Ceyuan haijing

From Wikipedia, the free encyclopedia
Jump to: navigation, search
The master figure in Sea mirror of circle measurements, that all the problems use. It shows a round town, inscribed in a right triangle and a square.

Ceyuan haijing (simplified Chinese: 测圆海镜; traditional Chinese: 測圓海鏡; pinyin: cè yuán hǎi jìng; literally: "sea mirror of circle measurements") is a treatise on solving geometry problems with the algebra of Tian yuan shu written by the mathematician Li Zhi in 1248 in the time of the Mongol Empire. It is a collection of 692 formula and 170 problems, all derived from the same master diagram of a round town inscribed in a right triangle and a square. They often involve two people who walk on straight lines until they can see each other, meet or reach a tree or pagoda in a certain spot. It is an algebraic geometry book, the purpose of book is to study intricated geometrical relations by algebra.

Majority of the geometry problems are solved by polynomial equations, which are represented using a method called tian yuan shu, "coefficient array method" or literally "method of the celestial unknown". Li Zhi is the earliest extant source of this method, though it was known before him in some form. It is a positional system of rod numerals to represent polynomial equations.

Ceyuan haijing was first introduced to the west by the British Protestant Christian missionary to China, Alexander Wylie in his book Notes on Chinese Literature, 1902. He wrote:

The first page has a diagram of a circle contained in a triangle, which is dissected into 15 figures;the definition and ratios of the several parts are then given, and there are followed by 170 problems, in which the principle of the new science are seen to advantage. There is an exposition and scholia throughout by the author.[1]

This treatise consists of 12 volumes.

Volume 1[edit]

Reconstructed Diagram of circular city in alphabets

Diagram of a Round Town[edit]

The monography begins with a master diagram called the Diagram of Round Town(圆城图式). It shows a circle inscribed in a right angle triangle and four horizontal lines, four vertical lines.

  • TLQ, the large right angle triangle, with horizontal line LQ, vertical line TQ and hypotenuse TL

C: Center of circle:

  • NCS: A vertical line through C, intersect the circle and line LQ at N(南north side of city wall), intersects south side of circle at S(南).
  • NCSR, Extension of line NCS to intersect hypotenuse TL at R(日)
  • WCE: a horizontal line passing center C, intersects circle and line TQ at W(西,west side of city wall) and circle at E(东,east side of city wall).
  • WCEB:extension of line WCE to intersect hypotenuse at B(川)
  • KSYV: a horizontal tangent at S, intersects line TQ at K(坤),hypotenuse TL at Y(月)。
  • HEMV: vertical tangent of circle at point E, intersects line LQ at H, hypotenuse at M(山,mountain)
  • HSYY,KSYV, HNQ,QSK form a square, with inscribed circle C.
  • Line YS, vertical line from Y intersects line LQ at S(泉, spring)
  • Line BJ, vertical line from point B, intersects line LQ at J(夕,night)
  • RD, a horizontal line from R, intersects line TQ at D(旦,day)

The North, South, East and West direction in Li Zhi's diagram are opposite to our present convention.

Triangles and their sides[edit]

There are a total of fifteen right angle triangles formed by the intersection between triangle TLQ, the four horizontal lines, and four vertical lines.

The names of these right angle triangles and their sides are summarized in the following table

