integer lengths. More generally, if we take

a1 = 4p2q(p + q), b1 = 2p(p − q)(p + q)2, c1 = 2p(p + q)(p2 + q2),

these lengths are all integers. There are generically 68 different lengths associated with the eleventriangles∆n for n = 1, 2, 3, g, p, 10, 11, 12, 13, 14 and15. Each of these 68 lengths is givenby a quartic form inp andq. The data for the triangles∆n are summarized in the table below.

Lengths of segments associated with the triangles∆n

n 1 2 3 g p 10 11 12 13 14 15

cn 30 11 63 59 20 26 50 66 48 55 39an 7 2 19 18 5 6 16 20 15 17 14bn 25 9 58 54 18 24 44 59 43 53 37an + bn 32 13 65 61 22 28 51 68 49 57 41bn − an 31 12 64 60 21 27 47 67 45 56 40an + cn 29 10 62 58 19 25 46 63 44 54 38cn − an 24 8 54 53 17 23 43 55 42 52 36bn + cn 4 1 10 9 2 3 7 11 6 8 5cn − bn 16 5 38 37 14 15 35 39 34 36 33xjh 2 · 9 3 25 24 6 2 · 8 2 · 18 26 2 · 17 23 15xjj 2 · 19 7 46 44 16 2 · 18 2 · 38 50 2 · 37 43 35xhh 2 · 10 4 29 25 7 2 · 9 2 · 19 30 2 · 18 24 16xhj 2 · 18 6 44 43 15 2 · 17 2 · 37 48 2 · 36 42 34

In this table, an entryk = 1,2, . . . ,68 stands for a quartic formfk given in the followingdictionary, which also shows the various elements realized byfk. These quartic forms are arrangedlexicographically according to

p, q, p − q, p + q, p2 + q2 p2 − 2pq − q2, p2 + 2pq − q2.

The quartic formsfk

1g refers togao. These triangles are calledshang gao andxia gao.2p refers topıng. These triangles are calledshang pıng andxia pıng.3Such a construction is explained in the solution of Problem IX.14 ofJiuzhang Suanshu. In the Chinese tradition, the

assumption thatgou being shorter thangu requiresp2 − 2pq − q2 > 0.

2.4 9

k fk

1 4p4 b2 + c2

2 4p3q a2 bp + cp

3 4p3(p − q) c2 + (b2 − a2) b10 + c104 4p3(p + q) b1 + c1 a2 + b2 + c2

5 4p2q2 c2 − b2 ap b15 + c156 4p2q(p − q) (a2 + b2) − c2 cp + (bp − ap) a10

b13 + c137 4p2q(p + q) a1 c2 − (b2 − a2) ap + bp + cp

b11 + c118 2p2(p − q)2 c2 − a2 b14 + c14

2 · 8 4p2(p − q)2 c10 + (b10 − a10)9 2p2(p − q)(p + q) b2 bg + cg

2 · 9 4p2(p − q)(p + q) c1 + (b1 − a1) a10 + b10 + c1010 2p2(p + q)2 a2 + c2 b3 + c3

2 · 10 4p2(p + q)2 a1 + b1 + c111 2p2(p2 + q2) c2 b12 + c1212 2p2(p2 − 2pq − q2) b2 − a2

13 2p2(p2 + 2pq − q2) a2 + b2

14 4pq3 cp − bp a15

15 4pq2(p − q) (ap + bp) − cp c10 − b10 a13

c15 + (b15 − a15)16 4pq2(p + q) c1 − b1 cp − (bp − ap) a11

a15 + b15 + c1517 2pq(p − q)2 cp − ap a14

2 · 17 4pq(p − q)2 (a10 + b10) − c10 c13 + (b13 − a13)18 2pq(p − q)(p + q) ag bp

2 · 18 4pq(p − q)(p + q) (a1 + b1) − c1 c10 − (b10 − a10) c11 + (b11 − a11)a13 + b13 + c13

19 2pq(p + q)2 a3 ap + cp

2 · 19 4pq(p + q)2 c1 − (b1 − a1) a11 + b11 + c1120 2pq(p2 + q2) cp a12

21 2pq(p2 − 2pq − q2) bp − ap

22 2pq(p2 + 2pq − q2) ap + bp


k fk

23 2p(p − q)3 c10 − a10 c14 + (b14 − a14)24 2p(p − q)2(p + q) c1 − a1 cg + (bg − ag) b10

a14 + b14 + c1425 2p(p − q)(p + q)2 b1 c3 + (b3 − a3) ag + bg + cg

a10 + c10

26 2p(p − q)(p2 + q2) c10 c12 + (b12 − a12)27 2p(p − q)(p2 − 2pq − q2) b10 − a10

28 2p(p − q)(p2 + 2pq − q2) a10 + b1029 2p(p + q)3 a1 + c1 a3 + b3 + c330 2p(p + q)(p2 + q2) c1 a12 + b12 + c1231 2p(p + q)(p2 − 2pq − q2) b1 − a1

32 2p(p + q)(p2 + 2pq − q2) a1 + b133 4q4 c15 − b1534 4q3(p − q) c13 − b13 (a15 + b15) − c1535 4q3(p + q) c11 − b11 c15 − (b15 − a15)36 2q2(p − q)2 c14 − b14 c15 − a15

2 · 36 4q2(p − q)2 (a13 + b13) − c1337 2q2(p − q)(p + q) cg − bg b15

2 · 37 4q2(p − q)(p + q) (a11 + b11) − c11 c13 − (b13 − a13)38 2q2(p + q)2 c3 − b3 a15 + c15

2 · 38 4q2(p + q)2 c11 − (b11 − a11)39 2q2(p2 + q2) c12 − b12 c1540 2q2(p2 − 2pq − q2) b15 − a15

41 2q2(p2 + 2pq − q2) a15 + b1542 2q(p − q)3 c13 − a13 (a14 + b14) − c1443 2q(p − q)2(p + q) (ag + bg) − cg c11 − a11 b13

c14 − (b14 − a14)44 2q(p − q)(p + q)2 (a3 + b3) − c3 cg − (bg − ag) b11

a13 + c1345 2q(p − q)(p2 − 2pq − q2) b13 − a13

46 2q(p + q)3 c3 − (b3 − a3) a11 + c1147 2q(p + q)(p2 − 2pq − q2) b11 − a11

48 2q(p − q)(p2 + q2) (a12 + b12) − c12 c1349 2q(p − q)(p2 + 2pq − q2) a13 + b1350 2q(p + q)(p2 + q2) c11 c12 − (b12 − a12)51 2q(p + q)(p2 + 2pq − q2) a11 + b11

52 (p − q)4 c14 − a14

53 (p − q)3(p + q) cg − ag b1454 (p − q)2(p + q)2 c3 − a3 bg a14 + c1455 (p − q)2(p2 + q2) c12 − a12 c1456 (p − q)2(p2 − 2pq − q2) b14 − a14

57 (p − q)2(p2 + 2pq − q2) a14 + b1458 (p − q)(p + q)3 b3 ag + cg

59 (p − q)(p + q)(p2 + q2) cg b1260 (p − q)(p + q)(p2 − 2pq − q2) bg − ag

61 (p − q)(p + q)(p2 + 2pq − q2) ag + bg

2.4 11

k fk

62 (p + q)4 a3 + c363 (p + q)2(p2 + q2) c3 a12 + c1264 (p + q)2(p2 − 2pq − q2) b3 − a3

65 (p + q)2(p2 + 2pq − q2) a3 + b366 (p2 + q2)2 c1267 (p2 + q2)(p2 − 2pq − q2) b12 − a12

68 (p2 + q2)(p2 + 2pq − q2) a12 + b1269 (p2 − 2pq − q2)2

70 (p2 − 2pq − q2)(p2 + 2pq − q2)71 (p2 + 2pq − q2)2


CHEN Weiqi (1889) introduced the notion of generalized product (f an jı) and discovered thatthe various quartic forms are integer combinations ofa g, bg, ap, a12, andb12. 4

ag bg ap a12 b12

a1 1 1 1b1 1 1 1c1 1 1 1 1

a2 1 1b2 1 1c2 1 1 1

a3 1 1b3 1 1c3 1 1 1

a10 1 −1 1b10 −1 1 1c10 1 1 −1 1

a11 −1 1 1b11 1 −1 1c11 1 1 1 −1

c12 1 1

a13 1 1 −1b13 1 1 −1c13 −1 −1 1 1

a14 −1 1b14 −1 1c14 1 1 −1

a15 −1 1b15 −1 1c15 1 1 −1

WANG Guitong: if x = ap andy = bg, everyfk is an integer combination ofx, y,√

xy,√x(x + y), and

√y(x + y) with coefficients0, ±1, ±2.

4See KONG Guoping’s Introduction to CYHJ,Compendium, volume I, p.728.

Chapter 4

Shıbie Zajı: Section I

4.1 SBZJ I.1-2: Zhu za ming mu

4.1.1 Text

1.1 Tian zhı yu rı yu rı zhi yu xın tong.1.2 Xın zhı yu chuan yu chuan zhı yu de tong.2.1.1 Rı zhı xın yu rı zhı yu shan tong,2.1.2 gu yı shan zhı chuan wei xiao cha.2.2.1 Chuan zhı xın yu chuan zhı yu yue tong,2.2.2 gu yi yue zhı rı wei da cha.

4.1.2 Translation

1.1 AK = KO, (c6 = b12).1.2 ON = NB, (a12 = c9).2.1.1 KO = KM , (b12 = c6).2.1.2 ThereforeMN = c15 = KN − KM = KN − KO and is calledxiao cha.2.2.1 NO = NL, (a12 = c8).2.2.2 ThereforeLK = c14 = NK − NL = NK − NO and is calledda cha.

4.1.3 Notes

(1.1) and (2.1.1)

The easiest explanation ofAK = KO is by joining AK and noting that∠KOA = ∠OAC =∠OAB = ∠OAK = ∠KAO. There is, however, no evidence that the ancient Chinese, beforethe translation of Euclid’sElements, reasoned with angle measurement in right triangles. We shallseek an explanation using the “out-in principle” which LIU Hui used in his commentary ofJiuzhangSuanshu.

The equalityAK = KO would follow from the congruence of the right trianglesAKP andPOT , whereT , though not indicated in the circular city diagram, is the point where the incircletouches the hypotenuseAB. This congruence may not be immediately obviously. If, however, weswap the positions of the hypotenuse andgu of triangleKOT , this will result in a right triangle withhypotenuse along the lineKB, gu alongKJ , and withgou equal toOT , the radius of the incircle.This must be the right triangleKMR, which clearly is also congruent toAKP . From this we have1.1 and2.1.

14 Shıbie Zajı: Section I

A

B C

K

O

A

B C

I

J

K

O

P

T

(a)

A

B C

I

J

K

M

O

P

R

(b)

(2.1.2), (2.2.2) and the use of da cha and xiao cha

The use of the termsda cha (big difference) andxiao cha (small difference) is somewhat ambiguousin CYHJ. Section II ofSBZJ is on the five sums and five differences (wu hu o wu jiao). The fivedifferences are

b − a c − a = b − d c − b = a − d (a + b) − c c − (b − a)

gou gu cha gou xian cha gu xian cha xian huo cha xian cha chacha da cha xiao cha

The use ofda andxiao is explained byc − a > c − b. These terms therefore apply to specificright triangles. This reference triangle is understood when it is the main one. (2.1.2) and (2.2.2)establishc15 = MN andc14 = LK as thexiao cha andda cha of ∆12, but simply name them asxiao cha andda cha, with implicit reference to∆1.

