卷上

卷上 第一問至第二十二問，全是關于正方形和圓形的問題。

例子：第八問：今有方田一段，內有園池水占之。外有地一十三畝七分半。只雲內外方圓周共相和，得三百步。問方圓周各多少？

答曰：外方周二百四十步，內圓周六十步。

法曰：立天元一為圓徑，以三之為圓周，以減共步，得

:::::: 太

:::::::

為方周，以自增乘，得

::: 太

::: （天元：相當于未知數 x）

:::::

十六段方田積于頭，再立天元圓徑以自之又十二之，得

::: 太

:::

::

為十六個圓池積以減頭位得

:::: 太

::::

:::::::

為十六段如積，寄左然後列真積一十三畝七分半，以畝法通之得五萬二千八百步，與左相消得

::::

::::

::::::

開平方為圓池徑，又三之為圓池周。

卷中

第二十三問至第四十二問，共20 問，解長方形和圓形的問題。

例題：第三十六問

今有圓田一段，中有直池水占之，外計地六千步。只雲從內池四角斜至四楞各一十七步半。其內池長闊共相和得八十五步。問三事各多少？

答曰外田徑一百步，池長六十步，闊二十五步。

法曰 （天元術）：立天元一位內池斜角，加二倍池角到圓池的距離為圓的直徑

圓直徑：

:::::::: 太 （常數項）

::::::::: （天元：相當于未知數 x）

圓直徑= x+35

將近似圓周率 3 乘圓直徑的平方 得圓面積的四倍= 3 (x+35)^2 = 3 (x^2+70x+1225) = 3x^2+210x+3675

四段圓面積：

:::::::: 太（常數項）

:::::::::: 元（x）

::::::::::: （x^2項）

四段圓面積減去四倍土地面積得池面積的四倍=3x^2+210x+3675- 4 x 6000 =3x^2+210x-20325

四段池積：

:::::::: 太

:::::::::: 元（x）

::::::::::: （x^2項）

池長闊之和 85的平方

::::::: （7225）

等于 四段池面積加 一段較（水池長與闊之差）冪（平方）

又二段池面積 加一段較冪 等于 長的平方加闊的平方 等于

水池對角線冪（對角線長度的平方）：

:::::::: 太

:::::::: 元 （x）

:::::::: （x^2）

（四池積 + 較冪） -(二池積+ 較冪) = 二池積 = 7225-x^2

::::::: （7225）

::::::::::

::::::::::

二段池積 乘二 = 四段池積:

::::::::

:::::::::::

:::::::::::

與上四段池積：

:::::::: 太

:::::::::: 元（x）

::::::::::: （x^2項）

合併得一元二次方程式 5x^2+210x-34775:

::::::: 太

::::::::::: 元

:::::::::::

解之得池對角線長度為65步

圓直徑= 65 + 2 * 17.5 = 65+35=100

較 = 長 - 闊 =35步

長 +闊 =85

由此 池長=60 步 池闊 =25 步

卷下

第四十三問至第六十四問，共22 問，是關于比較複雜的圖形。

第五十四問：今有方田一段，內有直池結角占之；外計地一千一百五十步只雲從田角至水兩邊各一十四步，一十九步；問三事各多少？

答曰：方四十五步，池長三十五步，闊二十五步。

法曰：設天元一為池闊：x

::::::::

:::::::: 元

池闊加田角到池邊距離的2倍 （38步）等于 方田對角線的長度：x+38

::::::::

::::::::: 元

取平方得展田（以方田對角線為邊的正方形）面積 x^2 +76x +1444

:::::::::

::::::::::: 元

::::::::::::

:::::又池長-池闊 = 2 (19-14) = 10

池長 = 池闊 +10：x+10

:::::::: 太

:::::::::

池面積 = 池闊 乘 池長：x(x+10) = x^2 +10x

:::::::: 太

::::::::

:::::::::

池面積 乘 1.96 ( \sqrt(2) =1.4 的平方 =1.96) 得 1.96x^2 +19.6x

:::::::: 太

:::::::::

展田面積 - 池面積 乘 1.96得地積乘1.96：

x^2 +76x +1444 - 1.96x^2 +19.6x = ： -0.96x^2 +56.4 x +1444

::::::::

