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HISTORIA
MATHEMATICA
Historia Mathematica 31 (2004) 148-162
www.elsevier.com/locatc/hm
Poles and walls in Mesopotamia and Egypt
Duncan J.
Melville
Department of Mathematics, St. Lawrence University, Canton, NY 13617, USA
Abstract
The problem of a pole or reed leaning against a wall is an example of a mathematical “riddle,” or exercise,
that occurs in many cultures, among other “Pythagorean”-type problems. There are a few known instances of the
problem in Mesopotamia and Egypt. We analyze the way the problem is treated in these cases, paying particular
attention to the structure of the algorithms used to find the solutions.
© 2003 Elsevier Inc. All rights reserved.
Résumé
Le probléme d’un poteau ou d’un roseau penchés contre un mur est une énigme mathématique, de type
«pythagoricien» comme on en trouve dans beaucoup de cultures. Il en existe quelques exemples en Mésopotamie et
en Egypte. Nous analysons la manière dont le problème est traité dans ces cas, en prétant une attention particulière
a la structure des algorithmes employés pour arriver aux solutions.
© 2003 Elsevier Inc. All rights reserved.
MSC: 01A16; 01A17
Keywords: Pythagorean problems; Old Babylonian; Seleucid; Demotic; Reed and wall
EN
‘DA
1. Introduction
Ar\4 TN, +Losa
on deciphering the texts themselves, learning the technical vocabulary and notation. Once the
M
In the early phases of the rediscovery of Mesopotamian mathematics, attention was first concentrated
terminology was understood, the emphasis moved to uncovering the level and extent of mathematical
knowledge implied by the texts. The search was to find how much was known, or how advanced Old
Babylonian mathematics was. More recently, there has been a shift in the field toward understanding
how mathematics was deployed within the contexts of the general culture, tracing the origins of certain
types of concerns, and unraveling the syllabus indicated by the texts, which, for the Old Babylonian
E-mail address: dmelville@stlawu.edu.
0315-0860/S — see front matter © 2003 Elsevier Inc. All rights reserved.
doi:10.1016/j.hm.2003.08.004
Pagina 2
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Poles and walls in Mesopotamia and Egypt
Duncan J. Melville
Department of Mathematics, St. Lawrence University, Canton, NY 13617, USA
Abstract
The problem of a pole or reed leaning against a wall is an example of a mathematical “riddle,” or exercise,
that occurs in many cultures, among other “Pythagorean”-type problems. There are a few known instances of the
problem in Mesopotamia and Egypt. We analyze the way the problem is treated in these cases, paying particular
attention to the structure of the algorithms used to find the solutions.
2003 Elsevier Inc. All rights reserved.
Résumé
Le problème d’un poteau ou d’un roseau penchés contre un mur est une énigme mathématique, de type
«pythagoricien» comme on en trouve dans beaucoup de cultures. Il en existe quelques exemples en Mésopotamie et
en Égypte. Nous analysons la manière dont le problème est traité dans ces cas, en prêtant une attention particulière
à la structure des algorithmes employés pour arriver aux solutions.
2003 Elsevier Inc. All rights reserved.
MSC: 01A16; 01A17
Keywords: Pythagorean problems; Old Babylonian; Seleucid; Demotic; Reed and wall
1. Introduction
In the early phases of the rediscovery of Mesopotamian mathematics, attention was first concentrated
on deciphering the texts themselves, learning the technical vocabulary and notation. Once the
terminology was understood, the emphasis moved to uncovering the level and extent of mathematical
knowledge implied by the texts. The search was to find how much was known, or how advanced Old
Babylonian mathematics was. More recently, there has been a shift in the field toward understanding
how mathematics was deployed within the contexts of the general culture, tracing the origins of certain
types of concerns, and unraveling the syllabus indicated by the texts, which, for the Old Babylonian
E-mail address: dmelville@stlawu.edu.
0315-0860/$ – see front matter 2003 Elsevier Inc. All rights reserved.
doi:10.1016/j.hm.2003.08.004
Pagina 3
Vedi nel PDF(si apre in una nuova finestra)period in particular, are mostly school texts (see, for example, Robson [1999, 2000, 2002], Veldhuis
[1997]). Høyrup admirably summed up these developments in the field of mathematical Assyriology up
to the early 1990s in Høyrup [1996a].
In the past 10 years or so, the themes or trends identified by Høyrup have continued and been joined by
others. Høyrup himself has devoted much concern to identifying and separating various strata in ancient
mathematics. In particular, he has been interested in what he has termed a “lay,” or “surveyors,” tradition
of problems or riddles, in contrast to, and connection with, the scholastic scribal tradition [Høyrup, 1990a,
1996b, 1997a, 1997b, 1997c, 1999, 2001, 2002a, 2002b]. Among the problems whose origins Høyrup
identifies in this lay tradition are a number based on Pythagorean triples, of which the most famous in
the problem of the pole or reed leaning on a wall. In [1999], Høyrup collected and discussed all instances
of the Pythagorean “rule” in Mesopotamian mathematics. Here, we consider only the pole and wall
problem and caution against making any wider Pythagorean conclusions without considering how the
problem compares with other Pythagorean-type problems.
