Commentary on Proposition VI.27¶
Back to Proposition VI.27.
We have already (note on I. 44) seen the significance, in Greek geometry, of the theory of “the application of areas, their exceeding and their falling-short.” In I. 44 it was a question of “applying to a given straight line (exactly, without ‘excess’ or ‘defect’) a parallelogram equal to a given rectilineal figure, in a given angle.” Here, in VI. 27—29, it is a question of parallelograms applied to a straight line but “deficient (or exceeding) by parallelograms similar and similarly situated to a given parallelogram.”
Apart from size, it is easy to construct any number of parallelograms “deficient” or “exceeding” in the manner described. Given the straight line AB to which the parallelogram has to be applied, we describe on the base CB, where C is on AB, or on BA produced beyond A, any parallelogram “similarly situated” and either equal or similar to the given parallelogram (Euclid takes the similar and similarly situated parallelogram on half the line), draw the diagonal BD, take on it (produced if necessary) any points as E, K, draw EF, or KL, parallel to CD to meet AB or AB produced and complete the parallelograms, as AH, ML.
If the point E is taken on BD or BD produced beyond D, it must be so taken that EF meets AB between A and B. Otherwise the parallelogram AE would not be applied to AB itself, as it is required to be.
The parallelograms BD, BE, being about the same diameter, are similar [VI. 24], and BE is the defect of the parallelogram AE relatively to AB. AE is then a parallelogram applied to AB but deficient by a parallelogram similar and similarly situated to BD.
If K is on DB produced, the parallelogram BK is similar to BD, but it is the excess of the parallelogram AK relatively to the base AB. AK is a parallelogram applied to AB but exceeding by a parallelogram similar and similarly situated to BD.
Thus it is seen that BD produced both ways is the locus of points, such as E or K, which determine, with the direction of CD, the position of A, and the direction of AB, parallelograms applied to AB and deficient or exceeding by parallelograms similar and similarly situated to the given parallelogram.
The importance of VI. 27—29 from a historical point of view cannot be overrated. They give the geometrical equivalent of the algebraical solution of the most general form of quadratic equation when that equation has a real and positive root. It will also enable us to find a real negative root of a quadratic equation ; for such an equation can, by altering the sign of x, be turned into another with a real positive root, when the geometrical method again becomes applicable. It will also, as we shall see, enable us to represent both roots when both are real and positive, and therefore to represent both roots when both are real but either positive or negative.
The method of these propositions was constantly used by the Greek geometers in the solution of problems, and they constitute the foundation of Book X. of the $Elements$ and of Apollonius’ treatment of the conic sections. Simson’s observation on the subject is entirely justified. He says namely on VI. 28, 29 : “These two problems, to the first of which the 27th Prop. is necessary, are the most general and useful of all in the Elements, and are most frequently made use of by the ancient geometers in the solution of other problems ; and therefore are very ignorantly left out by Tacquet and Dechales in their editions of the Elements, who pretend that they are scarce of any use.”
It is strange that, with this observation before him, even Todhunter should have written as follows. “We have omitted in the sixth Book Propositions 27, 28, 29 and the first solution which Euclid gives of Proposition 30, as they appear now to be never required, and have been condemned as useless by various modern commentators; see Austin, Walker and Lardner.”
27 contains the διορισμός, the condition for a real solution, of the problem contained in the proposition following it. The maximum of all the parallelograms having the given property which can be applied to a given straight line is that which is described upon half the line (τὸ ἀπὸ τῆς ἡμισείας ἀναγραφόμενον). This corresponds to the condition that an equation of the form
$$ax - px^2 = A$$
may have a real root. The correctness of the result may be seen by taking the case in which the parallelograms are rectangles, which enables us to leave out of account the sine of the angle of the parallelograms without any real loss of generality. Suppose the sides of the rectangle to which the defect is to be similar to be as $b$ to $c$, $b$ corresponding to the side of the defect which lies along AB. Suppose that AKFG is any parallelogram applied to AB having the given property, that AB $= a$, and that FK $= x$.
Then
KB $= \frac{b}{c}x$, and therefore AK $= a - \frac{b}{c}x$.
Hence $(a - \frac{b}{c}x)x = S$, where $S$ is the area of the rectangle AKFG.
Thus, given the equation
$$ax - \frac{b}{c}x^2 = S,$$
where $S$ is undetermined, VI. 27 tells us that, if $x$ is to have a real value, $S$ cannot be greater than the rectangle CE.
Now CB $= \frac{a}{2}$, and therefore CD $= \frac{c}{b}\cdot\frac{a}{2}$; whence
$$S \ngtr \frac{c}{b}\cdot\frac{a^2}{4},$$
which is just the same result as we obtain by the algebraical method.
In the particular case where the defect of the parallelogram is to be a square, the condition becomes the statement of the fact that, if a straight line be divided into two parts, the rectangle contained by the parts cannot exceed the square on half the line.
Now suppose that, instead of taking F on BD as in the figure of the proposition, we take F on BD produced beyond D but so that DF is less than BD.
Complete the figure, as shown, after the manner of the construction in the proposition.
Then the parallelogram FKBH is similar to the given parallelogram to which the defect is to be similar. Hence the parallelogram GAKF is also a parallelogram applied to AB and satisfying the given condition.
We can now prove that GAKF is less than CE or AD.
Let ED produced meet AG in O.
Now, since BF is the diagonal of the parallelogram KH, the complements KD, DH are equal.
But
DH = DG, and DG is greater than OF.
Therefore KD > OF.
Add OK to each; and
AD, or CE, > AF.
This other “case” of the proposition is found in all the mss., but Heiberg relegates it to the Appendix as being very obviously interpolated. The reasons for this course are that it is not in Euclid’s manner to give a separate demonstration of such a “case”; it is rather his habit to give one case only and to leave the student to satisfy himself about any others (cf. I. 7). Internal evidence is also against the genuineness of the separate proof. It is put after the conclusion of the proposition instead of before it, and, if Euclid had intended to discuss two cases, he would have distinguished them at the beginning of the proposition, as it was his invariable practice to do. Moreover the second “case” is the less worth giving because it can be so easily reduced to the first. For suppose F' to be taken on BD so that FD = F'D. Produce BF to meet AG produced in P. Complete the parallelogram BAPQ, and draw through F' straight lines parallel to and meeting its opposite sides.
Then the complement F'Q is equal to the complement AF'.
And it is at once seen that AF, F'Q are equal and similar. Hence the solution of the problem represented by AF or F'Q gives a parallelogram of the same size as AF' arrived at as in the first “case.”
It is worth noting that the actual difference between the parallelogram AF and the maximum area AD that it can possibly have is represented in the figure. The difference is the small parallelogram DF.