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[[File:益古演段.jpg|thumb|right|200px|Preface to Yigu yanduan, [[Siku Quanshu]]]]
 
'''''Yigu yanduan''''' (益古演段 old mathematics in expanded sections) is a 13th-century mathematical work by [[Yuan dynasty]] [[mathematician]] [[Li Zhi (mathematician)|Li Zhi]]. Yigu yanduan was based on North Song mathematician Jiang Zhou (蒋周) Yigu Ji (益古集 Collection of Old Mathematics)which was extinct.  However from fragments quoted in [[Yang Hui]]'s work  ''The Complete Algorithms of  Acreage(田亩比类算法大全)'', we know that this lost mathematical treatise Yigu Ji  was about solving  area problems  with geometry. Li Zhi used the examples of Yigu Ji to introduce  the art of  [[Tian yuan shu]] to new comers to this field. Although Li Zhi's previous monograph  ''Ceyuan haijing'' also used tian yuan shu, however it is harder to understand than Yigu yanduan.
 
Yigu yanduan was later collected into [[Siku Quanshu]].
Yigu yanduan consists of  three volumes with 64 problems solved with [[Tian yuan shu]] in parallel with geometrical method. Li Zhi intended to introduced students to the art of Tian yuan shu thru ancient geometry. Yigu yanduan together with Ceyuan haijing are considered  major contribution to [[Tian yuan shu]] by Li Zhi. These two works are also considered as the earliest extant documents on Tian yuans shu.
 
All the 64 problems  followed more or less the same format, it started with a question(问),  followed by an answer（答曰）, a diagram, then an algorithm（术）, in which Li Zhi explained step by step how to set up algebra equation with [[Tian yuan shu]], then followed by geometrical interpretation (Tiao duan shu). The order of arrangement of Tian yuan shu equation in Yigu yanduan is the reverse of that in Ceyuan haijing, i.e., here with the
constant term at top, followed by first order tian yuan, second order tian yuan, third order tian yuan etc. This later arrangement conformed
with contemporary convention of  algebra equation( for instance, Qin Jiushao's  [[Mathematical Treatise in Nine Sections]]), and later became a norm.
 
Yigu yanduan was first introduced to the western world by the British Protestant Christian missionary to China, [[Alexander Wylie (missionary)|Alexander Wylie]] who wrote:{{quote|Yi koo yen t'wan...written in 1282 onsists of 64 geometrical problem, illustrated the principle of Plane Measurement, Evolution and other rules, the whole being developed by means of T'een yuen}}.<ref>Alexander Wylie, Notes on Chinese Literature, p117, Shanghai 1902, reprinted by kessinger Publishing</ref>
 
In 1913  Van Hée translated all 64 problems in Yigu yanduan into French<ref>van Hée ''Li Yeh, Mathématicien Chinois du XIIIe siècle'', TP,1913,14,537</ref>
 
==Volume I==
[[File:益古第八问.jpg|thumb|right|200px|Problem 8 in ''Yigu yanduan'' solved by in line [[Tian yuan shu]]]]
Problem 1 to 22, all about the mathematics of a circle embedded in a square.
 
Example: problem  8
 
''There is a square field, with a circular pool in the middle, given that the land is  13.75 mu, and the sum of the circumferences of the square field and the circular pool equals to 300 steps, what is the circumferences of the square and circle respective ?
 
Anwwer: The circumference of the square is 240 steps, the circumference of the circle is 60 steps.''
 
Method: set up tian yuan one (celetial element 1) as the diameter of the circle,    x
:::::::[[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod v1.png]]
 
multiply it by 3 to get the circumference of the circle  3x (pi ~~3)
:::::::[[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod v3.png]]
 
subtract this from the sum of circumfereces  to obtain the circumference of the square  <math>300-3x</math>
 
:::::::[[File:Counting rod v3.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]] TAI
:::::::::[[File:Counting rod v-3.png]]
 
The square of it equals to 16 times the area of the square  <math>(300-3x)*(300-3x) = 900 -1800x +9x^2</math>
:::::::[[File:Counting rod v9.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod h1.png]][[File:Counting rod v-8.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]]
::::::::::[[File:Counting rod v9.png]]
 
Again set up tian yuan 1 as the diameter of circle,  square it up and multiplied by 12 to get
16 times the area of circle as
 