When the same, or similar, problem occurs in different cultures, the question of transmission of
ideas inevitably arises. Pythagorean problems, whether derived from triangles or rectangles, have a wide
diffusion and, given the relative simplicity of the underlying ideas, may have appeared independently in
many contexts. However, the closer the “clothing” of an abstract problem, that is, the greater the similarity
in dressing the problem in practical guise, the greater the suspicion of contact between the authors.
Unfortunately, scarcity of sources often militates against proving direct chains of transmission, especially
over large reaches of space and time. The question of transmission of mathematical ideas among various
ancient cultures is addressed at length in Dold-Samplonius et al. [2002], where, in particular, Høyrup
considers the Seleucid disjunction in techniques with older Mesopotamian practices and connections
with mathematics in Egyptian, Greco-Roman, Islamic, Indian, and Chinese societies. In his comments
on the Mesopotamian and Egyptian pole and wall problems Høyrup notes only that they are “solved
similarly” [Høyrup, 2002b, p. 20]. The purpose of this paper is to probe that similarity from a somewhat
different direction.
Mesopotamian and Egyptian mathematics have been interpreted in three primary ways: algebraically,
geometrically, and algorithmically. As algebra is so deeply embedded in modern culture, it is natural
for modern commentators to recast problems in algebraic symbolism and view the ancient procedure as
solving an equation. Algebra is a powerful and compact language for communicating mathematical ideas
and such an approach provides a convenient means for bridging the gap between ancient mathematics
and modern readers. However, the algebra itself is alien to the ancient cultures and obscures questions of
how people in those cultures viewed the mathematics they were doing.
As a closer approximation to the conceptual framework of ancient scribes, Høyrup in particular has
argued extensively [1990b, 1996a, 1996b, 1997a, 1997b, 1997c, 1999, 2001, 2002a] and persuasively for
a “naive,” or cut-and-paste, geometric underpinning for the recorded procedures. His analysis depends
on a close reading and technical interpretation of the vocabulary of particular texts and analyzing steps
of a procedure as geometric actions. The reconstructed diagrams are not attested in the original sources
and remain speculative in certain particulars, although the thrust of his interpretation is convincing.
A third approach, to view the written descriptions of mathematical procedures as algorithms and
akin to computer programs has a long history (see, for example, Knuth [1972]) and appeal in an age
of computerization. Such an approach focuses on the formal aspects of the procedures, rather than
the underlying conceptions. In the context of locating scribal mathematics in Egypt and Mesopotamia
within a domain of professionalization encompassing medicine, divination and jurisprudence, Ritter has
Pagina 4
Vedi nel PDF(si apre in una nuova finestra)developed techniques for exposing the rhetorical construction of the texts [1995, 1998]. This approach
has also been adopted by Imhausen in her explorations of the algorithms used in Egyptian mathematics
[2002]. Here, we adapt Ritter’s ideas to provide a formalism suited to describing the structure of
algorithms in a way that allows comparison between procedures from widely differing sources. To
provide focus to the technique, we concentrate on a single problem, and consider only its appearance
in Mesopotamia and Egypt.
The problem of a ladder, or pole, leaning against a wall has an extensive history and is still a staple
exercise in mathematics. Nowadays, it can be found in almost every calculus text, where it is typically
used as a related rates problem. Here is an example from a recent version of a calculus book by the
ubiquitous Stewart:
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the
top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? [Stewart, 1998, p. 269]
The exercise uses related rate techniques and the Pythagorean theorem. Note that since the ladder is 10 ft
long, and the base is 6 ft from the wall at the moment of interest, the resulting Pythagorean triple is
6–8–10, or twice a 3–4–5 triangle.
In the years B.C. (Before Calculus), the rate of motion was not of interest, just the position; ladders
were moved, but did not slide. In this form, the problem has a long and venerable tradition and appears
in various guises in many cultures. Here, we restrict ourselves to a discussion of the ways in which the
problem was used and solved in the mathematics of Mesopotamia and Egypt.1 The sources come from the
Old Babylonian period (ca. 2000–1600), Seleucid Mesopotamia (third or second century), and Ptolemaic
Egypt (early third century).