Number Name Vertices Hypotenuse0c Vertical0b Horizontal0a
1 通 TONG 天地乾 ΔTLQ 通弦(TL天地) 通股(TQ天乾) 通勾(LQ地乾)
2 边 BIAN 天西川 ΔTWB 边弦(TB天川) 边股(TW天西) 边勾(WB西川)
3 底 DI 日地北 ΔRDN 底弦(RL日地) 底股(RN日北) 底勾(LB地北)
4 黄广 HUANGGUANG 天山金 ΔTMJ 黄广弦(TM天山) 黄广股(TJ天金) 黄广勾(MJ山金)
5 黄长 HUANGCHANG 月地泉 ΔYLS 黄长弦(YL月地) 黄长股(YS月泉) 黄长勾(LS地泉)
6 上高 SHANGGAO 天日旦 ΔTRD 上高弦(TR天日) 上高股(TD天旦) 上高勾(RD日旦)
7 下高 XIAGAO 日山朱 ΔRMZ 下高弦(RM日山) 下高股(RZ日朱) 下高勾(MZ山朱)
8 上平 SHANGPING 月川青 ΔYSG 上平弦(YS月川) 上平股(YG月青) 上平勾(SG川青)
9 下平 XIAPING 川地夕 ΔBLJ 下平弦(BL川地) 下平股(BJ川夕) 下平勾(LJ地夕)
10 大差 DACHA 天月坤 ΔTYK 大差弦(TY天月) 大差股(TK天坤) 大差勾(YK月坤)
11 小差 XIAOCHA 山地艮 ΔMLH 小差弦(ML山地) 小差股(MH山艮) 小差勾(LH地艮)
12 皇极 HUANGJI 日川心 ΔRSC 皇极弦(RS日川) 皇极股(RC日心) 皇极勾(SC川心)
13 太虚 TAIXU 月山泛 ΔYMF 太虚弦(YM月山) 太虚股(YF月泛) 太虚勾(MF山泛)
14 明 MING 日月南 ΔRYS 明弦(RY日月) 明股(RS日南) 明勾(YS月南)
15 叀 ZHUAN 山川东 ΔMSE 叀弦(MS山川) 叀股(ME山东) 叀勾(SE川东)

In problems from Vol 2 to Vol 12, the names of these triangles are used in very terse terms. For instance

"明差","MING difference" refers to the "difference between the vertical side and horizontal side of MING triangle.
"叀差","ZHUANG difference" refers to the "difference between the vertical side and horizontal side of ZHUANG triangle."
"明差叀差并" means "the sum of MING difference and ZHUAN difference"(b_{14}-a_{14})+(b_{15}-a_{15})

Length of Line Segments[edit]

This section (今问正数)lists the length of line segments, the sum and difference and their combinations in the diagram of round town, given that the radius r of inscribe circle is  r=120 paces a_{1}=320,b_{1}=640.

The 13 segments of ith triangle(i=1 to 15) are:

  1. Hypoteneuse c_{i}
  2. Horizontal a_{i}
  3. Vertical b_{i}
  4.  :勾股和 :sum of horizontal and vertical a_{i}+b_{i}
  5.  :勾股校:difference of vertical and horizontal b_{i}-a_{i}
  6.  :勾弦和:sum of horizontal and hypotenuse a_{i}+c_{i}
  7.  :勾弦校:difference of hypotenuse and horizontal c_{i}-a_{i}
  8.  :股弦和:sum of hypotenuse and vertical b_{i}+c_{i}
  9.  :股弦校:difference of hypotenuse and vertical c_{i}-b_{i}
  10.  :弦校和:sum of the difference and the hypotenuse c_{i}+(b_{i}-a_{i})
  11.  :弦校校:difference of the hypotenuse and the difference c_{i}-(b_{i}-a_{i})
  12.  :弦和和:sum the hypotenuse and the sum of vertical and horizontal a_{i}+b_{i}+c_{i}
  13.  :弦和校:difference of the sum of horizontal and vertical with the hypotenuse a_{i}+b_{i}

Among the fifteen right angle triangles, there are two sets of identical triangles:

ΔTRD=ΔRMZ,
ΔYSG=ΔBLJ

that is

a_{6}=a_{7};
b_{6}=b_{7};
c_{6}=c_{7};
a_{8}=a_{9};
b_{8}=b_{9};
c_{8}=c_{9};

Segment numbers[edit]

There are 15 x 13 =195 terms, their values are shown in Table 1:[2]

Segment Table 1

Definitions and formula[edit]

Miscellaneous formula[edit]

[3]

  1. (c_{1}-a_{1})*(c_{1}*b_{1})= 1 \over 2*(d_{1})^2
  2. a_{10}*b_{11} = 1 \over 2(d_{1})^2
  3. a_{13}*b_{1} = 1 \over 2(d_{1})^2
  4. a_{1}*b_{13} = 1 \over 2(d_{1})^2
  5. b_{2}*b_{15} = (r_{1})^2
  6. a_{14}*a_{3} = (r_{1})^2
  7. a_{5}*b_{4} = (d_{1})^2
  8. a_{8}*b_{6} = a_{9}*b_{7}=(r_{1})^2
  9. (b_{14}*c_{14})*(a_{15}+c_{15}) = (r_{1})^2
  10. c_{6}*c_{8} = c_{7}*c_{9})=a_{13}*b_{13}

