Commentary on Proposition VI.28¶
Back to Proposition VI.28.
The second part of the enunciation of this proposition which states the διορισμός appears to have been considerably amplified, but not improved in the process, by Theon. His version would read as follows. “But the given rectilineal figure, that namely to which the applied parallelogram must be equal (ᾧ δεῖ ἴσον παραβαλεῖν), must not be greater than that applied to the half (παραβαλλομένου instead of ἀναγραφομένου), the defects being similar, (namely) that (of the parallelogram applied) to the half and that (of the required parallelogram) which must have a similar defect” (ὁμοίων ὄντων τῶν ἐλλειμμάτων τοῦ τε ἀπὸ τῆς ἡμισείας καὶ ᾧ δεῖ ὅμοιον ἐλλείπειν). The first amplification “that to which the applied parallelogram must be equal” is quite unnecessary, since “the given rectilineal figure” could mean nothing else. The above attempt at a translation will show how difficult it is to make sense of the words at the end; they speak of two defects apparently and, while one may well be the “defect on the half,” the other can hardly be the given parallelogram “to which the defect (of the required parallelogram) must be similar.” Clearly the reading given above (from $P$) is by far the better.
In this proposition and the next there occurs the tacit assumption (already alluded to in the note on VI. 22) that if, of two similar parallelograms, one is greater than the other, either side of the greater is greater than the corresponding side of the less. As already remarked, VI. 28 is the geometrical equivalent of the solution of the quadratic equation
$$ax - \frac{b}{c} x^2 = S,$$
subject to the condition necessary to admit of a real solution, namely that
$$S \ngtr \frac{c}{b} \cdot \frac{a^2}{4}.$$
The corresponding proposition in the Data is (Prop. 58), If a given (area) be applied (i.e. in the form of a parallelogram) to a given straight line and be deficient by a figure (i.e. a parallelogram) given in species, the breadths of the defect are given.
To exhibit the exact correspondence between Euclid’s geometrical and the ordinary algebraical method of solving the equation we will, as before (in order to avoid bringing in a constant dependent on the sine of the angle of the parallelograms), suppose the parallelograms to be rectangles. To solve the equation algebraically we change the signs and write it
$$\frac{b}{c} x^2 - ax = - S.$$
We may now complete the square by adding $\frac{c}{b} \cdot \frac{a^2}{4}$. Thus
$$\frac{b}{c} x^2 - ax + \frac{c}{b} \cdot \frac{a^2}{4} = \frac{c}{b} \cdot \frac{a^2}{4} - S;$$
and, extracting the square root, we have
$$\sqrt{\frac{b}{c}} \cdot x - \sqrt{\frac{c}{b}} \cdot \frac{a}{2} = \pm \sqrt{\frac{c}{b} \cdot \frac{a^2}{4} - S},$$
and
$$x = \frac{c}{b} \cdot \frac{a}{2} \pm \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} - S\right)}.$$
Now let us observe Euclid’s method.
He first describes GEBF on EB (half of AB) similar to the given parallelogram D. He then places in one angle FGE of GEBF a similar and similarly situated parallelogram GQ, equal to the difference between the parallelogram GB and the area C.
With our notation,
GO $:$ OQ $= c : b$,
whence
OQ $=$ GO $\cdot \frac{b}{c}$.
Similarly
$\frac{a}{2} =$ EB $=$ GE $\cdot \frac{b}{c}$,
so that
GE $= \frac{c}{b} \cdot \frac{a}{2}$.
Therefore the parallelogram GQ $=$ GO2 $\cdot \frac{b}{c}$, and the parallelogram GB $= \frac{c}{b} \cdot \frac{a^2}{4}$.
Thus, in taking the parallelogram GQ equal to $(GB - S)$, Euclid really finds GO from the equation
GO2 $\cdot \frac{b}{c} = \frac{c}{b} \cdot \frac{a^2}{4} - S$.
The value which he finds is
GO $= \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} - S\right)}$,
and he finds QS (or $x$) by subtracting GO from GE; whence
$$x = \frac{c}{b} \cdot \frac{a}{2} - \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} - S\right)}.$$
It will be observed that Euclid only gives one solution, that corresponding to the negative sign before the radical. But the reason must be the same as that for which he only gives one “case” in VI. 27. He cannot have failed to see how to add GO to GE would give another solution. As shown under the last proposition, the other solution can be arrived at
by placing the parallelogram GOQP in the angle vertically opposite to FGE so that GQ' lies along BG produced. The parallelogram AQ' then gives the second solution. The side of this parallelogram lying along AB is equal to SB. The other side is what we have called $x$, and in this case
$x =$ EG $+$ GO $= \frac{c}{b} \cdot \frac{a}{2} + \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} - S\right)}$.
A parallelogram similar and equal to AQ' can also be obtained by producing BG till it meets AT produced and completing the parallelogram B'ABA', whence it is seen that the complement QA' is equal to the complement AQ, besides being equal and similar and similarly situated to AQ'.
A particular case of this proposition, indicated in Prop. 85 of the Data, is that in which the sides of the defect are equal, so that the defect is a rhombus with a given angle. Prop. 85 proves that, If two straight lines contain a given area in a given angle, and the sum of the straight lines be given, each of them will be given also. AB, BC being the given straight lines “containing a given area AC in a given angle ABC,” one side CB is produced to D so that BD is equal to AB, and the parallelograms are completed. Then, by hypothesis, CD is of given length, and AC is a parallelo-
gram applied to CD falling short by a rhombus (AD) with a given angle EDB. The case is thus a particular case of Prop. 58 of the $Data$ quoted above (p. 263) as corresponding to VI. 28.
A particular case of the last, that namely in which the defect is a square, corresponding to the equation
$$ax - x^2 = b^2,$$
is important. This is the problem of applying to a given straight line a rectangle equal to a given area and falling short by a square; and it can be solved, without the aid of Book VI., as shown above under II. 5 (Vol. I. pp. 383—4).