These terms are also used for two of the right triangles.∆10 is namedda cha gougu. It is theright triangle whosegu is equal toda cha (c− a), i.e., b10 = b− d = c− a. Likewise,∆11 is namedxiao cha gougu, with a11 = a − d = c − b.

Remark. b−a is most often referred to simply ascha, but occasionallyzhong cha. See, for example,IV.2.7,8 and IV.5.2.

4.2 SBZJ I.3: Zhu za ming mu

3.1 Ming gou zhuan gu xiang de, mıng wei nei luu, qiu xu jı.3.2 Ming gu zhuan gou xiang de, mıng wei wai luu, qiu xu jı.3.3 Xu gou xu gu xiang de, mıng xu luu, qiu xu jı.

4.2 SBZJ I.3: Zhu za ming mu 15

4.2.1 Translation3.1 a14 · b15 is callednei luu, to findxu ji.3.2 b14 · a15 is calledwai luu, to findxu ji.3.3 a13 · b13 is calledxu luu, to findxu ji.

4.2.2 Notes

a14

b15

b13

a13

a15

b14



4.1 Fan gou gu huo jı xian huang huo.4.2 Fan da cha jı gu huang jiao.4.3 Fan xiao cha jı gou huang jiao.

4.3.1 Translation

4.1 a + b = c + d.4.2 c − a = b − d.4.3 c − b = a − d.

4.3.2 Notes

(1) Fan means “whenever”. A statement beginning withf an (or da fan) conveys universality. Theseformulae were given by LIU Hui in his commentary on Problem IX.16.1

A

B C

I

J

O

Figure 4.1:a + b = c + d

(2) Huang means “yellow”. Ancient manuscripts ofJiuzhang Suanshu andZhoubi Suanj ıngpresumably were accompanied by colored diagrams. Traditionally certain elements were named bycolors. The side length of a yellow square is calledhuang fang zhı mian, or simplyhuang fang.In the present context,huang refers to the diameter of the inscribed circle, which presumably wascolored yellow.

1LIU Hui gave at least three formulae: (i)b − (c − a) = d, (ii) (a + b) − c = d, (iii) 2(c − a)(c − b) = d2, which,along witha − (c − b) = d easily follow from the solution of Problem IX.12.



4.4.1 Text5.1.1 Gao gu pıng gou cha mıng jiao cha, you mıng yuan cha.5.1.2 Cı shu jı gao pıng er cha gong ye.5.1.3 You wei mıng huo zhuan huo jiao ye.5.2.1 Mıng zhuan er cha gong mıng cı cha, you mıng jın cha, you mıng lie huo.5.2.2 Cı shu you wei mıng da cha zhuan xiao cha jiao ye.5.3.1 Gou yuan cha zhı gu, gu yuan cha zhı gou xiang bıng mıng hun dong huo.5.3.2 Cı shu you wei yı jıng yı xu xian gong ye.5.4.1 Mıng zhuan er cha jiao mıng bang cha.5.4.2 Cı shu you wei gao pıng er cha jiao,5.4.3 you wei jı shuang cha nei jian xu huo,5.4.4 you wei jı xian nei jian cheng jıng ye.5.5.1 Xu cha bu jı bang cha mıng cuo cha.5.5.2 Cı shu you wei da cha cha nei qu jiao cha,5.5.3 you wei jı cha nei qu er zhı pıng cha,5.5.4 you wei cı cha nei qu xiao cha cha,5.5.5 you wei mıng qu zhuan gou gong qu er zhı mıng gou ye.5.6 Xu cha bang cha gong mıng cuo huo.

4.4.2 Translation5.1.1 Jiao cha (yuan cha) := bg − ap

5.1.2 = (bg − ag) + (bp − ap)5.1.3 = (a14 + b14) − (a15 + b15).5.2.1 Cı cha (jın cha, lie cha) := (b14 − a14) + (b15 − a15)5.2.2 = (c14 − a14) − (c15 − b15).5.3.1 Hun tong huo := b11 + a10

5.3.2 = d + c13.5.4.1 Bang cha := (b14 − a14) − (b15 − a15)5.4.2 = (bg − ag) − (bp − ap)5.4.3 = (c12 − a12) + (c12 − b12) − (a13 + b13)5.4.4 = c12 − d.5.5.1 Cuo cha := bang cha −(b13 − a13)5.5.2 = (b10 − a10)− jiao cha5.5.3 = (b12 − a12) − 2(bp − ap)5.5.4 = cı cha −(b11 − a11)5.5.5 = b14 + a15 − 2a14.5.6 Cuo huo := (b13 − a13)+ bang cha.


4.4.3 Notes

(5.1)

(1) The right triangles∆6 (shang gao) and∆7 (xia gao) being congruent, they are simply referredto as (gao) and denoted by∆g when the position are not important in the context. Likewise,a g fora6 = a7 etc. The same applies to the pair of congruent triangles∆8 (shang pıng) and∆9 (xia pıng).They are denoted by∆p with ap for a8 = a9 etc.

(2) The segmentsbg = b6 andap = a9 are situated at the corners of the circular city diagram.Their difference is namedyuan (distant) orjiao (corner). This is given by the quartic form

f70 = (p2 − 2pq − 2q2)(p2 + 2pq − q2).

(5.2): cı cha

See III.3.2.6.

(5.3)

(1) Yuan means “circle”. Here it is taken as the diameterd of the inscribed circle.Gou yuan chazhı gu is thegu on the right triangle which has the differencea − d (g ou yuan cha) asgou. Sincea − d = c − b = (xiao cha) = a11, this is b11. Similarly, gu yuan cha zhı gou is a10. The sumb11 + a10 is calledhun dong huo. This is also the sum ofd andc13.

(2) See IV.4.1.

(5.4)

(1) Shuang means “pair”.Jı shuang cha is the pair of differencesc12 − a12 andc12 − b12 .in ∆12.

(2) Bang cha is given by the quartic form

f69 = (p2 − 2pq − q2)2.

What is the relevance of these definitions?


This important section contains a list of 10 formulae of the formf i · fj = k · r2 for a constantk.The list is exhaustive except forf10 · f36 = r2 which can be written as

d13 =d2

a + b + c.

Is this somewhere in CYHJ?


4.5.1 Text

6.1 Fan da xiao cha xiang cheng wei ban duan jıng mı.6.1′ Da cha gou xiao cha gu xiang cheng yı tong.6.2 Xu gou cheng da gu de ban duan jıng mı.6.2′ Xu gu cheng da gou yı tong shang.6.3 Bian gu zhuan gu xiang cheng de ban jıng mı.6.3′ Mıng gu dı gou xiang cheng yı tong shang.6.4 Huang guang gu huang chang gou xiang cheng wei jıng mı.6.5 Gao gu pıng gou xiang cheng de ban jıng mı.6.6 Mıng xian mıng gu bıng yu zhuan xian zhuan gou bıng xiang cheng de ban jıng mı.6.6′ Mıng xian mıng gou bıng yu zhuan xian zhuan gu bıng xiang cheng yı tong shang.6.7 Gao xian bıng xian xiang cheng wei yı duan huang jı jı.6.8 Mıng gou zhuan gu xiang cheng bei zhı wei yı duan tai xu jı.6.8′ Mıng gu zhuan gou yı tong.

4.5.2 Translation

6.1 b10 · a11 = 12d2

6.1′ a10 · b11 = 12d2

6.2 a13 · b1 = 12d2

6.2′ b13 · a1 = 12d2

6.3 b2 · b15 = r2

6.3′ a14 · a3 = r2

6.4 b5 · a4 = d2

6.5 bg · ap = r2

6.6 (b14 + c14)(a15 + c15) = r2

6.6′ (a14 + c14)(b15 + c15) = r2

6.7 cg · cp = a12 · b12

6.8 2a14 · b15 = a13 · b13

6.8′ = 2b14 · a13

4.5.3 Exegetical notes

4.5.4 SBZJ I.6.1

The use offan suggests reading this as(c−a)(c− b) = 12d2 for an arbitrary right triangle.2 This, as

we mentioned, had been given by LIU Hui.3 However, in the circular city diagram,dacha asc − ais also equal todacha gu b10, 4 andxiaocha asc− b is also equal toxiaocha gou a11. 5 This explainsI.6.1. Along this line, RUAN Yuan remarked that the product ofDacha gou a10 andxiaocha gu b11

is also the same. This gives I.6.1′.


a10b11 = 1

2d2(c-a)(c-b) = b10a11 =

1

2d2

b13a1=1

2d2a13 b1 =

1

2d2


a14a3= r2 b2a15 = r2

bgap= r2 b4a5 = d2


4.5.5 SBZJ I.6.2

4.5.6 SBZJ I.6.3

4.5.7 SBZJ I.6.4

4.5.8 SBZJ I.6.5

4.5.9 SBZJ I.6.6

c14 + b14 = c2 − a2, c15 + a15 = c3 − b3. See the section onwu huo wu jiao. 6

Now, c2 − a2 = b2b · (c − a) andc3 − b3 = a3

a · (c − b). Therefore,(c14 + b14)(c15 + a15) =b2·a3

ab · (c − a)(c − b) = 12 · 2r2 = r2.

Note: It is easy to see thatb2 · a3 = 12ab since the complement ofb2 · a2 in ab is equal to

∆1 = 12ab.

b15+c15 = ap a14+c14= bg

6

14

9

15

2See notes on I.4.3See notes on SBZJ I.4.4b10 = b − d = c − a.5a11 = a − d = c − b.6b14 +c14 = b12 +c14−r = c6 +c14−r = c10−r = c10 +a12−(a12 +r) = c10 +a12−a2 = c10 +c8−a2 =

c2 − a2.


4.5.10 SBZJ I.6.7

b15

a15

b14

a14

cp

cg

cgcp =b12a12 2a14 b15 = a13b13 = 2a15b14

12

13

9

4.5.11 SBZJ I.6.8


Chapter 5

Shıbie Zajı: Section II

On the five sums and five differences of the various right triangles.

1. The right triangle∆1 on tong xian.

(a) a1 + b1 = d + c1.

(b) b1 − a1 = (b1 − d) − (a1 − d).

(c) a1 + c1 = 2a1 + (c1 − a1).

Gou xian huo ji er gou yi da cha.

(d) c1 − a1 is calledda cha.

Qi jiao ze da cha ye.

(e) b1 + c1 = 2b1 + (c1 − b1).

Gou xian huo ji er gu yi xiao cha.

(f) c1 − b1 is calledxiao cha.

Qi jiao ze xiao cha ye.

(g) c1 + (b1 − a1) = d + 3(b1 − a1).

Xian jiao huo wei yi jing san cha gong.

(h) c1 − (b1 − a1) = a1 + (c1 − b1).

Qi jiao ze da gou xiao cha gong ye.

(i) a1 + b1 + c1 = (a2 + b2 + c2) + a1 = (a3 + b3 + c3) + b1.

San shi huo ji bian xian san shi huo dai da gou ye, you wei di xian san shi huodai da gu ye.

