The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Menaechmus' First Solution

(Heiberg 78.14) Let Α and Ε be two given straight lines: thus it is necessary to find two mean proportionals between them. 

Analysis. Let it happen, and let the means be Β and Γ. Let the straight line ΔΗ, being terminated at Δ, be placed into position, and at Δ, let ΔΖ be made equal to Γ. Let ΖΘ be drawn at right angles, and let ΖΘ be made equal to Β. So since the three straight lines Α, Β, and Γ are proportional, \begin{equation} \tag{76} \text{ rect.(Α, Γ) = sq.(Β),} \end{equation} therefore the rectangle contained by the given lines Α and Γ, that is, the rectangle ΔΖ, is equal to the square on B, that is, ΖΘ. Therefore Θ is on a parabola drawn through Δ. Let parallels ΘΚ and ΔΚ be drawn [parallel to ΔΖ and ΘΖ, respectively]. And since the rectangle formed by Β and Γ is given (since it is equal to the rectangle formed by Α and Ε), therefore the rectangle ΚΘΖ is also given. Therefore Θ is on a hyperbola, contained in the asymptotes ΚΔ and ΔΖ. Therefore Θ is given, so that Ζ is, too. 

Above: Menaechmus’ First Diagram. First construct the two conic sections (see the next diagram for how). GeoGebra is unable to find the intersections of two loci, though. I chose 5 points at random on each locus, computed the conic that passed through each set of five points, and then found the intersection point of those conics. I then hid everything except the intersection point. Moving the bottom point of segments A and E will adjust the ratio of A to E; everything will be dynamically recalculated.

Synthesis. This will be synthesized thusly. Let Α and Ε be the two given straight lines, and let ΔΗ, being terminated at Δ, be put into position. Through Δ, let a parabola be drawn, the axis of which is ΔΗ, and the upright side of the figure is Α. Let the lines drawn at right angles to ΔΗ be equal in square to the [rectangular] areas applied along Α, having breadths the segments cut off by them in the direction of the point Δ.

Let ΔΘ be drawn, and let ΔΚ be at right angles (to it). In the asymptotes ΚΔ and ΔΖ, let a hyperbola be drawn, from which the lines drawn parallel to ΚΔ and ΔΖ make an area equal to the rectangle contained by Α and Ε: indeed, it (the hyperbola) will cut the parabola. Let it cut at Θ, and let perpendiculars ΘΚ and ΘΖ be drawn. 

So since [by Conics I.11] \begin{equation} \tag{77} \text{ sq.(ΖΘ) = rect.(A, ΔΖ),} \end{equation}therefore \begin{equation} \tag{78} \text{Α : ΖΘ = ΖΘ : ΖΔ. } \end{equation}

Above: Constructing the Conics in Menaechmus’ First Diagram. This diagram is not given by Eutocius, but shows how the two conics can be constructed. The two means Β and Γ are not shown here (see the previous figure for details about them). As the points Z1 (hyperbola) and Z2 (parabola) are moved on the line ΔΗ, the points K1 and K2 are recomputed. For the parabola (in red), the relationship is sq.(ΔΚ2) = rect.(ΔZ2, A). For the hyperbola (in blue), the relationship is rect.(ΔΖ1, ΔK1) = rect.(Α, Ε). For more details about the symptoms of conic sections in the Greek view, see Stoudt’s article here in Convergence [2015].

Again, since [by Conics I.12] \begin{equation} \tag{79} \text{rect.(Α, Ε) = rect.(ΘΔΖ),} \end{equation}therefore \begin{equation} \tag{80} \text{ Α : ΖΘ = ΖΔ : Ε. } \end{equation}But \begin{equation} \tag{81} \text{Α : ΖΘ = ΖΘ : ΖΔ, } \end{equation}and therefore \begin{equation} \tag{82} \text{ Α : ΖΘ = ΖΘ : ΖΔ = ΖΔ : Ε. } \end{equation}

Let Β be made equal to ΘΖ, and Γ made equal to ΖΔ; therefore, \begin{equation} \tag{83} \text{ Α : Β = Β : Γ = Γ : Ε.} \end{equation}Therefore the lines Α, Β, Γ, and Ε are in continuous proportion: the very thing it was required to find.