2. Old Babylonian poles
Only one example of this problem survives in the corpus of Old Babylonian mathematics, where it
occurs as Problem 9 on the tablet BM 85196, a large (19 × 12.3 cm) tablet containing 18 problems on
a variety of subjects. It is an example of what Friberg calls “recombination texts,” those compiled from
a collection of thematic problem texts [Friberg, 1996]. Bought by the British Museum, it does not have
a secure archaeological context. However, it shares a number of features with a group of tablets that
Robson suggests all come from Sippar and are the work of one Iškur-mansum.2 The tablet was originally
published by Thureau-Dangin [1935] in copy, photograph, transliteration, translation, and commentary.
It was then included by Neugebauer in Volume 2 of Mathematische Keilschrifttexte [Neugebauer, 1935–
1 Fibonacci (ca. 1200) included a version with a pole leaning against a tower as the first problem in the geometric portion
of the last chapter of Liber Abaci [Sigler, 2002, p. 543]. He used a 12–16–20 triple and presumably took the problem from an
Islamic source. In China, the Nine Chapters (ca. 100 B.C.) includes a number of similar problems, only one of which actually
features a pole and a wall. Another has a reed growing out of a pond that just touches the water when pulled to the bank,
a variant which also appears in India [Shen Kangshen et al., 1999, pp. 469–473]. Sesiano traces some of the European history
of the ladder and the wall in Sesiano [1987].
2 Høyrup also identifies Sippar as the source [2002a, p. 332], including BM 85196 in his Group 6A of groups of texts
organized by geographical origin; Friberg also suggests Sippar as the origin of the Group 6A texts, though with less force
[2000, p. 169]. The tablet has not yet been dated more precisely than Old Babylonian (ca. 2000–1600).
Pagina 5
Vedi nel PDF(si apre in una nuova finestra)1937] and commented on again by Thureau-Dangin in Textes Mathématiques Babyloniens [1938]. The
various problems on the tablet have been the subject of extensive commentary since. A recent translation
and transliteration appear in Høyrup [2002a, p. 275].
Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line,
but the structure of the problem can be safely restored. The problem reads as follows:
A pole. 30 (a reed). From its . . .
Above, it has descended 6, below, how far has it moved away?
You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.
Square 24: 9,36 you see. Subtract 9,36 from 15:
5,24 you see. What is the square root of 5,24? 18 is the square root. 18
on the ground it has moved away. If 18 on the ground
above, what did it descend? Square 18: 5,24 you see.
Subtract 5,24 from 15: 9,36 you see. Of 9,36,
what is the square root? 24 is the square root. Subtract 24 from 30:
6 you see descended. Such is the procedure.
The problem begins by stating its subject—a pole. Next, the length is given, with a gloss to provide scale.
The pole is 30 (nindan) long, that is a reed-length, and so the 30 should be read as 0;30, that is, half a
nindan, or approximately 3 m.3 As the rest of the first line is badly damaged, it is not possible to restore
completely the statement of the problem. Hence, we cannot be sure which elements were explicitly
stated, and which remained implicit. Certainly, the pole begins upright, but there is no explicit mention
of a wall.4 Much like Stewart, the problem’s author is interested in the point at which the top of the pole
has moved down 6 (units). The problem is to find how far the base of the pole has moved.
The exercise follows the conventions of Old Babylonian mathematics, where first the problem is stated
and then the instructor, addressing the student, “you,” explains the procedures to be followed. As the
structure is very formulaic, the terminology often becomes quite abbreviated. Here, all that is left of the
phrase, “You, in your workings, . . .” is the imperative, “You.”
Since Mesopotamian problems always use specific numbers for the various quantities, there are two
topics of interest: the general algorithm or procedure used, and the specific numbers chosen. We will take
these topics in turn.
The organization of the procedure is typical of such Old Babylonian texts. The task is divided into
steps. Each step is one simple calculation, usually using the result of the previous step, and the result
of each calculation is given. As the student is working with particular numbers, the procedure invokes
these data. However, to get a general understanding of the underlying algorithm, it is helpful for the
modern reader to have a more symbolic representation. Here, we follow an approach adapted from that
introduced by Ritter [1995]. The convention is the following. We denote by Di the ith piece of data
given in the statement of the problem, and by Rj the result of the j th step of the solution procedure.
This technique has the advantage of focusing on the steps of the procedure, rather than determining an
equation. This formalism represents how the algorithm proceeds, rather than presenting a proof of why
it works. We take for granted that the algorithms are “correct”; a concerned reader may easily construct
an algebraic proof from the algorithm as a check.
3 We follow Neugebauer’s convention for writing multiplace sexagesimal numbers: sexagesimal places are separated by a
comma, and, when absolute sizes are given, a semicolon represents the “sexagesimal point,” the position of units.
