Commentary on Proposition VI.29

Back to Proposition VI.29.

[Folio 0272 | p. 266]

The corresponding proposition in the $Data$ is (Prop. 59), If a given (area) be applied (i.e. in the form of a parallelogram) to a given straight line exceeding by a figure given in species, the breadths of the excess are given.

The problem of VI. 29 corresponds of course to the solution of the quadratic equation

$$ax + \frac{b}{c} x^2 = S.$$

The algebraical solution of this equation gives

$$x = - \frac{c}{b} \cdot \frac{a}{2} \pm \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} + S\right)}.$$

The exact correspondence of Euclid’s method to the algebraical solution may be seen, as in the case of VI. 28, by supposing the parallelograms to be rectangles. In this case Euclid’s construction on EB of the parallelogram EL similar to D is equivalent to finding that

FE = $\frac{c}{b} \cdot \frac{a}{2}$, and EL = $\frac{c}{b} \cdot \frac{a^2}{4}$.

[Folio 0273 | p. 267]

His determination of the similar parallelogram MN equal to the sum of EL and $S$ corresponds to proving that

$$FN^2 \cdot \frac{b}{c} = \frac{c}{b} \cdot \frac{a^2}{4} + S,$$

or

$$FN = \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} + S\right)},$$

whence $x$ is found as

$$x = FN - FE = \sqrt{\frac{c}{b} \left(\frac{c}{b} \cdot \frac{a^2}{4} + S\right)} - \frac{c}{b} \cdot \frac{a}{2}.$$

Euclid takes, in this case, the solution corresponding to the positive sign before the radical because, from his point of view, that would be the only solution.

No διορισμός is necessary because a real geometrical solution is always possible whatever be the size of $S$.

Again the $Data$ has a proposition indicating the particular case in which the excess is a rhombus with a given angle. Prop. 84 proves that, If two straight lines contain a given area in a given angle, and one of the straight lines is greater than the other by a given straight line, each of the two straight lines is given also. The proof reduces the proposition to a particular case of $Data$, Prop. 59, quoted above as corresponding to VI. 29.

Again there is an important particular case which can be solved by means of Book II. only, as shown under II. 6 above (Vol. I. pp. 386—8), the case namely in which the excess is a square, corresponding to the solution of the equation

$$ax + x^2 = b^2.$$

This is the problem of applying to a given straight line a rectangle equal to a given area and exceeding by a square.