3.5: BM 13901 #14
- Page ID
- 47605
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Obv. II
44 The surfaces of my two confrontations I have heaped: \(25^{\prime} 25^{\prime \prime}\).
45 The confrontation, two-thirds of the confrontation and \(5^{\prime}\), \(\mathrm{NINDAN}\).
46 1 and \(40^{\prime}\) and \(5^{\prime}\) over-going \(40^{\prime}\) you inscribe.
47 \(5^{\prime}\) and \(5^{\prime}\) you make hold, \(25^{\prime \prime}\) inside \(25^{\prime} 25^{\prime \prime}\) you tear out:
Rev. I
1 \(25^{\prime}\) you inscribe. 1 and 1 you make hold, 1. \(40^{\prime}\) and \(40^{\prime}\) you make hold,
2 \(26^{\prime} 40^{\prime \prime}\) to 1 you join: \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) to \(25^{\prime}\) you raise:
3 \(36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\) you inscribe. \(5^{\prime}\) to \(40^{\prime}\) you raise: \(3^{\prime} 20^{\prime \prime}\)
4 and \(3^{\prime} 20^{\prime \prime}\) you make hold, \(11^{\prime \prime} 6^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) to \(36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\) you join:
5 by \(36^{\prime} 17^{\prime \prime} 46^{\prime \prime \prime} 40^{\prime \prime \prime}\), \(46^{\prime} 40^{\prime \prime}\) is equal. \(3^{\prime} 20^{\prime \prime}\) which you have made hold
6 inside \(46^{\prime} 40^{\prime \prime}\) you tear out: \(43^{\prime} 20^{\prime \prime}\) you inscribe.
7 igi \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) is not detached. What to \(1^{\circ} 26^{\prime} 40^{\prime \prime}\)
8 may I posit which \(43^{\prime} 20^{\prime \prime}\) gives me? \(30^{\prime}\) its bandûm.
9 \(30^{\prime}\) to 1 you raise: \(30^{\prime}\) the first confrontation.
10 \(30^{\prime}\) to \(40^{\prime}\) you raise: \(20^{\prime}\), and \(5^{\prime}\) you join:
11 \(25^{\prime}\) the second confrontation.
Even this problem deals with two squares (lines Obv. II.44–45).6 The somewhat obscure formulation in line 45 means that the second “confrontation” equals two-thirds of the first, with additional \(5^{\prime} \mathrm{NINDAN}\). If \(c_{1}\) and \(c_{2}\) stands for the two “confrontations,” line 44 informs us that the sum of the areas is \(\square\left(c_{1}\right)+\square\left(c_{2}\right)=25^{\prime} 25^{\prime \prime}\), while line 45 states that \(c_{2}=40^{\prime} \cdot c_{1}+5^{\prime}\).
This problem cannot be solved by means of a simple false position in which a hypothetical number is provisionally assumed as the value of the unknown—that only works for homogeneous problems.7 The numbers 1 and \(40^{\prime}\) in line 46 show us the way that is actually chosen: \(c_{1}\) and \(c_{2}\) are expressed in terms of a new magnitude, which we may call \(c\):
\(c_{1}=1 \cdot c \quad, \quad c_{2}=40^{\prime} \cdot c+5^{\prime}\).
That corresponds to Figure 3.6. It shows how the problem is reduced to a simpler one dealing with a single square \(\square(c)\). It is clear that the area of the first of the two original squares ( \(\square\left(c_{1}\right)\)) equals \((1 \times 1) \square(c)\), but that calculation has to wait until line Rev. I.1. The text begins by considering \(\square\left(c_{2}\right)\), which is more complicated and gives rise to several contributions. First, the square \(\square\left(5^{\prime}\right)\) in the lower right corner: \(5^{\prime}\) and \(5^{\prime}\) you make hold, \(25^{\prime \prime}\). This contribution is eliminated from the sum \(25^{\prime} 25^{\prime \prime}\) of the two areas: \(25^{\prime \prime}\) inside \(25^{\prime} 25^{\prime \prime}\) you tear out: \(25^{\prime}\) you inscribe. The \(25^{\prime}\) that remains must now be explained in terms of the area and the side of the new square \(\square(c)\).
\(\square\left(c_{1}\right)\), as already said, is \(1 \times 1=1\) times the area \(\square(c)\): 1 and 1 you make hold, 1.8 After elimination of the corner \(5^{\prime} \times 5^{\prime}\) remains of \(\square\left(c_{2}\right)\), on one hand, a square \(\square\left(40^{\prime} c\right)\), on the other, two “wings” to which we shall return imminently. The area of the square \(\square\left(40^{\prime} c\right)\) is \(\left(40^{\prime} \times 40^{\prime}\right) \square(c)=26^{\prime} 40^{\prime \prime} \square(c)\): \(40^{\prime}\) and \(40^{\prime}\) you make hold, \(26^{\prime} 40^{\prime \prime}\). In total we thus have \(1+26^{\prime} 40^{\prime \prime}=1^{\circ} 26^{\prime} 40^{\prime \prime}\) times the square area \(\square(c)\): \(26^{\prime} 40^{\prime \prime}\) to 1 you join: \(1^{\circ} 26^{\prime} 40^{\prime \prime}\).
