4.7: TMS VIII #1

1 The surface \(10^{\prime}\). The 4th of the width to the width I have joined, to 3 I have gone … over

2 the length \(5^{\prime}\) went beyond. You, 4, of the fourth, as much as width posit. The fourth of 4 take, 1 you see.

3 1 to 3 go, 3 you see. 4 fourths of the width to 3 join, 7 you see.

4 7 as much as length posit. \(5^{\prime}\) the going-beyond to the to-be-torn-out of the length posit. 7, of the length, to 4, ^¿of the width^?, raise,

5 28 you see. 28, of the surfaces, to \(10^{\prime}\) the surface raise, \(4^{\circ}40^{\prime}\) you see.

6 \(5^{\prime}\), the to-be-torn-out of the length, to four, of the width, raise, \(20^{\prime}\) you see. \(\frac{1}{2}\) break, \(10^{\prime}\) you see. \(10^{\prime}\) make hold,

7 \(1^{\prime} 40^{\prime \prime}\) you see. \(1^{\prime} 40^{\prime \prime}\) to \(4^{\circ}40^{\prime}\) join, \(4^{\circ} 41^{\prime} 40^{\prime \prime}\) you see. What is equal? \(2^{\circ} 10^{\prime}\) you see.

8 \(10^{\prime}\) ^¿…^? to \(2^{\circ} 10^{\prime}\) join, \(2^{\circ} 20^{\prime}\) you see. What to 28, of the surfaces, may I posit which \(2^{\circ} 10^{\prime}\) gives me?

9 \(5^{\prime}\) posit. \(5^{\prime}\) to 7 raise, \(35^{\prime}\) you see. \(5^{\prime}\), the to-be-torn-out of the length, from \(35^{\prime}\) tear out,

10 \(30^{\prime}\) you see, \(30^{\prime}\) the length. \(5^{\prime}\) the length to 4 of the width raise, \(20^{\prime}\) you see, 20 the length (mistake for width).

In BM 13901 #12 we saw how a problem about squares could be reduced to a rectangle problem. Here, on the contrary, a problem about a rectangle is reduced to a problem about squares.

Translated into symbols, the problem is the following;

\(\frac{7}{4} w-\ell=5^{\prime}\), \((\ell, w)=10^{\prime}\)

(“to 3 I have gone” in line 1 means that the “joining” of \(\frac{1}{4} w\) in line 1 is repeated thrice). The problem could have been solved in agreement with the methods used in TMS IX #3 (page 57), that is, in the following way:

\(7 w-4 \ell=4 \cdot 5^{\prime}\), \((\ell, w)=10^{\prime}\)

\(7 w-4 \ell=20^{\prime}\), \((7 w, 4 \ell)=(7 \cdot 4) \cdot 10^{\prime}=28 \cdot 10^{\prime}=4^{\circ} 40^{\prime}\)

\(7 w=\sqrt{4^{\circ} 40^{\prime}+\left(\frac{20^{\prime}}{2}\right)^{2}}+\frac{20^{\prime}}{2}=2^{\circ} 20\),

\(\begin{array}{l}
4 \ell=\sqrt{4^{\circ} 40^{\prime}+\left(\frac{20^{\prime}}{2}\right)^{2}}-\frac{20^{\prime}}{2}=2 \\
w=20^{\prime}, \quad \ell=30^{\prime}
\end{array}\).

Figure \(4.13\): The method of TMS VIII #1.

However, once again the calculator shows that he has several strings on his bow, and that he can choose between them as he finds convenient. Here he builds his approach on a square whose side \((z)\) is \(\frac{1}{4}\) of the width (Figure 4.13). In that way, the width will equal 4, understood as 4 \(z\) (You, 4, of the fourth, as much as width posit), and the length prolonged by \(5^{\prime}\) will be equal to 7, understood as \(7z\) (7 as much as possible). Line 4 finds that the rectangle with sides \(7z\) and \(4z\)—in other words, the initial rectangle prolonged by \(5^{\prime}\)—consists of \(7 \cdot 4=28\) small squares \(\square(z)\).¹¹ These 28 squares exceed the area \(10^{\prime}\) by a certain number of sides (\(n \cdot z\)), the determination of which is postponed until later. As usual, indeed, the non-normalized problem

\(28 \square(z)-n \cdot z=10^{\prime}\)

is transformed into

\(\square(28 z)-n \cdot(28 z)=28 \cdot 10^{\prime}=4^{\circ} 40^{\prime}\).

Line 6 finds \(n=4 \cdot 5^{\prime}=20^{\prime}\), and from here onward everything follows the routine, as can be seen on Figure 4.14: 28 \(z\) will be equal to \(2^{\circ} 20\), and \(z\) hence to \(5^{\prime}\).¹² Therefore, the length \(\ell\) will be \(7 \cdot 5^{\prime}-5^{\prime}=30^{\prime}\), and the width \(w\) \(4 \cdot 5^{\prime}=20^{\prime}\).

Figure \(4.14\): Resolution of the normalized equation of TMS VIII #1.