第三問
今有方田一段, 內有圓池水占, 之外計地一萬一千三百二十八步. 只云 從外田角斜至內 池楞各五十二步.
問內徑
342外方各多少.
荅曰:外田方一百二十步. 內池徑六十四步.
343
法曰:立天元一為內池徑. 加倍至步得 為方斜. 以自增乘得 為 方斜幂, 於頭.
其方斜上本合身外減四. 今不及減便是寄一步四分為分母也. 今此方斜幂乃是變斜為方面. 以自乘之數, 又別得是展起之數也.
又立天元為池徑. 自之,又三因, 四而一為池積.
今為方田積, 既以展起, 則此池積亦須展起. 故又用ㄧ步九分六釐乘之, 得一 步四分七釐, 亦為一个展起底圓池積也.
以一步九分六釐乘之者, 葢為分母十四. 以自之得一步九分六釐也.
以池積減田積 餘
344為一段如
345積, 寄左. 然後列真積, 一萬
一千三百二十八步, 亦用分母幂, 一步九分六釐, 乘之
或兩度下加四亦同得二萬
342 面徑 is in WYG and WJG
343 A:方田. b: 圓池. c: 五十二步.
344 47 instead of 0.47 in last line in WYG and WJG.
345 虛 instead of 如 in WJG and WYG.
188
二千二百 0 二步八分八釐. 與左相消得
346. 平方開之得六十四步為內池 徑也. 倍至步, 加池徑, 身外除四,見方面也.
一法求所展池積. 以徑自之了,更不須三因,四除及,以一步九分六釐 乘之. 只於徑幂上以一步四分七釐 .
○案:
此即三因四除一步九分六釐之數.乘之,便為所 展之池積也。
依條段求之. 展積內減四段至步幂, 餘為實. 四之至步為從. 四分七釐 益隅.
347
義曰:凡言展積,者是於正積上,以一步九分六釐乘起之數. 元法本是方面上 寄一步四分. 分母自乘過, 於每步上, 得一步九分六釐. 故今命之為展起之數 也. 諸變斜為方面
348者皆準. 此所展之池積是, 於一步圓積上, 展出九分六釐.
346 47 instead of 0.47 in last line in WJG and WYG
347 J1-4:減 . c1-4:從 . abcd: 四分七釐. klmn:二分五釐.
Diagram is slightly different in WYG, the outer circle is passing through the corner of the inside square.
348 面 is not in WYG.
189
若以池徑上 取斜為外圓徑, 則一步上,止生得四分七釐也. 故以四分七釐為 虛常法. 又取方幂, 一步九分六釐, 四分之三亦得圓積, 一步四分七釐也.
銳案:此圖元本脫左右兩從字,今增.
案: 法內皆,以徑一周三方五斜七為率. 故各面積分數與密率不合.葢此書專為明理而作密率數.繁礙於講解.
故用古率以從簡且其法既明即用密率亦無不可.
Problem Three.
Suppose there is one piece of square field, inside of which there is a circular pond full of water, while outside a land of eleven thousand three hundred and twenty eight bu is counted. One only says that [the distances] from the angle of the outer square obliquely reaching the edge of the inside pond are 52 bu each.
One asks how long the inside349 diameter and the outer side are each.
350
The answer says: the side of the outer field is one hundred and twenty bu. The diameter of the inside pond is sixty four bu.
The method says: Set up one Celestial Source as the diameter of the inside pond. Adding twice the reaching bu yields 104
1
tai 351
as a diagonal of the square.
349 There is the character 面, mian, “side” instead of 内, nei, “inside”, in the WJG and WYG.
350 a: square field. b: circular pond. c: fifty two bu.
351 Starting from this problem there is no commentary by the editor of the Siku quanshu on polynomial expressions anymore.
190
as the square of the diagonal of the square352,
which is sent to the top.
From353 the diagonal of the square in the text above, one must remove out of the body its four [tenth]. Now, one does not carry out this diminution; one sends instead one bu four fen to make the denominator. Now, the square of the diagonal of the square354 is then the transformation of the diagonal into the side of a square. Besides, by self-multiplying this quantity, one obtains the expansion of this quantity355.
One sets up further the Celestial Source as the diameter of the pond. This times itself, and increased further by three, then divided by four makes the area of the pond. Now, if one has expanded the area of the square field, then the area of the pond also needs to be expanded. That is why one uses further one bu nine fen six li to multiply this, what yields one bu four fen seven li, which also makes one expanded area of the inside356 circular pond.