Each "wing" is a rectangle 
(\(5^{\prime}, 40^{\prime}c\), whose area can be written \(5^{\prime} \cdot 40^{\prime} c=3^{\prime} 20^{\prime \prime} c\): \(5^{\prime}\) to \(40^{\prime}\) you raise: \(3^{\prime} 20^{\prime \prime}\). All in all we thus have the equation
\(1^{\circ} 26^{\prime} 40^{\prime \prime} \square(c)+2 \cdot 3^{\prime} 20^{\prime \prime} c=25^{\prime}\)
This equation confronts us with a problem which the Old Babylonian author has already foreseen in line Rev. I.2, and which has caused him to postpone until later the calculation of the wings. In modern terms, the equation is not "normalized," that is, the coefficient of the second-degree term differs from 1. The Old Babylonian calculator might correspondingly have explained it by stating in the terminology of TMS XVI that "as much as (there is) of surfaces" is not one—see the left part of Figure 3.7, where we have a sum of \(\alpha\) square areas (the white rectangle (\(c, \alpha c\))) and \(\beta\) sides, that is, the shaded rectangle (\(c, \beta\)), corresponding to the equation
\(\alpha \square(c)+\beta c=\Sigma\)
(in the actual case, \(\alpha=1^{\circ} 26^{\prime} 40^{\prime \prime}\), \(\beta=2 \cdot 3^{\prime} 20^{\prime \prime}\), \(\Sigma=25^{\prime}\)). This prevents us from using directly our familiar cut-and-paste procedure. "Breaking" \(\beta\) and making the two "moieties" "hold" would not give us a gnomon.
The Babylonians got around the difficulty by means of a device shown in the right-hand side of figure 3.7: the scale of the configuration is changed in the vertical direction, in such a way that the vertical side becomes \(\alpha c\) instead of \(c\); in consequence the sum of the two areas is no longer \(\Sigma\left(=25^{\prime}\right)\) but \(\alpha \Sigma\left(=1^{\circ} 26^{\prime} 40^{\prime \prime} \cdot 25^{\prime}=36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\right)\): \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) to \(25^{\prime}\) you raise: \(36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\) you inscribe. As we see, the number \(\beta\) of sides is not changed in the operation, only the value of the side, namely from \(c\) into \(\alpha c\).9
In modern symbolic language, this transformation corresponds to a multiplication of the two sides of the equation
\(\alpha c^{2}+\beta c=\Sigma\)
by \(\alpha\), which gives us a normalized equation with the unknown \(\alpha c\):
\((\alpha c)^{2}+\beta \cdot(\alpha c)=\alpha \Sigma\),
an equation of the type we have encountered in BM 13901 #1. We have hence arrived to a point where we can apply the habitual method: "breaking" the shaded rectangle and make the two resulting "moieties" "hold" a quadratic complement (Figure 3.8); the outer “moiety” is lightly shaded in its original position and more heavily in the position to which it is brought). Now, and only now, does the calculator need to know the number of sides in the shaded rectangle of Figure 3.7 (that is, to determine \(\beta\)). As already said, each "wing" contributes \(5^{\prime} 40^{\prime \prime}=3^{\prime} 20^{\prime \prime}\) sides. If the calculator had worked mechanically, according to fixed algorithms, he would now have multiplied by 2 in order to find \(beta\). But he does not! He knows indeed that the two wings constitute the excess that has to be "broken" into two "moieties." He therefore directly makes \(3^{\prime} 20^{\prime \prime}\) and \(3^{\prime} 20^{\prime \prime}\) "hold,"which produces the quadratic complement, and "joins" the resulting area \(11^{\prime \prime} 6^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) to that of the gnomon \(36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\): \(3^{\prime} 20^{\prime \prime}\) and \(3^{\prime} 20^{\prime \prime}\) you make hold, \(11^{\prime \prime} 6^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) to \(36^{\prime} 6^{\prime \prime} 40^{\prime \prime \prime}\) you join: [...] \(36^{\prime} 17^{\prime \prime} 46^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\).
\(36^{\prime} 17^{\prime \prime} 46^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) is thus the area of the completed square, and its side \(\sqrt{36^{\prime} 17^{\prime \prime} 46^{\prime \prime \prime} 40^{\prime \prime \prime \prime}}=46^{\prime} 40^{\prime \prime}\): by \(36^{\prime} 17^{\prime \prime} 46^{\prime \prime \prime} 40^{\prime \prime \prime \prime}\) , \(46^{\prime} 40^{\prime \prime}\) is equal. This number represents \(1^{\circ} 26^{\prime} 40^{\prime \prime} \cdot c+3^{\prime} 20^{\prime \prime}\); therefore, \(1^{\circ} 26^{\prime} 40^{\prime \prime} c\) is \(46^{\prime} 40^{\prime \prime}-3^{\prime} 20^{\prime \prime}=43^{\prime} 20^{\prime \prime}\): \(3^{\prime} 20^{\prime \prime}\) which you have made hold inside \(46^{\prime} 40^{\prime \prime}\) you tear out: \(43^{\prime} 20^{\prime \prime}\) you inscribe. Next, we must find the value of \(c\). \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) is an irregular number, and the quotient \(46^{\prime} 40^{\prime \prime} / 1^{\circ} 26^{\prime} 40^{\prime \prime}\) is given directly as \(30^{\prime}\):10 igi \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) is not detached. What to \(1^{\circ} 26^{\prime} 40^{\prime \prime}\) may I posit which \(43^{\prime} 20^{\prime \prime}\) gives me? \(30^{\prime}\) its bandûm.
In the end, \(c_{1}\) and \(c_{2}\) are determined, \(c_{1}=1 \cdot c=30^{\prime}\), \(c_{2}=40^{\prime} \cdot c+5^{\prime}=25^{\prime}\):11 \(30^{\prime}\) to 1 you raise: \(30^{\prime}\) the first confrontation. \(30^{\prime}\) to \(40^{\prime}\) you raise: \(20^{\prime}\) and \(5^{\prime}\) you join: \(25^{\prime}\) the second confrontation. The problem is solved.