Chapter 15 Right Triangles in Zhoubi Suanjing In the ancient Chinese tradition, a right triangle is called a gougu. Gou and gu mean respectively the shorter and longer legs of a right triangle. The hypotenuse is called xian. 15.1 The right triangle theorem From the Zhou Dynasty Canon of Gnomonic Computations (Zhoubi suanjing) compiled around the second century, B.C.: The circle originates from the square, the square from the rectangle, and the rectangle from 9 9 = 81. Therefore, cutting up a rectangle, we have (a right triangle) of width 3, length 4, and distance between two corners 5. After erecting a square on this (diagonal), surround it with (four) half-rectangles to form a square. In this way, we find (that the sides of the right triangle are) 3-4-5: the two rectangles (that make up the corners) have to increase by 25 (to fill up the whole square). This is called piling up rectangles.
86 Right Triangles in Zhoubi Suanjing 15.2 A Commentary on the diagram of right triangle, square and circle 1 Multiply each of the legs 2 to itself, combine them to give the area [of the square] on the hypotenuse. 3 The square root is the hypotenuse. According to the diagram, 4 the product of the legs can be taken as twice the area of the red triangle. Doubling it yields four triangles. Multiply the difference of the legs to itself as the area of the yellow square in the middle. Adding this area [from the difference of the legs] to the four triangles also gives the area [of the square] on the hypotenuse. 5 Figure 15.1: The Hypotenuse diagram Subtract this area [of the square on the difference between the leg] from the area on the hypotenuse, half the difference, take the square root of this, with the 6 difference [between the legs] as middle coefficient. In this way we get the shorter leg again. To get the longer leg, add the difference to the shorter one. 7 Whenever the areas of [the squares on] the shorter and the longer legs are combined, they give the area of [the square on] the hypotenuse. Whether as a square inside, or as a gnomon outside, the shapes may appear singular, yet the quantities 8 1 gōu gǔfāng yuán túzhù, a collection of rules for solving problems on right triangles is appended to the Zhoubi Suanjing, attributed to ZHAO Shuang in the third century A.D. 2 In the annotations here, we shall use a and b to stand for the legs, and c for the hypotenuse. 3 This is the main theorem, known as the theorem of gōu gǔ. 4 The diagram for the hypotenuse (xian tu), associated with the text of Zhoubi, but was lost. There are various emandations for the diagram. The one we adopt here follows QIAN Baocong. QIAN also associates several supplementary diagrams. 5 c 2 =4 1 2 ab +(b a)2. 6 Given c and b a, one can find a by solving the equation x(x +(b a)) = 1 2 (c2 (b a) 2 ). But Zhao did not explain how to solve this. It is apparent that this process (called dai cong kai fang), as an extension of the method of extraction of square roots, was presumed to be well known to the reader. 7 This paragraph gives the solution of a right triangle with c and b a given. 8 area.
15.2 2 87 are uniform; the bodies may be different, yet the numbers are the same. 9 a c a c b c b c c + b c b Gnomon for the shorter leg The gnomon 10 for the area on the shorter leg has as its breadth the difference of the hypotenuse and the longer leg, and as its length their sum, and the area for this longer leg forms the square inside. Subtracting the area of this gnomon from the area of the square on the hypotenuse, extracting the square root, we get the longer leg. Twice the longer leg as middle coefficient, extracting [square root], the corner 11 of the gnomon is the difference between the hypotenuse and the longer leg. Adding the longer leg [to this difference] gives the hypotenuse. Dividing the area [of the square] on the shorter leg by the difference [between the hypotenuse and the longe leg], we get the sum of the hypotenuse and the longer leg. 12 Dividing the area [of the square] on the shorter leg by the sum [of the hypotenuse and the longer leg], we get the difference of the hypotenuse and the longer leg. 13 Multiply the sum [of the hypotenuse and the longer leg] by itself. Add this to the area on the shorter leg. With this sum as dividend, and double the sum 14 [of hypotenuse and the longer leg] as divisor, we get the hypotenuse again. Subtracting the area on the shorter leg from the square of the sum [of the hypotenuse and the longer leg], with [this as dividend and] the same divisor [as before], we get 15 16 the longer leg.... 9 The paragraph reiterates the theorem on areas as a basic principle. In what follows the square on the shorter leg is also construed as the area of a gnomon. 10 The Chinese word here is ju. This is also the same word for rectangle. But the context dictates that the shape is the carpenter s square, hence the translation gnomon. Eucl.II, Definition 2: And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon. 11 Given the longer leg b and the area A of this gnomon, one can find the difference c b as the solution of x(x +2b) =A. 12 c + b = a 2 (c b). 13 c b = a 2 (c + b). 14 c =[(c + b) 2 + a 2 ] 2(c + b), because c = c+b 15 b =(c + b) 2 a 2 ] 2(c + b), because b = c+b 2 + c b 2 = c+b 2 + a2 2(c+b). 2 c b 2 = c+b 2 a2 2(c+b). 16 This paragraph gives the solution of a right triangle with given a and one of c ± b.