The Five Sums and The Five Differences[edit]

  1. a_{2}+b_{2}+c_{2}=b_{1}+c_{1}[4]
  2. a_{3}+b_{3}+c_{3}=a_{1}+c_{1}
  3. a_{4}+b_{4}+c_{4}=2b_{1}
  4. a_{5}+b_{5}+c_{5}=2a_{1}
  5. a_{6}+b_{6}+c_{6}=b_{1}
  6. a_{7}+b_{7}+c_{7}=b_{1}
  7. a_{8}+b_{8}+c_{8}=a_{1}
  8. a_{9}+b_{9}+c_{9}=a_{1}
  9. a_{10}+b_{10}+c_{10}=b_{1}+c_{1}-a_{1}
  10. a_{11}+b_{11}+c_{11}=c_{1}-b_{1}+a_{1}
  11. a_{12}+b_{12}+c_{12}=c_{1}
  12. a_{13}+b_{13}+c_{13}=a_{1}+b_{1}-c_{1}
  13. a_{14}+b_{14}+c_{14}=c_{1}-a_{1}
  14. a_{15}+b_{15}+c_{15}=c_{1}-c_{1}
  • (b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15})=2*(b_{12}-a_{12})
  • a_{8}+(b_{7}-a_{7})+(b_{8}-a_{8})=b_{7}

……………………Etc.

Li Zhi derived a total of 692 formula in Ceyuan haijing. Eight of the formula are incorrect, the rest are all correct[5]

From vol 2 to vol 12, there are 170 problems, each problem utilizing a selected few from these formula to form 2nd order to 6th order polynomial equations. As a matter of fact, there are 21 problems yielding third order polynomial equation, 13 problem yielding 4th order polynomial equation and one problem yielding 6th order polynomial[6]

Volume 2[edit]

This volume begins with a general hypothesis[3]

Suppose there is a round town, with unknown diameter. This town has four gates, there are two WE direction roads and two NS direction roads outside the gates forming a square surrounding the round town. The NW corner of the square is point Q, the NE corner is point H, the SE corner is point V, the SW corner is K. All the various survey problems are described in this volume and the following volumes.

All subsequent 170 problems are about given several segments, or their sum or difference, to find the radius or diameter of the round town. All problems follow more or less the same format; it begins with a Question, followed by description of algorithm, occasionally followed by step by step description of the procedure.

Nine types of inscribed circle

The first ten problems were solved without the use of Tian yuan shu. These problems are related to various types of inscribed circle.

Question 1
Two men A and B start from corner Q. A walks eastward 320 paces and stands still. B walks southward 600 paces and see B. What is the diameter of the circular city ?
Answer: the diameter of the round town is 240 paces.
This is inscribed circle problem associated with ΔTLQ
Algorithm:d={2a_{1} \times b_{1} \over a_{1} + b_{1}+c_{1}}
={2 * 320 * 600 \over 320 +600 +\sqrt(320^2+600^2)}=240
Question 2
Two men A and B start from West gate. B walks eastward 256 paces, A walks south 480 paces and sees B. What is the diameter of the town ?
Answer 240 paces
This is inscribed circle problem associated with ΔTWB
From Table 1, 256 = a_{2}; 480 =b_{2}
Algorithm:
{2a_{2} \times b_{2} \over a_{2} + b_{2}+c_{2}}=d
={2*256 *480 \over 256+600+\sqrt(256^+600^2)}=240
Question 3
inscribed circle problem associated with ΔRDN

{2a_{3} \times b_{3} \over a_{3} + b_{3}+c_{3}}=d

Question 4:inscribed circle problem associated with ΔRSC

{2a_{12} \times b_{12} \over c_{12}}=d

Question 5:inscribed circle problem associated with ΔTWB

{2a \times b \over a+b}=d

Question 6

{2a_{10} \times b_{10} \over b_{10} - a_{10}+c_{10}}=d

Question 7

{2a_{11} \times b_{11} \over b_{11} - a_{11}+c_{11}}=d

Question 8

{2a_{13} \times b_{13} \over b_{13} + a_{13} -c_{13}}=d

Question 9

{2a_{14} \times b_{14} \over c_{14} - a_{14}}=d

Question 10

{2a_{15} \times b_{15} \over c_{15} - b_{15}}=d

Tian yuan shu[edit]

