4.3: VAT 7532

Figure \(4.5\): The diagram of VAT 7532. The "upper width" is to the left

Obv.

1 A trapezium. I have cut off a reed. I have taken the reed, by its integrity

2 1 sixty (along) the length I have gone. The 6th part

3 broke off for me: \(1‵12\) to the length I have made follow.

4 I turned back. The 3rd part and \(\frac{1}{3}\) kùš broke off for me:

5 3 sixty (along) the upper width I have gone.

6 With that which broke off for me I enlarged it:

7 36 (along) the width I went. 1 bùr the surface. The head (initial magnitude) of the reed what?

8 You, by your proceeding, (for) the reed which you do not know,

9 1 may you posit. Its 6th part make break off, \(50^{\prime}\) you leave.

10 igi \(50^{\prime}\) detach, \(1^{\circ} 12^{\prime}\) to 1 sixty raise:

11 \(1‵12\) to \(1‵12\) join: \(2‵24\) the false length it gives you.

12 (For) the reed which you do not know, 1 may you posit. Its 3rd part make break off,

13 \(40^{\prime}\) to 3 sixty of the upper width raise:

14 \(2‵\) it gives you. \(2‵\) and 36 the lower width heap,

15 \(2‵36\) to \(2‵24\) the false length raise, \(6``14`24\) the false surface.

16 The surface to 2 repeat, \(1``\) to \(6``14`24\) raise

17 \(6````14```24``\) it gives you. And \(\frac{1}{3}\) kùš which broke off

18 to 3 sixty raise: 5 to \(2‵24\), the false length,

19 raise: \(12‵\). \(\frac{1}{2}\) of \(12‵\) break, \(6‵\) make encounter,

Rev.

1 \(36``\) to \(6````14```24``\) join, \(6````15```\) it gives you.

2 By \(6````15```\), \(2``30^{\prime}\) is equal. \(6`\) which you have left

3 to \(2``30`\) join, \(2``36`\) it gives you. igi \(6``14`24\),

4 the false surface, I do not know. What to \(6``14`24\)

5 may I posit which \(2``36\) gives me? \(25^{\prime}\) posit.

6 Since the 6th part broke off before,

7 6 inscribe: 1 make go away, 5 you leave.

8 igi 5 detach, \(12^{\prime}\) to 25 raise, \(5^{\prime}\) it gives you . \(5^{\prime}\) to \(25^{\prime}\) join: \(\frac{1}{2} \mathrm{NINDAN}\), the head of the reed it gives you.

This problem also deals with a field—yet with a field which the surveyor would only encounter in dream (or rather, in a nightmare). “Real life” enters through the reference to the unit bùr, a unit belonging to practical agricultural administration, and through the reference to measuring by means of a reed cut for this purpose; its length ( \(\frac{1}{2} \mathrm{NINDAN}\)) corresponds indeed to a measuring unit often used in practical life and called precisely a “reed” (gi in Sumerian). One may also imagine that such reeds would easily break. Finally, the use of the numeral “sixty” shows us one of the ways to express numbers unambiguously.

Everything else, however—that is, that the area of the field is known before it is measured, and also the ways to indicate the measures of the pieces that break off from the reed—shows which ruses the Old Babylonian school masters had to make use of in order to produce second-degree problems having some taste of practical life.

For once, Figure 4.5 reproduces a diagram that is traced on the tablet itself. In general, as also here, diagrams are only drawn on the tablets when they serve to clarify the statement; they are never used to explain the procedure. On the other hand, Figure 4.5 shows once more that the solution is known in advance: the numbers \(1‵\), 45 and 15 are indeed the measures of the sides expressed in \(\mathrm{NINDAN}\).

We thus undertake to measure the trapezium by means of a reed of unknown length \(R\). We manage to measure \(1`\) reed lengths along the length of the trapezium before the reed loses a sixth of its length and is reduced to \(r=\frac{5}{6} R\). What remains of the length turns out to be \(1^{\prime} 12 r\) (lines Obv. 2-3).

Then the reed breaks for the second time. According to lines Obv. 4 and 5, the measure of the "upper width" (to the left)⁶ is \(3` z\), where \(z=\frac{2}{3} r-\frac{1}{3}\) kùš is the length of the reed after this second reduction.

The piece that broke off last is put back into place, and the "(lower) width" (evidently to the right) is paced out (line Obv. 7) as \(36 r\). Finally we learn that the area of the filed is 1 bùr = \(30`\) sar (1 sar = 1 \(\mathrm{NINDAN}^{2}\)), see page 17). We are asked to find the original length of the reed—its "head" in the sense of "beginning.".

Lines Obv. 9-11 determine the length in units \(r\) by means of a false position: if \(R\) had been equal to 1, then \(r\) would have been \(50^{\prime}\); conversely, \(R\) must correspond to \(r\) multiplied by igi \(50^{\prime}=1^{\circ} 12^{\prime}\). \(1`\) steps of \(R\) thus correspond to \(1`12 \cdot r\), and the complete length will be

\(1` 12 \cdot r+1` 12 \cdot r=2` 24 \cdot r\).

Figure \(4.6\): The doubled trapezium of VAT 7532.

The text speaks of \(2`24\) as the "false length," that is, the length expressed in units \(r\).

