The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Philon the Byzantine

Colin B. P. McKinney (Wabash College)

Biographical note: Philon (or Philo) the Byzantine (ca 280 BCE–ca 220 BCE). Other than his method for duplicating the cube, we know little about him, and only some of his writings survive. His interests are similar in nature to Heron’s; one extant work of his deals with the construction of ballistae. Read more about Philon at MacTutor.

(Heiberg 60.29) Let there be two straight lines ΑΒ and ΒΓ, between which it is necessary to find two mean proportionals. Let them be placed so as to contain a right angle at B. And when ΑΓ is joined, let the semicircle ΑΒΕΓ be drawn around it, and let ΑΔ be drawn at right angles to ΒΑ, and ΒΓ drawn at right angles to ΓΖ. And let a moving ruler be placed at Β, and let it be moved around B, cutting ΑΔ and ΓΖ, until positioned so that the line from Β to Δ is equal to the line from Ε to Ζ, that is, to the segment between the circumference of the circle and ΓΖ. 

Philon’s Diagram. The Greek text for the first paragraph doesn’t really make clear that ΔΖ should be one straight line, passing through Β and Ζ. The next paragraph, and Eutocius’ later comment about the utility of Philon’s approach, do make this clear. The red semicircle and the point G are not part of the manuscript tradition; I added them to make it visually clear when ΔΒ is equal to ΕΖ. The semicircle is defined here as having center G and radius GΔ, where G is the midpoint of ΒΕ. Moving the points Β or Δ results in the semicircle shrinking and/or moving. When the semi-circle passes through Ζ (and has diameter ΔΖ), the segments ΔΒ and ΕΖ are equal. By default, the length ΑΒ is half of ΑΓ; adjusting this ratio (by moving Β towards Α to adjust the length AB) will require then re-adjusting Δ to make ΔΒ equal to ΕΖ.


So let a ruler have been conceived, having the position which ΔΒΕΖ has, when, as has been said, ΔΒ is equal to ΕΖ. I say that the straight lines ΑΔ and ΓΖ are two mean proportionals between ΑΒ and ΒΓ.

For let ΔΑ and ΖΓ be imagined, having been projected and intersecting at Θ: it is clear that, since ΒΑ and ΖΘ are parallel, the angle at Θ is right, and the circle ΑΕΓ, being completed, will also pass through Θ.

So since ΔΒ is equal to ΕΖ, therefore also the rectangle ΕΔΒ is equal to the rectangle ΘΔΑ: for each is equal to the square on the tangent from Δ. But the rectangle ΒΖΕ is equal to the rectangle ΘΖΓ: for similarly, each is equal to the square on the tangent from Ζ: so that also the rectangle ΘΔΑ is equal to the rectangle ΘΖΓ. 

This results from several applications of Elements III.36. If we complete the semicircle ΑΒΕΓ into a full circle ΑΒΕΓΘ, then it passes through Θ since the angle at Θ is right (by construction) and ΑΓ is a diameter. Hence we have two lines, ΔΘ and ΔΕ, which cut the circle. We then draw the two tangents to the circle, one from Δ and one from Ζ, here ΔΨ and ΖΩ. 
Elements III.36 demonstrates that, for ΔΨ,  \begin{equation} \tag{20} \text{rect.(ΔE, ΔB) = sq.(ΔΨ) = rect.(ΔΘ, AΔ);} \end{equation} and for ΖΩ, \begin{equation} \tag{21} \text{rect.(ZB, ZE) = sq.(ΖΩ) = rect.(ZΘ, ZΓ).} \end{equation} Since, by construction, ΔΒ equals ΖΕ, and ΒΕ is common, we can conclude that ΔΕ equals ΖΒ. Therefore, the left hand sides of equation (20) and (21) are the same. Eutocius’ next claim then follows. 

And for this reason, \begin{equation} \tag{22} \text{ΔΘ : ΘZ = ΓZ : AΔ.} \end{equation} But \begin{equation} \tag{23} \text{ΘΔ : ΘZ = BΓ : ΓZ = AΔ : AB,}\end{equation}for ΒΓ has been drawn parallel to ΔΘ in the triangle ΔΘΖ. Therefore \begin{equation} \tag{24} \text{BΓ : ΓZ = ΓZ : AΔ = AΔ : AB,}\end{equation}which was set out to show. 

It is necessary to note, that this construction is nearly the same as that given by Heron: for the parallelogram ΒΘ is the same as that taken in Heron’s construction, the sides ΘΑ and ΑΓ being projected, and the straightedge moved about the point Β. They are different in one way only: that in Heron’s construction, we moved the straightedge around B until the segments from the midpoint of ΑΓ (that is, K), having been drawn to ΘΔ and ΘΖ, are cut equally by it [K] as the segments ΚΔ and ΚΖ. But here,[14] we moved the straightedge until ΔΒ became equal to ΕΖ. But each of the constructions proceeds in the same way.

The one mentioned now[15] is better suited to use, for it is possible to observe the equality of ΔB and ΕΖ when the ruler ΔΖ is divided into equal parts in a continuous manner. This is much easier than attempting to use a compass [to observe] the equality of the lines from Κ to Δ and Ζ. 

[14] In the Greek, Eutocius does not explicitly say the word "straightedge." The particles μέν and δέ are used here by Eutocius to set up a comparison between Heron’s and Philon’s methods. Often μέν and δέ are translated as "on the one hand" and "on the other hand" to emphasize a comparison.

[15] Philon’s version.