::::::::[[File:Counting rod 0.png]]  TAI
::::::::[[File:Counting rod 0.png]]
:::::::[[File:Counting rod h1.png]][[File:Counting rod v2.png]]
 
subtract from 16 time square area we have 16 times area of land
 
:::::::[[File:Counting rod v9.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod h1.png]][[File:Counting rod v-8.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]]
::::::::::[[File:Counting rod v-3.png]]
 
put it at right hand side
and put 16 times  13.75 mu = 16 * 13.75 *240 =52800 steps at left,
after cancellation, we get <math>-3x^2-1800x+37200</math>=0:
 
:::::::[[File:Counting rod v3.png]][[File:Counting rod h7.png]][[File:Counting rod v2.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod h1.png]][[File:Counting rod v-8.png]][[File:Counting rod 0.png]][[File:Counting rod 0.png]]
::::::::::[[File:Counting rod v-3.png]]
 
Solve this equation to get diameter of circle = 20 steps, circumference of circle = 60 steps
 
==Volume II==
[[File:益古第36问.jpg|right|250px]]
 
Problem 23 to 42,  20 problems in all solving  geometry of rectangle embedded in circle with tian yuan shu
 
Example, problem 35
 
Suppose we have a circular field with a rectangular water pool in the center, and the distance of  a corner to the circumference is 17.5 steps,
and the sum of length and width of the pool is 85 steps, what is the diameter of the circle, the length and width of the pool ?
 
Answer: The diameter of the circle is one hundred steps, the length of pool is 60 steps, and the width 25 steps.
Method: Let tian yuan one as the diagonal of rectangle, then the diameter of circle is tian yuan one plus 17.5*2
 
:                  <math>x+35</math>
multiply the square of diameter with  <math>\pi \approx 3</math> equals to four times the area of the circle:
 
:                  <math> 3 (x + 35)^2 =3x^2+ 210x + 3675</math>
 
subtracting four times the area of land to obtain:
 
:  four times the area of pool  = <math>3 x^2 + 210x + 3675- 4 x 6000 </math>= <math>3 x^2+210x -20325 </math>
now
 
The square of the sum of length and width of the pool =85*85 =7225
which is four times the pool area plus the square of the difference of its length and width (<math> (L-W)^2 </math>)
 
Further
double the pool area plus <math> (L-W)^2 </math> equals to  <math> L^2 +W^2</math> = the square of the diagonal of the pool
thus
 
( four time pool area + the square of its dimension difference ) - (twice the pool area + square if its dimension difference)
equals <math>7225 -x^2</math> = twice the pool area
 
so four times the area of pool = <math> 2 (7225 - x^2) </math>
 
equate this with the four times pool area obtained above
 
: <math> 2 (7225 - x^2) </math> =<math>3 x^2+210x -20325 </math>
 
we get a quadratic equation <math> 5 x^2 + 210x - 34775</math>=0
Solve this equation to get
*diagonal of pool =65 steps
*diameter of circle =65 +2*17.5 =100 steps
*Length - width =35 steps
*Length + width =85 steps
*Length =60 steps
*Width =25 steps
 
==Volume III==
[[File:益古演段下54.jpg|right|350px]]
 
Problem 42 to 64, altogether 22 questions about the mathematics of more complex diagrams
 
Q: fifty-fourth. There is a square field, with a rectangular water pool lying on its diagonal. The area outside the pool is one thousand one hundred fifty paces. Given that from the corners of the field to the straight sides of the pool are fourteen paces and nineteen paces.  What is the area of the square field, what is the length and width of the pool?
 
Answer: The area of the square field is 40 square paces, the  length of the pool is thirty five paces, and the width is twenty five paces.  
 
Let the width of the pool be Tianyuan 1.
 