4 However, Muroi has suggested a restoration including the wall [1991].
Pagina 6
Vedi nel PDF(si apre in una nuova finestra)In this particular case, the problem gives two pieces of numerical data: the length of the pole, and thus
the original height, 30, which we label D1 , and the distance the tip of the pole has moved down, namely 6,
which we denote D2 . The first instruction is to compute with the first piece of data given, although the
result of that computation is not immediately used. The second instruction utilizes the second piece
of data given. This is a common approach in such algorithms. The layout of the algorithm is given in
Table 1. At the conclusion, it is noted that the 18 of R5 is the quantity sought. The remainder of the
problem reverses the procedure in a somewhat abbreviated form, beginning with the 18 and determining
that, indeed, the pole has been moved down by 6 from its original position.
It has been common to render the results of these algorithms as algebraic formulas, a practice which
obscures the proceduralnature of the mathematics. However, tracing through the problem gives the
induced equation R5 = D12 − (D1 − D2 )2 . In the notation of Fig. 1, recalling that D1 = d, the length of
the pole and D2 = b, the amount moved down, then
R5 = d 2 − (d − b)2 = d 2 − l 2 = a.
Two points are important. The first is that the algorithm produces the correct result. The second is that the
steps of the algorithm lead towards that solution in a natural order. Applications of the Pythagorean rule
in Old Babylonian mathematics occur in a variety of problems. The entire corpus of such problems has
recently been analyzed by Høyrup [1999], whose conclusion is that the rule was discovered sometime
“between 2025 . . . and 1825 B.C.” [Høyrup, 1999, p. 407].
As to the numbers chosen, the length of the pole is the very common 30 (one-half in the sexagesimal
system) and the resulting Pythagorean triple is 18–24–30, or a multiple of the basic 3–4–5. It is not,
however, the favorite 45–1–1,15 version of the 3–4–5 triple that often appears, perhaps because of a
desire to use a more realistic length for the pole.
Table 1
BM 85196, No. 9
Computation
Step 1
Step 2
Step 3
Step 4
Step 5
Symbolic instruction
Square 30, result 15
Subtract 6 from 30, result 24
Square 24, result 9,36
Subtract 9,36 from 15, result 5,24
Find the square root of 5,24, result 18
Fig. 1. The pole and the wall.
R1 = D12
R2 = D1 − D2
R3 = R22
Pagina 7
Vedi nel PDF(si apre in una nuova finestra)3. Seleucid walls
Although Old Babylonian mathematics is represented by a fair number of texts, the same cannot be
said of subsequent periods. In particular, the dearth of texts means that it is impossible to delineate
transmission of ideas through the next 1500 years that separate the Old Babylonian and Seleucid sources.
All we can do when faced with a recurrence of similar problems in the Seleucid period is tentatively to
decide whether to join the dots of these isolated pieces of evidence. While we have large numbers of
Seleucid (and Late Babylonian) astronomical tablets utilizing intricate calculations, there are very few
mathematical tablets of the Old Babylonian school type. We are fortunate in having a tablet with an
instance of our favorite problem. The date of the tablet is not known; it is perhaps second century.
The example is Problem 12 on BM 34568 (see Tables 2 and 3), a tablet featuring a collection of
about twenty problems mostly to do with rectangles; Problem 12 is something of an anomaly (the tablet
is broken and the later problems are very damaged). The tablet was originally published by Neugebauer
[1935–1937] in Volume III; an up-to-date transliteration and technical translation can be found in Høyrup
[2002a, p. 394]. The translation given below is somewhat freer and thus disguises some important features
of the text. One is that the vocabulary used is very different from that of Old Babylonian mathematics.
While this is perhaps not surprising over a gap of 1500 or so years, the lack of continuity of vocabulary
is one strand in Høyrup’s argument for a lack of continuity in mathematical traditions. Besides questions
of grammar and word choice, there are several other fascinating features of this text. Normalizing the
vocabulary, the problem reads as follows:
A reed and a wall I erected. 3 kuš I have gone down,
9 kuš it has moved away. What is the reed? What is the wall?
Since you do not know,
3 times 3: 9. 9 times 9: 1,21. Add 9 to 1,21.
1,30 times 30 you go: 45. The reciprocal of 3 is 20. 20 times 45 you go:
15, the reed. What is the wall? 15 times 15: 3,45. 9 times 9:
1,21. Subtract 1,21 from 3,45, leaving 2,24.
What times what may I go to get 2,24?
12 times 12: 2,24. 12 the wall.
Although not entirely clear from the statement of the problem, the set-up is that the reed was moved
from a vertical position so that the top descended 3 cubits, the base moved out by 9, and the tip of the
reed was then resting on the top of the wall, and here there really is a wall. After the set-up, the questions
are given: What is the length of the reed? and What is the height of the wall? The solutions are found in
that order. That is, first the length of the reed is found and from that, the height of the wall determined.
Interestingly, the height of the wall is computed using the Pythagorean rule, rather than as a subtraction.