Ciyuan haijing vol II Problem 14 detail procedure(草曰)
From problem 14 onwards, Li Zhi introduced "Tian yuan one" as unknown variable, and set up two expressions according to Section Definition and formula, then equate these two tian yuan shu expressions. He then solved the problem and obtained the answer.
Question 14:“Suppose a man walking out from West gate and heading south for 480 paces and encountered a tree. He then walked out from the North gate heading east for 200 paces and saw the same tree. What is the radius of the round own?"。
Algorithm: Set up the radius as Tian yuan one, place the counting rods representing southward 480 paces on the floor, subtract the tian yuan radius to obtain

480-x

Counting rod v-1.png
Counting rod v-4.pngCounting rod h8.pngCounting rod 0.png

Then subtract tian yuan from eastward paces 200 to obtain:

200-x

Counting rod v-1.png
Counting rod v2.pngCounting rod 0.pngCounting rod 0.png
multiply these two expressions to get:x^2-680x+96000
Counting rod v1.png
Counting rod h6.pngCounting rod h-8.pngCounting rod 0.png
Counting rod v9.pngCounting rod h6.pngCounting rod 0.pngCounting rod 0.pngCounting rod 0.png
Counting rod v2.png
Counting rod 0.png

that isx^2-680x+96000=2x^2

thus:-x^2-680x+96000=0

Counting rod v-1.png
Counting rod h6.pngCounting rod h-8.pngCounting rod 0.png
Counting rod v9.pngCounting rod h6.pngCounting rod 0.pngCounting rod 0.pngCounting rod 0.png

Solve the equation and obtain  r= 120

Volume 3[edit]

17 problems associated with segment b_{2}i.e TW in ΔTWB[7]

The a_{10} pairs with b_{11},a_{11} pairs with b_{10} and a_{15} pairs with b_{14} in problems with same number of volume 4. In other words, for example, change a_{11} of problem 2 in vol 3 into b_{10} turns it into problem 2 of Vol 4.[8]

Problem # GIVEN x Equation
1 b_{2}c_{4} direct calculation without tian yuan
2 b_{2}a_{11} d x^2+a_{11}x-2b_{2}a_{11}=0
3 b_{2}b_{11} r x^2+b_{2}x-b_{2}b_{11}=0
4 b_{2}a_{15} d x^3+a_{15}x^2-4a_{15}b_{2}^2=0
5 b_{2}a_{14} d x^3-(b_{2}-2a_{14})x^2+a_{14}^2*x+a_{14}^2*b_{2}=0
6 b_{2}a_{10} r x^2+(b_{2}-(b_{2}-c_{10}))x+b_{2}(b_{2}-c_{10})=0
7 b_{2}c_{2} r ((1/2)*c_{2}-(1/2)*b_{2}+b_{2})*x^2-(1/2)*(c_{2}-b_{2})b_{2}^2=0
8 b_{2}c_{1} r 2x^2+((c_{1}+b_{2})+(c_{1}-b_{2}))x-((c_{1}+b_{2})(c_{1}-b_{2})-(c_{1}-b_{2})^2))=0
9 b_{2}c_{6} r 2x^2-2(b_{2}-2(b_{2}-c_{5}))b_{2}=0
10 b_{2}b_{14} r x^2-2b_{2}x+((b_{2}-b_{14})^2-b_{14}^2=0
11 b_{2}a_{10} r (2b_{2}-a_{10})x-b_{2}a_{10}=0
12 b_{2}c_{15} b_{15} x^2+(b_{2}+c_{15})x-b_{2}c_{15}=0
13 b_{2}c_{14} a_{14} x^4-2(b_{2}-c_{14})x^3+(b_{2}-c_{14})^2x^2+2b_{2}c_{14}^2x-(2(b_{2}-c_{14})-b_{2}))b_{2}c_{14}^2=0
14 b_{2}c_{6} r=\sqrt((2c_{6}-b_{2})b_{2})
15 b_{2}c_{8} r -x^3-c_{8}x^2-b_{2}^2x+c_{8}b_{2}^2=0
16 b_{2}b_{14}+c_{14} calculate with formula for inscribed circle
17 b_{2}a_{15}+c_{15} Calculate with formula forinscribed circle