:::::::[[File:Counting rod 0.png]] TAI
:::::::[[File:Counting rod v1.png]]
 
Add the width of the pool to the twice the distance from field corner to short long side of pool equals to the length of diagonal of the field  x+38
 
::::::::[[File:Counting rod h3.png]][[File:Counting rod v8.png]]
:::::::::[[File:Counting rod v1.png]]  TAI
 
Square it to obtain  the area of square with the length of the pool diagonal as its sides
 
:::::::  <math>x^2 +76x +1444 </math>
 
:::::::::[[File:Counting rod h1.png]][[File:Counting rod v4.png]][[File:Counting rod h4.png]][[File:Counting rod v4.png]]
:::::::::::[[File:Counting rod h7.png]][[File:Counting rod v6.png]] TAI
::::::::::::[[File:Counting rod v1.png]]
 
:::::The length of pool minus the width of pool multiplied by 2 =  2 (19-14)  = 10
 
Pool length =  pool width +10：x+10
 
::::::::[[File:Counting rod h1.png]][[File:Counting rod 0.png]] TAI
:::::::::[[File:Counting rod v1.png]] 
 
Pool area  =  pool with times pool length ：x(x+10) =<math> x^2 +10x </math>
 
:::::::::[[File:Counting rod 0.png]] TAI
::::::::[[File:Counting rod h1.png]][[File:Counting rod 0.png]]
:::::::::[[File:Counting rod v1.png]]
 
Area of pool times  乘  1.96 ( the square root of 2) =1.4 ) 
 
we have    <math>1.96x^2 +19.6x </math>
 
::::::::[[File:Counting rod h1.png]][[File:Counting rod v9.png]][[File:Counting rod h6.png]] tai
:::::::::[[File:Counting rod v1.png]][[File:Counting rod h9.png]][[File:Counting rod v6.png]]
 
Area of diagonal square subtract area of pool multiplied 1.96 equals to area of land times 1.96：
:      <math>x^2 +76x +1444 </math> - <math>1.96x^2 +19.6x = </math>：<math> -0.96x^2 +56.4 x +1444</math>
 
::::::::[[File:Counting rod h1.png]][[File:Counting rod v4.png]][[File:Counting rod h4.png]][[File:Counting rod v4.png]]
::::::::::[[File:Counting rod h5.png]][[File:Counting rod v6.png]][[File:Counting rod h4.png]] TAI
:::::::::::[[File:Counting rod 0.png]][[File:Counting rod h9.png]][[File:Counting rod v-6.png]]
 
Occupied plot times  1.96  =1150 *  1.96  =2254=<math> -0.96x^2 +56.4 x +1444</math>
 
hence  =<math> -0.96x^2 +56.4 x -810</math>：
 
:::::::::[[File:Counting rod v8.png]][[File:Counting rod v-1.png]][[File:Counting rod 0.png]]
::::::::::[[File:Counting rod h5.png]][[File:Counting rod v6.png]][[File:Counting rod h4.png]] TAI
:::::::::::[[File:Counting rod 0.png]][[File:Counting rod h9.png]][[File:Counting rod v-6.png]]
 
Solve this equation and we obtain
 
width of pooll  25 paces
therefore pool length  =pool width +10  =35 paces
length of pool =45 paces
 
==References==
 
{{reflist}}
*[[Yoshio Mikami]] The Development of Mathematics in China and Japan, p81
*Annotated Yigu yanduan by  Qing dynasty mathematician  Li Rui.
 
{{DEFAULTSORT:Yigu Yanduan}}
[[Category:Chinese mathematics]]
[[Category:13th-century works]]

Revision as of 22:36, 24 June 2013

File:益古演段.jpg
Preface to Yigu yanduan, Siku Quanshu

Yigu yanduan (益古演段 old mathematics in expanded sections) is a 13th-century mathematical work by Yuan dynasty mathematician Li Zhi. Yigu yanduan was based on North Song mathematician Jiang Zhou (蒋周) Yigu Ji (益古集 Collection of Old Mathematics)which was extinct. However from fragments quoted in Yang Hui's work The Complete Algorithms of Acreage(田亩比类算法大全), we know that this lost mathematical treatise Yigu Ji was about solving area problems with geometry. Li Zhi used the examples of Yigu Ji to introduce the art of Tian yuan shu to new comers to this field. Although Li Zhi's previous monograph Ceyuan haijing also used tian yuan shu, however it is harder to understand than Yigu yanduan.

Yigu yanduan was later collected into Siku Quanshu.

Yigu yanduan consists of three volumes with 64 problems solved with Tian yuan shu in parallel with geometrical method. Li Zhi intended to introduced students to the art of Tian yuan shu thru ancient geometry. Yigu yanduan together with Ceyuan haijing are considered major contribution to Tian yuan shu by Li Zhi. These two works are also considered as the earliest extant documents on Tian yuans shu.

