*Biographical note: Pappus of Alexandria (ca. 290 CE – ca. 350 CE). Pappus’ chief work was the *Collection*, a series of eight books on a wide variety of topics in geometry. Only some of the books survive. Book III, which does survive, contains a collection of solutions to the cube duplication problem, but not as many as Eutocius’ collection presented here. Read more about Pappus at MacTutor.*

(Heiberg 70.7) Pappus proposed to find a cube having a given ratio to a given cube. His argument proceeds towards this goal, but it is clear that, if the following thing is found, then what is proposed will also be found: for when two straight lines are given, if the first of the two means required to be found is determined, then the second one is also determined.^{20}

For as he himself says in his work, let the semicircle ΑΒΓ be drawn, and let ΔΒ be drawn from the center Δ at right angles,^{21} and let a ruler be placed at the point Α, so that one of its ends remains fixed by a peg at the point Α, and the other end moves around another peg centered between Β and Γ. After constructing these things, let it be required to find two cubes having to one another the prescribed ratio.

Above: Pappus’ Diagram. The red circle is my addition, and makes it easy to determine when ΚΘ and ΘΗ are equal. The points Ε and Θ are movable. Moving E changes the given ratio (ΓΔ : ΔΕ), which by default is 2:1. Once a ratio is changed, Θ should be moved until the red circle passes through both Κ and Η. Also, notice how when the diagram is in its final configuration, the lines ΗΑ, ΑΛ, ΛΓ, and ΛΗ are configured relative to each other just as in Eutocius’ description of Plato’s device.

And let the ratio of ΒΔ to ΔΕ be made the same as the prescribed ratio, and having joined ΓΕ, let it be produced to Ζ. Let the ruler be moved slightly between Β and Γ, until the part it cuts off between the straight lines ΖΕ and ΕΒ becomes equal to the part between the straight line ΒΕ and the arc ΒΚΓ. Making an attempt and adjusting the ruler as needed,^{22} this will be easily found. Let this happen when the ruler is in the position ΑΚ, so that ΗΘ and ΘΚ are equal. I say that the cube on ΒΔ has to the cube on ΔΘ the prescribed ratio, that is, the ratio which ΒΔ has to ΔΕ. Let the circle be imagined as completed,^{23} and having joined ΚΔ, let ΚΔ be extended to Λ, and let ΛΗ be joined. Therefore ΛΗ is parallel to ΒΔ, since ΚΘ is equal to ΗΘ, and ΚΔ is equal to ΔΛ. Let ΑΛ and ΛΓ both be joined. So since the angle ΑΛΓ is right (since it is in a semicircle) and ΛΜ is the height, \begin{equation} \tag{32} \text{sq.(ΛM) : sq.(MA) :: ΓM : MA :: sq.(AM) : sq.(MH).} \end{equation}

Two parts of this are relatively simple: we have two right angles, at Λ and Α. In right triangle ΓΛΑ, we have that ΛΜ is a mean proportional between ΓΜ and ΜΑ. In the right triangle ΗΛΑ, we have that ΜΑ is a mean proportional between ΗΜ and ΜΛ. This yields \begin{equation} \tag{33} \text{ΓM : MΛ :: ΛM : MA (for triangle ΓΛΑ) } \end{equation} and \begin{equation} \tag{34} \text{HM : MA :: MA : MΛ (for triangle ΗΛΑ). } \end{equation} Then by Elements V def. 9, the ratio ΑΜ : ΜΓ is the duplicate ratio of the ratio ΑΜ : ΜΛ, so \begin{equation} \tag{35} \text{AM : MΓ :: sq.(AM) : sq.(MΛ).} \end{equation} Similarly, \begin{equation} \tag{36} \text{sq.(AM) : sq.(MH) :: ΓM : MA.} \end{equation}Pappus' claim then follows. |

Let the ratio of ΑΜ to ΜΗ be placed in common: therefore \begin{equation} \tag{37} \text{ (ΓM : MA) comp. (AM : MH) :: ΓΜ : ΜΗ :: (sq.(ΑΜ) : sq.(ΜΗ)) comp. (AM : ΜΗ). } \end{equation}