88 Right Triangles in Zhoubi Suanjing Multiply the two differences 17, double it and extract the square root. To the result add the difference between the hypotenuse and the longer leg, we get the shorter leg. 18 Added to the difference between the hypotenuse and the shorter leg, this gives the longer leg. 19 Added to the two differences this gives the hypotenuse. 20 21 a a a + b c a + b c c b c a b c a c b b From twice the square on the hypotenuse, remove the square on the difference between the longer and the shorter legs. This is the square on the sum of the legs, as is clear from the [hypotenuse] diagram (xián tú). Indeed, the double of the area on the hypotenuse fills up the large, outside square with a surplus. 22 This surplus is indeed the area on the difference between the longer and the shorter legs. Subtracting from this [double of the area on the hypotenuse] the area on the difference, extracting [square root], we get the side of the outer large square. The side of this large square is the sum of the longer and the shorter legs. Multiply this sum by itself, double, and subtract from the double of the area on the hypotenuse. Extracting [the square root of] the difference, we get the yellow area in the middle. The side of this area is the difference between the longer and the shorter legs. 23 Subtract the difference from the sum, half it; we get the shorter leg. 24 25 Add the difference to the sum, half it; we get the longer leg. 17 Between the hypotenuse and each of the two legs. 18 a = 2(c a)(c b)+(c b). 19 b = 2(c a)(c b)+(c a). 20 c = 2(c a)(c b)+(c a)+(c b). 21 The reconstruction of the associated diagram is explained in LIU Hui s commentary on JZSS 9.12. 22 the yellow area. 2c 2 =(a + b) 2 +(b a) 2. 23 b a = 2c 2 (a + b) 2. 24 a = 1 2 (a + b) 1 2 (b a); b = 1 2 (a + b)+ 1 2 (b a). 25 The paragraph gives the solution of right triangle with c and one of b ± a given.
15.2 2 89 [Suppose we have a rectangle.] A segment 26 is its breadth and width combined. Its area is the short leg and the long leg [of a right triangle] multiplied together. 27 Subtract four times the area from the area [of the square on the hypotenuse]. Extract [the square root] of the remainder to get the difference [between the breadth and the width]. 28 Subtract this difference from the sum, half the remainder; and we get the breadth. 29 Subtract the breadth from the segment, and we get what we want. 30 26 Twice the hypotenuse in the text. The Chinese character bèi (for twice ) apparently is redundant. It is inconsistent with the last sentence of the paragraph: subtract the breadth from the hypotenuse and we get what we want, namely, the width. The present paragraph presents the method of breadth and width; it amounts to finding two unknown given their sum and product. Putting x + y =2c and xy = a 2 for a right triangle with one leg a and hypotenuse 2c trivializes the issue here, for then clearly, x y =2b and y = c b, x = c + b. Incidentally, a slight variant of this method leads to the extraction of square root with middle coefficient : finding two unknown lengths with given product and difference. Perhaps it is for this reason that the method of dai cong kai fang is never explicitly explained in the ancient Chinese texts. 27 To find the breadth, say y, and the width, say x, of the rectangle, imagine you have the diagram for the right triangle with these unknowns lengths as legs. 28 x y = (x + y) 2 4xy = l 2 4ab, where l is the length of the given segment. 29 y =[(x + y) (x y)] 2. 30 x = l y.