All the 64 problems followed more or less the same format, it started with a question(问), followed by an answer（答曰）, a diagram, then an algorithm（术）, in which Li Zhi explained step by step how to set up algebra equation with Tian yuan shu, then followed by geometrical interpretation (Tiao duan shu). The order of arrangement of Tian yuan shu equation in Yigu yanduan is the reverse of that in Ceyuan haijing, i.e., here with the constant term at top, followed by first order tian yuan, second order tian yuan, third order tian yuan etc. This later arrangement conformed with contemporary convention of algebra equation( for instance, Qin Jiushao's Mathematical Treatise in Nine Sections), and later became a norm.

Yigu yanduan was first introduced to the western world by the British Protestant Christian missionary to China, Alexander Wylie who 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..[1]

In 1913 Van Hée translated all 64 problems in Yigu yanduan into French[2]

Volume I

File:益古第八问.jpg
Problem 8 in Yigu yanduan solved by in line Tian yuan shu

Problem 1 to 22, all about the mathematics of a circle embedded in a square.

Example: problem 8

There is a square field, with a circular pool in the middle, given that the land is 13.75 mu, and the sum of the circumferences of the square field and the circular pool equals to 300 steps, what is the circumferences of the square and circle respective ?

Anwwer: The circumference of the square is 240 steps, the circumference of the circle is 60 steps.

Method: set up tian yuan one (celetial element 1) as the diameter of the circle, x

File:Counting rod 0.png TAI
File:Counting rod v1.png

multiply it by 3 to get the circumference of the circle 3x (pi ~~3)

File:Counting rod 0.png TAI
File:Counting rod v3.png

subtract this from the sum of circumfereces to obtain the circumference of the square 300−3x

File:Counting rod v3.pngFile:Counting rod 0.pngFile:Counting rod 0.png TAI
File:Counting rod v-3.png

The square of it equals to 16 times the area of the square (300−3x)∗(300−3x)=900−1800x+9x2

File:Counting rod v9.pngFile:Counting rod 0.pngFile:Counting rod 0.pngFile:Counting rod 0.pngFile:Counting rod 0.png TAI
File:Counting rod h1.pngFile:Counting rod v-8.pngFile:Counting rod 0.pngFile:Counting rod 0.png
File:Counting rod v9.png

Again set up tian yuan 1 as the diameter of circle, square it up and multiplied by 12 to get 16 times the area of circle as

File:Counting rod 0.png TAI
File:Counting rod 0.png
File:Counting rod h1.pngFile:Counting rod v2.png

subtract from 16 time square area we have 16 times area of land

File:Counting rod v9.pngFile:Counting rod 0.pngFile:Counting rod 0.pngFile:Counting rod 0.pngFile:Counting rod 0.png TAI
File:Counting rod h1.pngFile:Counting rod v-8.pngFile:Counting rod 0.pngFile:Counting rod 0.png
File:Counting rod v-3.png

put it at right hand side and put 16 times 13.75 mu = 16 * 13.75 *240 =52800 steps at left, after cancellation, we get −3x2−1800x+37200=0:

File:Counting rod v3.pngFile:Counting rod h7.pngFile:Counting rod v2.pngFile:Counting rod 0.pngFile:Counting rod 0.png TAI
File:Counting rod h1.pngFile:Counting rod v-8.pngFile:Counting rod 0.pngFile:Counting rod 0.png
File:Counting rod v-3.png

Solve this equation to get diameter of circle = 20 steps, circumference of circle = 60 steps

Volume II

File:益古第36问.jpg

Problem 23 to 42, 20 problems in all solving geometry of rectangle embedded in circle with tian yuan shu

Example, problem 35

Suppose we have a circular field with a rectangular water pool in the center, and the distance of a corner to the circumference is 17.5 steps, and the sum of length and width of the pool is 85 steps, what is the diameter of the circle, the length and width of the pool ?