This suggests that the extended problem is built up from a simpler original. The steps for this subproblem
can be summarized in Table 2, where we set D1 to be the length of the reed, that is, the hypotenuse of
the triangle, and D2 the distance the base moved.
In terms of numbers, the wall is 12 kuš, the reed 15, and the base of the triangle 9 kuš. Hence, the
problem uses a 9–12–15 Pythagorean triple, or, once again, a multiple of the basic 3–4–5.
In form, the procedure is similar to that followed in the Old Babylonian text considered above
and shows the same facility with “Pythagorean” problems. The more difficult part of the problem is
determining the length of the reed in the first place. The given data are D1 = 3, the distance moved down,
and D2 = 9, the distance moved from the base of the wall. The procedure is given in Table 3.
Pagina 8
Vedi nel PDF(si apre in una nuova finestra)Table 3
BM 34568, No. 12, Part 1
Table 2
BM 34568, No. 12, Part 2
Step 1
Step 2
Step 3
Step 4
Computation
Symbolic instruction
15 times 15: 3,45
9 times 9: 1,21
Subtract 1,21 from 3,45,
remainder 2,24
Find square root of 2,24 as 12
R1 = D12
R2 = D22
R3 = R2 − R1
R4 =
√
R3
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Computation
Symbolic instruction
3 times 3: 9
9 times 9: 1,21
9 plus 1,21: 1,30
1,30 times 30: 45
Reciprocal of 3: 20
20 times 45: 15
R1 = D12
R2 = D22
R3 = R1 + R2
R4 = 12 R3
R5 = D1
1
R6 = R5 × R4
The final result, R6 , is the length sought. The algorithm conforms well to the rhetoric of Old
Babylonian mathematics we saw above: it begins by using the given data, and then most steps use as
input the result of the previous step. Steps 4 and 5 introduce a certain amount of complexity into the
procedure and should alert us to a closer study. The solution to the problem is really rather elegant, and
I suspect that anyone stumbling through the algorithm without additional explanation would have little
feeling for why the obtained result is correct. For the benefit of the algebraically minded, unravelling the
procedure and setting d to be length of the reed gives the induced formula:
1 1 2
D1 + D22 .
D1 2
The steps in the procedure are the same as would be taken in computing this formula from the given data.
Why is this formula correct? In fact, it is exactly the formula that would be obtained by modern algebraic
reasoning. Using the notation in Fig. 1, set d to be the length of the reed and l the height of the wall; then
l = d − D1 , and the Pythagorean theorem gives
d = R6 =
d 2 = (d − D1 )2 + D22 = d 2 − 2dD1 + D12 + D22 ,
which naturally (algebraically) simplifies to
2dD1 = D12 + D22 ,
(1)
and so
1 2
D1 + D22 .
2D1
Given the data D1 and D2 , d would be computed by much the same procedure as that detailed in the
tablet. The challenge is to provide a geometrical argument that makes this formula natural.
Høyrup rises to this challenge by noting that D1 = d − l, and so Eq. (1) indicates that the sum of the
square on (d − l) and the square on D2 (the base of the triangle) is equal to twice the rectangle on d
and d − l (see Høyrup [2002a, p. 399, 2002b, pp. 15–17]). This connection between the rectangles and
squares is easily obtainable by a “naive geometrical” argument and may well underlie the construction
of the problem. It also accounts for the order of Steps 5 and 6 of the algorithm. However, Høyrup notes
in concluding his analysis that, “the method that is used . . . resembles nothing we have encountered in
the Old Babylonian corpus (or the pre-Seleucid Late Babylonian texts), and is therefore highly unlikely
to be old” [Høyrup, 2002a, p. 399]. The problem is a curious mixture of old and new. The problem is
given the dress of pole and wall as in the older source. While it uses a multiple of the standard 3–4–5
Pagina 9
Vedi nel PDF(si apre in una nuova finestra)triple, it is not the same multiple as the earlier case and the fact that two sources use multiples of the
simplest Pythagorean triple does not provide evidence for contact between them. The organization of the
procedure conforms to expectations, but the algorithm used is not old.
4. Egyptian poles
Although there is no evidence of interest in leaning poles against walls in the meagre sources we have
on Pharaonic Egyptian mathematics, the problem does appear in a demotic papyrus from the early third
century B.C., roughly contemporary with the Seleucid tablet studied above, probably slightly older.
A collection of mathematical papyri was published by Parker [1972]. One of these, P. Cairo J.E.