90 Right Triangles in Zhoubi Suanjing
Chapter 16 Right triangles: Chapter 9 of JZSS The gougu rule and JZSS 9.1-3 JZSS 9.1 Now, given a right triangle, the lengths of its gou and gu are 3 chi and 4 chi respectively. Question: What is the length of the hypotenuse? Answer: 5 chi. JZSS 9.2 Now, given a right triangle, the lengths of its hypotenuse and gou are 5 chi and 3 chi respectively. Question: What is the length of its gu? Answer: 4 chi. JZSS 9.3 Now, given a right triangle, the lengths of its hypotenuse and gu are 5 chi and 4 chi respectively. Question: What is the length of its gou? Answer: 3 chi. The right triangle (Gougu) L1 rule Add the squares of gou and gu, take the square root [of the sum] giving the hypotenuse. L2 Further, the square of gu is subtracted from the square on the hypotenuse. The square root of the remainder is the gou. Further, the square of the
92 Right triangles: Chapter 9 of JZSS you is subtracted from the square on the hypotenuse. The square root of the remainder is gu. L3 L1: The shorter side is gou, and the longer side gu. The side opposite to the right angle is called the hypotenuse (xian). Gou is shorter than gu, and gu is shorter than the hypotenuse. These apply in all rates (standards), and so are explicated here. L2: Let the square on gou be red in colour, the square on gu be blue. Let the deficit and excess parts be mutually substituted into corresponding positions, the other parts remain unchanged. They are combined to form the square on the hypotenuse. Extract the square root to obtain the hypotenuse. Figure 16.1: LI Huang s proof of the gougu theorem L3: The sum of the squares on gou and the gu is the square on the hypotenuse. [If] one of them is missing, the remaining one can be obtained from the rest. JZSS 9.4 Now given a circular log, 2 chi 5 cun in diameter. Assume it is turned into a rectangular plank 7 cun thick. Question: What is the width? Answer: 2 chi 4 cun. Method: Square the diameter 2 chi 5 cun, subtract the square of 7 cun from it, extract the square root of the remainder. This is the width. L L: Here, take the diameter 2 chi 5 cun as the hypotenuse [of a right triangle], the thickness of the plank 7 cun as gou. So its width is gu.
93 JZSS 9.5 Now given a tree 2 zhang high, and 3 chi in circumference. A kudzu vine winds around it 7 times from its root to its top. Question: What is the length of the vine? Answer: 2 zhang 9 chi. Method: Multiply the circumference by 7 circuits as gu, let the height of the tree be gou. Find the hypotenuse as the length of the vine. L L: Obtain the length of the vine according to the circumference and height of the tree. The vine winds up the tree. Take a blue thread winding round a pen. Straighten the thread and observe. Every circuit gives a right triangle, with the segment as gu, the circumference as gou and the length of the vine as the hypotenuse. Multiply the circumference by 7 circuits means combining all gou into one [large] gou. Here gou is, unexpectedly, longer than the gu, so in the Method we consider reversely the height of the tree as the gou and the circumference as gu. To find the hypotenuse from gou and gu also refers to the first figure given under the [ordinary] no treecircumference condition. It is obvious that the sum of the squares on the gou and gu is the square on the hypotenuse. However, the squares on gu or gou may be put inside the square on the hypotenuse, either in the interior or on the exterior. If put in the interior, it is a square, if put on the exterior it becomes a gnomon, in these cases they are different in form but equal in area. In the figure, the red gnomon, whose area is equal to the square on gou, is put on the exterior, its area is also equal to a rectangle with length the sum of gu and the hypotenuse, and width the difference between gu and the hypotenuse. The square on the gu is put in the interior. If the blue gnomon, whose area is equal to the square on gu, is put in the exterior, its area is also equal to a rectangle with length the sum of gou and the hypotenuse, and width the difference between gou and the hypotenuse. The square on gu is put in the interior. Therefore divide [the square on gou or gu] by the sum or difference [of gu or gou and the hypotenuse] or multiply the sort [difference] by the long [sum to find th unknown gou and gu].
94 Right triangles: Chapter 9 of JZSS JZSS 9.6 A pond is a 1 zhang (= 10 chi) square. A reed grows at its center and extends 1 chi out of the water. If the reed is pulled to the side of the pond, it reaches the side precisely. Question: What are the depth of the water, and the length of the reed? Answer: water 12 chi deep; reed 13 chi long. Method: Square half the side of the pond. L1 From it we subtract the square of 1 chi, L2 the height above the water. Divide the remainder by twice the height above the water to obtain the depth of the water. L3 The sum of the result and the height above the water is the length of the reed. L1: Here take half the side of the pond, 5 chi, asgou, the depth of the water as gu, and the length of the reed as the hypotenuse. Obtain gu and the hypotenuse from gou and the difference between gu and the hypotenuse. Therefore, square gou for the area of the gnomon. L2: The height above the water is the difference between gu and the hypotenuse. Subtract the square of this difference from that of the area of the gnomon; take the remainder. L3: Let the difference between the width of the gnomon and the depth of the water be gu. Therefore construct [a rectangle] with a width of 2 chi, twice the height above the water. Its length is the depth of water to be found.