(j) (a1 + b1) − c1 = d.

Qi jiao ze cheng jing ye.

26 Shıbie Zajı: Section II

2. The right triangle∆2 onbian xian.

(a) a2 + b2 = b1 + cp.

bian xian shang gou gu huo wei tong gu ping xian gong.

(b) b2 − a2 = b10 − cp.

Qi jiao ze da cha gu nei qu ping xian ye.

(c) a2 + c2 = b1 + a3.

gou xian huo ji tong gu di gou gong.

(d) c2 − a2 = b14 + c14.

Qi jiao ze ming gu ming xian gong ye.

(e) b2 + c2 = (b1 + c1) − a2.

Gu xian huo ji tong gu tong xian huo nei shao ge bian gou ye.

(f) c2 − b2 = ap.

Qi jiao ze ping gou ye.

(g) c2 + (b2 − a2) = b10 + c10.

Xian jiao gong wei da cha shang gu xian huo.

(h) c2 − (b2 − a2) = a1.

Qi jiao ze da gou ye.

(i) a2 + b2 + c2 = b1 + c1 = (a4 + b4 + c4) + (a1 − d). 1

San shi huo ji tong xian shang gu xian huo, you wei huang guang san shi huoshang dai gou yuan cha ye.

(j) (a2 + b2) − c2 = a4 = cp + (bp − ap) = b13 + c13.

Qi jiao ze da cha gou ye, you wei ping xian shang xian jiao huo, you wei tai xuxian shang gu xian huo ye.

1BAI Shangshu [p.32] has writtend1 − a1 instead.

27

3. The right triangle∆3 ondı xian.

(a) a3 + b3 = a1 + cg.

dı xian shang gou gu huo wei tong gou gao xian gong.

(b) b3 − a3 = cg − a11.

Qi jiao ze gao xian nei qu xiao cha gou ye.

(c) a3 + c3 = (c1 − (b1 − a1)) + bg.

gou xian huo wei tong gou shang xian jiao jiao yu gao gu gong.

(d) c3 − a3 = bg.

Qi jiao ze gao gu ye.

(e) b3 + c3 = 12 (a1 + b1 + c1).

Gu xian huo wei ban ge tong xian shang san shi huo.

(f) c3 − b3 = c15 + a15.

Qi jiao ze zhuan xian shang gou xian huo ye.

(g) c3 + (b3 − a3) = a10 + c10.

Xian jiao huo wei da cha shang gou xian huo ye.

(h) c3 − (b3 − a3) = a11 + c11.

Qi jiao ze xiao cha shang gou xian huo ye.

(i) a3 + b3 + c3 = a1 + c1 = (a5 + b5 + c5) + (b1 − d).

San shi huo ji tong xian shang gou xian huo, you wei huang chang san shi huoshang dai gu yuan cha.

(j) (a3 + b3) − c3 = b11 = cg − (bg − ag) = a13 + c13.

Qi jiao ze xiao cha gu ye, you wei gao xian shang xian jiao jiao, you wei tai xuxian shang gou xian huo.


4. The right triangle∆4 onhuang guang xian.

(a) a4 + b4 = b1 + b13 = (a1 + b1) − (a11 + b11).

Huang guang xian shang gou gu huo wei da gu xu gu gong, you wei tong gugong nei shao ge xiao cha shang gou gu huo.

(b) b4 − a4 = 2(bg − ag).

Qi jiao ze liang ge gao cha ye.

(c) a4 + c4 = 2cg + d.

gou xian huo wei liang gao xian yi yuan jing gong.

(d) c4 − a4 = 2b14.

Qi jiao ze liang ming gu ye.

(e) b4 + c4 = c1 + (b1 − a1).

Gu xian huo wei tong xian shang xian jiao huo.

(f) c4 − b4 = 2b15.

Qi jiao ze er zhuan gu ye.

(g) c4 + (b4 − a4) = 2b10.

Xian jiao huo wei liang ge da cha gu ye.

(h) c4 − (b4 − a4) = 2b11.

Qi jiao ze liang ge xiao cha gu ye.

(i) a4 + b4 + c4 = 2b1.

San shi huo wei liang ge da gu ye.

(j) (a4 + b4) − c4 = 2b13.

Qi jiao ze liang xu gu ye.

29

5. The right triangle∆5 onhuang chang xian.

(a) a5 + b5 = a1 + a13 = (a1 + b1) − (a10 + b10).

Huang chang xian shang gou gu huo wei da gou xu gou gong, you wei tonghuo nei shao ge da cha shang gou gu huo ye.

(b) b5 − a5 = 2(bp − ap).

Qi jiao ze liang ge ping cha ye.

(c) a5 + c5 = c1 − (b1 − a1).

gou xian huo wei tong xian shang xian jiao jiao.

(d) c5 − a5 = 2a14.

Qi jiao ze liang ge ming gou ye.

(e) b5 + c5 = 2d + 2a15.

Gu xian huo wei er yuan jing er zhuan gou.

(f) c5 − b5 = 2a15.

Qi jiao ze er zhuan gou ye.

(g) c5 + (b5 − a5) = 2a10.

Xian jiao huo wei liang ge da cha gou ye.

(h) c5 − (b5 − a5) = 2a11.

Qi jiao ze liang ge xiao cha gou ye.

(i) a5 + b5 + c5 = 2a1.

San shi huo wei liang da gou.

(j) (a5 + b5) − c5 = 2a13.

Qi jiao ze liang xu gou ye.


6. The right triangles∆g ongao xian.

(a) ag + bg = cg + b13 = d + (bg − ag).

Gao xian shang gou gu huo wei gao xian xu gu gong, you wei yi jing ji gao gougao gu cha ye.

(b) bg − ag = c3 − a1 = b2 − b3.

Qi jiao ze dı xian nei jian da gou ye, you wei bian gu nei jian dı gu ye.

(c) ag + cg = b3.

gou xian gong ze dı gu.

(d) cg − ag = b14.

Qi jiao ze ming gu ye.

(e) bg + cg = b2.

Gu xian gong ji bian gu ye.

(f) cg − bg = b15.

Qi cha ze zhuan gu ye.

(g) cg + (bg − ag) = b10.

Xian jiao gong ze da [gu] cha.

(h) cg − (bg − ag) = b11.

Qi jiao ze xiao cha gu ye.

(i) ag + bg + cg = b1.

San shi huo ji da gu.

(j) (ag + bg) − cg = b13 = c11 − a11 = c14 − (b14 − a14).

Qi jiao ze xu gu ye, you wei xiao cha shang gou xian jiao, you wei ming xianshang xian jiao jiao.

31

7. The right triangles∆p onping xian.

(a) ap + bp = cp + a13.

Ping xian shang gou gu gong ji ping xian xu gou gong ye.

(b) bp − ap = b1 − c2.

Qi jiao ze da gu nei jian bian xian ye.

(c) ap + cp = a3.

gou xian gong ji dı gou.

(d) cp − ap = a14.

Qi jiao ze ming gou ye.

(e) bp + cp = a2.

Gu xian gogn ji bian gou.

(f) cp − bp = a15.

Qi jiao ze zhuan gou ye.

(g) cp + (bp − ap) = a10.

Xian jiao gong da cha gou.

(h) cp − (bp − ap) = a11.

Qi jiao ze xiao cha [gou] ye.

(i) ap + bp + cp = a1.

San shi huo ji da gou.

(j) (ap + bp) − cp = a13 = c10 − b10 = c15 + (b15 − a15).

Qi jiao ze xu gou ye, you wei da cha shang gu xian jiao, you wei zhuan xianshang xian jiao huo.


8. The right triangle∆10 onda cha xian.

(a) a10 + b10 = b1 − a13.

Da cha shang gou gu huo ji da gu nei qu xu gou.

(b) b10 − a10 = c10 − d. 2

Qi jiao ze da cha xian nei qu yuan jing ye.

(c) a10 + c10 = b1.

gou xian gong ji da gu.

(d) c10 − a10 = b10 − 2a14.

Qi jiao ze da cha gu nei qu er zhi ming gou ye.

(e) b10 + c10 = b1 + (b10 − a10).

Gu xian huo wei da gu shang jia ge da zhong cha 3 ye.

(f) c10 − b10 = a13.

Qi jiao ze xu gou ye.

(g) c10 + (b10 − a10) = 2(c2 − a2).

Xian jiao huo wei liang ge bian xian shang gou xian jiao.

(h) c10 − (b10 − a10) = d.

Qi jiao ze cheng jing ye.

(i) a10 + b10 + c10 = b1 + (b1 − d) = c1 + (b1 − a1) = 2b2.

San shi huo ji gu yu gu yuan cha gong, you wei da xian da jiao gong, 4 you weier bian gu.

(j) (a10 + b10) − c10 = c13 + (b13 − a13).

Qi jiao ze tai xu shang xian jiao huo ye.

2Check this. BAI Shangshu [p.37] takesyuan jing as the diameter of the incircle of∆10 here and in (h) below. This isclearly wrong. The formlae in (b) and (h) are equivalent, and in (h) the text clearly meansd, the diameter of the city.

3LI Rui corrected an earlier remark (which asserted thatda zhong cha is the difference ofb + c of ming and the radius)by saying thatzhong cha meansb − a, andda zhong cha is b − a of ∆10 (da cha).

4The meaning ofda jiao follows BAI Shangshu, [p.37].

33

9. The right triangle∆11 onxiao cha.

(a) a11 + b11 = a1 − b13.

Xiao cha shang gou gu huo ji da gou nei qu xu gu ye.

(b) b11 − a11 = d − c11.

Qi jiao ze yuan jing nei qu xiao cha xian ye.

(c) a11 + c11 = a1 − (b11 − a11). [cf. BAI [p.38].]

gou xian huo wei da gou shang jian ge xiao zhong cha. 5

(d) c11 − a11 = b13.

Qi jiao ze xu gu ye.

(e) b11 + c11 = a1.

Gu xian gong ji da gou.

(f) c11 − b11 = a11 − 2b15.

Qi jiao ze xiao cha gou nei qu liang ge zhuan gu ye.

(g) c11 + (b11 − a11) = d.?

Xian jiao huo wei yuan jing.

(h) c11 − (b11 − a11) =.?

Qi jiao ze liang ge dı xian shang gu xian jiao, you wei liang ge zhuan xianshang gou xian huo ye.

(i) a11 + b11 + c11 = a1 + (a1 − d) = c1 − (b1 − a1) = 2a3. ?? See BAI’s footnote 49.

San shi huo ji gou yu gou yuan cha gong ye, you wei da xi an da jiao jiao, 6

you wei er dı gou.

(j) (a11 + b11) − c11 = c13 − (b13 − a13).

Qi jiao ze tai xu shang xian jiao jiao ye.

5Again, LI Rui corrected an earlier remark by asserting thatxiao zhong cha is b − a of ∆11 (xiao cha).6Remark: This isxian jiao jiao of tong xian


10. The right triangle∆12 onhuang jı.

(a) a12 + b12 = cg + cp.

Huang jı gou gu huo ji gao xian ping xian gong.

(b) b12 − a12 = b14 − a15.

Qi jiao ze ming gu nei qu zhuan gou ye.

(c) a12 + c12 = c3.

gou xian gong ji dı xian.

(d) c12 − a12 = c14.

Qi jiao ze ming xian ye.

