On the pages D22 and enlarged dolmens the plan of the chamber of some dolmens are compared with the geometrical figures below. These figures can be formed as follows: In a grid a diagonal will be used with the proportion 1:2. Three of those diagonals are joined together in a loop so that afterwards an integral right angled triangle evolves.
From the proportion 1:2 and the rules for congruence can be explained, that we deal with a 3:4:5-triangle. The triangles CDE and ACE are congruent with BEF and therefore CD:CE:AC = 1:2:4. By means of simple calculations the sides can be found. Because of the construction it's clear that AD and BE cross each other at a right angle.
The step from a proportion 1:2 to an arbritary proportion 1:N is rather obviously. By extending diagonal BE upward, another congruent triangle ACG emerges. It's clear that CE:AC:CG = 1:N:N^2. For a double diagonal this gives the length EG = N^2 + 1, so that for a single diagonal counts AB = 1/2 (N^2 + 1). This method corresponds with the Pythagorean algoritm of finding the triples for the right angled triangles . For the sides near the right angle counts BC = 1/2 (N^2 - 1) and AC = N.
For the next step towards the general proportion M:N some understanding is needed in how the proportion evolves in the figure. Below the proportion 3:7 is used as an example, which comes from the Jiuzhang suanshu (Nine Chapters on Mathematical Art - 200 BC) .
From P via B to D the length gets increasingly enlarged by factor M/N. For D this leads to the fraction D = P x N^2/M^2. This fraction can be smoothed out by taking a length of M^2 for P. Then we have B = N/M x P = M x N and D = N/M x B = N^2. One can find the rates of the enscribed triangle from the construction. The steps in the first column should be followed to do so. The method finds its parallel in the solution to problem 9.14 from the Jiuzhang suanshu (the other columns). In problem 9.14 the distances of a walk are found accordingly to the proportion of the walking speed. Strikingly Kangshen and v.d. Waerden calculate side A correctly in their translation, but still both differ from each other.
|Pattern||Problem 9.14||Problem 9.14|
|C = 1/2 (P + D)||C = 1/2 (M^2 + N^2)||C = 1/2 (7*7 + 3*3) = 29|
|A = D – C||A = N^2 – C||A = 7*7 – 29 = 20||(Kangshen )|
|A = C – P||A = C – M^2||A = 29 – 3*3 = 20||(v.d. Waerden )|
|B = N/M x P||B = N x M||B = 7 x 3 = 21|
Plato's method in finding a right angled triangle , can be deduced from the method above quite easily. Therefore we need to fill in diagonal C with A = C - M^2 and all sides need to be doubled. This gives B = 2MN, C = M^2 + N^2 and A = N^2 - M^2. Consequently Plato's method forms a generalization of that of Pythagoras.
Although the proportional approach is a handy tool for manipulations with right angled triangles, it has one backdraw. It isn't clear which triangles emerge from which proportional numbers. It could be, that ancient man has tried to find a solution for this problem by choosing the rates of the triangle sides as the proportional numbers. In that case the figure needs to be used differently.
From the old Babylonian times (2000-1600 Other websites) a clay tablet remains with problems about right angled triangles (coded VAT 6598) . Here the students are asked to find diagonal size within a given rectangle. What follows isn't a real calculation but a rough approach: C = B + 1/2 (A^2 / B). On basis of the proportion pattern this approach can be understood. In the figure left one sees D:A = A:B, so that D = A^2 / B. After circling around (green) it appears that C can be looked at as C = B + 1/2 D more or less. The figure aside is based on the proportion 3:4, which gives a rather big fault. Because in most of the integral right angled triangles side A is small according to sides B and C, this gives an even better result.
It's generally accepted that the geometrical thinking has spread from Mesopotamia towards Egypt and Greece. In that light tablet VAT 6598 can be seen as a stepping stone for the theorem of Phytagoras. At least it shows the interest to work things out from the sides in stead of the proportion numbers. Also the theorem can be found via the proportion pattern, but then D must be enlarged as in the figure for the proportion M:N (see above). As with the clay tablet the pattern is approached from the triangle which has the proportion as it's sides. In the figure left one sees D:B = B:A, so that D = B^2 / A. Furhermore its clear that C:B = (D + B):C, which makes C^2 = (B^2/A + A) x A. This results in the theorem of Pythagoras C^2 = A^2 + B^2.
