Plimpton 322


 

Abstract

The ancient Babylonian clay tablet known as Plimpton 322 is a set of solutions to a problem.  The problem from which these solutions are derived can be stated simply and the values on the Tablet can be found using Euclidean geometry and algebra.  The Tablet is too short not to have limits imposed upon its values.  These limits are found by imposing geometric/algebraic constraints on the problem, but not in terms of angular measure or trigonometry.

I first learned of the Babylonian clay tablet called Plimpton 322 while reading the textbook written by H. L. Resnikoff and R. O. Wells, entitled Mathematics in Civilization.  The subject intrigued me greatly and I began a search for ways to understand this Tablet.  I formulated several questions and/or observations:  First, I felt that there had to be a problem that was being solved.   Second, what method or methods were available in the time of the Tablets’ construction to solve the problem?  Third, how are we permitted to interpret the numbers that resulted from their calculations?   Finally, what limits were placed upon the tabular results of the problem?

Today, this tablet resides in a display case in the Rare Book & Manuscript Library located within the Butler Library of Columbia University in New York City.

 

Outline of the paper:

I    The statement of the problem

II   Methods of solution

III Interpreting the calculated results on the Tablet

IV Limits on the Tablet values

 

The setting of the problem:  Suppose two travelers are walking along a road some three thousand years ago.  Noticing that there is an ample supply of reeds growing nearby, one of the two proceeds to pull up one of the reeds.  Noting the approximate center of the reed, he makes a bend in the reed, not at its center, and forms a right angle at the bend.  Handing the bent reed to his companion, he says, “ Here is a problem for you:  make a bend on the long side at a point so that the end point just touches the short side end point, all the while keeping all the sides straight and the right angle that I gave to you remains fixed.”

We formalize this problem, using modern notation as follows:

 

Let the reed end points be labeled A and B.  Bend the reed at any point C, between A and B, forming a right angle at C and, without loss of generality, let AC < BC (a right angle is constructible by the methods of Euclidean Geometry).  See Figure 1.

Figure 1- The bent reed.

Part I Statement of the Problem

 

Formal statement of the problem:  Find the point D on BC at which AD = DB.  As a further explanation of this statement, the point D may be thought of as that point about which the line segment  DB  is rotated so that the point at B may be brought to A in a sweeping circular arc with D as center.  See Figure 2.

Figure 2- With D as center, the circular arc from B to A.

Part II Solutions

Solution by geometric methods:

 

From Figure 1, draw AB.  Find the midpoint E, of AB.  Construct the perpendicular bisector of AB at E and extend the perpendicular through E, meeting BC at D.  Then AD = DB because D lies on the perpendicular bisector of AB and is therefore equidistant from the two points A and B .  See Figure 3.

Figure 3- The geometric solution.

Solution using geometry, algebra, and line segment lengths in terms of  \mathit{p} and  \mathit{q} :

Starting with Figure 1, we construct Figure 4 as follows:

let  \mathit{p=BC},  \mathit{q=AC} and  \beta =\angle ABC .

Find the midpoint F of BC.  At F, construct FE perpendicular to BC, making FE parallel to AC, and meeting AB at E.   E is the midpoint of AB and the length of EF=\frac{1}{2}q.  At E, construct ED perpendicular to AB, meeting BC at D.  Then, as in the geometric solution, D is that point at which AD = DB.

From  rt\Delta DFE,   it follows that   \angle DEF=\beta.  Now  \angle DEF=\angle ABC=\beta.  Therefore

rt\Delta ACB\approx rt\Delta DFE.  It follows that

\frac{AC}{BC}=\frac{DF}{EF},   or    \frac{q}{p}=\frac{DF}{\frac{1}{2}q},   so that  DF=\frac{q^{2}}{2p}.

Then

AD=BD=BF+DF=\frac{p}{2}+\frac{q^{2}}{2p}=\frac{p^{2}+q^{2}}{2p}.

Finally,

DC=FC-FD=\frac{p}{2}-\frac{q^{2}}{2p}=\frac{p^{2}-q^{2}}{2p}    and  AC=q.

 

The above derived line segment lengths are illustrated in Figure 4 below.

Figure 4- Geometric and algebraic solution.