(e) b12 + c12 = c2.

Gu xian gong ji bian xian.

(f) c12 − b12 = c15.

Qi jiao ze zhuan xian ye.

(g) c12 + (b12 − a12) = cg + c14 = b1 − a10 = c10.

Xian jiao huo wei gao xian ming xian gong, you wei da gu nei jian da cha gou,you wei da cha xian.

(h) c12 − (b12 − a12) = c11.

Qi jiao ze xiao cha xian ye.

(i) a12 + b12 + c12 = c1.

San shi huo ji tong xian.

(j) (a12 + b12) − c12 = c13 = a14 + b15 = cg − c14 = cp − c15 = a10 − b13 = a11 − a13.

Qi jiao ze tai xu xian ye, you wei ming gou zhuan gu gong, you wei gao xiannı jian ming xian, you wei ping xian nei jian zhuan xian, you wei da cha goushang jian xu gu, you wei xiao cha gu shang jian xu gou ye.

35

11. The right triangle∆13 on tai xu.

(a) a13 + b13 = d − c13 = c13 + d13 = c12 − (b14 + a15). ??

Tai xu gou gu huo ji yuan jing jian xu xian, you wei xi an xian huang fang gong,you wei huang ji nei jian qu ming gu zhuan gong.

(b) b13 − a13 = a10 − b11.

Qi jiao ze da cha gou nei jian ge xiao cha gu ye.

(c) a13 + c13 = b11.

gou xian gong ji xiao cha gu ye.

(d) c13 − a13 = b13 − d13.

Qi jiao ze xu gu nei jian ge xiao huang fang ye.

(e) b13 + c13 = a10.

Gu xian gong wei da cha gou.

(f) c13 − b13 = a13 − d13.

Qi jiao ze xu gou nıe jian ge xiao huang fang ye.

(g) c13 + (b13 − a13) = (a10 + b10) − c10 = c5 − a5 = 2a14.

Xian jiao huo wei da cha xian shang xian huo jiao, you wei huang chang xianshang gou xian jiao, you wei liang ge ming gou.

(h) c13 − (b13 − a13) = d11.

Qi jiao ze xiao cha shang huang fang mian ye.

(i) a13 + b13 + c13 = d.

San shi huo ji da huang fang.

(j) (a13 + b13) − c13 = 2(c14 − b14) = 2(c15 − a15) = (c14 − b14) + (c15 − a15).

Qi jiao ze liang ge ming xian shang gu xian jiao, you wei zhuan xian shangliang ge gou xian jiao, you wei ming xian shang xiao cha yu zhuan xian shangda cha gong ye.


12. The right triangle∆14 onming xian.

(a) a14 + b14 = (c1 − a1) − c14.

Ming xian shang gou gu huo ji da cha nei jian ming xian.

(b) b14 − a14 = c14 − b13.

Qi jiao ze ming xian nei jian xu gu ye.

(c) a14 + c14 = bg.

gou xian pian ji gao gu.

(d) c14 − a14 = bg − 2a14.

Qi jiao ze gao gu nei shao er zhi ming gou ye.

(e) b14 + c14 = b2 − a10 = c2 − a2.

Gu xian huo ji bian gu nei jian da cha gou, you wei bian gou bian xian cha.

(f) c14 − b14 = 12d13.

Qi jiao ze ban ge xu huang fang ye ye.

(g) c14 + (b14 − a14) = c10 − a10.

Xian jiao huo ji da cha shang gou xian jiao.

(h) c14 − (b14 − a14) = b13.

Qi jiao ze xu gu ye.

(i) a14 + b14 + c14 = b1 − d.

San shi huo wei gu yuan cha.

(j) (a14 + b14) − c14 = c13 − a13 = b13 − d13.

Qi jiao ze tai xu shang gou xian jiao, you wei xu gu nei jian xu huang fang ye.

37

13. The right triangle∆15 on zhuan xian.

(a) a15 + b15 = (c1 − b1) − c15.

zhuan xian shang gou gu huo ji xiao cha nei jian zhuan xian.

(b) b15 − a15 = a13 − c15.

Qi jiao ze xu gou nei jian zhuan xian ye.

[BAI’s edition omittedgou].

(c) a15 + c15 = a3 − b11 = c3 − b3.

gou xian huo ji dı gou nei jian xiao cha gu, you wei dı gu dı xian cha.

(d) c15 − a15 = 12d13.

Qi jiao ze ban ge xu huang fang ye.

(e) b15 + c15 = ap.

Gu xian huo ji ping gou.

(f) c15 − b15 = ap − 2b15.

Qi jiao ze ping gou nei shao liang ge zhuan gu ye.

(g) c15 + (b15 − a15) = a13.

Xian jiao huo ji xu gou.

(h) c15 − (b15 − a15) = c11 − b11.

Qi jiao ze xiao cha shang gu xian jiao ye.

(i) a15 + b15 + c15 = a1 − d.

San shi huo ji gou yuan cha.

(j) (a15 + b15) − c15 = c13 − b13 = a13 − d13.

Qi jiao ze tai xu shang gu xian cha, you wei xu gou nei jian xu huang fang ye.


14. Addendum: for the right triangle∆4 huang guang above.

(a) b4 − a4 = (c1 − a1) − b11 = (b1 − a1) − (b11 − a11) = b4 − d.

Qi gou gu jiao you wei da cha shang shao ge xiao cha gu, you wei zhong chanei shao ge xiao cha jiao, you wei huang guang g u nei shao yi jing.

(b) a4 + c4 = 2b3 = b2 + b11.

gou xian gong you wei liang dı gu, you wei da gu yu xiao cha gu gong.

(c) b4 + c4 = c1 + (b1 − a1) = 2b2.

Gu xian huo you wei da xian zhong cha gong, you wei liang ge bian gu.

(d) c4 − b4 = d11.

Gu xian cha you wei xiao cha shang huang fang mian.

15. Addendum: for the right triangle∆5 huang chang above .

(a) b5 − a5 = a10 − (c1 − b1) = d − a5.

Qi gou gu jiao you wei da cha gou shang shao ge xiao cha ye, you wei yuanjing nei shao ge huang chang gou.

(b) a5 + c5 = 2a3 = a1 + a11.

Gou xian gong you wei liang ge dı gou, you wei da gou yu xiao cha gou gong.

(c) c5 − a5 = d10.

Gou xian jiao you wei da cha shang huang fang mian.

(d) b5 + c5 = 2a2.

Gu xian gong you wei liang ge bian gou.

40 Shıbie Zajı Section III: The hypotenuses

Chapter 6

Shıbie Zajı Section III: Thehypotenuses

6.1 SBZJ III.1

6.1.1 Text1.1.1 Da xian wei da gou yu gu yuan cha gong,1.1.2 you wei da gu yu gu yuan cha gong.1.2.1 Bian xian nai bian gu pıng gou gong,1.2.2 you wei da gu nei jian pıng xian shang gou gu jiao.1.3.1 Dı xian nai dı gou gao gu gong,1.3.2 you wei da gou jia yı ge gao cha.1.4.1 Huang guang xian wei da gu nei jian xu gu,1.4.2 you wei bian gu zhuan gu gong.1.5.1 Huang chang xian nai da gou nei jian xu gou,1.5.2 you wei dı gou mıng gou gong1.g.1 Gao xian nai da cha xian nei jian mıng xian,1.g.2 you wei mıng xian xu xian gong.1.p.2 Pıng xian nai xiao cha xian nei jian zhuan xian,1.p.2 you wei zhuan xian xu xian gong.1.10.1 Da cha xian nai da gu nei jian da cha gou,1.10.2 you wei gao xian mıng xian gong,1.10.3 you wei da xian nei qu huang chang xian.1.11.1 Xiao cha xian wei da gou nei jian xiao cha gu,1.11.2 you wei pıng xian zhuan xian gong,1.11.3 you wei da xian nei qu huang guang xian.1.12.1 Jı xian nai gao gu pıng gou gong,1.12.2 you wei pıng xian mıng xian gong,1.12.3 you wei gao xian zhuan xian gong,1.12.4 you wei da cha xian nei jian gao pıng er xian jiao,1.12.5 you wei xiao cha xian nei jia gao pıng er xian jiao.1.13.1 Xu xian nai huang ji huang fang mian,1.13.2 you wei mıng gou zhuan gu gong,1.13.3 you wei gao xian nei jian mıng xian,1.13.4 you wei pıng xian nei jian zhuan xian.1.14 Mıng xian nai gao xian nei jian xu xian.1.15 Zhuan xian nai pıng xian nei jian xu xian.

6.1 SBZJ III.1 41

Translation

1.1.1 c1 = a1 + (b1 − d)1.1.2 = b1 + (a1 − d).1.2.1 c2 = b2 + ap

1.2.2 = b1 − (bp − ap).1.3.1 c3 = a3 + bg

1.3.2 = a1 + (bg − ag)1.4.1 c4 = b1 − b13

1.4.2 = b2 + b15.1.5.1 c5 = a1 − a13

1.5.2 = a3 + a14.1.g.1 cg = c10 − c14

1.g.2 = c14 + c13.1.p.1 cp = c11 − c15

1.p.2 = c15 + c13.1.10.1 c10 = b1 − a10

1.10.2 = cg + c14

1.10.3 = c1 − c5.1.11.1 c11 = a1 − b11

1.11.2 = cp + c15

1.11.3 = c1 − c4.1.12.1 c12 = bg + ap

1.12.2 = cp + c14

1.12.3 = cg + c15

= c10 − (cg − cp)= c11 + (cg − cp).

1.13.1 c13 = d12

1.13.1 = a14 + b15

1.13.2 = cg − c14

1.13.3 = cp − c15.1.14 c14 = cg − c13.1.15 c15 = cp − c13.

6.1.2 Notes

1.3gao cha, the “dıfference (cha)” of ∆g refers to the dˇıfference ofgou andgu.

1.g.2 BAI’s dısplay has “c6 − c14”.

1.13.huang ji huang fang mian, the side of the yˇellow of the right triangle∆12.


6.2 SBZJ III.2

6.2.1 Text

2.1 Huang guang xian huang chang xian xiang bıngwei da xian xu xian gong ye.

2.2 Yı cı shu jian yu da huo yu jı xu huo.2.3 Ruo yı er xian xiang jian jı xu xian pıng xian gong ye.2.3′ Cı shu ou he, dang wei er jı cha.2.4 Huang guang xian you wei da cha xian xu xian gong.2.5 Huang chang xian you wei xiao cha xian xu xian gong.2.6 Yı huang chang xian jian yu da gou yu jı xu gou.2.7 Yı huang guang xian jian yu da gu yu jı xu gu.

6.2.2 Translation

2.1 c4 + c5 = c1 + c13.2.2 (a1 + b1) − (c4 + c5) = a13 + b13.2.3 c4 − c5 = c13 + cp. not correct2.3′ c4 − c5 = 2(b12 − a12).2.4 c4 = c10 + c13.2.5 c5 = c11 + c13.2.6 a1 − c5 = a13. same as1.5.2.7 b1 − c4 = b13. same as1.4.

6.2.3 Notes

(2.3) and (2.3′)

The Qing commentators pointed out that (2.3.1) was incorrect; it was a numerical coincidence:510 − 272 = 238 = 102 + 136. LI Rui remarked equality fails for each of the four new sets. Forexample, with the first new set,c4 − c5 = 100, butcp + c13 = 150 + 100 = 250.