89127-30, 89137-43, in a very fragmentary state, has a legal code on the recto and some 40 mathematical
problems on the verso. It was dated on paleographic grounds from a study of the legal text.5 Both
the beginning and end of the papyrus are lost and some of the remaining problems are missing too
much content to be restored, but among them is a series of eight problems to do with a pole (literally,
a tree) whose top moves down or base moves out. As in the Old Babylonian problem, and against the
Seleucid version, the Egyptian pole appears unaccompanied by explicit mention of a wall.6 While the
pole problems on this papyrus are in certain cases badly damaged, textual parallelism allows us to restore
the mathematical content with confidence even if there linger some technical philological questions.
As is common in texts with series of problems, here some basic sets of parameters are chosen and then
the problems ring a variety of changes on which parameters are given and which are computed. The first
underlying Pythagorean triple, accounting for three of the eight problems, is 6–8–10 (units are cubits),
which, as we saw, was still going strong in Stewart. Both the Old Babylonian and Seleucid pole problems
also used multiples of the 3–4–5 triple. Another two problems use the same triple, but with the role of
the 6 and 8 reversed. Interestingly, however, the triple used in P. Cairo for the other three problems is
10–10 12 –14 12 . Presumably this was chosen so that in the one case the largest member of the triple was 10,
and in the other case, the smallest was 10, thereby linking the two sets of parameters by the number 10.
The problems are numbered 24–31 in Parker’s text. We retain that numbering here for convenience.
The translation is based on Parker’s, but simplified. For original transliterations and translations see
Parker [1972, pp. 35–40].
Problem 24 (see Table 4)
A pole is 10 cubits when erect. If its foot is moved out
6 cubits, how much is the top lowered? The working.
5 The recto of the papyrus fragments contains the “Legal Code of Hermopolis West.” The papyrus was found “in the debris
of a ruined building opposite the place of mummification . . . believed to be one of the temple archives” [Mattha, 1975, p. xi].
Mattha continues that the papyrus “can be safely assigned to the reign of Philadelphus, possibly earlier,” that is, probably early
third century. In light of Ritter’s arguments about common domains of knowledge and professionalization among mathematical
and legal material, it is remarkable that the mathematics occurs on the same papyrus as a legal code. Although much of the code
deals with leasing of land and property, and inheritance, there is nothing particularly mathematical about the legal procedures
described. The source clearly locates the mathematics within a temple-based scribal milieu.
6 The apparent absence of the wall should not be given too much emphasis. The context of the problem would have been
understood by its intended users and it is only the Seleucid problem that computes the height of the wall, and thus forces its
explicit mention.
Pagina 10
Vedi nel PDF(si apre in una nuova finestra)Table 4
P. Cairo, No. 24
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
10 times 10, result 100
6 times 6, result 36
Subtract 36 from 100, remainder 64
Find square root of 64, result 8
Subtract 8 from 10, remainder 2
R1 = D12
R2 = D22
R3 = R2 − R1
√
R4 = R3
R5 = R1 − R4
You compute 10 times 10: result 100. You compute 6 times 6: result 36.
Subtract it from 100: remainder 64. Reduce 64 to its square root: result 8.
Subtract 8 from 10: remainder 2.
You shall say, “Its top is lowered 2 cubits.”
Instead of finding the height of the (missing) wall, or the height of the top of the pole from the ground,
the goal is to find by how much the top has lowered when the base of the pole is moved out a specified
distance. The problem is solved by a straightforward use of the Pythagorean theorem. The given data are
D1 = 10, the length of the pole, and D2 = 6, the distance the base is moved.
As in Mesopotamia, the procedure first utilizes the data, in the order given in the statement of the
problem and then makes single arithmetical steps, with the output from one step usually forming the
input of the next. The final step requires recalling the initial data. The formal structure, then, of the
algorithm is very similar to that of the Mesopotamian problems.
Problem 25 (see Table 5)
The second problem is formally the same as the first, but with different values for the data.
A pole is 14 12 cubits when erect.
If its foot is moved out 10 cubits, how much is the top lowered?
You compute 14 12 times 14 12 : result 210 14 .
You compute 10 times 10: result 100. Subtract it from 210 14 :
remainder 110 14 . Reduce 110 14 to its square root: result 10 12 .
Subtract it from, 14 12 : remainder 4.
You shall say, “Its top is lowered 4 cubits.”
Table 5
P. Cairo, No. 25
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
14 12 times 14 12 , result 210 14
R1 = D12
R2 = D22
R3 = R2 − R1
√
R4 = R3
R5 = R1 − R4
10 times 10, result 100
Subtract 100 from 210 14 , remainder 110 14
Find square root of 110 14 , result 10 12
Subtract 10 12 from 14 12 , remainder 4
Pagina 11
Vedi nel PDF(si apre in una nuova finestra)Table 6
P. Cairo, No. 26
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
10 times 10, result 100
6 times 6, result 64
Subtract 64 from 100, remainder 36
Find square root of 36, result 6
Subtract 6 from 10, remainder 4
R1 = D12
R2 = D22
R3 = R2 − R1
√
R4 = R3
R5 = R1 − R4
This time the given data are D1 = 14 12 , the length of the pole, and D2 = 10, the distance the base is
moved.