If each line segment length of Figure 4 is multiplied by 2p then

DC=p^{2}-q^{2},AC=2pq,AD=p^{2}+q^{2},BF=p^{2},FD=q^{2} \textup{and}\mathit{EF=pq}

 

 

Part III

Interpreting the calculated results on the Tablet

 

Looking at the results for DC, AC, and AD above, one might argue that we have succeeded in finding a method of producing Pythagorean Triples.  However, we lack an adequate hypothesis stating that we are seeking a method of finding right triangle lengths satisfying

(DC)^{2}+(AC)^{2}=(AD)^{2} .

Furthermore, we do not know that the sum of the squares of the two legs is equal to the square on the hypotenuse for a right triangle.  That is, we have obtained the three sides of a right triangle without knowing anything about a relation between the legs and the hypotenuse.  In other words, our hypothesis is empty.  Therefore in the context of this paper, we may not conclude that we have produced formulas to generate Pythagorean Triples nor, for that matter, have we proven the Theorem of Pythagoras.

 

It may be observed from Figure 4, with

\beta =\angle ABC ,  that \angle ADC=2\beta because   \angle ADC   is the sum of the two opposite interior angles,

\angle DAB  and  \angle DBA   of the isosceles   \Delta ABD .  This result will be used later in the paper.

 

Part IV

The Limits

The values that appear in column I of the Tablet are equal to the square of the ratio of the hypotenuse  AD  to the leg  AC   and will be called

R^{2}=\left ( \frac{p^{2}+q^{2}}{2pq} \right )^2.                                                                     (1)

The values of Column I on the tablet Plimpton 322, when written in modern Hindu – Arabic numerals, are equivalent to a set of values which range from 1:23:13:46:40 to 1:59:00:15.

The Tablet end values of this range strongly suggest that there are limiting values between which the Tablet values lie.  In this paper we assume that these lower and upper values are 1:20 or 4/3 in Hindu-Arabic notation, and 2 respectively.   This pair of lower and upper limiting values for the entries in Column I may be obtained by considering the following two geometric problems.

 

Refer to Figure 4.

The upper limit.   Let us ask the following question:  What conditions must exist on sides  AC \textup{and}\mathit{BC}  so that  AC=DC? That is, under what conditions is  \Delta ACD  a right isosceles triangle?  This question is motivated by the fact that we know that the values of  DC=p^{2}-q^{2}   and AC=2pq   are respectively DC=119   and AC=120,

(p = 12 and q = 5).   Only the leg value,  DC=119,   appears on the Tablet  in Line 3, Column II.  We observe that the two legs, DC  and AC   are very nearly equal for these p and q, hence the question above.  For these two legs to be equal, we impose the condition upon them that AC=DC  or  p^2-q^2=2pq.   We solve this quadratic for pin terms of qand obtain p=(1+\sqrt{2})q, choosing the solution from the quadratic so that p > 0.   Substituting the right hand side of this equation into equation (1) above and simplifying, we obtain R^2=2.  This value is only slightly larger than the value of 1:59:00:15 of Line 3, Column I of the Tablet.

 

Background and motivation for the lower limit of the Tablet

 

Earlier we commented that in any solution to the problem, we always have  \angle ADC=2\beta .

Furthermore, we always have  \beta =\angle BAD=\angle ABD.   If, in  \Delta ADC,   we impose the condition that   \angle DAC=\beta, then this causes  \angle BAC=2\beta.  By asking for these conditions to exist, we create the condition that

rt\Delta DCA\approx rt\Delta ACB.  From this similarity constraint within the figure, we will obtain the proposed lower limit of 4/3 on the Tablet values of Column I as follows:

With the conditions holding as outlined in the above paragraph, we then have  \frac{AC}{CD}=\frac{BC}{AC}   or  \frac{2pq}{p^2-q^2}=\frac{2p^2}{2pq}  , so that p^2=3q^2 or  p=\sqrt{3}q  .  Substituting this last expression into the equation for R^2 , we have R^2=\frac{4}{3} .  This modern Hindu-Arabic numerical ratio is slightly smaller than the Tablet value in line 17, column I, which is 1:23:13:46:40.

A more elegant presentation of these upper and lower limits is to call them   \frac{2}{1}  and  \frac{4}{3} , which ratios contain the first four counting numbers when written in Hindu-Arabic numerals.