95 JZSS 9.7 (Hanging rope) When one end of a rope is tied to the top of a (vertical) pole, 3 chi rests on the ground. When stretched, the other end of (the rope) is 8 chi from the pole. Question: How long is the rope? Answer: 1 zhang 2 1 6 chi. Method: Square the distance from the foot, L1 divide by the length of the end lying on the ground, L2 add the result to the length of the end. Halve it, giving the length of the rope. L3 L1: Here consider the distance from the foot, 8 chi, asgou, the length of the string as the hypotenuse. The Problem of the string tightly stretched is the same type as that of the gate away from the threshold, i.e. given the gou and the difference between gu and the hypotenuse, find gu and the hypotenuse. Square the distance to get the area of the gnomon. L2: The end of the string lying on the ground is the difference between the hypotenuse and gu; take it to divide the area of the gnomon to obtain the sum of the hypotenuse and the gu. L3: If the numerator cannot be halved, double the denominator. The sum added to the difference is twice the length of the string. So halve the result. Subtract the difference from the sum and halve it to get the height of the pole. JZSS 9.8 Now give a wall 1 zhang high. A pole leans against the wall so that its top is even with the top of the wall. If the foot of the pole is moved 1 chi further from the wall, the pole will lie flat on the ground. Question: What is the length of the pole? Answer: 5 zhang 5 cun. Method: Square the height of the wall 10 chi and divide it by the distance moved by the foot of the pole; halve the sum to obtain the length of the pole. L
96 Right triangles: Chapter 9 of JZSS L: Here take 1 zhang, the height of the wall, as the gou, the length of the pole as the hypotenuse, and 1 chi, the distance moed from the wall as the difference between gu and the hypotenuse. The Problem is the same type as that of a string hanging from the top of a pole. JZSS 9.9 Now given a circular log of unknown size buried in a wall. When sawn 1 cun deep, it shows a breadth of 1 chi. Question: What is the diameter of the log? Answer: 2 chi 6 cun in diameter. Method: Square half of the breadth sawn, L1 divide by its depth. Add the result to the depth to obtain the diameter. L2 L1: In the Method, take the breadth sawn as gou, the diameter as the hypotenuse. The depth, 1 cun, is half the difference between gu and the hypotenuse. So the breadth is also halved. L2: The depth added is also half the difference. In the Problem [7] above one should halve the distance to the foot of the pole. Here all the data have been halved, so the difference is no more halved. JZSS 9.10 Now given a gate, partially opened, is 1 chi away from the threshold. There is a gap of 2 cun between the halves [of the gate]. Question: What is the width of the gate? Answer: 1 zhang 1 cun. Method: Square 1 chi, the distance from the threshold, divide by half the distance between the halves to obtain the width of the gate. L L: Here, take the distance from the threshold as the gou, half the width of the gate as the hypotenuse, and half the distance between the halves as the difference between the gu and the hypotenuse. To find the hypotenuse, the result should be halved. However, the width of the gate is twice the hypotenuse, so it is no longer halved.
97 JZSS 9.11 Now given a door, whose height exceeds the width by 6 chi 8 cun. Two [opposite] corners are 1 zhang apart. Question: What are the height and the width of the door? Answer: The width 2 chi 8 cun; height 9 chi 6 cun. Method: Let the square of 1 zhang be shi. Halve the difference; square it, and double. Subtract this from shi. Halve the remainder and extract its square root. Subtract from this half the given difference, giving the the width of the door. Add [to this square] half the difference, giving the height of the door. L, L: Let the width of the door be gou, the height be gu and the distance between two [opposite] corners be the hypotenuse. The height exceeds the width by 6 chi 8 cun; this is the difference between gou and the gu. Calculate according to the diagram. The square of the hypotenuse is exactly 10000 [square] cun. Double it, and subtract from it the square of the difference between the gou and gu. Extract the square root of the remainder to obtain the sum of the width and the height. Subtract the difference from the sum and halve it to obtain the width of the door, to [the width] add this difference to obtain the height of the door. The Method here finds half the dimensions of the door first. For there are 4 red square areas and 1 yellow square area in a 1 zhang square. Twice the square of half the difference between the width and height is 2 yellow square area. Half the 4 difference between the shi and half the yellow square area is 2 red square areas and 1 4 yellow square, which is just 1 of the big square. So its square root is half the sum of 4 the width and height of the door. From that we subtract half the difference between the height and width to obtain the width, and adding that to half the difference we obtain the height of the door. In the same diagram, multiply gou and gu, double the product, and add the square of the difference (of gou and gu). This also yields the square on the hypotenuse. [If] the product of gou and gu is known, then the hypotenuse can be determined. 1 1 This translation follows LI Jimin s critical edition, which is preferable in view of the subsequent
