 # Hunebedden (Dutch dolmens) in proportion

## About orientation patterns with hunebedden (in European context)

### Geometry in ancient times

On the pages Orientation grid the plan of the chamber of some hunebedden and 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 similar to 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 [10.1]. It’s clear that CE:AC:CG = 1:N:N2. For a double diagonal this gives the length EG = N2 + 1, so that for a single diagonal counts AB = ½ (N2 + 1). This method corresponds with the Pythagorean algoritm of finding the triples for the right angled triangles [10.2]. For the sides near the right angle counts BC = ½ (N2 - 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) [10.3]. From P via B to D the length gets increasingly enlarged by factor M/N. For D this leads to the fraction D = P × N2/M2. This fraction can be smoothed out by taking a length of M2 for P. Then we have B = N/M × P = M × N and D = N/M × B = N2. 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 Construction Method Working out C = ½ (P + D) C = ½ (M2 + N2) C = ½ (72 + 32) = 29 A = D � C A = N2 � C A = 72 � 29 = 20 (Kangshen [10.3]) A = C � P A = C � M2 A = 29 � 32 = 20 (v.d. Waerden [10.4]) B = N/M × P B = N × M B = 7 × 3 = 21

Plato’s method of finding a right angled triangle [10.5], can be deduced from the method above. Therefore we need to fill in diagonal C with A = C - M2 and all sides need to be doubled. This gives B = 2MN, C = M2 + N2 and A = N2 - M2. Consequently Plato’s method forms a generalization of that of Pythagoras. But, as a matter of fact it weren’t the Greece who discovered this method. The much older clay tablet Plimpton 322 from the old Babylonian times (2000-1600 BC) can be put in this tradition also. This tablet holds four columns with numbers and probably two columns have broken off at the beginning. The remaining colums suggest that it’s about the sizes of right anlged triangles. Amongst many speculations about the broken columns, the ideas of Bruins fit best with the train of thoughts in Mesopotamia [10.6]. He demonstrates how the table can be setup, starting with pairs of reciprocals, which deminish regularly. (Reciprocals are numbers that make a power of 60 when they are multiplied with each other - in this case 600 = 1.) When taking these reciprocals as the proportion M:N, the following situation emerges: B = 2 × r × 1/r = 2, C = r2 + 1/r2 and A = r2 - 1/r2. Since squared reciprocals of 1 are reciprocals for themselves again, one can choose any pair of reciprocals and take them as a square of another pair. Furthermore on the clay tablet B is taken to be 1, so that by the rules of similarity A and C should be halved also. This gives B = 1, C = (r + 1/r) / 2 and A = (r - 1/r) / 2, which is the formula used by Bruins to find the widths and diagonals in the tablet. [10.7]

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. Also from the old Babylonian times a tablet with problems about right angled triangles remains (coded VAT 6598) [10.8]. Here (problem XVIII) the students are asked to find the diagonal size within a given rectangle. What follows isn’t a real calculation but a rough approach: C = B + ½ (A2 / 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 = A2 / B. After circling around (green) it appears, that C can be looked at as C = B + ½ 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 proportional numbers. Also the theorem can be found via the proportion pattern, but then D should be enlarged as in the figure above for the proportion M:N. 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 = B2 / A. Furhermore its clear that C:A = (D + A):C, which makes C2 = (B2/A + A) × A. This results in the theorem of Pythagoras C2 = A2 + B2.

The working out directly from the sides on the right angle is algebraic partly. Their sizes cannot be read from the figure directly. In that 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) the next statements are read [10.9]:

1.9
The diagonal of a square becomes the side of a square twice as large. 1.10
When the breadth is as wide as the side of a square and
the length is as wide of the side of a square twice as large,
then its diagonal is as wide of a square three times as large.

1.11
The other way around, it counts for the side of a square,
which is three times smaller as another square,
that this side makes one ninth of that other square.

1.12
The areas produced seperately by the length and the breadth of a rectangle together are equal to the area produced by the diagonal.

1.13
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.

The statements in the text become more and more general. Statement 1.9 can be proofed geometrically quite easily. This is not the case with statement 1.10, for the sides don’t form integral values compared to each other. Although unmistakable text 1.12 describes the theorem, the proof is lacking. Apparently this text explains the role of the right angled triangles in statement 1.13. It doesn’t pretend to be a math lesson, but rather it was used as a guide to manipulate shapes.

The grid proofs to be a good tool to solve geometrical 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 concerns diagonal in a rectangle also, but it is solved via the manipulation of areas. Below the method is rendered sec without calculations. The old Babylonians used sexadeciamal numbers, which can’t be transfered to simple decimal values all the time. The numbers are replaced by indications in square brackets.

[10.10]
The diagonal is 1¼ and the area is ¾. How long are the length and the breadth?
You, as you operate:

l.1 - Draw the diagonal and its counter-side and join them together [= square them] to be the ’base’.
l.2 - Copy the area twice and remove [it from the base]. This becomes the ’remainder’.
l.3 - Take the square-side [of the remainder] and break it midst. [This is the ’averager’.]
l.4 - Square [the averager] and take it in your hand.
l.5 - [Put] the area [= ¾] ’over’ your hand and take the square-side.
l.6 - Draw this [side] and its counter-side.
l.7 - Combine one together [with the averager] and remove [the averager] from the other.
l.8 - This makes length 1 and breadth ¾.