Volume 4[edit]

17 problems, given a_{3}and a second segment, find diameter of circular city.[9]

Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
second line segment c_{5} b_{10} a_{10} b_{14} b_{15} c_{11} c_{13} c_{1} c_{9} a_{15} b_{11} c_{14} c_{15} c_{9} c_{7} a_{15}+c_{15} b_{14}+c_{14}

Volume 5[edit]

18 problems, givenb_{1}[9]

Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
second line segment b_{14} a_{14} a_{15} b_{15} b_{11} a_{11} c_{10} c_{4} c_{2} c_{1} c_{6} c_{9}-a_{11} c_{15} c_{14} c_{9} c_{12} a_{15}+b_{14} c_{13}

Volume 6[edit]

18 problems.

Q1-11,13-19 givena_{1},and a seond line segment, find diameter d.[9]
Q12:given a_{1}+c_{3}and another line segment, find diameter d.
Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Given a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1}+c_{3} a_{1} a_{1} a_{1} a_{1} a_{1} a_{1}
Second line segment a_{15} b_{15} b_{14} a_{14} a_{10} b_{10} c_{11} c_{5} c_{3} c_{1} c_{9} b_{10}-c_{6} c_{14} c_{15} c_{6} c_{12} a_{15}+b_{14} a_{13}

Volume 7[edit]

18 problems, given two line segments find the diameter of round town[10]

Q Given
1 a_{14}b_{15}
2 a_{15}b_{14}
3 a_{14}a_{15}
4 b_{14}b_{15}
5 a_{14}c_{8}
6 b_{15}c_{7}
7 b_{15}c_{13}
8 a_{14}c_{13}
9 d-a_{14}d-b_{15}
10 d-b_{14}d-a_{15}
11 c_{12}a_{15}+b_{14}
12 a_{15}+b_{14}c_{13}
13 b_{15}+c_{13}c_{13}-b_{15}
14 a_{14}+b_{15}+c_{13}a_{14}+b_{15}+c_{13}-a_{14}
15 c_{14}d-b_{15}
16 c_{5}d-a_{14}
17 a_{1}-a_{14}b_{1}-b_{15}
18 a_{1}+a_{14}b_{1}-b_{15}

Volume 8[edit]

17 problems, given three to eight segments or their sum or difference, find dimeter of round city.[10]

Q Given
1 a_{14}+b_{14}a_{15}+b_{15}c_{12}
2 a_{14}+b_{14}a_{15}+b_{15}c_{13}
3 (c_{12}-a_{12})+(c_{12}-b_{12})d_{14}+d_{15}
4 c_{15}c_{14}
5 b_{14}+c_{14}c_{15}+b_{15}
6 a_{15}+c_{15}a_{14}+c_{14}
7 a_{1}+c_{1}a_{15}+c_{15}
8 a_{1}+c_{1}a_{14}+c_{14}
9 b_{1}+c_{1}b_{15}+c_{15}
10 b_{1}+c_{1}b_{14}+c_{14}
11 b_{14}+c_{14}a_{15}+c_{15}b_{14}+a_{15}-c_{13}
12 (b_{8}-a_{8})+(b_{2}-a_{2})b_{14}+a_{15}-c_{13}
13 b_{7}-a_{8}(b_{14}-a_{14})+(b_{15}-a_{15})c_{12}-d
14 (b_{7}-a_{7})+(b_{8}-a_{8})(b_{14}-a_{14})+(b_{15}-a_{15})
15 a_{14}+b_{14}a_{15}+b_{15}
16 a_{14}+a_{15}b_{14}+b_{15}

Problem 14[edit]

Given the sum of GAO difference and MING difference is 161 paces and the sum of MING difference and ZHUAN difference is 77 paces. What is the diameter of the round city?
Answer: 120 paces.