As is clear from Tables 4 and 5, precisely the same algorithm is used in the two problems. The
problems are nicely chosen so that the answer in the first case is 2 cubits, and in the second case, 4 cubits.
It is also interesting that the problem of finding the square root of 110 14 should be considered sufficiently
elementary not to distract from the main purpose of the problem.
Problem 26 (see Table 6)
The next problem gives the same two pieces of data, the length of the pole and the distance the foot is
moved out. It reverts to the first 6–8–10 triple, but now with the 6 and the 8 swapped around.
A pole is 10 cubits when erect. If
its foot is moved out 8 cubits, how much is the top lowered?
You compute 10 times 10: result 100.
You compute 8 times 8: result 64. Subtract it from 100: remainder 36.
Reduce 36 to its square root: result 6. Subtract it from 10: remainder 4.
You shall say, “Its top is lowered 4 cubits.”
Hence, the given data are D1 = 10, the length of the pole, and D2 = 8, the distance the base is moved.
Certainly there is a clear idea of the relationships between the elements of Pythagorean triples, and
the dressing of the problem, as well as the order of the elements of the triple, merely provide concrete
examples.
Problem 27 (see Table 7)
Problem 27 is the inverse to the first problem. The relevant triple is again 6–8–10 and the length of
the pole is given. However, this time, the distance the top moves down is given and the distance the foot
moves out is sought.
A pole is 10 cubits when erect.
The distance its foot is moved out will not be stated
Lower its top 2 cubits. How much is the foot moved out?
You compute 10 times 10: result 100. Subtract it, 2, from 10:
remainder 8. You compute 8 times 8: result 64.
Subtract it from 100: remainder 36. Reduce it to its square root: result 6.
You shall say, “Its foot is moved out 6 cubits.”
Pagina 12
Vedi nel PDF(si apre in una nuova finestra)Table 7
P. Cairo, No. 27
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
10 times 10, result 100
Subtract 2 from 10, remainder 8
8 times 8, result 64
Subtract 64 from 100, remainder 36
Find square root of 36, result 6
R1 = D12
R2 = D1 − D2
R3 = R22
R4 = R
√1 − R3
R5 = R4
The given data are D1 = 10, the length of the pole, and D2 = 2, the distance the top is lowered.
Whereas the first group of problems, where the distance the pole is moved out does not appear in the
Mesopotamian sources, this group treats the same situation as in BM 85196. A comparison of Tables 7
and 1 shows that the Egyptian algorithm is exactly the same as the Old Babylonian one.
Problem 28
Problem 28 is parallel to Problem 27, but using the second set of data, so that it is the inverse of
Problem 25. The length of the pole is given as 14 12 cubits and the top is lowered by 4 cubits. The distance
the foot is moved outwards is sought and determined to be 10 cubits. The structure of the algorithm is
precisely that of Problem 27.
Problem 29
Problem 29 is the third similar problem, being parallel to Problem 27, but using the third set of data,
that is, the 6–8–10 data of Problem 26, where the foot is moved out 8 cubits. The given data are the length
of the pole, 10 cubits, and how far the top is lowered, 4 cubits. Despite slight differences in wording (for
example, this problem does not tell you what will not be stated), the solution follows exactly the same
steps as in Problems 28 and 27.
Problem 30 (see Table 8)
The last two problems in the series, Problems 30 and 31, are both very heavily damaged.7 The
reconstruction of the problems depends on assumed parallels between the two problems and requirements
of the problem. Despite the extensive restoration, one can have reasonable confidence in the algorithms
used.
These two problems ask for the length of the pole, when the amount the top is lowered and the distance
the foot is moved out are known.
A pole. Its foot is moved out 6 cubits and
its height will not be stated. If the top is lowered
2 cubits, what was its height?
7 The expected ninth problem, using the third set of data, does not appear on the papyrus. Instead, the next problem is one of
determining the diameter of a circle given its area.
Pagina 13
Vedi nel PDF(si apre in una nuova finestra)Table 8
P. Cairo, No. 30
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
6 times 6, result 36
2 times 2, result 4
Add 36 to 4, result 40
Add 2 to 2, result 4
Divide 40 by 4, result 10
R1 = D12
R2 = D22
R3 = R1 + R2
R4 = D2 + D2
R5 = R3 /R4
You compute 6 times 6: result 36. You compute 2 times 2: result 4.
You add 36 to 4: result 40. You add 2 to 2: result 4. Take it to 40: result 10.
The height of the pole was 10.
The given data are D1 = 6, the distance the foot is moved out, and D2 = 2, the distance the top was
lowered. The procedure is a new one.
