4. Description of the Art of the Celestial Source, Tian Yuan Shu, 天元術 118
4.1.1 Descriptive examples 133
Here follows as illustration the translation of the first and one of the last problems –the first being basic and the other being more elaborated- and a transcription of those into modern mathematical terms:
Problem One.
Suppose there is one piece of square field, inside of which there is a circular pond full of water, while outside a land of thirteen mu seven fen and a half is counted. One does not record the diameter of the inside circle and the side of the outer square. One only says that [the distances] from the edge134 of the outer field reaching the edge of the inside pond [made] on the four sides are twenty bu each.
One asks how long the diameter of the inside circle and the sides of the outer square are.
The answer says: The side of the outer square field is sixty bu. The diameter of the inside pond is twenty bu.
The method says: Set up one Celestial Source as the diameter of the inside pond. Adding twice the reaching bu yields 40
1
tai as the side of the field. Augmenting this by
self-multiplying yields 1600
80 1
tai
as the area of the square, which is sent to the top.
133 See complete translation of the problem and Chinese text in supplement.
134 楞, leng.
60
135
Set up again one Celestial Source as the diameter of the inside pond. This times itself and
increased further by three then divided by four yields 0 0 0.75
tai
as the area of the pond.
Subtracting this from the top position yields 1600
80 0.25
tai
as a piece of the empty area,
which is sent to the left.
After, place the genuine area. With the divisor of mu, making this communicate yields three thousand and three hundred bu.
With what is on the left, eliminating from one another yields 1700
80 0.25
Opening the square yields twenty bu as diameter of the circular pond. Adding twice the reaching bu to the diameter of the pond gives the side of the outer square.
135 a: distance to the water, 20 bu. b: side of the square field, 60 bu.
61 Description in modern terms of problem one:
Statement: Let a be the distance from the middle of the side of the square to the pond, 20bu; let A be the area of the square field (S) less the area of the circular pond (C), 13mu 17fen, or 3300bu; and d be the diameter of the pond. One asks the side of the square and diameter.
x = d
Side of the square = 2a + x = 40 + x
S = (2a + x)² = 4a² + 4ax + x² = 1600 + 80x + x² C = 43x2= 0.75x², with π=3.
S – C = 4a² + 4ax + x² - 43x2= 1600 + 80x + 0.25x² = 3300 = A A- (4a² + 4ax + x² - 43 x2) = 1700 - 80x - 0.25x² = 0
62 Problem Sixty three
Suppose there is one piece of a big circular field and two pieces of a small and a big square field. Inside of the small square field there is a circular pond full of water. The sum of the outer areas, sixty one thousand three hundred bu is counted. One only says [the distance]
from the side of the small square field reaching the edge of the pond is thirty bu. The side of the big square field exceeds the side of the small field by fifty bu. The diameter of the circular field also exceeds the side of the big square field by fifty bu.
One asks how long the four136 things each are.
The answer says: the side of the small square field is one hundred bu. The diameter of the pond is forty bu. The side of the big square field is one hundred fifty bu. The diameter of the circular field is two hundred bu.
The method says: set up one Celestial Source as the diameter of the inside pond. Adding twice what reaches the water, sixty bu, makes the side of the small square field.
On the side of the small square, adding further the difference of the sides of the small and the big squares, fifty bu, gives the side of the big square.
On the side of the big square, add further the difference between the diameter of the big circle and the side of the big square, fifty bu, gives the diameter of the big circle.
137
A diagram138 is provided on the left.
One diameter of the inside circle: 0 1
tai
136 “four” in the siku quanshu; “three” in Li Rui edition.
137 a: big circular field. b: big square field. c: small square field. d: pond
138 Here the word 圖, tu, “diagram”, refers to the four dispositions on the surface for computing. These
dispositions which are on the right in the original edition are presented immediately following in my translation.
This term could also be translated by plural “diagrams”. I do not known if the character refer to the four polynomials placed in two columns in the Chinese text, or to each of the polynomial independently.
63
yuanas four pieces of the area of the circle pond, which on the
above [position]139.
Put further the side of the small square, 60 1
Quadrupling this yields the following pattern:
48400 880 4
as140 four pieces of the area of the
small square, which is on the second [position].
Put further the side of the big square. This times itself yields
12100
139 The dispositions on the surface for computation are different than in the other problem. One of the specifity of this problem is that it introduce different names for the position on the counting support. As one has four areas to compute, those areas are placed on different positions: the above [position] (上), the second [position]
(次) , the bottom [position] (下) and the position under the bottom (下位之次). In other problem polynomial are placed on the top (頭) and on the left (左) only.
140 The character 太 tai is not in the three following polynomials. Many of the polynomials of this problem lack of the character tai or yuan. I tend to think that this is a mistake of the copyist.
64
as141 the square of the diameter of the big circle.
Tripling this yields the following pattern:
76800 960 3
as four pieces of the area of the big circle,
which is on the position under the bottom.
Combining what is on the three last positions yields the following pattern:
139600 2320 11
, which is
on the right.
Subtracting the four areas of the pond, 0 3
yuanfrom what is on the right yields
139600 2320 8
as
four pieces of the equal area, which is sent on the left.
After, place the genuine area, sixty one thousand three hundred bu. Multiplying by four because of the distribution yields two hundred forty five thousand two hundred bu. With
what is on the left eliminating them from one another yields:
105600 2320 8
What yields from opening the square is forty bu as the diameter of the inside pond. Adding each difference of the bu gives each side of the squares and the diameter of the circle.
141 The character 太 tai is not in the three following polynomials.
65 Transcription in modern term of Problem sixty four:
Let a be the distance of 30 bu from the side of the small field to the circular pond; let A be the area of the big square field (S) plus the area of the circular (C) and the area of the small square field (B) less the area of the circular pond (P), 61300 bu; and x be the diameter of the pond. And let d be the side of the small field, b, the side of the big square field and c, the diameter of the circular field knowing that: d + 50 = b and b + 50 = c. Let call i= 50
The procedure of the Celestial Source:
d = x + 2a = x +60
b = d + 50 = x + 110 or b = x + 2a + i c = b + 50 = x + 160 or c = x + 2a + 2i 4P = 3x²
B = d² = (x + 2a)² = 4a² + 4ax + x² = 3600 + 120x + x² 4B = 16a² + 16ax + 4x² = 14400 + 480x + x²
S = b² = (x + 2a + i)² = x² + 2(2a + i)x + (2a + i)²= (x + 110)² = 12100 + 220x + x² 4S = 4x² +8(2a + i)x + 4(2a +i)² = 48400 + 880x + 4x²
c² = (x + 2a + 2i)² = x² + 2(2a + 2i)x + (2a + 2i)² = 25600 + 320x + x² 4C = 3c² = 3x² + 6(2a + 2i)x + 3(2a + 2i)² = 76800 + 960x + 3x²
4B + 4S + 4C = 4(2a + i)² + 3(2a + 2i)² + 16a² + x[16a + 8(2a + i) + 6(2a + 2i)] + 11x²
= 139600 + 2320x + 11x²
4B + 4S + 4C – 4P = 4(2a + i)² + 3(2a + 2i)² + 16a² + x[16a + 8(2a + i) + 6(2a + 2i)] + 8x² = 4A
= 139600 + 2320x + 8x² = 245200
66 We have the following equation: 4A – [4(2a + i)² + 3(2a + 2i)² + 16a²] - x[16a + 8(2a + i) + 6(2a + 2i)] - 8x² = 105600 - 2320x - 8x² = 0