Alexandru Proca

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 (Template:Zh) 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:31 year-old Systems Analyst Bud from Deep River, spends time with pursuits for instance r/c cars, property developers new condo in singapore singapore and books. Last month just traveled to Orkhon Valley Cultural Landscape.

This treatise consists of 12 volumes.

Volume 1

Reconstructed Diagram of circular city in alphabets

Diagram of a Round Town

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

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	HypotenuseTemplate:0c	VerticalTemplate:0b	HorizontalTemplate:0a
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

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

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

Segment Table 1

Definitions and formula

Miscellaneous formula

1. $(c_1-a_1)\ast (c_1\ast b_1)$= $\frac{1}{2}$*$(d_1{)}^2$
2. $a_{10}\ast b_{11}$= $\frac{1}{2}$$(d_1{)}^2$
3. $a_{13}\ast b_1$= $\frac{1}{2}$$(d_1{)}^2$
4. $a_1\ast b_{13}$= $\frac{1}{2}$$(d_1{)}^2$
5. $b_2\ast b_{15}$= ${\displaystyle }$$(r_1{)}^2$
6. $a_{14}\ast a_3$= $(r_1{)}^2$
7. $a_5\ast b_4$= $(d_1{)}^2$
8. $a_8\ast b_6$= $a_9\ast b_7$$=(r_1{)}^2$
9. $(b_{14}\ast c_{14})\ast (a_{15}+c_{15})$= $(r_1{)}^2$
10. $c_6\ast c_8$= $c_7\ast c_9)$=$a_{13}\ast b_{13}$
11. ${\displaystyle }$

The Five Sums and The Five Differences

1. $a_2+b_2+c_2=b_1+c_1$
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. $a_{15}+b_{15}+c_{15}=c_1-c_1$

• $(b_7-a_7)+(b_8-a_8)+(b_{14}-a_{14})+(b_{15}-a_{15})=2\ast (b_{12}-a_{12})$
• $a_8+(b_7-a_7)+(b_8-a_8)=b_7$

……………………Etc.

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

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

Volume 2

This volume begins with a general hypothesis

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:$d=\frac{2a_1\times b_1}{a_1+b_1+c_1}$

$=\frac{2\ast 320\ast 600}{320+600+\sqrt{(}320^2+600^2)}=240$

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:

$\frac{2a_2\times b_2}{a_2+b_2+c_2}=d$

$=\frac{2\ast 256\ast 480}{256+600+\sqrt{(}256^{+}600^2)}=240$

Question 3

inscribed circle problem associated with ΔRDN

$\frac{2a_3\times b_3}{a_3+b_3+c_3}=d$

Question 4：inscribed circle problem associated with ΔRSC

$\frac{2a_{12}\times b_{12}}{c_{12}}=d$

Question 5：inscribed circle problem associated with ΔTWB

$\frac{2a\times b}{a+b}=d$

Question 6

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

Question 7

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

Question 8

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

Question 9

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

Question 10

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

Tian yuan shu

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

：$480-x$

Template:V-1元

Template:V-4Template:H8Template:Rod0。

Then subtract tian yuan from eastward paces 200 to obtain:

$200-x$

Template:V-1元

Template:Rod2Template:Rod0Template:Rod0

multiply these two expressions to get：$x^2-680x+96000$

Template:V1

Template:H6Template:H-8Template:Rod0元

Template:V9Template:H6Template:Rod0Template:Rod0Template:Rod0

Template:Rod2

Template:Rod0元

that is$x^2-680x+96000=2x^2$

thus：$-x^2-680x+96000=0$

Template:V-1

Template:H6Template:H-8Template:Rod0元

Template:V9Template:H6Template:Rod0Template:Rod0Template:Rod0

Solve the equation and obtain $r=120$

Volume 3

17 problems associated with segment $b_2$i.e TW in ΔTWB

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.

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	$x^3-(b_2-2a_{14})x^2+a_{14}^2\ast x+a_{14}^2\ast b_2=0$
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	$((1/2)\ast c_2-(1/2)\ast b_2+b_2)\ast x^2-(1/2)\ast (c_2-b_2)b_2^2=0$
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	$b_2$，$c_{14}$	$a_{14}$	$x^4-2(b_2-c_{14})x^3+(b_2-c_{14}{)}^2{x}^2+2b_2{c}_{14}^{2}x-(2(b_2-c_{14})-b_2))b_2{c}_{14}^2=0$
14	$b_2$，$c_6$		$r=\sqrt{(}(2c_6-b_2)b_2)$
15	$b_2$，$c_8$	r	$-x^3-c_8{x}^2-b_2^{2}x+c_8{b}_2^2=0$
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

17 problems, given $a_3$and a second segment, find diameter of circular city.

。

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

18 problems, given$b_1$。

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

18 problems.

Q1-11，13-19 given$a_1$，and a seond line segment, find diameter d.