(2.6)

(1) Jian yu means “to subtract from”. Therefore,ji a jian yu yı meansyı−jia. In comtemporaryChinese,a jian b meansa − b. The usage ofjian was different in ancient times. For example, inJiuzhang Suanshu, jia jian yı meansyı − jia

(2) Yu jı means “the remainder is”.

6.3 SBZJ III.3 43

6.3 SBZJ III.3

6.3.1 Text

3.1.1 Bian xian dı xian xiang bıng wei da xian huang jı xian gong ye.3.1.2 Yı cı bıng shu nei jian da huo yu wei huang jı xian nei jian yuan jıng ye.3.2.1 Ruo yı er xian xiang jian jı huang jı cha ye.3.2.2 Cı shu tong zhe zuı duo, gu you wei huang jı xian nei shao ge xiao cha ixan,3.2.3 you wei gao xian pıng xian jiao,3.2.4 you wei mıng gu nei shao zhuan gou,3.2.5 you wei da cha xian nei shao huang jı xian3.2.6 you wei cı cha xu cha gong ye.3.3.1 Bian xian you wei huang jı gu xian gong,3.3.2 you wei huang guang xian zhuan xian gong.3.4.1 Dı xian you wei huang jı gou xian gong,3.4.2 you wei huang guang xian zhuan xian gong ye.3.5.1 Yı bian xian jian da gu yu wei ban jıng nei jian pıng gou3.5.2 you wei pıng xian nei jian xiao cha.3.6.1 Dı xian nei da gou yu wei gao gu nei ban jıng,3.6.2 you wei da cha nei jian gao xian ye.

6.3.2 Translation

3.1.1 c2 + c3 = c1 + c12

3.1.2 (c2 + c3) − (a1 + b1) = c12 − d.3.2.1 c2 − c3 = b12 − a12

3.2.2 = c12 − c11

3.2.3 = cp − cg

3.2.4 = b14 − a15

3.2.5 = c10 − c12

3.2.6 = ((b14 − a14) + (b15 − a15)) + b13 − a13 see I.5.2.1.3.3.1 c2 = b12 + c12

3.3.2 = c4 + c15.3.4.1 c3 = a12 + c12

3.4.2 = c5 + c14.3.5.1 b1 − c2 = r − ap

3.5.2 = cp − (c1 − b1).3.6.1 c3 − a1 = bg − r3.6.2 = (c1 − a1) − cg.

6.3.3 Notes

(3.1)

(c2 + c3) − (a1 + b1) = (c1 + c12) − (a1 + b1) = c12 − (a1 + b1 − c1) = c12 − d.

(3.5)

Yı jia jian yı meansyı − jia.


6.4 SBZJ III.4–8

6.4.1 Text

4.1 Huang guang xian nei jian bian gu jı zhuan gu.4.2 Huang chang xian nei jian diı gou jı mıng gou ye.5.1 Gao xian gao gu gong jı bian gu.5.2 Pıng xian pıng gou gong jı dı gou.5.3 Gao xian gao gou gong jı dı gu.5.4 Pıng xian pıng gu gong jı bian gou.6.1 Shang gao xian jian yu tong gu yu jı bian gu nei jian mıng gu ye.6.2 Xia pıng xian jian yu tong gou yu jı bian gou nei jian mıng gou ye.6.3.1 Gao xian pıng xian xiang bıng jı da xian nei shao ge huang jı xian ye.6.3.2 Ruo yı xiang bıng shu jian yu da huo yu wei huang jı xian yuan jıng gong ye.6.3.3 Gao xian pıng xian xiang jian jı huang jı cha ye,6.3.4 you wei huang jı xian shang jian xiao cha xian ye.6.3.5 Ruo yı xiang jian shu que jia yxiang bıng shu jı huang guang xian ye.7.1 Gao xian nei jian mıng gu de ban jıng.7.2 Pıng xian nei jian zhuan gou yı tong shang.8.1 Huang jı gou shang jia mıng xian wei huang jı xian.8.2 Huang jı gu shang jia zhuan xian yı tong shang.

6.4.2 Translation

4.1 c4 − b2 = b15.4.2 c5 − a3 = a14.5.1 cg + bg = b2.5.2 cp + ap = a3.5.3 cg + ag = b3.5.4 cp + bp = a2.6.1 b1 − c5 = b2 − b14.6.2 a1 − c9 = a2 − a14.6.3.1 cg + cp = c1 − c12.6.3.2 (a1 + b1) − (cg + cp) = c12 + d.6.3.3 cg − cp = b12 − a12

6.3.4 = c12 − c11.6.3.5 (cg + cp) + (cg − cp) = c4.7.1 cg − b14 = r.7.2 cp − a15 = r.8.1 a12 + c14 = c12.8.2 b12 + c15 = c12.

6.4.3 Notes

(6.1) and (6.2)

c5 can be replaced bycg andc9 can be replaced bycp.

6.5 SBZJ III.9 45

6.5 SBZJ III.9

6.5.1 TextHuang jı xian

9.1 de jı gou jı dı gu,9.2 de jı gu jı bian xian,9.3 nei qu jı gou jı mıng xian,9.4 qu jı gu jı zhuan xian,9.5 jian yu tong xian jı j’ı huo,9.6 de xu xian yı tong shang,9.7 nei qu xu xian jı mıng xian zhuan xian gong,9.8 qu xu huang jı mıng huo zhuan huo gong ge,9.9 qu cheng jıg jı bang cha,9.10 nei jia jı cha jı da cha xian,9.11 qu jı cha jı xiao cha xian,9.12 jia jiao cha jı liang ge gao gu,9.13 jian jiao cha jı er pıng gou.

6.5.2 Translation

9.1 c12 + a12 = c3.9.2 c12 + b12 = c2.9.3 c12 − a12 = c14.9.4 c12 − b12 = c15.9.5 c1 − c12 = a12 + b12.9.6 c12 + c13 = a12 + b12.9.7 c12 − c13 = c14 + c15.9.8 c12 − d13 = (a14 + b14) + (a15 + b15).9.9 c12 − d = bang cha.9.10 c12 + (b12 − a12) = c10.9.11 c12 − (b12 − a12) = c11.9.12 c12 + jiao cha = c12 + (bg − ap) = 2bg.9.13 c12 − jiao cha = c12 − (bg − ap) = 2ap.

6.5.3 Notes

(9.1)

de means “gain”.

(9.8)

Xu huang means the diameter of the incircle of∆13.

(9.9)

See the definition ofbang cha in (5.4.4).

(9.12) and (9.13)

See the definition ofjiao cha in (5.1.1).


6.6 SBZJ III.10

6.6.1 TextTai xu xian

10.1 jia ru jı xian wei jı huo,10.2 jı xian nei qu zhı jı mıng da cha zhuan xiao cha bıng ye,10.3 zai qu zhı jı gou jı mıng xian,10.4 jia yu da cha xian jı huang guang xian,10.5 jia yu xiao cha xian jı huang chang xian,10.6 nei qu mıng gou jı zhuan gu,10.7 jia mıng gou wei yuan jıng nei shao xu huang zhuan gu gong,10.8 jia ru mıng gu wei mıng huo zhuan gu gong,10.9 jian yu mıng gu jı mıng jiao nei qu zhuan gu.10.10 jia ru mıng xian wei jı gu,10.11 jian yu mıng xian wei mıng da cha zhuan xiao cha nei shao ge zhuan xian,10.12 jia yu mıng huo jı liang ge xu xian yı ge gaocha gong ye,10.13 jian yu mıng huo jı gao cha ye,10.141 nei qu zhuan gou jı mıng gou zhuan jiao gong,10.142 you wei zhuan gu pıng cha gong,10.15 jia yu zhuan gou jı zhuan huo mıng gou gong ,10.161 jia yu zhuan gu wei er xu xian nei shao mıng gou,10.162 you wei yuan jıng nei shao xu huang mıng gou gong,10.17 nei jian zhuan gu jı mıng gou,10.18 nei jia zhuan xian jı jı gou,10.19 nei jian zhuan xian wei mıng gou nei shao ge zhuan xiao cha,10.20 jia ru zhuan huo jı liang ge xu xian nei shao ge pıng cha ye,10.21 nei jian zhuan huo jı pıng cha ye,10.22 jia ru mıng zhuan er huo gong jı jı huo nei shao ge xu huang ye,10.23 ruo jian yu mıng zhuan er huo gong jı mıng gu zhuan gou gong,10.24 jian yu gao xian jı mıng xian,10.25 jian yu pıng xian jı zhuan xian,10.26 jia yu jiao cha jı er mıng gou yı jı cha gong,10.27 jian yu jiao cha jı yı jı cha er zhuan gu jiao ye ,10.28 de bang cha jı mıng gu zhuan gou gong,10.29 nei jian bang cha jı xu san shı huo nei qu le jı shuang cha ye,10.30 nei jia xu cha jı er mıng gou,10.31 nei jian xu cha jı er zhuan gu,10.32 nei jia xu huang fang jı xu huo,10.33 nei jian xu huang fang jı xu jı da xiao cha bıng ye.

You zhu xian.

6.6 SBZJ III.10 47

6.6.2 Translation

10.1 c13 + c12 = a12 + b12.10.2 c12 − c13 = c14 + c15.10.3 c12 − c13 − c13 = (c14 − b14) + (c15 − b15).10.4 c13 + c10 = c4.10.5 c13 + c11 = c5.10.6 c13 − a14 = b15.10.7 c13 + a14 = d − (d13 + b15).10.8 c13 + b14 = (a14 + b14) + b15.10.9 b14 − c13 = (b14 − a14) − b15.10.10 c13 + c14 = b12.10.11 c14 − c13 = (c14 − a14) + (c15 − b15) − c15.10.12 c13 + (a14 + b14) = 2c13 + (bg − ag).10.13 (a14 + b14) − c13 = bg − ag.10.141 c13 − a15 = a14 + (b15 − a15),10.142 = b15 + (bp − ap),10.15 c13 + a15 = (a15 + b15) + a14,10.161 c13 + b15 = 2c13 − a14

10.162 = d − (d13 + a14).10.17 c13 − b15 = a14.10.18 c13 + c15 = a12.10.19 c13 − c15 = a14 − (c15 − b15).10.20 c13 + (a15 + b15) = 2c13 − (bp − ap).10.21 c13 − (a15 + b15) = bp − ap.10.22 c13 + ((a14 + b14) + (a15 + b15)) = a12 + b12 − d13.10.23 ((a14 + b14) + (a15 + b15)) − c13 = b14 + a15.10.24 cg − c13 = c14.10.25 cp − c13 = c15.10.26 c13 + jiao cha = 2a14 + (b12 − a12).10.27 jiao cha − c13 = (b12 − a12) − 2b15.10.28 c13 + bang cha = b14 + a15.10.29 c13 − bang cha = (a13 + b13 + c13) − ((c12 − a12) + (c12 − b12)).10.30 c13 + (b13 − a13) = 2a14.10.31 c13 − (b13 − a13) = 2b15.10.32 c13 + d13 = a13 + b13.10.33 c13 − d13 = (c13 − a13) + (c13 − b13).


6.6.3 Notes

(10.11)

This is the same asc13 = a14 + b15.