98 Right triangles: Chapter 9 of JZSS [A digression on the case of equal gou and gu] Ifgou and gu are equal, each multiplied to itself gives a square, and their sum is the square on the hypotenuse. Added to the double of the square of half of the difference (between gou and gu). [In this form] the difference still appears. All these multiplied to themselves or to each other as dividend and divisors, and the case of long gu and short gou, are merely different streams from the same source. If gou and gu are each 5, the square on the hypotenuse is 50. Its square root is 7 chi, with remainder 1, not exact. If the hypotenuse is 10, its square is 100, whose half is the square on gou or gu, i.e. 50 each. The extraction of its square root is also not exact. That is why one speaks of circumference 3 and diameter 1, and square side 5 and diagonal 7. Though these are not exact, it is still possible to speak of their approximations. [In the general case, consider] the square of the sum of gou and gu, and also the product of gou and gu. Subtract from the square 4 times the [latter] area. Extract the square root. This gives the difference of gou and gu. Add to the sum and halve to give gu. Subtract the difference from the sum, and halve to give gou. The gou, gu and hypotenuse are the height, width, and diagonal. All these follow from this diagram. [Take] twice of the hypotenuse as the sum of width and length, and the area of gou gnomon. The width is the difference of gu and hypotenuse. [Take] the area of gou gnomon (ju gou zhi mi). With twice of gu as congfa, extract the square root. This also gives the difference between the hypotenuse and the gu. Here is one [more] method: subtract the square on the difference of gou and gu from the square of the hypotenuse. Halve the remainder [as shi]. With the difference as congfa, extract the square root. This gives gou. JZSS 9.12 There are a door of unknown height and width, and a pole of unknown length. Horizontally the pole is 4 chi too long, and 2 chi vertically. But it can just be carried through the door diagonally. Question: What are the height, width, and the diagonal of the door? Answer: The width is 6 chi, height 8 ci, and diagonal 1 zhang. Method: Double the product of the two excesses [of the bamboo pole] and extract its square root. The result plus the excess [of the pole] over the height is the width.a digression on the case of equal gou and gu. Here is an alternative translation following standard editions. In the same diagram, the square of the sum of gou and gu, added to the square of their difference, is twice the square on the hypotenuse. Halve this, and extract the square root to obtain the hypotenuse. [Furthermore], subtract from twice the square on the hypotenuse the square of the difference of gou and gu [and extract square root] to find the sum of gu and gou. Knowing the hypotenuse [and the difference], it is possible to determine gou and gu.
99 The result plus the excess [of the pole] over the width is the height. The result plus both excesses is the diagonal of the door. L L: Here consider the width, height and diagonal of the door respectively as gou, gu and hypotenuse of a right triangle. The squares on gou and gu may be put with the squares on the hypotenuse as an interior square or an exterior gnomon. Observe two ends of the exterior gnomon outside the square on gou. The square on gou is [partly] covered by a blue gnomon leaving a yellow square. The area of the two overlapping rectangles between the two gnomons is equal to that of the yellow square. Each of the rectangles has the difference between the hypotenuse and gou as its length, and the difference between the hypotenuse and gu as its width. So twice the product of the two differences is the area of the yellow square. Extract its square root to obtain the side. The width of the blue gnomon outside is the difference between the hypotenuse and the gu, to which one adds [the side of the yellow square] to obtain gou. JZSS 9.13 A1zhang bamboo breaks and its top reaches ground, 3 chi from the bamboo. Question: How tall is the broken bamboo? Answer: 4 11 20 chi.
100 Right triangles: Chapter 9 of JZSS Method: Square the distance from the base L1 and divide this by its height. L2 Subtract the result from the height and halve the difference to obtain the height of the break. L3 L1: Here we consider 3 chi, the distance from the base as the gou, the height left standing as gu, [and] the length of the fallen tip as the hypotenuse. Find gu using the gou and the sum of gu and the hypotenuse. Hence first square gou to obtain the area of a gnomon [which has the area of gou square]. L2: The total height of the bamboo, 1 zhang, is the sum of gu and the hypotenuse. Divide the square of the gou by that to obtain the difference [between gu and the hypotenuse]. L3: The Method is the same type as that for hanging rope. 2 It is an alternative problem. Similar to the above Method: square the height, i.e. the sum of the area of the square on gu plus the hypotenuse. From the square [of the total height] subtract the square of the distance from the base. Consider the remainder as dividend, twice the height as divisor. Divide, giving the height of the break. JZSS 9.14 Now two men stand at the same point. A and B walk with rates 7 and 3 respectively. B walks [eastward]. A walks 10 bu south, then walks in a northeasterly direction till the two men meet. Question: What is the distance covered by each? Answer: B walks 10 1 bu eastward and A walks 14 1 bu diagonally to meet him. 2 2 Method: Halve the sum of the square of 7 and the square of 3, which gives the rate of A walking diagonally. Subtract this from the square of 7. The remainder is the rate of A walking south. The rate of B walking eastwards is 3 times 7. L1 Lay down the 10 bu southwards. Multiply this by the rate of A walking diagonally and again by the rate of B walking east as individual dividends. Divide each by the rate of A walking south to obtain both the required distances. L2 L1: Here, take the southward distance as gou, the eastward as gu, and the diagonal distance as gu and the hypotenuse. The rate for gu is 3. The rate for the sum of gou and the hypotenuse is 7. In order to obtain the rate for the hypotenuse divide 2 JZSS 9.7.