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

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

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

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

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}$${\displaystyle }$
8	$a_1+c_1$，$a_{14}+c_{14}$${\displaystyle }$
9	$b_1+c_1$，$b_{15}+c_{15}$${\displaystyle }$
10	$b_1+c_1$，$b_{14}+c_{14}$，${\displaystyle }$
11	$b_{14}+c_{14}$，$a_{15}+c_{15}$，$b_{14}+a_{15}-c_{13}$
12	$(b_8-a_8)+(b_2-a_2)$，$b_{14}+a_{15}-c_{13}$${\displaystyle }$
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})$${\displaystyle }$
15	$a_{14}+b_{14}$，$a_{15}+b_{15}$${\displaystyle }$
16	$a_{14}+a_{15}$，$b_{14}+b_{15}$${\displaystyle }$

Problem 14

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:

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:

$\frac{(b_7-a_7)+(b_8-a_8)+(b_{14}-a_{14})+(b_{15}-a_{15})}{2}$ $=(b_{12}-a_{12})$

$b_{12}-a_{12}=$$\frac{161+77}{2}$$=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)

$c_{12}=x+b_7=2\ast x+(b_7-a_7)+(b_8-a_8)=2\ast x+161$

$c_{12}^2=(x+b_7{)}^2=(2\ast x+161{)}^2=4\ast x^2+644\ast x+25921$

$c_{12}^2-(b_{12}-a_{12}{)}^2$

$=4\ast x^2+644\ast x+25921-$$\frac{((b_7-a_7)+(b_8-a_8)+(b_{14}-a_{14})+(b_{15}-a_{15}){)}^2}{4}$

$=4\ast x^2+644\ast x+11760=d$(diameter of round town),

$d^2=(4\ast x^2+644\ast x+11760{)}^2=16\ast x^4+5152\ast x^3+508816\ast x^2+15146880\ast x+138297600$

Now, multiply the length of RZ by $4\ast x$

$4\ast x\ast b_7=4\ast x\ast (x+(b_7-a_7)+(b_8-a_8))=4\ast x\ast (x+161)=4\ast x^2+644\ast x$

multiply it with the square of RS:

$d^2=4\ast x\ast b_7\ast c_{12}^2=$$(4\ast x^2+644\ast x)\ast (4\ast x^2+644\ast x+25921)=$$16\ast x^4+5152\ast x^3+518420\ast x^2+16693124$

equate the expressions for the two $d^2$

thus

$16\ast x^4+5152\ast x^3+518420\ast x^2+16693124=$$16\ast x^4+5152\ast x^3+508816\ast x^2+15146880\ast x+138297600$

We obtain:

$9604\ast x^2+1546244\ast x-138297600=0$

solve it and we obtain $x=a_8=64$;

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

Volume 9

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

8 problems

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

：Miscellaneous 18 problems：

Q	GIVEN
1	$c_2$，$c_3$${\displaystyle }$
2	$c_5$，$c_4$${\displaystyle }$
3	$b_{11}$，$c_4$${\displaystyle }$
4	$a_{10}$，$c_3$${\displaystyle }$
5	$a_{10}$，$b_{11}$${\displaystyle }$
6	$b_7$，$a_8$${\displaystyle }$
7	$b_1-b_{11}$，$a_1-a_{10}$
8	$b_{10}-a_{10}$，$b_{11}-a11$${\displaystyle }$
9	$c_{13}$，$a_{10}-b_{11}$${\displaystyle }$
10	$a_{12}+b_{12}$，$a_{13}+b_{13}$${\displaystyle }$
11	$c_1$，$\frac{b_1}{a_1}$${\displaystyle }$
12	$d_{10}-d_{11}$，$d_{12}-d_{13}$${\displaystyle }$
13	$c_{12}-[c_{10}-(b_{10}-a_{10})]$，$c_{11}+(b_{11}-a_{11})-c_{13}$，$b_{12}-a_{12}$
14	$c_8-(c_1-b_1)$，$(c_1-a_1)-c_7$${\displaystyle }$
15	$a_1+c_1$，$(c_1-a_1)+(c_1-b_1)$${\displaystyle }$
16	$a_{12}+b_{12}+c_{12}$，$(a_{13}+b_{13})-c_{13}$${\displaystyle }$
17	From the book Dongyuan jiurong
18	From Dongyuan jiurong

Volume 12

14 problems on fractions

Problem	given
1	$b_1+c_1$，$a_1$= $\frac{8}{15}$$b_1$
2	$a_1+c_1$，$a_1$= $\frac{8}{15}$$b_1$
3	$a_1=(1-5/9)\ast 3d$，$b_1-a_1$${\displaystyle }$
4	$a_3=(5/6)\ast d$，$b_2-a_3$
5	$(15/16)b_1=d$，$a_1+b_1$${\displaystyle }$
6	$a_{12}=(8/15)\ast b_{12}$，$c_{12}-b_{12}$，$c_{12}-a_{12}$
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$${\displaystyle }$
10	$a_{11}+(1/3)b_{10}$，$b_{10}-(3/4)a_{11}$${\displaystyle }$
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

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

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

References

42 year old Obstetrician and Gynaecologist Merle from Komoka, has interests including entertaining, property developers house in singapore singapore and train collecting. Just had a family trip to China Danxia.

• 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》