(10.33)

Xu jı da xiao cha bıng. This apparently has nothing to do with thex u jı in Section I.3. It should readXu zhı da xiao cha bıng.

Chapter 5

SBZJ Section IV: Daxiao cha

5.1 SBZJ Section IV.1

5.1.1 Text

1.1 Da cha xian xiao cha xian gong jı liang ge jı xian ye,1.2 yı liang ge jı cha wei zhı jiao.1.3 Da cha cha xiao cha cha gong jı liang ge jı cha ye,1.4 yı liang ge bang cha wei zhi jiao.1.5 Da cha shang da cha, xiao cha shang da cha gong jı liang ge mıng xian ye,1.6 yı liang ge mıng cha wei zhi jiao.1.7 Da cha shang xiao cha, xiao cha shang xiao cha gong jı liang ge zhuan xian ye,1.8 yı liang ge zhuan cha wei zhi jiao.1.9 Da cha huang xiao cha huang shu gong jı liang ge jı huang ye,1.10 yı liang ge xu cha wei zhi jiao.1.11 Da cha gou xiao cha gou gong jı liang ge jı gou ye,1.12 yı liang ge pıng cha wei zhi jiao.1.13 Da cha gu xiao cha gu gong jı liang ge jı gu ye,1.14 yı liang ge gao cha wei zhi jiao.1.15 Liang huo gong wei er jı huo,1.16 yı er jiao cha wei zhi jiao.

50 SBZJ Section IV: Daxiao cha

5.1.2 Translation

1.1 c10 + c11 = 2c12.1.2 c10 − c11 = 2(b12 − a12).1.3 (b10 − a10) + (b11 − a11) = 2(b12 − a12).1.4 (b10 − a10) − (b11 − a11) = 2 · bang cha.1.5 (c10 − a10) + (c11 − a11) = 2c14.1.6 (c10 − a10) − (c11 − a11) = 2(b14 − a14).1.7 (c10 − b10) + (c11 − b11) = 2c15.1.8 (c10 − b10) − (c11 − b11) = 2(b15 − a15).1.9 d10 + d11 = 2d12.1.10 d10 − d11 = 2(b13 − a13).1.11 a10 + a11 = 2a12.1.12 a10 − a11 = 2(bp − ap).1.13 b10 + b11 = 2b12.1.14 b10 − b11 = 2(bg − ag).1.15 (a10 + b10) + (a11 + b11) = 2(a12 + b12).1.16 (a10 + b10) − (a11 + b11) = 2 · jiao cha.

5.2 SBZJ Section IV.2

5.2.1 Text

2.1 Da cha xian shang xian jiao jiao jı yuan jıng.2.2 Xiao cha xian shang xian jiao huo yı tong.2.3 Da cha xiang xiao cha jı xu gou.2.4 Xiao cha xiang da cha jı xu gu ye.2.5 Da cha xian yu mıng gou gong jı bian gu.2.6 Xiao cha xian yu zhuan gu gong jı dı gou ye.2.7 Da cha xian nei jian zhong cha jı huang chang gou.2.8 Xiao cha xian nei jia zhong cha jı huang guang gu ye.2.9 Da gu nei jian xiao cha gu jı huang guang gu.2.10 Da gou nei jian da cha gou jı huang chang gou ye.2.11 Xu xian de xu gu jı da cha gou.2.12 Xu xian de xu gou jı xiao cha gu ye.2.13

5.3 SBZJ Section IV.3–6 51

5.2.2 Translation

2.1 c10 − (b10 − a10) = d.2.2 c11 + (b11 − a11) = d.2.3 c10 − b10 = a13.2.4 c11 − a11 = b13.2.5 c10 + a14 = b2.2.6 c11 + b15 = a3.2.7 c10 − (b − a) = a5.2.8 c11 − (b − a) = b4.2.9 b1 − b11 = b4.2.10 a1 − a10 = a5.2.11 c13 + b13 = a10.2.12 c13 + a13 = b11.

5.3 SBZJ Section IV.3–6

5.3.1 Text

3.1 Da xian nei jian da xiao cha gong jı yuan jıng,3.2 san shı huo nei jian er zhı da xiao cha gong jı san ge yuan jıng ye.4.1 Da cha gou xiao cha gu xiang bıng mıng hun tong huo,4.2 ruo yı xiaang jian j’ı xu cha ye.5.1 Da cha huo xiao cha huo er shu xiang bıng jı da xian xu xian gong ye,5.2 er shu xiaang jian j’ı zhong cha xu cha gong ye,5.3.1 you ban zhı bıng shu jı wei jı xian xu xian gong ye,5.3.2 you wei gao xian pıng xian gong,5.3.3 you wei huang jı gou gu gong ye.6.1.1 Da cha cha xiao cha cha er shu xiang bıng jı liang ge huang jı cha,6.1.2 you wei da cha xian nei jian xiao cha xian ye.6.2 Liang shu xiaang jian er ban zhı jı shı huan jı xian shang jian yuan jıng ye.

You da xiao cha.

5.3.2 Translation

3.1 c10 − ((c1 − a1) + (c1 − b1)) = d.3.2 (a1 + b1 + c1) − 2((c1 − a1) + (c1 − b1)) = 3d.4.1 hun tong huo := a10 + b11.4.2 a10 − b11 = b13 − a13.5.1 (a10 + b10) + (a11 + b11) = c10 + c13.5.2 (a10 + b10) − (a11 + b11) = (b1 − a1) + (b13 − a13).5.3.1 1

2 ((a10 + b10) + (a11 + b11)) = c12 + c13

5.3.2 = cg + cp

5.3.3 = a12 + b12.6.1.1 (b10 − a10) + (b11 − a11) = 2(b12 − a12)6.1.2 = c10 − c11.6.2 1

2 ((b10 − a10) − (b11 − a11)) = c12 − d.

52 SBZJ Section IV: Daxiao cha

5.3.3 Notes

(4.1)

Hun tong huo see I.5.3.1,2.

Chapter 7

Construction Problems

Construction fromai andaj

i/j 15 14 13 12 11 10 p 3 21 2 2 2 2 1 1 1 1 12 1 2 2 1 2 1 1 23 2 2• 2 2• 1 2 2•

p 1 2• 1 2• 1 110 2 1 1 2 211 1 2 1 212 1 2• 213 1 114 2

Construction fromci andcj

c/j 15 14 13 12 11 10 p g 3 21 2 2 1 1 1 1 1 1 1 12 2• 2 2 2• 2 1 1 2• 23 2 2• 2 2• 1 2 2• 1g 2• 1 1 2• 1 1 2p 1 2• 1 2• 1 110 2 1 1 1 111 1 2 1 112 2• 2• 113 1 114 2

54 Construction Problems

Construction fromai andcj

a/c 15 14 13 12 11 10 g 3 2 11 3 4 3 4 2 3 3 3 2 22 3 4 3 4 3 3 4 4 2 33 3 4 3 4 2 3 3 2• 3 2p 2 4 3 2• 3 3 4∗ 4 2 310 3 3 2 4 3 2 3 4 2 311 2 4 3 4 2 3 3 3 3 213 2 3 2 4 3 2 3 4 3 314 3 2• 2 4∗ 3 2 3 4∗ 3 315 2 4 3 4 3 3 4 4 3 3

a12 = cp.

Construction fromai andbj

a/b 15 14 13 12 11 10 g 3 2 11 2 3 2• 3 2 2 3 3 1 22 3 4∗ 3 4 3 3 4 4∗ 2 33 2 3 2 3 1 2 2 2 2 1p 2• 4 3 4∗ 3 3 2• 4 2 310 2 3 2 3 2• 2 3 3 1 211 1 3 2 3 2 2• 3 3 2 212 3 4 3 2• 3 3 4∗ 4 3 313 1 3 2 3 2 2 3 3 2 2•

14 2 2 1 3 2 1 2• 3 2 215 2 4∗ 3 4 3 3 4 4∗ 3 3

Construction from ai and aj

1. Trivial constructions

• (a1, a3): r = a1 − a3; (a3, a11), (a2, a12), (a10, a1), (a12, a15),

• (a1, a11): r = 12 (a1 − a11); (a2, a15),

• (a10, a13): r = 12 (a10 + a13).

• (a13, a14): r = a13 + a14.

2. (a1, a10): 1r = 1

a1+ 1

a10.

• (a1, a2) follows from(a1, a10) sincea10 = 2a2 − a1

• (a2, a10) also follows from(a1, a10) sincea1 = 2a2 − a10.

• (a9, a10) follows from (a1, a10) sincea1 = 2a9 + a10.

3. (ap, a15): r =a2

p−a215

2a15; 1

d = 1ap−a15

− 1ap+a15

. It is easier to rewrite this as

ap

r=

2apa15

a2p − a2

15

55

• (a9, a13) follows froma15 = a9 − a13.

• (a9, a11) follows froma15 = a11 − a9.

• (a11, a13) from a9 = 12 (a11 + a13) anda15 = 1

2 (a11 − a13).

• (a13, a15) follows fromap = a13 + a15.

Constructions from ci and cj

Examples

(1) If c8 = x, c15 = y, thenr = x2−y2

x2+y2 · x. This also leads to the constructions from(c8, cj) forj = 13, 11, 10.

(2) Note thata12 = c8.(3) If c10 = x andc11 = y, thenr = xy

x+y . It is still not easy to position the circle.c13 = w

satisfiesw2 + (x + y)w = xy. b2 is a root of

t(t − x) =(

r

y

)2

· xy = 0.

From ai and bj

(1) (a1, b2) and(a3, b1): r = a1b2a1+2b2

.

(2) (a3, b11) and(a10, b2): r = a3b112a3−b11

.

(3) (a11, b15) and(a14, b10): r = a11b15a11−2b15

.

(4) (a13, b15) and(a14, b13): r = a14b132a14−b13

.

Construction by square roots

(a3, a14): r2 = a3a14

(a8, a14) also follows from(a3, a14) sincea3 = 2a8 + a14.(a12, a14) follows from(a3, a14) sincea3 = 2a12 − a14.

(a8, a12): r2 = a212 − a2

8

Given a segmentON with a pointS on it, to construct a right triangleABC such that the parallel toBC through the incenterO intersects the hypotenuse atN , and the perpenicular toBC atS intersectthe hypotenuse at a pointL on the upper edge of the square cirumscribing the incircle.

Construction: (1) Construct the perpendicular toBC atS.(2) Construct the circleN(O) to intersect the line in (1) atL.(3) Complete the rectangleOSLH .(4) The circleO(H) is the incircle. The hypotenuse is along the lineNL.

(a3, a8): r2 = a3(a3 − 2a8)

Given a segmentBJ with a pointQ on it such thatBQ < 12 ·BJ , to construct a right triangleABC

such that the incircle touchesBC at J and the parallel toBC from the incenterO intersects theperpendicular toBC atQ on the hypotenuse.

Construction: (1) MarkY onBC such thatBY = 2 · BJ .(2) Construct the semicircle with diameterBJ .


(3) Construct the perpendicular toBC atY to intersect the semicircle atO ′.(4) MarkO′ on the perpendicular toBC atJ such thatJO = JO ′.(5) The circleO(J) is the incircle of the triangle. LetN be the intersection of the parallel toBCthroughO and the perpendicular toBC atQ. Then the hypotenuse is along the lineBN .