:::::::::: 元

:::::::::::

占地 乘 1.96 =1150 * 1.96 =2254= -0.96x^2 +56.4 x +1444

由此得 = -0.96x^2 +56.4 x -810：

:::::::::

:::::::::: 元

:::::::::::

解方程得 池闊 25步

由此得 池長 =池闊 +10 =35步

放田邊長 =45 步

李冶在出版《測圓海鏡》之後，撰寫《益古演段》，將抽象的代數方法（天元術）和直觀性強的幾何方法（條段法）相結合來闡述問題，圖文並茂，便于學習天元術，使此書成為當時受人們歡迎的數學教材。蔣周的《益古集》已失傳，《益古演段》成為保留最詳盡的條段法資料的文獻。

1902年英國倫敦會傳教士偉烈亞力在1902年上海出版的《中國典籍扎記》簡略地介紹了《益古演段》.

1913年法國赫師慎（van Hée）將《益古演段》64問翻譯成法文，發表在通報上。

1984年新加坡大學藍麗蓉和馬來西亞的洪天賜發表英文論文題為《李冶及其益古演段》。

Overview

Yigu yanduan was based on Northern Song mathematician Jiang Zhou's (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), 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 newcomers to this field. Although Li Zhi's previous monograph Ceyuan haijing also used Tian yuan shu, it is harder to understand than Yigu yanduan.

Yigu yanduan was later collected into Siku Quanshu. pp :D

Yigu yanduan consists of three volumes with 64 problems solved using Tian yuan sh] in parallel with the geometrical method. Li Zhi intended to introduce students to the art of Tian yuan shu through ancient geometry. Yigu yanduan together with Ceyuan haijing are considered major contributions 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, starting 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 English readers by the British Protestant Christian missionary to China, Alexander Wylie who wrote:

In 1913 Van Hée translated all 64 problems in Yigu yanduan into French.

Volume I

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 (celestial element 1) as the diameter of the circle, x

::::::: TAI

:::::::

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

::::::: TAI

:::::::

subtract this from the sum of circumferences to obtain the circumference of the square 300-3x

::::::: TAI

:::::::::

The square of it equals to 16 times the area of the square (300-3x)*(300-3x) = 90000 -1800x +9x^2

::::::: TAI

:::::::

::::::::::

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

:::::::: TAI

::::::::

:::::::

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

::::::: TAI

:::::::

::::::::::

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 -3x^2-1800x+37200=0:

::::::: TAI

:::::::

::::::::::

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

Volume II

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 \pi \approx 3 equals to four times the area of the circle:

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

subtracting four times the area of land to obtain:

four times the area of pool = 3 x^2 + 210x + 3675- 4 x 6000 = 3 x^2+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 L^2 +W^2 = 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 -x^2 = twice the pool area

so four times the area of pool = 2 (7225 - x^2)

equate this with the four times pool area obtained above

2 (7225 - x^2) =3 x^2+210x -20325

we get a quadratic equation 5 x^2 + 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

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.

::::::: TAI

:::::::

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

::::::::

::::::::: TAI

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

::::::: x^2 +76x +1444

:::::::::

::::::::::: TAI

::::::::::::

:::::The length of pool minus the width of pool multiplied by 2 = 2 (19-14) = 10

Pool length = pool width +10: x+10

:::::::: TAI

:::::::::

Pool area = pool with times pool length : x(x+10) = x^2 +10x

::::::::: TAI

::::::::

:::::::::

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

one has 1.96x^2 +19.6x

:::::::: tai

:::::::::

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

x^2 +76x +1444 - 1.96x^2 +19.6x = ： -0.96x^2 +56.4 x +1444

::::::::

:::::::::: TAI

:::::::::::

Occupied plot times 1.96 =1150 * 1.96 =2254= -0.96x^2 +56.4 x +1444

hence = -0.96x^2 +56.4 x -810:

:::::::::

:::::::::: TAI

:::::::::::

Solve this equation and we obtain

width of pool 25 paces

therefore pool length =pool width +10 =35 paces

length of pool =45 paces