Another false position is applied in line Obv. 12. The text posits 1 for the length \(r\) of the reed once shortened, and deducts that what remains after the loss of \(\frac{1}{3}\) must be equal to \(40^{\prime}\). Leaving aside the extra loss of \(\frac{1}{3}\) kùš, the false upper width (the upper width measured in units \(r\)) is thus \(40^{\prime}\) times 3 sixties, that is, \(40^{\prime} \cdot 3^{\prime}=2`\). In other words, the upper width measures \(2` r\)—still leaving aside the missing piece of \(\frac{1}{3}\) kùš.

Since line Obv. 7 indicates that the false (lower) width is 36, we thus know—with the same reserve concerning the missing \(\frac{1}{3}\) kùš—the three sides that will allow us to determine the area of the trapezium in units \(\square(r)\).

Yet the text does not calculate this area: The surface of 2 repeat. Instead it doubles the trapezium so as to form a rectangle (see the left part of Figure 4.6), and the lines Obv. 14–16 calculate the area of this rectangle (the “false surface”), finding \(6``14`24\) (in the implicit unit \(\square(r)\)).

If the reed had not lost an ulterior piece of \(\frac{1}{3}\) kùš, we might now have found the solution by means of a final false position similar to that of BM 13901 #10 (see page 46): according to line Obv. 7, the area of the field is 1 bùr, the doubled area hence 2 bùr = \(1`` \mathrm{NINDAN}^{2}\) (Obv. 16: The surface to 2 repeat, \(1``\)). However, things are more complicated here. For each of the \(3`\) steps made by the twice shortened reed a piece of \(\frac{1}{3}\) kùš is missing from our calculation, in total thus \(3` \cdot \frac{1}{3} \mathrm{kùš}=1` \mathrm{kùš}=5 \mathrm{NINDAN} \left(1 \mathrm{kùš}=\frac{1}{12} \mathrm{NINDAN}\right)\): And \(\frac{1}{3}\) kùš which broke off to 3 sixty raise: 5(Obv. 17-18). Therefore the area of the real field does not correspond to what we see to the left in Figure 4.6 but to that which remains after elimination of the shaded strip to the right. The area of this strip is \(5 \cdot 2` 24 r=12` r\): \(5\) to \(2`24\), the false length, raise: \(12`\). The relation between the "false surface" and that of the doubled real trapezium can now be expressed by the equation

\(6``14` 24 \square(r)-12` r=1``\).

This non-normalized equation is solved in the usual way. First it is multiplied by \(6``14` 24\): \(1``\) to \(6``14` 24\) raise \(6````14```24``\) it gives you (Obv. 16-17). That leads to the normalized equation

\(\square\left(6`` 14` 24 r\right)-12` \cdot\left(6`` 14` 24 r\right)=6```` 14``` 24``\)

or, with \(s=6`` 14` 24 r\) \(r\) as unknown,

\(\square(s)-12 s=6```` 14``` 24``\).

From here onward, the procedure coincides with that of BM 13901 #2 (page 43), with a small variation in the end. the calculations can be followed in Figure 4.7.

The area \(6```` 14``` 24``\) corresponds to the rectangle of (height) \(s\) and breadth \(s-12`\). Half of the excess of the height over the breadth is "broken" and repositioned as seen in the diagram: lightly shaded in the original positions, heavily shaded where it is moved to. The construction of the completing square is described with one of the synonyms of "making hold," namely "to make encounter" (Obv. 19).

Figure \(4.7\)

After the usual operations we find that \(s=6`` 14` 24 r=2`` 36`\), and in line Rev. 5 that \(r=25^{\prime}\). We observe, however, that the "moiety" that was moved around is not put back into its original position, which would have reconstituted \(s\) in the vertical direction. Instead, the other "moiety," originally left in place, is also moved, which allows a horizontal reconstitution \(s=6`` 14` 24 r=2`` 36\): \(6`\) which you have left to \(2``30`\) join, \(2``36`\) it gives you.⁷

In the lines Rev. 6-8, the calculator introduces a third false position: if \(R\) had been equal to 6, then \(r\) would be 5. The difference of 1 between \(R\) and \(r\) is \(\frac{1}{5}\) of \(r\) or \(12^{\prime}\) times \(r\). Now the true value of \(r\) is \(25^{\prime}\); in order to obtain \(R\) we must hence "join" \(12^{\prime} \cdot 25^{\prime}=5^{\prime}\) to it. Therefore \(R=25^{\prime}+5^{\prime}=30^{\prime}=\frac{1}{2} \mathrm{NINDAN}\).

One might believe this problem type to be one of the absolute favorites of the Old Babylonian teachers of sophisticated mathematics. We know four variants of it differing in the choice of numerical parameters. However, they all belong on only two tablets sharing a number of terminological particularities—for instance, the use of the logogram \(\frac{1}{2}\) for the "moiety," and the habit that results are "given," not (for example) "seen" or "coming up." Both tablets are certainly products of the same locality and local tradition (according to the orthography based in Uruk), and probably come from the same school or even the same hand. A simpler variant with a rectangular field, however, is found in an earlier text of northern origin, and also in a text belonging together with the trapezium variants; if not the favorite, the broken reed was probably a favorite.