101 the square on gu by the sum of gou and hypotenuse to obtain the difference between the gou and the hypotenuse. Add this to the sum and halve it to get the hypotenuse. Subtract the difference between gou and the hypotenuse from the hypotenuse, to get gou. Fractions, if any, should be reduced to a common denominator, and simplified before the required rates are determined. The Rule considered the square on the sum of gou and the hypotenuse being taken as the red and yellow squares joined so as to avoid denominators appearing. Turn the blue gnomon with the area of gu squared into a rectangle, whose length is the sum of gou and the hypotenuse and width is the difference between gou and the hypotenuse. As a whole, the rectangle is twice the hypotenuse in length, and the sum of gou and the hypotenuse in width. Cut it horizontally half and half, with one half as the hypotenuse rate. The square of 7 is considered as the rate for the sum of the gou and the hypotenuse, from which subtract the hypotenuse rate and the remainder is gou rate. The point where the two men stand is the start of the walk. Both the hypotenuse rate and gou rate are [rectangles] with the sum of the hypotenuse and gou as their length [and with the hypotenuse and the gou respectively as their width]; then gu rate will also be a rectangle with the sum of the hypotenuse and the gou as its length [and with the gu as its width]. L2: The southward distance, 10 bu, is known as the given gou (jian gou). It should be multiplied by the hypotenuse rate and gu rate respectively and divided by gou rate [to obtain the required hypotenuse and gu]. JZSS 9.15 Now given [a right triagle] whose gou and gu are 5 bu and 12 bu respectively. Question: What is the side of the inscribed square? Answer: The side is 3 9 bu. 17 Method: The product of the gou and gu is the dividend, the sum of the gou and gu the divisor. Divide, giving the side of the square. L L: The product of gou by gu comprises three pairs of figures, red, blue and yellow. Put the yellow figures at the bottom, and combine the red and blue figures as rectangles at the top, with the side of the central yellow square as the width, and the sum of gou and gu as the length. That is why one adds the sum of gou and the
102 Right triangles: Chapter 9 of JZSS gu as divisor. In the figure, the square is contained in the given right triangle. To the top and to the right of the square there appear respective smaller right triangles. The relations between their sides retain the same rates as in the original triangle. The sum of the gou and gu of the small triangle [blue] on gu [side] is equal to the gu. Let gu of the original triangle be the sought rate, and the given gou,5bu,bethe given number. Apply the Rule of Three, giving the side of the inscribed square. Alternatively, let gou of the original triangle be the sought rate, the sum of gou and gu the given rate, and gu, 12bu, the given number. Apply the Rule of Three giving the side of the inscribed square. Although the statement lacks derivation, one arrives at the Method using the divisor and dividend. In the next Problem about an inscribed circle, one can see how this is done using the Rule of Three and the Proportional Distribution Rule. JZSS 9.16 Now given [a right triangle] whose gou and gu are 8 bu and 15 bu respectively. Question: What is the diameter of the inscribed circle? Answer: The side is 6 bu. Method: 8 bu is the gou, 15bu is gu. Find the hypotenuse. Add these three as divisor. Take gou to multiply the gu. Take twice [the product] as dividend. Divide, giving the diameter. L L: The product of gou by gu is the chief subject in the figure, in which there are three pairs of figures, red, blue and yellow. Doubling them, get four of each. Copy them onto a small piece of paper, and cut them out. Arrange them upright, slant or upside down by attaching the equal sides together so as to form a rectangle with the diameter as the width and the sum of gou, gu and the hypotenuse as the length. That is why we add gou, go and hypotenuse as divisor. From the figure, [the tip of] the blue triangle on the gu is equidistant from gou, gu and the hypotenuse. Measuring
103 this distance on gou and gu by a compass horizontally and vertically, one surely gets a small square. Further, draw a middle hypotenuse [parallel to the hypotenuse] with a compass. 3 In the middle of [each of] gou and gu there are small right triangles. The
104 Right triangles: Chapter 9 of JZSS diameter gou gu hypotenuse smaller gu [of the right triangle] on gou [of the original triangle] and gou [of the right triangle] on gu [of the given triangle] are both sides of the small square, i.e. half the diameter of the [inscribed] circle. Therefore the [relevant] data are all proportional. Let the gou, gu and the hypotenuse be the rates (lie cui), and take their sum as divisor. Multiply gou by the various summands as dividend, and divide. The gu of the right triangle on gou [of the given triangle] can be found. Multiply gu by the corresponding rate as dividend. 4 This gives the gou [of the right triangle] on gu [of the given triangle, as the product of gou and gu, divided by the sum of gou, gu, and hypotenuse. The diameter is twice of this]. Though differently expressed, the ways to form the dividend and divisor are actually the same. It is also possible to make use of the result of the previous problem on the inscribed square. Consider the right triangle with the middle hypotenuse. Its perimeter is the sum of gou and gu. 3 The recent critical edition of LI Jimin has yi gui tu hui, though DAI Zhen s edition had yi gui qi hui. 4 This means that the sum of gou, gu, hypotenuse of the right triangle on gou is equal to gou of the given triangle; similarly for gu. This is clear by considering the bisectors of the acute angles of the given right triangle.
