# The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Sporus

Author(s):
Colin B. P. McKinney (Wabash College)

Biographical note: Sporus (ca. 240 CE – ca. 300 CE). Other than his cube duplication, Sporus also worked on the problem of squaring a circle. He also criticized one attempted method to solve the problem, due to Hippias, as being logically circular (pun intended). Read more about Sporus at MacTutor

(Heiberg 76.2) Let ΑΒ and ΒΓ be two given unequal straight lines: thus it is necessary to find two mean proportionals in continued proportion.

Let ΔΒΕ be drawn from Β at right angles to ΑΒ, and with center Β and radius ΒΑ, let the semicircle ΔΑΕ be drawn. Let a straight line joining Ε to Γ be drawn through to Ζ, and let some straight line be drawn from Δ so that it makes ΗΘ equal to ΘΚ (for this is possible). From the points Η and K, let perpendiculars ΗΛ and ΚΝΜ be drawn to ΔΕ.

Above: Sporus’ Diagram. Notice how this diagram is mostly a rotation of Diocles’ and Pappus’. The red circle is my addition.

So since $$\tag{48} \text{ΚΘ : ΘΗ :: ΜΒ : ΒΛ,}$$

 This can be seen by imagining a horizontal line drawn through Η, perpendicular to ΗΛ.  The horizontal segments would be equal to ΛΒ and ΒΜ, respectively.

and ΚΘ is equal to ΘΗ, therefore ΜΒ is equal to ΒΛ. Therefore also the remainder ΜΕ is equal to the remainder ΛΔ. And therefore the whole ΔΜ is equal to the whole ΛΕ, and because of this, $$\tag{49}\text{ΜΔ : ΔΛ :: ΛΕ : ΕΜ.}$$But by repeated use of Elements VI.8 we have $$\tag{50} \text{ ΜΔ : ΔΛ :: ΚΜ : ΗΛ [using tri.(ΜΔΚ)] }$$and $$\tag{51} \text{ ΛΕ : ΕΜ :: ΗΛ : ΝΜ [using tri.(ΕΛΗ)]. }$$Again, since $$\tag{52} \text{ΔΜ : ΜΚ :: ΚΜ : ΜΕ [using tri.(ΔΚΕ)], }$$therefore by Elements VI.19 cor. $$\tag{53} \text{ ΔΜ : ΜΕ :: sq.(ΔΜ) : sq.(ΜΚ) }$$and $$\tag{54} \text{sq.(ΔΜ) : sq.(ΜΚ) :: sq.(ΔΒ) : sq.(ΒΘ), }$$so $$\tag{55} \text{sq.(ΔΜ) : sq.(ΜΚ) :: sq.(AΒ) : sq.(ΒΘ), }$$for ΔΒ is equal to ΒΑ. Again, since $$\tag{56} \text{ ΜΔ : ΔΒ :: ΛΕ : EB [by (49)], }$$and by using Elements VI.2 twice  $$\tag{57} \text{ΜΔ : ΔΒ :: ΚΜ : ΘΒ [using tri.(ΔΚΜ)] }$$and $$\tag{58} \text{ ΛΕ : ΕΒ :: ΗΛ : ΓΒ [using tri.(ΗΛΕ)], }$$therefore $$\tag{59} \text{ ΚΜ : ΘΒ :: ΗΛ : ΓΒ [combining (56) through (58)].}$$Therefore alternando [Elements V, def. 12] $$\tag{60} \text{ ΚΜ : ΗΛ :: ΘΒ : ΓΒ.}$$But $$\tag{61} \text{ ΚΜ : ΗΛ :: ΜΔ : ΔΛ [by (50)], }$$ $$\tag{62} \text{ ΜΔ : ΔΛ : : ΔΜ : ΜΕ,}$$and $$\tag{63} \text{ ΔΜ : ΜΕ :: sq.(ΑΒ) : sq.(ΘΒ) [combining (53) through (55)];}$$therefore $$\tag{64} \text{ sq.(ΑΒ) : sq.(ΘΒ) :: ΒΘ : ΒΓ.}$$

 This last claim takes a bit more work. By (55), $$\tag{65} \text{sq.(ΔΜ) : sq.(ΜΚ) :: sq.(ΑΒ) : sq.(ΒΘ). }$$Also, by (52), $$\tag{66} \text{ sq.(ΔΜ) : sq.(ΜΚ) :: ΔΜ : ΜΕ. }$$By construction, ΔΛ = ΜΕ, so $$\tag{67} \text{ ΔΜ : ΜΕ :: ΔΜ : ΔΛ. }$$But by (50), we have $$\tag{68} \text{ ΔΜ : ΔΛ :: ΚΜ : ΗΛ, }$$and by (60), we have $$\tag{69} \text{ΒΘ : ΒΓ :: ΚΜ : ΗΛ. }$$Putting these together, we get the desired result.

Let Ξ be taken as the mean proportional between ΘΒ and ΒΓ: $$\tag{70} \text{ ΘΒ : Ξ :: Ξ : ΒΓ.}$$So since $$\tag{71} \text{ sq.(ΑΒ) : sq.(ΒΘ) :: ΒΘ : ΒΓ, }$$but $$\tag{72} \text{ sq.(ΑΒ) : sq.(ΒΘ) :: duplicate ratio of (ΑΒ : ΒΘ), }$$ $$\tag{73} \text{ ΘΒ : ΒΓ :: duplicate ratio of (ΘΒ : Ξ), }$$and $$\tag{74} \text{ ΑΒ : ΒΘ :: Ξ : ΒΓ, }$$ therefore $$\tag{75} \text{ ΑΒ : ΒΘ :: ΘΒ :Ξ :: Ξ : ΒΓ. }$$But it is clear that this construction is the same as those given by both Pappus and Diocles.

Colin B. P. McKinney (Wabash College), "The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Sporus," Convergence (May 2016)