Literally, “added” in common, but with the sense of “multiplied.” This is a common source of headache when studying Greek ratio theory. The middle ratio is the result of “cancelling” the common factor ΜΑ. |

But \begin{equation} \tag{38} \text{(sq.(ΑΜ) : sq.(ΜΗ)) comp. (AM : ΜΗ) :: cube(ΑΜ) : cube(ΜΗ); } \end{equation} and therefore \begin{equation} \tag{39} \text{ ΓΜ : ΜΗ :: cube(ΑΜ) : cube(AH). } \end{equation} But [using *Elements* VI.8 twice] \begin{equation} \tag{40} \text{ΓΜ : ΜΗ :: ΓΔ : ΔΕ [use tri.(ΓΜΗ)]} \end{equation} and \begin{equation} \tag{41} \text{ΑΜ : ΜΗ :: ΑΔ : ΔΘ [use tri.(ΑΔΘ)];} \end{equation}therefore \begin{equation} \tag{42} \text{ΒΔ : ΔΕ :: the given ratio = cube(ΒΔ) : cube(ΔΘ) [since ΓΔ=ΒΔ=ΑΔ].} \end{equation}

Therefore ΔΘ is the first^{24} of the two mean proportionals required to have been found; and if we suppose that [using *Elements* VI.11] \begin{equation} \tag{43} \text{ΒΔ : ΔΘ :: ΘΔ : some other length,} \end{equation}the second^{25} will also have been found.

But it is necessary to note that this construction is essentially the same as the one given by Diocles, differing in this way only: in Diocles’ construction, a certain curve is drawn through a succession of points between Α and Β, and on this curve, the point Η is taken where the line ΓΕ cuts the aforementioned curve. But in Pappus’ construction, the point Η is found by moving a ruler around Α. For that the point Η is the same, whether it be taken by a ruler, as in Pappus’ construction, or as Diocles said, we can see as follows. Having extended ΜΗ to Ν, let ΚΝ be joined. So since ΚΘ is equal to ΘΗ, and ΗΝ is parallel to ΘΒ, ΚΞ is also equal to ΞΝ. And ΞΒ is common and perpendicular, for ΚΝ is bisected at right angles by the line through the center. And therefore the base is equal to the base, and through this, the arc ΚΒ is equal to the arc ΒΝ. Therefore the point Η is on Diocles’ curve.

Above: Diagram Comparing Pappus’ and Diocles’ Constructions. In this figure, the ratio of ΓΔ to ΔΕ is held fixed at 2:1. Diocles’ curve is constructed in the manner of the previous section, by making the arc ΚΒ equal to the arc ΒΝ. The curve therefore passes through the point Η, with ΚΘ equal to ΘΗ, as in Pappus’ construction.

The proof is also the same. For Diocles asserted that

\begin{equation} \tag{44} \text{ ΓΜ : ΜΝ :: ΜΝ : ΜΑ :: ΑΜ : ΜΗ.} \end{equation}But NM is equal to MΛ, since the diameter cuts it at right angles. Therefore

\begin{equation} \tag{45} \text{ ΓΜ : ΜΛ :: ΛΜ : ΜΑ :: ΑΜ : MH.} \end{equation}Therefore ΛΜ and ΜΑ are two mean proportionals between ΓΜ and ΜΗ. But

\begin{equation} \tag{46} \text{ ΓΜ : ΜΗ :: ΓΔ : ΔΕ } \end{equation} and

\begin{equation} \tag{47} \text{ ΓΜ : ΜΛ :: ΑΜ : ΜΗ :: ΓΔ : ΔΘ.} \end{equation}

Therefore ΔΘ is the first^{26} mean proportional between ΓΔ and ΔΕ, as Pappus found also.

20. In the Greek, Eutocius refers to these as the second and third means. They are the second and third *terms* in a proportion, but the first and the second *means*.

21. Presumably, at right angles to a tangent.

22. Netz suggests “trial and error."

23. i.e. the semicircle ΑΒΓ is completed to form a full circle.

24. The Greek has “second”, and a literal reading would be that ΔΘ is the second mean proportional. What Eutocius means, though, is that it is the second *term* in the proportion, even though it is the first *mean*.

25. Again, the Greek has “third.” See the previous note.

26. Again, literally, “second.”