New Angles on Ancient Babylonian Geometry (Part 2)

The tablet shown above was found by a French archaeological expedition in 1934 in the ruins of the Royal Archives of Susa, in modern day Iran. It has been dated to the period of King Hammurabi of the Old Babylonian Empire, around the 17th century BC. The photograph above appears in the official expedition report published in 1961 Quelques Textes Mathématiques de la Mission de Suse par E.M.Bruin (available online here). The tablet displays a very remarkable piece of geometry which has attracted the attention of scholars and mathematicians since it was discovered.

I cited this tablet in The Map and the Manuscript because the same geometry can be found marked at scale in the landscape of the Languedoc. In this blog post, I'd like to revisit the subject, and in particular, to suggest an alternative interpretation to the geometry shown than the one conventionally adopted by scholars. I'd also like to take the opportunity to correct a small but significant error in my book.

In Part 1 of this series, I discussed the Babylonian tablet known as Plimpton 322. It comprises a table of values relating to angles, similar to a modern trigonometric table but with some important differences. The first relates to the manner in which the Babylonians understood the concept of the angle itself.

In modern geometry, an angle can be formed by the intersection of any two straight lines, and can take any value between 0° and 360°.

For the Babylonians, however, only those angles were considered which were formed by the diagonals of rectangles whose sides could be expressed as whole numbers. I call these grid angles, because the process is equivalent to joining points on a regular grid. So, the angles of interest to the Babylonians can all be obtained by joining points separated by whole numbers a and b taken as co-ordinates on a grid.

Of all such rectangles, the most highly prized were those for which the hypotenuse was also a whole number measure. Such triangles are now known as Pythagorean (though Pythagoras himself would not be born for another thousand years after this tablet was created).

To be clear, all right-angled triangles with sides of lengths a, b, and c satisfy the following relationship:

In particular, when a, b and c are all whole numbers, we call this a Pythagorean triangle and the set of three numbers (a,b,c) a Pythagorean triple.

Plimpton 322 comprises a list of such Pythagorean triples, including some examples with surprising, even astonishingly high values, for example (12,709; 13,500; 18,541) and (4,061; 8,100; 9,061). How did they manage to derive such sets of extremely large, whole numbers that satisfied the Pythagorean equation?

The answer is that they had a very convenient formula by the use of which they could generate virtually unlimited examples of Pythagorean triples on demand. All that it required was a pair of whole numbers to begin the process. The formula dictates that, for any pair of whole numbers a and b, then the following equations create a Pythagorean triple:

By the use of this formula, it is straightforward to generate Pythagorean triples as large, or as small, as you like, by simple choice of an initial starting pair. For example, if we take a = 1, and b = 2, then the result is the familiar (3,4,5) right-angled triangle.

There is another crucial piece of information that the Babylonian geometers understood about the relationship between the starting pair, and the resulting Pythagorean triple. It relates to the angles generated in each case. I will put it in bold to signal its importance:

If the initial pair of whole numbers is taken as a grid angle, then one of the angles in the right-angled triangle formed by the Pythagorean triple will be exactly double the size of that initial grid angle.

In the case of the example above, in contemporary notation, the grid angle [1:2] corresponds to 26.56° (that is, the inverse tangent of (1/2)). Twice this angle is 53.13°, and this is indeed one of the angles of the [3:4:5] triangle.

With this background in place, we can now turn our attention to the Susa tablet. The images below are taken from the original paper. First, an accurate drawing of the geometry and the cuneiform digits in sexagesimal (base-60) notation.

Next, a reconstruction of the full geometry with a transcription of the figures into conventional numerals. Note that these are still in base-60 as shown. That is to say, 31 15 should be read as 31 15/60, or 31 1/4, or 31.25 (in decimal, or base 10). Similarly, 8 45 may be read as 8.75 (decimal). Letters in red have been added for convenience of discussion below.

What are we looking at with this geometry? Consider first the triangle ABC. The hypotenuse AB has a measure of 50, and the lower side of 30. Obviously, this is a [3:4:5] Pythagorean triangle and we immediately recognise that the height, CB must be 40.

The geometry of the initial [3:4:5] triangle has been reflected to the right hand side, so that there is a second [3:4:5] triangle BCE.

A point D has been determined and marked, such that AD is the same length as DB and DE.

Notice the two new triangles that have been created: ACD, and its reflection, DCE.

A circle has been drawn with centre at D, which passes through A, B, and E.

The dimensions that are shown on the tablet are the sides of the new smaller right-angled triangles, given (in decimal) as [8.75: 30 : 31.25]. There is something very interesting about this right-angled triangle. I'll reserve pointing out what this is until further down the page, to give the curious reader the opportunity to figure it out first.

Now that we have a handle on the geometry, let's look at the description of it given in the original 1961 paper reporting the discoveries Quelques Textes Mathématiques de la Mission de Suse par E.M.Bruin. Here is the excerpt in question.

The first line may be translated: "This tablet (Figure 1) evidently contains the calculation of the radius of a circle circumscribing an isosceles triangle." Notice that "evidently". Is it evident?

(Also notice that a typo has crept into the paper. In the last line, x should be 31.15 (base60) and 40-x 8.45 (base60), not 31.45 and 8.15 as shown.)

The algebraic method that Bruin proposes that the scribe used is the straightforward and correct method that would be used today to solve it, that is, by constructing and solving a quadratic equation. But is it the only approach? And can we be sure that this is how the scribe thought about this problem?

There's a strong clue that something else might be going on here.

It has been noted by one commentator on this tablet, (I cannot now find the link, but the observation stands regardless of the source), that the diagram illustrating the geometry on the tablet happens to be very accurately drawn.