The working out directly from the sides near the right angle is partly algabraic, because it cannot be read from the figure directly. In a sence one could say that mathematical understanding stands on a 'higher' plan already. For a mather of fact also the Sulba sutra has pointed out as the initial source of the theorem. In the part Baudhayana (about 700 BC) next two utterances are read:
The areas produced seperately by the length and the breadth of a rectangle together are equal to the area produced by the diagonal.
This can be observed in the rectangles having 3 and 4, 12 and 5, 15 and 8, 7 and 24, 12 and 35, 15 and 36 as their sides.
Although unmistakable text 1.12 describes the theorem, it doesn't prove it either. Instead text 1.13 takes it as a matter of experience afterwards, which makes it less probable that 1.12 flows from a given prove.
The grid proofs to be a good tool to solve math problems via proportions an congruences. Via a few stepping stones a continuous development appears from the chamber pattern of some enlarged dolmens till the theorem of Pythagoras. On the other hand it's unlikely that such a development has taken place in one smooth stream of extention of ideas. Possibly different 'schools' have existed, trying to come to solutions using their own methods. For example both the before mentioned clay tablet VAT 6598 and another one, Db2-146, belong to the old Babylonian period, but Db2-146 follows a different strategy. The problem of this tablet also concerns diagonal in a rectangle, but it is solved via the manipulation of areas. (Below the method is rendered sec without calculation. The old Babylonians used sexadeciamal numbers, which can't be transfered to simple decimal values all the time.)
The diagonal is 1 1/4 and the area is 3/4. How long are the length and the breadth?
You, as you operate:
r.1 - Draw the diagonal and its counter-side and join them together [= square them] to be the 'base'.
r.2 - Copy the area twice and remove [it from the base]. This becomes the 'remainder'.
r.3 - Take the square-side [of the remainder] and break it midst. [This is the 'averager'.]
r.4 - Square [the averager] so you get the (???).
r.5 - [Put] the area [= 3/4] 'over' the (???) and take the square-side.
r.6 - Draw this [side] and its counter-side.
r.7 - Combine one together [with the averager] and remove [the averager] from the other.
r.8 - This makes length 1 and breadth 3/4.
Although the text doesn't have illustrations, the hypotenusa-diagram of Zhao (see the page 'Zhou bi' cosmology) can be put into service here successfully. This diagram can be seen as based in the pattern found with some dolmens.
The yellow area of the left diagram evolves by squaring the diagonal of the rectangle of the assignment.
In the text this is called the joining together with its counter-side.
(NB. Nowadays we're used to see an area as the outcome of to lines at a right angle,
but here the square is represented as the area between two lines opposite each other.)
Two rectangle areas should be taken away from the diagonal area.
This will not succeed without cutting at least one rectangle into pieces.
The hypotenuse-diagram delivers an elegant solution, which appears to be usefull in the rest of the text.
The four triangles (two pale yellow and two vivid) together form the two areas that have to be removed from the base area.
The 'remainder' is located between the triangles (orange).
From the hypotenuse-diagram two things can be understood now:
(1) Four rectangles together with the remainder form the complete field of the hypotenuse-diagram. The area of one rectangle with only a quarter of the remainder (the ??? of line 4) results in quarter of the field then (the blue square - line 5 - the central figure).
(2) Half a side of the remainder (the averager of line 3) forms the difference in length between a side of the blue square and the short side as well as the long side of the rectangle (line 7).
The area of the rectangle was given and that of the remainder could be calculated, so that the area of the blue square can be known too. This makes it possible to take the square-side of it. By adding or removing the 'averager' from it (the red arrows in the right diagram), the length and the breadth are found.
Summarizing the above, the 'intellectual distance' between the pattern of some dolmen chambers and the earliest written math sources appears not to be that huge after all. One can appoint phases in the solution finding, which support a gradual development of the knowledge, although this development probably has elapsed quite whimsically. On the page D22 and enlarged dolmens it is put forward, that the pattern rather should be seen as the by-product of the construction of the chamber instead of being intentional. It's unlikely that Neolithic man intended to solve math problems in their dolmens. They tried to reach 'something' in the plan of the dolmens and used the state of science at that moment. On the page 'Zhou bi' cosmology that 'something' gets a possible meaning. To get even better results (not only with their dolmens) they must have increased their knowledge step by step.