 

In summary of Part IV, in the first instance of setting limits, the condition requires the creation of a right isosceles triangle; that is, its two legs must be equal in length, and in the second, two right triangles must be similar.   See Figures 5 and 6 respectively.  Furthermore, if in Figure 5 we let   q=AC=1  then p=BC=1+\sqrt{2}  and if in Figure 6 we let  q=AC=1  then p=BC=\sqrt{3}.

 

For the purpose of comparing to a modern, trigonometry-based development of an interpretation of Plimpton 322, the angle \beta =\angle ABC=22.5^{\circ} in Figure 5 and in Figure 6, \beta =\angle ABC=30^{\circ}.

 

Figure 5- R^2=2 .

Figure 6- R^2=\frac{4}{3}.

 

If we accept the conditions on the values of  p  and  q , that p\leq 125, q< p  with both p  and  q  regular (meaning that both p and q are multiples of integral powers of 2, 3 and 5 only), then there are three unlisted pair of (p,q) values which would fit within the above supposed limits of the Tablet, namely (125, 72), (16, 9) and (125, 64).  That is, these three pair of values will generate R^2 values such that    \frac{4}{3}<R^2<2 .

 

 

Summary remarks on the paper:

 

Part I:  The problem was to bend an existing right-angled reed configuration on its longer leg so that the end point on the long leg meets the end point of the short leg, while maintaining the right angle.

 

Part II:  The methods of solution consisted of the use of Euclidean Geometry, and algebra.

 

Part III:  The values listed on the Tablet do not imply that the author of the Tablet knew that the derived values formed two the three values of Pythagorean Triples, much less to have established any knowledge of the Theorem of Pythagoras.

 

Part IV:  The limiting values of Column I of the Tablet were calculated on the assumption that a) the author of the Tablet placed a pair of geometric constraints on the constructs of the values, b) the values of the legs and the diagonal (hypotenuse) were calculated using only regular numbers which were deemed appropriate based upon a predetermined set of upper and lower limits, and finally c) the limits of the Tablet were calculated in such a way as to avoid using either the square root of 2 or the square root of 3.

 

One may only speculate at this point if the author of the Tablet sensed an appreciation of a limit approach to the ultimate calculation of the square roots of 2 and of 3.

 

Several references are made within this paper to isosceles triangles.  May I be permitted, in closing, to quote my favorite lines from one who spoke of isosceles triangles?

 

Scarecrow:  The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side.  Oh joy, rapture! I’ve got a brain!  How can I ever thank you enough?

Wizard:  Well, you can’t.

The Wizard of Oz (Victor Fleming, 1939

 

Bibliography

Bruins, E. M., 1949, On Plimpton 322, Pythagorean numbers in Babylonian mathematics, Akad. V. Wetenshappen, Amsterdam, Afdeling Natuurkunde, Proc. 52, 629-632.

Neugebaur, O. and A. Sachs, 1945, Mathematical Cuneiform Texts, American Oriental Series, Vol. 29, American Oriental Society, New Haven, Conn.

Resnikoff, H. L. and R. O. Wells Jr., 1973. Mathematics in Civilization, Holt, Rinehart and Winston, Inc., pages 73-76.

Robson, E., Neither Sherlock Holmes nor Babylon:  A Reassessment of Plimpton 322, Historia Mathematica 28 (2001), 167 – 206

Vogel, Kurt, 1959, Vorgriechische Mathematik, 2 vols., Hermann Schroedel  Verlag, Hannover, Germany

 

Classification Codes:  01A17, 14-03, 51M04

Keywords

Plimpton 322

Right triangle

Isosceles triangle

Pythagorean Triple

Similar triangles

 

Biographical Sketch

Joseph Mathis (b.1941) is a retired emeritus professor of Mathematics from William Jewell College, Liberty MO (37 years).  He received his B. A. in Mathematics from Howard Payne College and M. S. in Mathematics from Texas Christian University.   He spent his entire academic career teaching undergraduate Mathematics at a total of five institutions.

He enjoys working on motor scooters, listening to both straight-ahead jazz and classical music, and working on his classic Saab 900.

 

mathisj@william.jewell.edu