(a3, a12) follows from above.

(c8, c12): z2 =c28(c

212−c28)

c212

c8 = a12.Construction: (1) Construct a right triangleKNO with KN = c12 andNO = c8, and right

angle atO.(2) Construct the perpendicular fromO to KN and let the projection beT(3) The circleO(T ) is the incircle. The hypotenuse is along the lineNT .

(c8, c14) follows fromc12 = c8 + c14.(c12, c14) follows fromc8 = c12 − c14.

(c12, c15): r2 = (c12−c15)2

c212· c15(2c12 − c15)

2c12 − c15 = c2.Given a segmentNK with a pointM on it, extend it to a pointA such thatMK = KA.

(1) Construct the circleK(M), and a line throughN tangent to the circle atO.(2) The circle with centerO, tangent to the lineNK, is the incircle. The hypotenuse is along theline NK. A is the point such thatMK = KA.

(c3, a3): r2 = x−yx+y

· y2

(c14, a14): r2 = x+yx−y

· y2

(ap, c12): r2 = a8(c12 − a8)

(ap, bg): r2 = apbg

Note: Trivial caseag = bp = r.

(a1, b13): d2 = 2a1b13 = 2a13b1

(1) Given perpendicular segmentsAP = bg andQ′P = ap, extendAP to Q1 such thatPQ1 =PQ′.(2) Construct a semicircle with diameterAQ1 to intersect the rayPQ′ atK.(3) construct the incircle(O ′) of the right triangleAKP . (4) ExtendAO ′ to intersect the perpen-dicular toKP atK. Let O be the intersection, andI its projection on the lineAP .(5) O(I) is the incircle of the right triangleABC. The hypotenuse is along the lineAK.

(a8, b15) and (a14, b7) : r2 = a8

a8−2b15· b2

15 =(

b15a8−2b15

)2

· a8(a8 − 2b15)

(a10, b11) and (a11, b10)

r2 = 12a10b11 = 1

2a11b10

57

(a12, b12): 1r2 = 1

a212

+ 1b212

. Easy construction.

Constructions from ai and bi

i

12 1r2 = 1

x2 + 1y2

2 x · r2 + 2y2 · r − xy2 = 015 x · r2 − 2y2 · r − xy2 = 03 y · r2 + 2x2 · r − x2y = 014 y · r2 − 2x2 · r − x2y = 010 2r2 − 2(x − y)r − xy = 011 2r2 + 2(x − y)r − xy = 01 2r2 − 2(x + y)r + xy = 013 2r2 − 2(x + y)r + xy = 0

i

3 r2 = x2(y−x)x+y

14 y2 = x2(x+y)y−x

12 r2 = x2(y2−x2)y2

1 2r2 + 2(y − x)r − x(y − x) = 011 2r2 − 2(y − x)r − x(y − x) = 013 2r2 − 2(x + y)r + x(x + y) = 010 2r2 − 2(x + y)r + x(x + y) = 02 x · r2 + 2(y2 − x2)r − x(y2 − x2) = 015 x · r2 − 2(y2 − x2)r − x(y2 − x2) = 0


Chapter 8

CYHJ Book II:Zheng luu yi shi si wen

8.1 Introduction

Book II consists of 14 problems. The first ten problems are on the diameters of the various circlesassociated with a right triangle.

Problem Topics Diameter

II.1 gou gu rong yuan incircle 2a1b1a1+b1+c1

II.2 gou shang rong yuan tangent togu andxian andwith center ongou 2a2b2

b2+c2II.3 gu shang rong yuan tangent togou andxian and

with center ongu 2a3b3a3+c3

II.4 gou gu shang rong yuan tangent toxian and centerat right angle vertex 2a12b12

c12II.5 xian shang rong yuan tangent togou andgu and

with center onxian 2a′b′a′+b′

II.6 gou wai rong yuan excircle ongou 2a10b10c10+(b10−a10)

II.7 gu wai rong yuan excircle ongu 2a11b11c11−(b11−a11)

II.8 xian wai rong yuan excirle onxian 2a13b13(a13+b13)−c13

II.9 gou wai rong yuan ban tangent to extensions ofgou andxian and with center ongu 2a14b14

c14−a14II.10 gu wai rong yuan ban tangent to extensions ofgou and

xian and with center ongou 2a15b15c15−b15

LI Ye explained in his preface that this work was inspired by his study ofD ong yuan jiu rong,nine circles ofDong yuan. It was not known whetherDong yuan was a name or the title of a book.LI Di 1 explains that in the northernSong period, there was a Daoist of this name who might havebeen the originator of the tradition ofji urong, the nine circles associated with a right triangle.

Which one in Book II was not among the nine indongyuan jiurong? LI Shanlan said it wasthe first one,gougu rongyuan, since this already known in ancient times. It is Problem IX.16 ofJiuzhang Suanshu. However, LIU Yueyun 2 thought it wasxian shang rong yuan, since the numbersinvolved are not of the same type, and the diagram does not have the line for the right triangle. KˇongGuoping, in hisLI Ye, ZHU Shiji, yu Jin Yuan Shuxue, agrees with LIU.

1Zhongguo shuxue tongshi, Song Yuan juan, 1999,Jiangsu; p.200.21896.


Apart from the circle in II.5 (xian shang rong yuan), the other nine circles naturally form asystem. They are associated with similar right triangles whose vertices are the vertices, center, andmidpoints of edges of the square circumscribing the incircle. The right triangle in II.5 in generalis not similar to any of them.3 In fact, every line through the center of the incircle will serve thepurpose. Aftergou shang rong yuan andgu shang rong yuan in II.3 and II.4, very natuarlly LI Yuconsiders this circle here.

A

B C(1)

D(10)E(13)

F (11)

G(15)

H(14)

I(2)

J(3)

O(12)

Figure 8.1:

3In the circular city diagram, these points are respectively180 bu east and360 bu south ofC.

8.2 The problems in Book II 61

8.2 The problems in Book II

Suppose there is a circular city of unknown diameter. On each of the four sides there is a gate.Through each gate there is a crossroad running west to east and south to north. LetC be the intersection of the roads through the west and north gates,4

F be the intersection of the roads through the east and north gates,E be the intersection of the roads through the east and south gates, andD be the intersection of the roads through the west and south gates.A miscellany of surveying methods are considered through the problems below.

north

east

A

B C(1)

D(10)E

F (11)

G(15)

H(14)

I(2)

J(3)

K

L

M

NO(12)

P(6)

Q(9)

R(7)

S(8)

W (4)X(13)

Y (5)

4Qi xi bei shi zhi dou tou ting wei qian di.


CYHJ II.1: d = 2a1b1a1+b1+c1

, c21 = a2

1 + b21

Someone asks:Jia andYı are both atC. Yı walks eastward320 bu and stops.Jia walks southward600 bu and seesYı.Question: What is the diameter of the city?

north

east

a1

b1

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.2:gou gu rong yuan

Method. This is gou gu rong yuan. 5 Multiply gou andgu, double it as dividend (shı). Add thesquares ofgou andgu to find the hypotenuse. Further add [to the hypotenuse] the sum ofg ou andgu as divisor (fa).Solution.

5The problem of finding the diameter of the incircle of a right triangle was solved inJiuzhang Suanshu, Problem IX.16.


CYHJ II.2: d = 2a2b2b2+c2

, c22 = a2

2 + b22

Someone asks:Jia andYı are both at the west gate.Yı walks eastward256 bu. Jia walks southward480 bu and is able to seeYı.Question and answer as before.

north

east

b2

a2

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.3:gou shang rong yuan

Method. This is the inscribed circle with center ongou (gou shang rong yuan). Multiply gou andgu, double it as dividend. And add the squares ofg ou andgu to find the hypotenuse. Addgu [to thehypotenuse] as divisor.Solution.


CYHJ II.3: d = 2a3b3a3+c3

, c23 = a2

3 + b23

Someone asks:Jia andYı are both at the north gate.Yı walks eastward200 bu. Jia walks southward375 bu and is able to seeYı.Question and answer as before.

north

east

b3

a3

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.4:gu shang rong yuan

Method. This is the inscribed circle with ongu (gu shang rong yuan). Multiply gou and gu,double it as dividend. And add the squares ofg ou andgu to find the hypotenuse. Addgou [to thehypotenuse] as divisor.Solution.


CYHJ II.4: d = 2a12b12c12

, c212 = a2

12 + b212

Someone asks:Jia andYı are both at the center of the city.Yı walks eastward136 bu straight throughthe gate.Jia walks southward225 bu and is able to seeYı.Question and answer as before.

north

east

a12

b12

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.5:gou gu shang rong yuan

Method. This is the inscribed circle with center ongou andgu (gou gu shang rong yuan). Multiplygou andgu, double it as dividend. Add the squares ofgou andgu to find the hypotenuse6 as divisor.Solution.

Remark. BAI Shangshu erroneously took(x, y) to be(a3, b3). These should be(b12, a12).

6ru fa qiu xian.


CYHJ II.5: d = 2a′b′a′+b′

Someone asks:Jia andYı are both atC. Yı walks eastward180 bu and stops at a tower.Jia walkssouthward360 bu and sees the tower aligning with the middle of the city diameter.Question and answer as before.

north

east

a′

b′

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.6:xian shang rong yuan

Method. This is the inscribed circle with center on the hypotenuse (xi an shang rong yuan). Mul-tiply gou andgu, double it as dividend. Take the sum ofgou andgu as divisor.Solution.


CYHJ II.6: d = 2a10b10c10+(b10−a10)

Someone asks:Jia andYı are both atD. Yı walks eastward192 bu. Jia walks southward360 bu andis able to seeYı.Question and answer as before.

north

east

a10

b10

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.7:gou wai rong yuan

Method. This is the escribed circle tangent togou (gou wai rong yuan). Multiply gou andgu,double it as dividend. Take the sum of the hypotenuse and the difference ofg ou andgu 7 as divisor.Solution.

7xian jiao gong.


CYHJ II.7: d = 2a11b11c11−(b11−a11)

Someone asks:Jia andYı are both atF . Jia 8 walks southward150 bu. Yı walks eastward80 bu andseesJia aligning with the city.Question and answer as before.

north

east

a11

b11

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

T

Figure 8.8:gu wai rong yuan

Method. This is the escribed circle tangent togu (gu wai rong yuan). Multiply gou andgu, doubleit as dividend. Take the difference of the hypotenuse and the difference ofg ou andgu 9 as divisor.Solution.

8BAI Shangshu writesYong instead ofJia.9xian jiao jiao.


CYHJ II.8: d = 2a13b13(a13+b13)−c13

Someone asks:Jia andYı are both atE. Yı walks westward48 bu. Jia walks northward90 bu andseesYı aligning with the city.Question and answer as before.

north

east

a13

b13

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

T

Figure 8.9:xian wai rong yuan

Method. This is the escribed circle tangent to the hypotenuse (xi an wai rong yuan). Multiply gouandgu, double it as dividend. Take the difference of the hypotenuse and the sum ofg ou andgu asdivisor.Solution.


CYHJ II.9: d = 2a14b14c14−a14

Someone asks:Jia andYı are both at the south gate.Yı walks eastward72 bu. Jia walks southward135 bu and seesYı aligning with the city.Question and answer as before.

north

east

a14

b14

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

T

Figure 8.10:gou wai rong yuan ban

Method. This is the escribed semicircle tangent togou [at the vertex of the right angle] (gou wairong yuan ban). Multiply gou andgu, double it as dividend. Takeda cha as divisor.Solution.