One multiplies by one bu nine fen six li, because one takes fourteen as the denominator357, which, self-multiplied, yields one bu nine fen six li.
Subtracting the area of the pond from the area of the field remains 10816
353 Usually commentaries found in the siku quanshu are introduced by the character 案, an, “commentary”; and the commentaries by Li Rui are introduced by 銳案, Rui an. Here, this commentary has no title; it is attributed to Li Ye himself.
354方斜幂, fang ke mi.
355 Let a be the side of the square and d, the diagonal. The purpose stated in the beginning of the problem is to find a. The first polynomial gives the diagonal. To find the side of the square from its diagonal, one just has to reduce the diagonal by 2 . Here 21.4. “To diminish the diagonal by its four [tenth]” is, in modern terms:
1.4
d da or 1.4
d a. Instead of reducing the diagonal, Li Ye proposes another procedure. One keeps the diagonal to make an expanded square whose side is this diagonal and works with the values of this new square.
1.4 will be placed as denominator to reduce thereafter all the values. In the case of area, that is to use 1.4 × 1.4
= 1.96, which is “the expansion of this quantity”.
356 底, di.
357 分母十四, fen mu shi si. The denominator is 1.4, which self-multiplied gives 1.96. I do not understand why one read 14 in all the different editions. Either the denominator was moved of one place on the counting board what makes a multiplication by 10, or it is a mistake made by the copyists which was not noticed by Li Rui and the editor of the Siku quanshu.
358 The zero of 0.47 is missing in the WJG.
359 The expression 虛積, xuji, “empty area” is used in the WJG instead of 如積, ruji, “equal area”.
191 After, one places the genuine area, eleven thousand three hundred and twenty eight bu. One uses also the square of the denominator360, one bu nine fen six li, to multiply this.
Or to augment by four [tenth] at the second degree361 is also the same
It yields twenty two thousand two hundred and two bu eight fen eight li. With what is on the left,
eliminating from one another yields
Opening the square of this yields sixty four bu as the diameter of the inside pond. One adds twice the reaching bu to the diameter of the pond, and reduces out of the body its four [tenth], there appears the side of the square.
One looks for the expanded area of the pond according to another method. Once the diameter is self multiplied, it is not necessary to increase by three, to divide by four, and to multiply this by one bu nine fen six li. One just directly multiplies the square of the diameter by one bu four fen seven li,
Commentary: this is to multiply the quantity of one bu nine fen six li by three and to divide it by four.
what makes the expanded area of the pond362.
One looks for this according to the section of pieces [of area]. From the expanded area, one subtracts four pieces of the square of the reaching bu of the diagonal, what remains makes the dividend. Four times the reaching bu makes the joint. Four fen seven li is the augmented corner.
360 分母羃. Fen mu mi.
361 两度加四, liangdu jia si. The main discourse follows the procedure: a² × (1.4)². The commentary suggests another possibility that could be: [a + a × 0.4]²; a being the side of square. On each of the side, one adds 1 and then adds 0.4. That represents a lag on the counting table.
362 Let x be the diameter of the circular pond. In the previous method, the expanded area of the pond was:
2 2
. The alternative method suggests to combine the coefficients in order to have only one coefficient, that is to directly multiply x² by 1.47, the latter being the result of 3
41.96.
192
363
Commentary by Li Rui: the diagram on the original edition lacks of two characters “joint” on the left and on the right. Here they are added364.
The meaning says: Each time one talks about expanded areas, one means a quantity which emerges from a multiplication of a real area365 by one bu nine fen six li. The original method basically consisted in sending one bu four fen on the side of the square. Once the denominator is self-multiplied, for each bu, it yields one bu nine fen six li. Therefore, now, one names this “to make the expanded value”. Every transformations of a diagonal into a side of the square follow the same standard. The area of the pond which is expanded is that for each bu of the circular area, one expands by nine fen six li. If, on the diameter of the pond, one takes the diagonal to make the diameter of the outer circle, hence, on one bu, it produces only four fen seven li. Therefore, four fen seven li makes the empty constant divisor.
[According to the other method] one takes further the surface of the square366, three quarters of one bu nine fen six li, which also yields the circular area of one bu four fen seven li.
