4.1: TMS IX #3

19 Surface, length, and width I have heaped, 1 the surface. 3 lengths, 4 widths heaped,

20 its 17th to the width joined, \(30^{\prime}\).

21 You, \(30^{\prime}\) to 17 go: \(8^{\circ} 30^{\prime}\) you see.

22 To 17 widths 4 widths join, 21 you see.

23 21 as much as of widths posit. 3, of three lengths,

24 3, as much as of lengths posit. \(8^{\circ} 30^{\prime}\), what is its name?

25 3 lengths and 21 widths heaped.

26 \(8^{\circ} 30^{\prime}\) you see,

27 3 lengths and 21 widths heaped.

28 Since 1 to the length is joined and 1 to the width is joined, make hold:

29 1 to the heap of surface, length, and width join, 2 you see,

30 2 the surface. Since the length and the width of 2 the surface,

31 \(1^{\circ} 30^{\prime}\), the length, together with \(1^{\circ} 20^{\prime}\), the width, are made hold,

32 1 the joined of the length and 1 the joined of the width,

33 make hold, ^¿1 you see.^? 1 and 1, the various (things), heap, 2 you see.

34 3…, 21…, and \(8^{\circ} 30^{\prime}\) heap, \(32^{\circ} 30^{\prime}\) you see;

35 so you ask.

36 … of widths, to 21, that heap:

37 … to 3, lengths, raise,

38 \(1^{\prime} 3\) you see. \(1^{\prime} 3\) to 2, the surface, raise:

39 \(2^{\prime} 6\) you see, ^¿\(2^{\prime} 6\) the surface^?. \(32^{\circ} 30^{\prime}\) the heap break, \(16^{\circ} 15^{\prime}\) you see .

40 {…}. \(16^{\circ} 15^{\prime}\) the counterpart posit, make hold,

41 \(4` 24^{\circ} 3^{\prime} 45^{\prime \prime}\) you see. \(2^{\prime} 6\) ^¿erasure^?

42 from \(4` 24^{\circ} 3^{\prime} 45^{\prime \prime}\) tear out, \(2` 18^{\circ} 3^{\prime} 45^{\prime \prime}\) you see.

43 What is equal? \(11^{\circ} 45^{\prime}\) is equal, \(11^{\circ} 45^{\prime}\) to \(16^{\circ} 15^{\prime}\) join,

44 28 you see. From the 2nd tear out, \(4^{\circ} 30^{\prime}\) you see.

45 igi 3, of the lengths, detach, \(20^{\prime}\) you see. \(20^{\prime}\) to \(4^{\circ} 30^{\prime}\)

46 {…} raise: \(1^{\circ} 30^{\prime}\) you see,

47 \(1^{\circ} 30^{\prime}\) the length of 2 the surface. What to 21, the widths, may I posit

48 which 28 gives me? \(1^{\circ} 20^{\prime}\) posit, \(1^{\circ} 20^{\prime}\) the width

49 of 2 the surface. Turn back. 1 from \(1^{\circ} 30^{\prime}\) tear out,

50 \(30^{\prime}\) you see. 1 from \(1^{\circ} 20^{\prime}\) tear out,

51 \(20^{\prime}\) you see.

Lines 19 and 20 present a system of two equations about a rectangle, one of the first and one of the second degree. The former is of the same type as the one explained in TMS XVI #1 (see page 27). The second coincides with the one that was examined in section #2 of the present text (see page 54). In symbolic translation, the equation system can be written

\(\frac{1}{17}(3 \ell+4 w)+w=30^{\prime}\), \((\ell, w)+\ell+w=1\).

In agreement with what we have seen elsewhere, the text multiplies the first-degree equation by 17 (using the Akkadian verb “to go,” see page 19), thus obtaining integer coefficients (as much as):

\(3 \ell+(4+17) w=3 \ell+21 w=17 \cdot 30^{\prime}=8^{\circ} 30^{\prime}\).

This is done in the lines 21–25, while the lines 26 and 27 summarize the result.

Lines 28–30 repeat the trick used in section #2 of the text (Figure 3.10): the length and the width are prolonged by 1, and the square that is produced when that which the two “joined”¹ “hold” is “joined” to the “heap” \((\ell, w)+\ell+w\); out of this comes a "surface 2," the meaning of which is again explained in lines 30-33.

The lines 34-37 are very damaged, too damaged to be safely reconstructed as far as their words are concerned. However, the numbers suffice to see how the calculations proceed. Let us introduce the magnitudes \(\lambda=\ell+1\) and \(\phi=w+1\). The text refers to them as the length and width "of the surface 2"—in other words, \((\lambda, \phi)=2\). Further,

\(\begin{aligned}
3 \lambda+21 \phi &=3 \cdot(\ell+1)+21 \cdot(w+1) \\
&=3+21+3 t+21 w \\
&=3+21+8^{\circ} 30^{\prime} \\
&=32^{\circ} 30^{\prime}
\end{aligned}\).

In order to facilitate the understanding of what now follows we may further introduce the variables

\(L=3 \lambda \quad, \quad W=21 \phi\)

(but we must remember that the text has no particular names for these—in contrast to \(\lambda\) and \(\phi\) which do have names; we now speak about, not \(with\) the Babylonian author). Lines 36-39 find that

\((L, W)=(21 \cdot 3) \cdot 2=1^{\prime} 3^{\circ} 2^{\prime}=2^{\prime} 6^{\circ}

summing up we thus have

\(L+W=32^{\circ} 30^{\prime}\), \((L, W)=2^{\prime} 6^{\circ}\)

We have now come to line 39, and arrived at a problem type which we had not seen so far: A rectangle for which we know the area and the sum of the two sides.

\((L, W)\), traced in full and a square \(\square(L)\) to its right, drawn with a dotted line. Next, we let the two "moieties" of this segment "hold" a square (lines 39-40). As we see that part of the original rectangle \((L, W)\) which falls outside the new square can just be fitted into it so as to form a gnomon together with that part which stays in place. In its original position, this piece appears in light shading, whereas it is darkly shaded in its new position.

Figure22.png" />

Figure \(4.1\): The cut-and-pste method of TMS IX #3.

One part of the new square \(\square\left(16^{\circ} 15^{\prime}\right)\) is constituted by the gnomon, whose area results from recombination of the original rectangle \((L, W)\); this area is hence \(2^{\prime}6\). We also know the area of the outer square, \(16^{\circ} 15^{\prime} \times 16^{\circ} 15=4^{\prime} 24^{\circ} 3^{\prime} 45^{\prime \prime}\) (lines 40 and 41). When the gnomon is "torn out" (lines 41 and 42), \(2^{\prime} 18^{\circ} 3^{\prime} 45^{\prime \prime}\) remains for the square contained by the gnomon. Its side (that which "is equal") is \(11^{\circ} 45^{\circ}\), which must now be "joined" to one of the pieces \(16^{\circ} 15^{\prime}\) (which gives us \(W\)) and "torn out" from the other, its "counterpart" (which gives us \(L\)). This time, however, it is not the same piece that is "joined" and "torn out"; there is hence no reason to "tear out" before "joining," as in YBC 6967 (page 46), and the normal priority of addition can prevail. Lines 43-44 find \(W=28\) and \(L=4^{\circ} 30^{\prime}\). Finally, the text determines first \(\lambda\) and \(\phi\) and then \(\ell\) and \(w\)—we remember that \(L=3 \lambda\), \(\lambda=\ell+1\), \(W=21 \phi\), \(\phi=w+1\). Since 28 has no igi, line 48 explains \(21 \cdot 1^{\circ} 20^{\prime}=28\).