The problem of doubling the cube was henceforth tried exclusively in the form of the problem of the two mean proportionals. - Thomas Little Heath

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 in which he may have discovered the conic sections and their fundamental properties. It seems to me much better to give the complete story of the origin and development of the geometry of the conic sections in one place, and this has been done in the chapter on conic sections associated with the name of Apollonius of Perga. Similarly a chapter has been devoted to algebra (in connexion with Diophantus) and another to trigonometry (under Hipparchus, Menelaus and Ptolemy).

The outstanding personalities of Euclid and Archimedes demand chapters to themselves. Euclid, the author of the incomparable Elements, wrote on almost all the other branches of mathematics known in his day. Archimedes's work, all original and set forth in treatises which are models of scientific exposition, perfect in form and style, was even wider in its range of subjects. The imperishable and unique monuments of the genius of these two men must be detached from their surroundings and seen as a whole if we would appreciate to the full the pre-eminent place which they occupy, and will hold for all time, in the history of science.

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>