Answer: The diameter of the circle is one hundred steps, the length of pool is 60 steps, and the width 25 steps. Method: Let tian yuan one as the diagonal of rectangle, then the diameter of circle is tian yuan one plus 17.5*2

x+35

multiply the square of diameter with π≈3 equals to four times the area of the circle:

3(x+35)2=3x2+210x+3675

subtracting four times the area of land to obtain:

four times the area of pool = 3x2+210x+3675−4x6000= 3x2+210x−20325

now

The square of the sum of length and width of the pool =85*85 =7225 which is four times the pool area plus the square of the difference of its length and width ((L−W)2)

Further double the pool area plus (L−W)2 equals to L2+W2 = the square of the diagonal of the pool thus

( four time pool area + the square of its dimension difference ) - (twice the pool area + square if its dimension difference) equals 7225−x2 = twice the pool area

so four times the area of pool = 2(7225−x2)

equate this with the four times pool area obtained above

2(7225−x2) =3x2+210x−20325

we get a quadratic equation 5x2+210x−34775=0 Solve this equation to get

  • diagonal of pool =65 steps
  • diameter of circle =65 +2*17.5 =100 steps
  • Length - width =35 steps
  • Length + width =85 steps
  • Length =60 steps
  • Width =25 steps

Volume III

File:益古演段下54.jpg

Problem 42 to 64, altogether 22 questions about the mathematics of more complex diagrams

Q: fifty-fourth. There is a square field, with a rectangular water pool lying on its diagonal. The area outside the pool is one thousand one hundred fifty paces. Given that from the corners of the field to the straight sides of the pool are fourteen paces and nineteen paces. What is the area of the square field, what is the length and width of the pool?

Answer: The area of the square field is 40 square paces, the length of the pool is thirty five paces, and the width is twenty five paces.

Let the width of the pool be Tianyuan 1.

File:Counting rod 0.png TAI
File:Counting rod v1.png

Add the width of the pool to the twice the distance from field corner to short long side of pool equals to the length of diagonal of the field x+38

File:Counting rod h3.pngFile:Counting rod v8.png
File:Counting rod v1.png TAI

Square it to obtain the area of square with the length of the pool diagonal as its sides

x2+76x+1444
File:Counting rod h1.pngFile:Counting rod v4.pngFile:Counting rod h4.pngFile:Counting rod v4.png
File:Counting rod h7.pngFile:Counting rod v6.png TAI
File:Counting rod v1.png
The length of pool minus the width of pool multiplied by 2 = 2 (19-14) = 10

Pool length = pool width +10：x+10

File:Counting rod h1.pngFile:Counting rod 0.png TAI
File:Counting rod v1.png

Pool area = pool with times pool length ：x(x+10) =x2+10x

File:Counting rod 0.png TAI
File:Counting rod h1.pngFile:Counting rod 0.png
File:Counting rod v1.png

Area of pool times 乘 1.96 ( the square root of 2) =1.4 )

we have 1.96x2+19.6x

File:Counting rod h1.pngFile:Counting rod v9.pngFile:Counting rod h6.png tai
File:Counting rod v1.pngFile:Counting rod h9.pngFile:Counting rod v6.png

Area of diagonal square subtract area of pool multiplied 1.96 equals to area of land times 1.96：

x2+76x+1444 - 1.96x2+19.6x=：−0.96x2+56.4x+1444
File:Counting rod h1.pngFile:Counting rod v4.pngFile:Counting rod h4.pngFile:Counting rod v4.png
File:Counting rod h5.pngFile:Counting rod v6.pngFile:Counting rod h4.png TAI
File:Counting rod 0.pngFile:Counting rod h9.pngFile:Counting rod v-6.png

Occupied plot times 1.96 =1150 * 1.96 =2254=−0.96x2+56.4x+1444

hence =−0.96x2+56.4x−810：

File:Counting rod v8.pngFile:Counting rod v-1.pngFile:Counting rod 0.png
File:Counting rod h5.pngFile:Counting rod v6.pngFile:Counting rod h4.png TAI
File:Counting rod 0.pngFile:Counting rod h9.pngFile:Counting rod v-6.png

Solve this equation and we obtain

width of pooll 25 paces therefore pool length =pool width +10 =35 paces length of pool =45 paces

References

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.

  • Yoshio Mikami The Development of Mathematics in China and Japan, p81
  • Annotated Yigu yanduan by Qing dynasty mathematician Li Rui.
  1. ↑ Alexander Wylie, Notes on Chinese Literature, p117, Shanghai 1902, reprinted by kessinger Publishing
  2. ↑ van Hée Li Yeh, Mathématicien Chinois du XIIIe siècle, TP,1913,14,537
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