It would be inconvenient to interrupt the account of Menaechmus's solution of the problem of the two mean proportionals in order to consider the way … - Thomas Little Heath

Take the case of a famous problem which plays a great part in the history of Greek geometry, the doubling of the cube, or its equivalent, the finding of two mean proportionals in continued proportion between two given straight lines. ...if all the recorded solutions are collected together, it is much easier to see the relations, amounting in some cases to substantial identity, between them, and to get a comprehensive view of the history of the problem. I have therefore dealt with this problem in a separate section of the chapter devoted to 'Special Problems,' and I have followed the same course with the other famous problems of squaring the circle and trisecting any angle.

The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}</math>it follows that<math>(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}</math>and if <math>a = 2b</math>, [then <math>(\frac{a}{x})^3 = 2</math>, and]<math>a^3 = 2x^3</math>.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)<math>x^2 = ay, y^2 = bx, xy = ab</math>[or equivalently...<math>y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x}</math> ]and the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i.e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i.e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have <math>\begin{cases}y^2 = b.ON = b.PM = bx\\ and\\ xy = PM.PN = ab\end{cases}</math>whence<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math>
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a. The point P where the two parabolas intersect is given by<math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math>whence, as before,<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math>

While then for a long time everyone was at a loss, Hippocrates of Chios was the first to observe that, if between two straight lines of which the greater is double of the less it were discovered how to find two mean proportionals in continued proportion, the cube would be doubled; and thus he turned the difficulty in the original problem into another difficulty no less than the former. Afterwards, they say, some Delians attempting, in accordance with an oracle, to double one of the altars fell into the same difficulty. And they sent and begged the geometers who were with Plato in the Academy to find for them the required solution. And while they set themselves energetically to work and sought to find two means between two given straight lines, Archytas of Tarentum is said to have discovered them by means of half-cylinders, and Eudoxus by means of the so-called curved lines. It is, however, characteristic of them all that they indeed gave demonstrations, but were unable to make the actual construction or to reach the point of practical application, except to a small extent Menaechmus and that with difficulty.