363 j1-4: subtract. c1-4: joint. abcd: four fen seven li. klmn: two fen five li.
The diagram is slightly different in the WJG and WYG. The outer circler is passing through the corner of the inside square.
364 These characters are not lacking in WJG and WYG.
365正積, zheng ji.
366 方幂. Fang mi. Usually, I translate mi by “square”, here, I translate it by “surface”.
193 Commentary: All the methods uses the diameter one [with] the
circumference three, and the side five [with] the diagonal seven as lü.
Hence, none of the fraction of the sides of the area corresponds to the mi lü. The reason is that this book concentrates on explaining principles, and to make some complex mi lü quantities would impede the explanation.
That is why one uses the ancient lü in order to facilitate and clarify. Once the method is clear, nothing prevents from using also the mi lü367.
367 The commentator of the siku quanshu associates the values 1 with 3 and 5 with 7. In my translation, there are four quantities, but in fact these are the expressions of two ratios. To use mi lü or ancient lü makes the ratio changes. The term lü express the fact that quantities can vary in the same way together. See [Chemla. 2994.
pp.956-9]. The ratio “diameter one, circumference three” is equal to 3, which is the approximate value of π here;
and the ratio “side five, diagonal seven” is equal to 1.4, which is the approximate value of 2 here. The mi lü is a technical expression for the ratio 22 for the circumference and 7 for the diameter, what give a more accurate value of π. According to the commentator, Li Ye chooses to use ancient lü for didactical reasons. There are also practical reasons. To use π=3 is the simplest way to construct ratio between squares and circles: one can easily transform four circles into three squares whose side is equal to the diameter, as one will see in other problems.
194 Problem Three, description.
Let a be the distance from the angle of the square field, to the circle, 52 bu ; let A be the area of the square field (S) less the area of the circular pond (C), 11328 bu; and x be the diameter of the pond.
The procedure of the Celestial Source:
Diagonal of the square = 2a + x = 104 + x
Expanded area of S: S × (1.4)² = (2a + x)² = 4a² + 4ax + x² = 10816 + 208x + x² C = 43x2= 0.75x², since π=3
Expanded area of C: 1.96C = 43x2× (1.4)² = 1.47x² 1.96S – 1.96C = 4a² + 4ax + x² - (43x2× (1.4)²) = A × (1.4)²
= 10816 + 208x + x² - 1.47x² = 10816 + 208x - 0.47x² = 11328 × 1.96 = 22202.88bu.
We have the following equation:
(4a²- 1.96A) + 4ax + x² -1.47x² = -11386.88 +208x -0.47 x² = 0
The procedure by section of pieces of area:
1.96A = 4a² + 4ax + x² - 0.47x²
The equation: 1.96A – 4a² = 4ax – 0.47x²
A piece of the diagonal, a, is given with the area represented in pink in figure 3.1, and one looks for the diameter of the inner circle. Li Ye, by expressing the diagonal of the square in term of the unknown (2a + x), is transforming the diagonal into the side of a square. Instead of dividing by √2 (value of diagonal = side of the square with denominator √2), he is multiplying every length by 1.4.
(“The diagonal of the square (…) corresponds to an outer body diminished by four. Now, one does not
195 carry out this diminution; one places instead one bu four fen as denominator”). In the case of area, that means to multiply by 1.96, and “one names this “to make the expanded value”. Once, one has transformed the side of the square into a diagonal by using √2, and expanded all the dimensions (figure 3.2), one can, hence, proceeds exactly like in the problem one (figure 3.3, see problem One).
First one removes the four squares of side a which are in the corners. It remains four rectangles of length a and width x. Thus 4ax makes the joint divisor. Then to find the coefficient of the term in x², one has to consider the central square area of side x. From this area, one removes the expanded area of the circle. That is -1.47x² + x². To find the expanded area of the circle, one looks for the diagonal of the central square and use it as a diameter. “on the diameter of the pond, one takes the diagonal (figure 3.4) to make the diameter of the outer circle (figure 3.5)”. The difference between these two areas is the constant divisor: - 0.47x². “hence, on each bu, it produces only four fen seven li. Therefore, four fen seven li is the empty constant divisor (figure 3.6)”.
Figure 3.1
Figure 3.2
196
Figure 3.3
Figure 3.4
Figure 3.5
197
Figure 3.6
198