The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Archytas, According to Eudemus’ History

Colin B. P. McKinney (Wabash College)

Biographical notes: Archytas of Tarentum (ca. 428 BCE – ca. 350 BCE). Apparently the first mathematician to solve the cube duplication, he is known to have been an associate of Plato. Fragments of his works survive, including a brief quotation by Eutocius in his commentary on the Conics. Read more about Archytas at MacTutor.

Eudemus of Rhodes (ca. 350 BCE – ca. 290 BCE). An associate of Aristotle, Eudemus might well have been the first historian of mathematics. We know of his works only through reference or quotation (such as by Eutocius). Read more about Eudemus at MacTutor.

(Heiberg 84.13) Let ΑΔ and Γ be two given straight lines: thus it is necessary to find two mean proportionals between ΑΔ and Γ. 

For let the circle ΑΒΔΖ be drawn around the larger segment ΑΔ, and let ΑΒ be fitted [into the circle] equal to Γ (see Elements IV.1), and having extended ΑΒ, let it cut the tangent from Δ at the point Π. Let ΒΕΖ be drawn parallel to ΠΔΟ, and let a half-cylinder be imagined at right angles to the [plane of] the semicircle ΑΒΔ, and let [another] semicircle ΑΔ be imagined at right angles to the first semicircle, this second semicircle being placed in the parallelogram of the half-cylinder. Therefore this semicircle, being rotated from Δ to Β, with the endpoint Α of its diameter remaining fixed, will intersect the surface of the half-cylinder by its [the semicircle’s] rotation, and will draw a certain curve on it.

Above: Archytas’ First Curve. This is the 3D view of the rotating semicircle ΑΔΚ, which is currently in the position ΑΔ2Κ2. The intersection of this semicircle and the cylinder is the point Κ, which starts at the point Δ and moves upwards as Δ moves towards the position Δ2. As Δ2 rotates, Κ2 traces a curve along the surface of the cylinder. To trace out the curve, move Δ2 toward Δ and back


Below: Archytas’ First Curve, planar view. This is the base plane of the above diagram.


Again, when the segment ΑΔ is fixed, if the triangle ΑΠΔ is rotated, moving in the direction opposite of the semicircle, its motion will make a conical surface with the straight line ΑΠ, which, having been rotated, will meet the line on the cylinder at some point; at the same time [as the rotation], the point Β will also draw a semicircle on the surface of the cone.

Above: Archytas’ Second Curve. This is the 3D view of the rotating triangle ΑΔΠ1. Its intersection with the cylinder is at Κ1. As the triangle rotates, the point Κ1 sweeps out another curved line on the surface of the cylinder. To trace out the curve, move Π1 between Π and O.


Below: Archytas’ Second Curve, planar view. This is the base plane of the above diagram.


Let the rotating semicircle have its position at the place where the lines on the cylinder intersect, as ΔΚΑ, and the counter-rotating triangle have as its position ΔΛΑ, and let the point of the aforementioned intersection be Κ, and let there be the semicircle ΒΜΖ, being drawn through the point Β, and let the common section of it and the circle ΒΔΖΑ be the line ΒΖ.

Above: Archytas: Intersection configuration. The two curves (not shown) intersect at the point Κ, when the triangle and semicircle are in this position.


From Κ, let a perpendicular be drawn to the plane of the circle ΒΔΑ: indeed it will fall on the perimeter of the circle since the cylinder was set up at right angles. Let it fall as ΚΙ, and from Ι, extending it to Α, let it [ΑΙ] meet ΒΖ at Θ. Let ΑΛ meet the semicircle ΒΜΖ at Μ. Let the lines ΚΔ, ΜΙ, and ΜΘ be joined.

Above: Archytas’ Finished Diagram. The additional lines are drawn, and the diagram is now complete.


So since each of the semicircles ΔΚΑ and ΒΜΖ are perpendicular to the base plane, therefore the common section of them, ΜΘ, is also perpendicular to the plane of the circle: so that ΜΘ is also perpendicular to ΒΖ. Therefore, by Elements III.35, \begin{equation} \tag{94} \text{ rect.(ΒΘΖ) = rect.(ΑΘΙ) = sq.(ΜΘ).} \end{equation}

Therefore the triangle ΑΜΙ is similar to each of the triangles ΜΙΘ and ΜΑΘ, and the angle ΙΜΑ is right. But the angle ΔΚΑ is also right: therefore the lines ΚΔ and ΜΙ are parallel. And by similarity of the triangles, 

\begin{equation} \tag{95} \text{ΔΑ : ΑΚ = ΚΑ : ΑΙ } \end{equation}  \begin{equation} \tag{96} \text{ΔΑ : ΑΚ = ΙΑ : ΑΜ. } \end{equation}

Therefore the four segments ΔΑ, ΑΚ, ΑΙ, and ΑΜ are continuously proportional. And ΑΜ is equal to Γ, since it is also equal to ΑΒ; therefore ΑΚ and ΑΙ have been found as two mean proportionals between the given lines ΑΔ and Γ.