As Problems 30 and 31 use the same algorithm, we will discuss them together.
Problem 31 (see Table 9)
This problem sets up the same situation as the previous one, but using the second set of data, the
10–14–14 12 triple.
A pole. If its foot is moved out 10 cubits
and its top is lowered 4 cubits, what was its height?
You compute 10 times 10: result 100. You compute 4 times 4: result 16.
You add it to 16: result 116. You add 4 to 4: result 8.
Take it to 116: result 14 12 . The height of the pole was 14 12 .
The given data are D1 = 10, the distance the foot is moved out, and D2 = 4, the distance the top is
lowered. The procedure is as before.
As can be seen from the “Symbolic Instruction” columns of the two problems, the steps followed are
the same. One needs to be a little cautious, however, due to the extensive restoration (see Parker [1972, pp.
39–40] for lacunae and commentary on the restoration). Speaking in favor of the restoration, disregarding
minor textual issues, is the fact that although the precise wording changes in equivalent problems using
Table 9
P. Cairo, No. 31
Step 1
Step 2
Step 3
Step 4
Step 5
Computation
Symbolic instruction
10 times 10, result 100
4 times 4, result 16
Add 100 and 16, result 116
Add 4 to 4, result 8
Divide 116 by 8, result 14 12
R1 = D12
R2 = D22
Pagina 14
Vedi nel PDF(si apre in una nuova finestra)alternative data in the series of problems above, the procedures in the algorithms are identical in each
case.
This problem, in which the length of the pole is to be determined, given the distance the base is moved
out and the distance the top is moved down, is precisely the situation in the first half of the Seleucid
problem discussed above. In view of Høyrup’s declaration that the procedure followed in the Seleucid
problem is of recent discovery, it is interesting to compare the two algorithms.
First, note that although the same pieces of data are given, their order differs in the two texts. In BM
34568, the distance the top has moved down is given first, followed by the distance the foot has moved
away from the wall. In both Problems 30 and 31, the distance the foot has moved out is given first,
followed by the amount the top is lowered. The numbers used, the values of the data, are also different in
the two sources.
Each of the three algorithms begins by finding the squares of the two initial pieces of data, in the order
given, and then summing the squares. The remainder of the algorithms appears superficially different.
In the Seleucid text, the result of the initial steps, R3 , is first multiplied by 30 (i.e., divided by 2), and
then multiplied by the reciprocal of D1 . That is, R3 is divided by 2D1 , but in two steps, and of course
division in Mesopotamia is accomplished by multiplication by the reciprocal. Recalling that the order of
the initial data is reversed in P. Cairo, the two steps of the algorithm show that R3 is divided by 2D2 ,
where the doubling of D2 is accomplished by adding it to itself. Thus the two approaches to the problem
feature two multiplications in Mesopotamia, and an addition and division in Egypt (instead of the two
divisions that would have occurred if R3 had been first halved before being divided by D2 ). The two
algorithms are thus quite similar, given that multiplication by a reciprocal was not a standard part of
Egyptian arithmetic. Certainly, both approaches utilize the crucial identity of the sum of the squares and
twice the rectangle.
There is no doubt that both algorithms are derived from the same conceptual map of the problem,
whether or not it is the precise geometrical interpretation suggested by Høyrup. Despite differences in
data and ordering of initial data, the underlying structural similarities imply a common source for the
algorithms, whether in Mesopotamia, Egypt, or in some third location. Certainly there were extensive
contacts between Mesopotamia, Egypt and other regions during the Hellenistic period.
The fact that the Egyptian source may well be older than the Seleucid one might be taken to support an
argument for transmission from Egypt to Mesopotamia, but it seems unlikely that either source is original
in documenting the problems it contains. If both sources are compilations or copies of earlier documents,
little can be said about the relative ages of the origins of the problem in the two contexts. Similarly, the
fact that Problems 27, 28, and 29 of P. Cairo use precisely the same algorithm as that in BM 85196, while
using different data, suggests some form of unrecorded transmission over the intervening 1500 years, but
as yet we can not speculate on the means of that transmission.
5. Summary of the problems
Table 10 indicates the given data and data sought in the problems discussed above. Note that only BM
34568 actually has a wall.
We let d denote the length of the pole or reed, l the height of the wall, a the distance the foot is moved
out, and b the distance the top is moved down.
Pagina 15
Vedi nel PDF(si apre in una nuova finestra)Table 10
Data summary
Source
Date
BM 85196
BM 34568
P. Cairo Nos. 24, 25, 26
P. Cairo Nos. 27, 28, 29
P. Cairo Nos. 30, 31
ca. 2000–1600
ca. 300–100
ca. 300–250
Given data
Sought data
d, b
b, a
d, a
d, b
a, b
a
d, l
b
a
d
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