This is curious because knowing the length of the radius does not assist in the task of drawing the geometry. It is not permitted under the rules of geometry to to measure out the 31.25 length. This implies that the scribe must have had at their disposal a method of solving the problem geometrically, that is of constructing the required radius accurately with just compass and straightedge.

In light of this observation, we can conclude that, at the very least, if the scribe solved the problem using the quadratic method as suggested by Bruin, then he must have also had another geometric solution to the problem at his disposal.

But now, why would there be two solutions? If the problem was intended to be merely algebraic, then there was no need for the geometry to be drawn. Surely it was not provided as a mere illustration, with its construction unconnected to the problem under discussion.

A single approach to the problem which provided both the understanding to draw it geometrically, and to calculate the required values, would be a much better solution.

I now suggest such an approach. Recall the concept of grid angles. Another way to think of the [3:4:5] triangle on the tablet is as a depiction of the [3:4] grid angle. We can calculate this angle easily, of course, but for what follows it is not necessary to do so. It is the small angle of the triangle, the angle at the point B marked α.

A moment's glance at the diagram below will quickly convince the reader that the same angle appears twice at the point A, as shown, to give the slope of the line AE. And that the angle at E as shown is equal to twice the angle, or 2α. Please examine the diagram below to confirm this.

To summarise: we have grid angle [3:4], at point B. And we have right angled triangle AEC with an angle exactly twice this angle. Where have we heard this before?

This is of course the distinct feature of the Babylonian method of generating Pythagorean triangles discussed above. This tells us that the triangle AEC must be in the proportions of the Pythagorean triangle generated by the algorithm based on the input values [3,4].

Let's check. To refresh, here is the formula to generate the Pythagorean triple from the initial pair:

Now if a=3, and b= 4, as in this case, then the output of the formula is [7, 24, 25].

So [7, 24, 25] are the sides of a right-angled triangle. Our triangle AEC, [8.75: 30 : 31.25], should have sides in this proportion. And they do. AEC is a Pythagorean triangle, with sides of lengths that are whole number ratios [7,24,25].

Of course this triangle is scaled as it appears in the tablet, so these are not the final measures, just the proportions. However, it is straightforward to arrive at the values shown. We observe that the middle side of the generated Pythagorean triangle must have a length equal to a, the length of AC in the diagram above. This corresponds to the value 2ab in the formula.

This tells us that all we need to do to derive the required formula is divide throughout by 2b to adjust the three values correctly. The formula for deriving the lengths of the sides of the Pythagorean triangle is therefore:

If we test this using a=30 and b=40, we get [ 2500/80, 30, 700/80] which reduces to [31.25, 30, 8.75]. These are the decimal values, equal to the sexagesimal equivalents on the tablet: [31 15, 30 8 45].

This demonstrates that the scribe could have easily derived the values shown for the small right-angled triangle by recognising that the diagram was illustrating the relationship between a grid-angle and its corresponding Pythagorean triangle with an angle equal to twice the grid-angle.

Armed with this understanding, it is now straightforward to draw the geometry accurately.

We can see that all we need to do is generate the triangle ADB which is a reflection of ACE, and the same size.

So we open the compass to the length AC, and then with the compass point at B, swing an arc of this length. Then we open the compass to length BC, move the point to A, and swing a second arc.

The intersection is marked D in the diagram above.

AE=EB so a circle on E of this radius passes through the vertices of the triangles.

Hence, this method of understanding the diagram yields a simple method of drawing the geometry with accuracy, together with the calculation of the sides of the new small right-angled triangles generated.

It also reveals that this small triangle is a Pythagorean triangle, whenever the original upright triangle has a height and width which resolve to whole number measure, as in this case, 40 and 30 respectively. This is equivalent to saying that the geometry illustrates the algorithm by which the Babylonians could generate Pythagorean triangles, or triples, of any size required, based on any selected initial pair of whole numbers.

In summary, the method of reading the geometry of the Susa tablet that I have described offers two distinct advantages over the method given by Bruin, and repeated by commentators since. Both methods provide correct values for the radius of the circle, however only the method described above also leads directly to being able to construct the geometrical diagram. The second advantage is that it when this geometry is drawn for any whole number lengths a and b, the small right angled triangle created is revealed as a Pythagorean triangle, with sides in whole number ratio. The Bruin method does not of itself provide this insight.

Which leads me directly to the topic of the error in The Map and the Manuscript. It occurs in Chapter 14, on page 320, where I wrote, in regards to the geometry discussed in detail above, that the combination of the [3,4,5] triangle and the [7, 24, 25] triangle was unique, and that no other pair of Pythagorean triangles satisfied the geometry displayed on the Susa tablet.

In light of what I have learned since writing the book, and as laid out above, I could not have been more wrong with this statement. So now I take the opportunity to correct it.

In fact, the complete opposite is true: for every triangle for which the height and width are whole numbers, that is, for every single grid angle, including all Pythagorean triangles, the smaller triangle as shown in the geometry of the Susa tablet will be a Pythagorean triangle.

I've corrected the error in new printings of the book, with a footnote acknowledging the earlier mis-statement.

Finally, here is Figure 121 from The Map and the Manuscript, showing the Susa tablet superimposed on the Sunrise Line geometry in the landscape. The tablet geometry is a perfect 1:10 000 inch scale map of the geometry in the landscape. The existence of the Susa tablet, dated to 1700BC, and its extraordinary geometry, offers startling independent witness to the reality of the geometry of these alignments in the landscape of the Languedoc. Further details in the book.