105 The side of its inscribed square is therefore the product of the side of the inscribed square of the given triangle and the sum of gou and gu, divided by the sum of gou, gu and hypotenuse. This is the product of gou and gu, divided by the sum of gou, gu, and hypotenuse. Doubling this, we obtain the diameter of the circle inscribed in the given right triangle. Alternatively, subtract the difference between gu and the hypotenuse from the gou; subtract the difference between gou and the hypotenuse from gu; or subtract the hypotenuse from the sum of the gou and gu, to obtain the required diameter of the inscribed circle. Double the product of the difference between gou and the hypotenuse by the difference between gu and the hypotenuse. Extract the square root, giving the diameter. This is JZSS 9.12. JZSS 9.17 Now a given city 200 bu square, with gates opening in the middle of each side. 15 bu from the east gate there is a tree. Question: At how many bu from the south gate with one see the tree? Answer: 666 2 3 bu. Method: The number of bu from the east gate is the divisor. L1 Square half the side of the city as dividend. Divide giving the number of bu. L2 L1: The gou rate is the divisor. L2: Here take 15 bu, the distance from the east gate, as gou rate; 100 bu, the distance between the east gate and the southeast corner, as gu rate; and 100 bu, the distance between the south gate and the southeast corner as the given gou From the last, find gu as the number of bu from the south gate. The square of half the side is equivalent to the product of gu rate by the given gou. The two are equal. JZSS 9.18 Now given a city which is 7 li from east two west, 9 li from south to north, with gates opening in the middle of each side. There is a tree 15 li from the east gate. Question: At how many bu from the south gate with one see the tree? Answer: 315 bu. 5 Method: Take the distance between the east gate and the [southeast] corner to multiply the distance between the south gate and the [southeast] corner as dividend. The distance between the tree and the [east] gate as divisor. Divide, giving the distance. L 5 1 li = 300 bu
106 Right triangles: Chapter 9 of JZSS tree tree L: Here take the distance between the east gate and the [southeast] corner, 4 li, as gou rate, the distance from the east gate, 15 li, asgu rate. The distance between the south gate and the [southeast] corner, 3 li, as the given gu. From these find gou; i.e. the required number of bu from the south gate. The Method is the same as in the previous problem.
107 JZSS 9.19 Now given a square city to be measured. It has gates opening in the middle of each side. 30 bu from the north gate there is a tree. At 750 bu from the west gate one can see the tree. Question: What is the length of each side? Answer: 1 li. Method: Quadruple the product (shi) of the two distances from the gates. Extract the square root of the shi to obtain the length of each side. L tree L: According to the previous Method. square half the side; divide it by the distance from the east gate to obtain the distance from the south gate. Here the product of the two distances is the square of half the side, i.e. one-fourth of the area of the city. Therefore quadruple it to obtain the whole area. Extract its square root to obtain the side. JZSS 9.20 Now given a square city of unknown side, with gates opening in the middle. 20 bu from the north gate there is a tree, which is visible when one goes 14 bu from the south gate and then 1775 bu westward. Question: What is the length of each side? Answer: 250 bu. Method: Take the distance from the north gate to multiply the westward distance. Double the product as the shi. L1 Take the sum of the distance from the north and south gates as the linear coefficient (congfa). Extract the root to obtain the side of the city. L2 tree
108 Right triangles: Chapter 9 of JZSS L1: Here take the westward distance [1775 bu] asthegu; the distance from the tree to the south gate, 14 bu, asthegou. Take the distance from the north gate, 20 bu,asgou rate; the distance from the north gate to the western corner as gu rate; i.e. half the side of the city. Then take the gou rate, the distance from the north gate, to multiply the gu, the distance west. The product is the area of a rectangle whose width and length are gu rate, half the side of the city, and gou respectively. It is, however, only the area of the western half of the city. Doubling it means adding the eastern half to it for the whole. L2: In the Method the width from east to west of the rectangle is the same as the side of the city; the length from south to north is the distance from the tree up to 14 bu from the south gate. The two [small] rectangles have the sum of the two north and south distances as width and the city side as length. Then the sum of the two north-south distances [i.e. the difference between the length and width] is considered as the linear coefficient. The area is the surplus outside the city square. JZSS 9.21 A city is a 10 li 6 square. At the center of each side is a gate. A and B part from the center of the city. B walks eastwards and A begins southwards for some distance and then turns (in some northeast direction) through a corner of the city to meet B. Their speeds are in the ratio 5:3. How long has each walked? Answer: A walks 800 bu from the south gate and then northeastward 4887 1 bu until 2 he meets B. B walks 4312 1 bu eastwards. 2 Method: Let 5 be squared and 3 squared also, halve their sum to get the rate of the northeast advance. Subtract this rate from 5 squared. The remainder is the 6 1 li = 300 bu.