Algorithm:[3]

Given

(b_{7}-a_{7})+(b_{8}-a_{8})=161
(b_{14}-a_{14})+(b_{15}-a_{15})=77

:Add these two items, and divide by 2; according to #Definitions and formula, this equals to HUANGJI difference:

(b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}) \over 2 =(b_{12}-a_{12})
b_{12}-a_{12}=161 + 77 \over 2=119
Let Tian yuan one as the horizontal of SHANGPING (SG):
x=a_{8}
 x+ 161 =x+(b_{7}-a_{7})+(b_{8}-a_{8})=a_{8}+(b_{7}-a_{7})+(b_{8}-a_{8})
=b_{7} (#Definition and fomula)
Since a_{8}+b_{7}=c_{12} (Definition and formula)
c_{12}=x+b_{7}=2*x+(b_{7}-a_{7})+(b_{8}-a_{8})=2*x+161
c_{12}^2=(x+b_{7})^2=(2*x+161)^2=4*x^2+644*x+25921
c_{12}^2-(b_{12}-a_{12})^2
=4*x^2+644*x+25921-((b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}))^2 \over 4
=4*x^2+644*x+11760=d(diameter of round town),
d^2=(4*x^2+644*x+11760)^2=16*x^4+5152*x^3+508816*x^2+15146880*x+138297600
Now, multiply the length of RZ by 4*x
4*x*b_{7}=4*x*(x+(b_{7}-a_{7})+(b_{8}-a_{8}))=4*x*( x+ 161)=4*x^2+644*x
multiply it with the square of RS:
d^2=4*x*b_{7}*c_{12}^2=(4*x^2+644*x)*(4*x^2+644*x+25921)=16*x^4+5152*x^3+518420*x^2+16693124
equate the expressions for the two d^2
thus
16*x^4+5152*x^3+518420*x^2+16693124=16*x^4+5152*x^3+508816*x^2+15146880*x+138297600
We obtain:

9604*x^2+1546244*x-138297600=0

solve it and we obtain x=a_{8}=64;

This matches the horizontal of SHANGPING 8th triangle in #Segment numbers.[11]

Volume 9[edit]

Part I
Problems given
1 a_{12}+b_{12}+c_{12}b_{12}-a_{12}
2 c_{1}b_{1}-a_{1}
3 c_{1}a_{10}+b_{11}
4 c_{1}a_{2}+b_{3}
Part II
Problems given
1 a_{1}+b_{1}a_{2}b_{3}
2 a_{1}+b_{1}c_{13}+b_{13}-a_{13}c_{13}-b_{13}+a_{13}
3 a_{1}+b_{1}a_{11}+b_{11}a_{10}+b_{10}
4 a_{1}+b_{1}c_{10}-a_{10}c_{11}-b_{11}
5 a_{1}+b_{1}c_{6}+c_{8}c_{6}-c_{8}
6 a_{1}+b_{1}c_{10}c_{11}
7 a_{1}+b_{1}c_{4}c_{5}
8 a_{1}+b_{1}c_{2}c_{3}

Volume 10[edit]

8 problems[10]

Problem Given
1 a_{1}+b_{1}+c_{1}c_{1}-b_{1}
2 a_{1}+b_{1}+c_{1}c_{1}-a_{1}
3 a_{1}+b_{1}+c_{1}b_{1}-a_{1}
4 a_{1}+b_{1}+c_{1}(c_{1}-b_{1})+(c_{1}-a_{1})
5 a_{1}+b_{1}+c_{1}(c_{1}-b_{1})+(b_{1}-a_{1})+(c_{1}-a_{1})
6 a_{1}+b_{1}+c_{1}d_{14}+d_{15}
7 a_{1}+b_{1}+c_{1}c_{12}
8 a_{1}+b_{1}+c_{1}c_{13}

Volume 11[edit]

:Miscellaneous 18 problems:[10]

Q GIVEN
1 c_{2}c_{3}
2 c_{5}c_{4}
3 b_{11}c_{4}
4 a_{10}c_{3}
5 a_{10}b_{11}
6 b_{7}a_{8}
7 b_{1}-b_{11}a_{1}-a_{10}
8 b_{10}-a_{10}b_{11}-a{11}
9 c_{13}a_{10}-b_{11}
10 a_{12}+b_{12}a_{13}+b_{13}
11 c_{1}b_{1} \over a_{1}
12 d_{10}-d_{11}d_{12}-d_{13}
13 c_{12}-[c_{10}-(b_{10}-a_{10})]c_{11}+(b_{11}-a_{11})-c_{13}b_{12}-a_{12}
14 c_{8}-(c_{1}-b_{1})(c_{1}-a_{1})-c_{7}
15 a_{1}+c_{1}(c_{1}-a_{1})+(c_{1}-b_{1})
16 a_{12}+b_{12}+c_{12}(a_{13}+b_{13})-c_{13}
17 From the book Dongyuan jiurong
18 From Dongyuan jiurong