Although the text doesn’t have illustrations, the hypotenusa-diagram of Zhao can be put into service here successfully [10.11]. See the page ’Zhou bi’ cosmology on how the diagram fits to the pattern above found with some dolmens. Left: line 1 and 2 - the ’base’ is yellow, the ’remainder’ is the dark orange square in the middle
Midst: line 3 till 6 - the little dark orange square is taken in the hand - in the blue square the ’area’ is merged with (put ’over’) it
Right: line 7 and 8 - the calculation of the sides

Line 1 and 2. By squaring the diagonal of the ’area’ of this problem, the yellow base of the left figure emerges. This is called merging the diagonal with its opposite one. Nowadays we tend to see an area as the result of two square lines, but here it’s presented as the span between two opposite lines. Next two ’areas’ should be taken away from the square. This is not possible without cutting the ’area’ in pieces. The hypotenuse diagram offers an elegant solution to the problem, which stays usefull for the rest of the text. The four triangles in the field (two pale yellow and two vivid) together form the two ’areas’, which should be taken away from the base. The ’remainder’ resides between the four right angled triangles (dark orange).
Line 3 till 6. Squaring the half side of the ’remainder’ makes the little dark orange square in the middle figure - what is in the hand. Next the ’area’ has to be placed over ’the hand’. We should not takes this literally, for then the computation is no longer correct. Instead the word over can be understood as a merging - the modern mathematical phrase union of. By a slight adaptation of the ’area’ it’s possible to join it with ’the hand’ into a square (blue) indeed. Afterwards this is the side with its counter-side.
Line 7 and 8. The requested sizes of the sides of the ’area’ are found by substraction and addition (red arlines in the figure at right) of the ’averager’ to the sides of the blue square. Finally we deal with a cut-back of the 3:4:5-triangle: ¾:1:1¼.

The strategy of Db2-146 is worked out further in a method, which is sometimes called ’geometric algebra’. algebraic problems are solved by handling them as if they were geometric problems. Tablet YBC 6967 stands in this tradition also. Here two reciprocals are searched for having a mutual difference of 7. An area of a rectangle having reciprocals as its sides amounts for example 1 or 60 per definition. The solution of YBC 6967 makes clever use of this (line 3 below). From the strategy of Db2-146 only the manipulation of the rectangle and the blue square remains. In square brackets you will find cross-references to the solution of Db2-146.

[10.12]
A reciprocal exceeds the other by 7. Which are the reciprocals?
You:
l.1 - Break 7 in two pieces of 3½ [the avarager].
l.2 - Join 3½ together (=square) so that 12¼ emerges [the hand].
l.3 - Merge the area 60 with 12¼ so that you see 72¼ [the blue square].
l.4 - Which square makes 72¼? This is 8½ [blue side].
l.5 - Draw 8½ with its counter-side [the blue square].
l.6 - Then take 3½ - which made 12¼ [the hand] - from the one and add it to the other.
l.7 - The one becomes 12 and the other 5 [red arrows in the figure at right].
l.8 - The reciprocals are 12 and 5.

In particular line 5 and 7 are interesting, because they are superfluous for a pure algebraic solution. When left out, the next line is still comprehensible. Also the addition in line 6, that 3½ comes from 12¼, isn’t an explanation in algebraic terms. On the other hand these remarks do clarify the way the geometrical shapes are manipulated. Line 5 starts off with the drawing of the blue square. Next the averager (3½) is used to set up the rectangle.

The geometric algebra of Mesopotamia surpasses the operations above. One can see this on tablet BM13901 clearly. Again the first two problems can be understood in the light of the method above. One can think of them as looking for one of the sides of a rectangle [10.13]. But they form a stepping stone to the much more complex problems that follow. Then various adaptions in the construction are needed in order to find the solutions. Nevertheless the blue square and the ’remainder’ from the working out of Db2-146 remain important parts of the solution.

#### Conclusion

From the above it appears, that the ’intellectual distance’ between the pattern of some dolmen chambers and the earliest written math sources can be bridged via a systematical approach. The point is the proportional elongation of one of the right angled sides of the triangle. One can appoint phases in the solution finding, which supports a gradual development of the knowledge, although this development probably has elapsed quite whimsically. Meanwhile the geometrical understandings succeed each other in a theoretically and chronologically correct order. Furthermore this development seems not to be restricted to one specific region. Without going into who acted upon whom, the hypotenuse diagram of Zhao appears to be very usefull to understand math problems in Mesopotamia. On the page ’Zhou Bi’ cosmology the diagram appears to be interrelated with the mentioned pattern of the dolmens.

On the page Orientation grid it is put forward, that the pattern rather should be seen as the by-product of the construction of the chamber as 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.