109 rate of the southward advance. The product of 3 by 5 is the rate of B walking eastward. L1 Lay down half the side, multiply [it] by the rate of the southward advance; divide it by the rate of the eastward advance to obtain the number of bu from the south gate. L2 Add this to the half side to obtain the number of bu southward for A. L3 Lay down the number of bu southward; to find the hypotenuse, multiply by the rates of the northeastward; to find B multiply the rate of the eastward advance. Take the products as dividends. Respectively divide by the rate of the southward advance to obtain the required numbers of bu. L4 L1: The Method to find the three rates is the same as in the above Problem(4) in finding the number of bu for A and B. L2: 5li, the distance from the south gate to the southeast corner, i.e. half the side of the city, is considered a small gu. Required is the number of bu from the south gate, so multiply half the side of the city by gou rate of the southward advance and divide it by the gu rate. L3: To take half the side means starting from the centre of the city. L4: This Method is the same as that of the previous Problem in finding the number of bu for A and B. JZSS 9.22 Now a tree is an unknown distance away from a person. Stand four poles, 1 zhang apart, making the left two aligned with the observed object. Observe from the rightrear pole, [the line of sight] cuts 3 cun into the left of the front-right pole. Question: How far away is the tree from the person? Answer: 33 zhang 3 chi 3 1 3 cun. Method: Consider 1 zhang squared as dividend, 3 cun as divisor; and divide. L L: Here take 3 cun, the length cut by the line of sight into the left of the frontright pole, as gou rate; the distance between the two right poles as gu rate; the distance between the left and right poles as the given you. The distance between the tree and the person is the gu corresponding to the given gou. The gu rate is multiplied by the given gou, both of which are 1 zhang, so the Method says square. Divide by 3 cun to obtain the number of cun. JZSS 9.23 Now to the west of a tree there is a hill whose height is unknown. The distance between the tree and the hill is 53 li and the tree is 9 zhang 5 chi high. A person standing 3 li to the east of the tree observes that the summit of the hill and the treetop are aligned. Assume his eyes are at the height 7 chi. Question: What is the height of the hill? Answer: 164 zhang 9 chi 6 2 3 cun.
110 Right triangles: Chapter 9 of JZSS Method: Lay down the height of the tree [95 chi]; subtract the height of the eyes 7 chi. Take the remainder to multiply 53 li as dividend. Take the distance, 3 li, from the person to the tree as divisor. Divide, and add the height of the tree to the quotient to obtain the height of the hill. L L: Bythegougu Rule. from the height of the tree subtract that of the eyes. The remainder 8 zhang 8 chi is considered as gou rate, 3 li, the distance from the surveyor to the tree as gu rate; and 53 li, the distance from the tree to the hill, as the
111 given gu. Find gou, to which add the height of the tree to get the height of the hill. JZSS 9.24 Now given a well 5 chi in diameter with unknown depth. Stand a pole of 5 chi high on the mouth of the well. When one looks down from the tip of the pole to the edge of the water, the line of sight cuts the diameter 4 cun. Question: What is the depth of the well? Answer: 5 zhang 7 chi 5 cun. Method: Lay down 5 chi, the diameter of the well; subtract 4 cun, the length cut by the line of sight. Take the remainder to multiply 5 chi; take the height of the pole as dividend; 4 cun, the length cut by the line of sight, as divisor. Divide, giving the depth of the well. L L: Here take the length cut by the line of sight as the gou rate, the height of the pole as gu rate, and 4 chi 6 cun, the remainder, as the given you. The required depth is gu corresponding to the given gou.