CYHJ II.10: d = 2a15b15c15−b15

Someone asks:Jia andYı are both at the east gate.Jia walks southward30 bu. Yı walks eastward116 bu and seesJia aligning with the city.Question and answer as before.

north

east

a15

b15

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

T

Figure 8.11:gu wai rong yuan ban

Method. This is the escribed semicircle tangent togu [at the vertex of the right angle] (gu wairong yuan ban). Multiply gou andgu, double it as dividend. Takexiao cha as divisor.Solution.


CYHJ II.11: r =√

b2b15

Someone asks:Jia exits the west gate and walks480 bu southward.Yı exits the east gate and walkssouthward for30 bu and seesJia.Question and answer as before.

north

east

b2

b15

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.12:ban ai tı

Method. This is the inscribed circle of half of an isosceles trapezoid (ban ai tı). 10 Multiply thetwo bu and extract the square root,11 getting half of the diameter.Solution.

10Ban ai tı, half of a short ladder.11Yi liang bu xiang cheng wei shi, ru ping fang er yi.


CYHJ II.11b: d =√

4a14a3

Also asks: There are two personsJia andYı. Yı exits the south gate, turns and walks eastward for72 bu. Jia exits the north gate and turns eastward for200 bu and seesYı.Question and answer as before.

north

east

a3

a14

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.13: 11b

Method. Multiply the two bu. Quadruple the product and extract the square root,12 getting thediameter.

12De shu si zhi wei shi, ru ping fang er yi.


CYHJ II.11c: d =√

4a′a′′

And suppose:Yı exits the south gate, turns and walks eastward for20 bu. Jia exits the north gateand turns eastward for720 bu [and seesJia].For problems like this, use the same method.

north

east

a′

a′′

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.14: 11c


CYHJ II.12: d =√

2a11b10

Someone asks: There are two personsJia andYı. Yı walks eastward fromF for 80 bu. Jia walkssouthward fromD for 360 bu and seesJia.Question and answer as before.

north

east

b10

a11

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.15:

Method. This is finding the “yellow side” from the two differences.13 Multiply the twobu, doubleit asshı, and find its square root for the city diameter.

13Liang cha qiu huang fang is finding the yellow side from the two differences. In Problem IX.12 ofJiuzhang Suanshu,(a + b − c)2 = 2(c − a)(c − b). Here,d = a + b − c andc − a = b − d = b10, c − b = a − d = a11 . SeeSBZJ 1.6.1.


CYHJ II.13 Fang wu xie qı

Someone asks:Jia exits the east gate and walks48 bu. Yı exits the south gate and walks48 bu toseeJia.Question and answer as before.

north

east

A

B C

D

E

F

G

H

I

J

K

L

M

N

O

P

Q

R

S

WX

Y

Jia

Yı

Figure 8.16:Fang wu xie qı

Method. This should be solved by the method of side and diagonal of a square.14 Every2 bucorreponds to10 bu in the diameter.

14Fang wu xie qı, side5 and diagonal7.


CYHJ II.14: r2 + (a3 + b2)r = a3b2

Someone asks:480 bu south of the west gate there is a tree. Exit the north gate and walk200 b ueastward. The same tree can be seen.Question and answer as before.

north

east

a3

b2

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 8.17:

Method. With the product of the twobu asshı, the sum of the twobu ascong, 1 bu as constant(chang fa), get the radius.


Chapter 9

CYHJ Book III:Bian gu yi shi qi wen

Problem Jia YıIII.1 b2 c4 d2 = 2b2 · 2(c4 − b2)III.2 b2 a11 d(d + a11) = 2b2 · a11

III.3 b2 b11 r(r + b2) = b2 · b11

III.4 b2 a15 d2(d + a15) = 4b22 · a15

III.5 b2 a14 r3 − (b2 − 2a14)r2 + a214r + a2

14b2 = 0r2 − (b2 − a14)r + a14 · b2 = 0

III.6 b2 c10 d2 − c10d + b2(b2 − c10) = 0III.7 b2 c2 gou gu rong yuanIII.8 b2 c1 d2 + c1 · d − (c1 − b2)b2 = 0III.9 b2 c6 r2 = (2c6 − b2)b2

III.10 b2 b14 r2 − 2b2 · r + b22 − 2b2 · b14 = 0

III.11 b2 a10 (2b2 − a10)d = a10 · b2

III.12 b2 c15 x2 + (c15 + b2)x − c15 · b2 = 0; x = b15

III.13 b2 c14 d3 − 2b2d2 + 4b2(b2 + 2c14)d − 8b3

2 = 0.III.14 b2 c6 r2 = b2(2c6 − b2)III.15 b2 c8 r2(c8 + r) = b2

2(c8 − r)III.16 b2 b14 + c14 r = b2(b2−(b14+c14)

b2+(b14+c14)

III.17 b2 a15 + c15 r3 + b2r2 + b2(a15 + c15)r − b2

2(a15 + c15) = 0.


CYHJ III.1

Someone asks:Yı exits the east gate, walks southward in an unknown number ofb u, and stays there.Jia exits the west gate, walks southward480 bu and seesYı. Then he walks towardsYı 510 bu tomeet him.Question and answer as before.

north

east

A

B C

D

E

F

G(15)

H

I(2)

J

K

L

M

NO

P

Q

R

S

W (4)X

Y

Figure 9.1:

Method. Double the difference ofbu. Multiply to twice Jia’s southward distance aspingfang shi,[extracting the square root] to get the diameter of the city.Solution. From SBZJ,1 the difference of the two distances, namely, 30bu, is the same asYı’ssouthward distance from the east gate. Doubling the difference gives 60b u. Multiplying this totwice Jia’s southward distance, [which is] 960bu, we get 57600bu aspingfang shı. Extract [thesquare root]2 by the [standard] method, getting 240b u, which is the city diameter. Q.E.F.3

1c4 − b2 = a15.2From SBZJ I.6(c):215 · 2b2 = d2.3Quad erat faciendum, that which was to have been done, translatesHe wen.

CYHJ Book III 81

CYHJ III.2

Someone asks:Jia exits the west gate, walks480 bu southward and stays there. From positionF ,Yı walks80 bu eastward and seesJia.Question and answer as before.

north

east

A

B C

D

E

F (11)

G

H

I(2)

J

K

L

M

NO

P

Q

R

S

W (4)X

Y

Figure 9.2:

Method. Double the southward distance and multiply by the eastward distance as constant (sh ı),the eastward distance ascongfang, 1 aschangfa. Get the diameter.Solution. Let d (tian yuan yı) be the diameter of the circle. Subtract from twiceJi a’s southward

distance and get−1 yuan960 as twiceda cha.4 Multiply by Yı’s eastward distance, getting

−80 yuan76800

asd2 (yuan jıng mı). 5 [Place on the left]. Then cancellation withd2 and the left gives

−1−80

76800.

Extract square root,6 getting240 bu for the city diameter. Q.E.F.

4Da cha is the same asDacha gu, b10 = b2 − r. Therefore,2b2 − d = 2b10 .52b10 · a11 = d2 from SBJZ 1.6.1(a).6Dai cong kai fang.


CYHJ III.3

Someone asks:Jia exits the west gate, walks480 bu southward and stays there.Yı, starting fromF ,also walks southward, for 150bu and seesJia.Question and answer as before.

north

east

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 9.3:

Method. Multiply the two bu as constant, [Jia’s] southward distance7 ascongfang, 1 asyu. Getthe radius.Solution. Let r (tian yuan yi) be the radius of the city. Subtract fromY i’s southward distance

and get−1 yuan150 asban tı tou. 8 Multiply by Jia’s southward distance as [ban] tı dı, 9 getting

−480 yuan72000 as the square of the radius.10 [Place on left]. Then cancellation withr2 and [the

expression on] the left gives−1

−48072000

.

Extracting its square root,11 we get 120bu. Doubling it gives [the diameter of] the city. Q.E.F.

7There are twoNan xing bu (southward distances). The one in question isJia’s, b2.8According to BAI Shangshu [p.201], a trapeziodal field with two right angles is axie tian in JZSS. In Song dynasty, it

is called atı tian (ladder field). This, however, does not explain the use ofban (half). If we seeGMAI as one half of asymmetric trapezoid, thenGM is ban tı tou (half of the upper width) andIA ban tı dı, half of the lower width.

9See the above footnote.10SBZJ 1.6(c).11by the method ofDai cong kai fang.

CYHJ Book III 83

CYHJ III.4

Someone asks:Jia exits the west gate, walks480 bu southwards.Yı exits the east gate, walks16bu straight through and seesJia.Question and answer as before.

north

east

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 9.4:

Method. With 4 times the number of eastwardbu multiply the square of the southward (bu) asconstant (shı), then a blank space, the eastwardbu as (lian), 1 bu asyu fa. Get the diameter.

Solution. Let d (tian yuan yı) be the diameter. AddYı’s eastwardbu to get1 yuan

16 aszhong

gou, 12 Jia’s southward distance iszhong gu. Put the eastwardbu asxiao gou, multiply by zhonggu, getting 7680tai. This should be divided byzhong gou. Since it is not divisible, it is taken asxiao gu (b15). (Note that it containszhong gou a2 in the denominator). Multiplying it byzhong gu,get3, 686, 400. Further multiplying by 4, get14, 745, 600. This is the square of diameter, (y ı duanyuan jıng mı), and it includes in the denominatora2, place it on the left. Then multiplyd by itself,

and then byzhong gou, get1

160 yuan

as the same number. Cancellation with the left gives

−1−16

0147456000

.

Extracting the cube root,13 get240 as the diameter of the city. Q.E.F.

12This is thegou of thebian gougu.13Dai cong kai li fang.


CYHJ III.5

Someone asks:Yı exits the south gate, walks 72bu and stops.Jia exits the west gate, walks 480busouthward, and findsYı aligning with the city wall.Question and answer as before.

north

east

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 9.5:

Method. Multiply the square ofYı’s eastward distance byJia’s southward distance as constant(shı), Yı’s eastward distance as (congfang), Jia’s southward distance minus twiceYı’s eastward dis-tance as negative (yı lian), 1 bu as (changfa). Get the radius.Solution.Alternative Method. With the product of the given numbers (y un shu xiang cheng) as constant,their difference ascong, 1 as negative leading coefficient (xu fa). Extract square root (ping kai) toget the radius.Solution. From SBZJ,14 the sum of the given numbers isb1 − a13, and their difference isc10. Letr (tian yuan yı) be the radius.

14Bie de.

CYHJ Book III 85

CYHJ III.6

north

east

A

B C

D

E

F

G

H

I

J

K

L

M

NO

P

Q

R

S

WX

Y

Figure 9.6:


CYHJ III.12. r4 + b2(b2 + c15)r2 − b3

2c15 = 0

Someone asks: Givenb2 = 480 andc15 = 34.Question and answer as before.Method. Multiply c15 andb2. Half the product as constant, the sum of half ofb2 and half ofc15

ascong, half bu asyu. Extract the square and getb15 equal to 30.Solution. Let x (tian yuan yi) be thezhuan gu b15.

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