Volume 12[edit]

14 problems on fractions[10]

Problem given
1 b_{1}+c_{1}a_{1}= 8 \over 15 b_{1}
2 a_{1}+c_{1}a_{1}= 8 \over 15 b_{1}
3 a_{1}=(1-5/9)*3db_{1}-a_{1}
4 a_{3}=(5/6)*db_{2}-a_{3}
5 (15/16)b_{1}=da_{1}+b_{1}
6 a_{12}=(8/15)*b_{12}c_{12}-b_{12}c_{12}-a_{12}
7 c_{1}d=(1/2)b_{2}a_{3}=(5/6)d
8 b_{2}+a_{3}+c_{2}b_{2}=(12/17)c_{1}a_{3}=(5/17)c_{1}
9 a_{3}+(5/6)b_{2}b_{2}+(3/5)a_{3}
10 a_{11}+(1/3)b_{10}b_{10}-(3/4)a_{11}
11 b_{1}-d=(3/5)b_{1}a_{1}-d=(1/4)a_{1}(b_{1}-d)-(a_{1}-d)
12 b_{1}-d=(3/5)b_{1}a_{1}-d=(1/4)a_{1}(1/5)b_{1}-(1/4)a_{1}
13 b_{14}=(1-(15/24)b_{10})a_{15}=(1-(4/5))a_{11}b_{14}-a_{15},b_{10}-a_{11}
14 a_{1}+b_{1}+c_{1}(b_{1}/a_{1})=8(1/3)(a_{1}/b_{15})=10(2/3)a_{14}-a_{13}b_{13}-b_{15}

Research[edit]

In 1913, French mathematician L. van Hoe wrote an article about Ceyuan haijing. In 1982, K. Chema Ph.D thesis Etude du Livre Reflects des Mesuers du Cercle sur la mer de Li Ye. 1983, University of Singapore Mathematics Professor Lam Lay Yong: Chinese Polynomial Equations in the Thirteenth Century。

Footnotes[edit]

  1. ^ Alexander Wylie, Notes on Chinese Literature, Shanghai, p116, reprinted by Kessinger Publishing
  2. ^ Compiled from Kong Guoping p 62-66
  3. ^ a b c Bai Shangshu p24-25.
  4. ^ Wu Wenjun Chapter II p80
  5. ^ Bai Shangshu, p3, Preface
  6. ^ Wu Wenjun, p87
  7. ^ Li Yan p75-88
  8. ^ Martzloff, p147
  9. ^ a b c Li Yan p88-101
  10. ^ a b c d e Kong Guoping p169-184
  11. ^ Footnote:In Vol 8 problem 14, Li Zhi stop short at x=64. However the answer is evident, as from No 8 formular in #Miscellaneous formula: a_{9}*b_{7}=r^2, and from #Length of Line Segmentsa_{8}=a_{9}, thus a_{8}*b_{7}=r^2, radius of round town can be readily obtain. As a matter of fact, problem 6 of vol 11 is just such a question of given a_{8}andb_{7}, to find the radius of the round town.

References[edit]

  • Jean-Claude Martzloff, A History of Chinese Mathematics, Springer 1997 ISBN 3-540-33782-2
  • Kong Guoping, Guide to Ceyuan haijing, Hubei Education Press 1966 孔国平. 《测圆海镜今导读》 《今问正数》 湖北教育出版社. 1995
  • Bai Shangshu: A Modern Chinese Translation of Li Yeh Ceyuan haijing. Shandong Education Press 1985李冶 著 白尚恕 译 钟善基 校. 《测圆海镜今译》 山东教育出版社. 1985
  • Wu Wenjun The Grand Series of History of Chinese Mathematics Vol 6 吴文俊主编 《中国数学史大系》 第六卷
  • Li Yan, A Historic Study of Ceyuan haijing, collected works of Li Yan and Qian Baocong vol 8《李俨.钱宝琮科学史全集》卷8,李俨《测圆海镜研究历程考》