The title of this paper is a nod to Underwood Dudley’s book The Trisectors [1994], but the purpose is rather different. Whereas Dudley exposes the myriad false trisections, my paper shows the myriad true cube duplications. This has one major benefit to me over Dudley: since the authors of the cube-duplications are long dead, and their methods are sound, I do not expect to be sued [Dilworth v. Dudley 1996]. 

It is well known to mathematicians and students with knowledge of abstract algebra that both of these problems are impossible with straightedge and compass, as is the problem of squaring a circle. This, I reckon, was at least suspected by the Greeks: see, for example, the humorous reference to squaring the circle in Aristophanes’ Birds [2000, lines 999–1012]. By allowing tools beyond just a straightedge and compass, the problems of trisecting the angle and duplicating the cube are solvable. (Squaring the circle, is, alas, not.) 

Fortunately for us as historians, Greek mathematicians found a number of solutions to the problem by instead solving a related one of finding two mean proportionals; I’ll detail the relationship in the next section. These solutions are incredibly diverse: some make use of mechanical instruments, others involve motion of figures in three dimensions, still others make interesting uses of the properties of conic sections. These are recorded by two ancient authors: Pappus, in Book III of the Collection, and Eutocius, in his commentary on Archimedes’ work On the Sphere and the Cylinder. While Eutocius lived later than Pappus, his collection contains more of the duplications, and so I have chosen to start this work with him. In future work, I plan to translate the relevant passages of Pappus and remark on the relationship between Eutocius’ and Pappus’ presentations of the methods that they both present. 

The text of this paper has two major parts: a translation of the relevant passages in Eutocius, along with my own notes, commentary, and diagrams. The Greek text is that of Heiberg’s edition [Eutocius 1891], recently reprinted by Cambridge [2013], and available in machine-readable format in the Thesaurus Linguae Graecae [Pantelia 2014]. Page and line numbers from Heiberg’s edition are given in parentheses at the start of some paragraphs, for those readers wishing to look at the corresponding Greek text. Selections of this text (and also many others) are also available in a two-volume set from the Loeb Classical Library [Thomas 1939a, 1939b]. These versions are especially helpful for those with limited (or no) proficiency with Greek, but who still want to see the Greek: all Loeb volumes have facing page style translation along with the original language text (Greek in our case, but the collection also features a vast array of Latin works). 

Netz has done a translation of both Archimedes’ text [2004] and Eutocius’ commentary [2004]. In particular, his work also has copious commentary about the diagrams in the manuscript tradition. The diagrams I give here are heavily influenced by Netz and Heiberg, though I have made some changes in order to make use of GeoGebra or to improve clarity. For example, the manuscript diagrams represent conic sections as circular arcs, but I have represented them in true conic-section form. Also, I have added parts to diagrams when it seemed helpful; for example, a circle mentioned in an argument but not drawn in a figure. The captions to each figure detail what changes or additions I have made. 

Readers of Convergence will recognize the relation of this paper with a recent one by Cawthorne and Green, Cubes, Conic Sections, and Crockett Johnson [2014]. In it, the authors give an overview of the cube duplication problem, discuss some of Johnson’s biography, and provide an analysis of Johnson's relevant painting. Readers might also recall that Albrecht Dürer knew of both the problem and at least some of the ancient solutions. His work The Painter’s Manual [1977] contains adaptations of the solutions to the problem due to Sporus and Plato, along with a variation on the tale Eratosthenes gives for the history of the problem. He also gives a method for drawing conchoid lines, similar in spirit to the method Eutocius ascribes to Nicomedes, but does not give the cube duplication that makes use of the conchoid line. The 1538 edition adds another method which resembles in some way both Descartes’ mesolabe and Eratosthenes’ sliding plates. Ishizu [2009] gives a very interesting discussion of Dürer’s motivation for solving the problem as it relates to a particular engraving, Melancolia (1514). 

As we will see in the History of the Problem section, solutions to the cube duplication problem were relevant also in engineering. Eratosthenes explicitly mentions the need for the solution if one is constructing artillery. Two other solutions appear in artillery manuals written by Heron and Philon; in the case of Heron, Eutocius cites Heron’s artillery manual as his source (though, it should be noted, the texts are not identical). 

For readability, I’ve “translated” some of the mathematical statements into what I think is an abbreviated form. Table 1.1 gives a brief list. 

Table 1.1: Summary of Notation

tri.(ABΓ)	The triangle with vertices A, B, Γ.
sq.(AB)	The square with side AB.
rect.(AB,ΓΔ)	The rectangle with sides AB, ΓΔ.
rect.(AΓ)	The rectangle about the diagonal AΓ.
rect.(ABΓ)	The rectangle with sides AB and BΓ.
quad.(AΓ)	The quadrilateral about the diagonal AΓ.
pllg.(AB,BΔ)	The parallelogram with sides AB, BΔ.
pllg.(AB,ΓΔ)	The parallelogram with sides equal to AB, ΓΔ.
pllg.(AΓ)	The parallelogram about the diagonal AΓ.
A : B :: Γ : Δ	As A is to B, so too is Γ to Δ.
dup.(A : B)	The duplicate ratio of A : B.
(A : B) comp. (Γ : Δ)	The ratio compounded from the ratios A : B and Γ : Δ
arc(AB)	The arc from A to B (or through both).

One downside to this approach is that it does over-distill some of the Greek phrasing, but it makes for more readable English. For example, Eutocius often makes statements within statements: he might say something like “Therefore A is to B, that is C, as D is to E.” This can be cumbersome to translate, especially when trying to use equation-style formatting for readability. In a situation such as this, I have tended to overly notate, and render the statement as \begin{align} \text{A : B} & = \text{A : C} \\\\ & = \text{D : E.} \end{align}

A general point: Greek texts do not label points in the same way that we tend to today. By this I mean that Greek texts freely interchange ΑΒ and ΒΑ, or seemingly randomly order the vertices of a triangle ΑΒΓ, ΒΑΓ, ΒΓΑ, etc. As I work frequently with the Greek texts, the practice carries over both to my own translations (where it is justified) and to my supplementary notes and commentary (where it is not). This, I think, points out how critical it is for a reader, be he ancient or modern, to have a diagram at hand. 

References to propositions in the Elements [2003] or the Conics [2013] are given as I found helpful; for the Elements, I have also hyperlinked to David Joyce’s online edition ([8]). The reader should note, however, that these references are almost never explicitly made in the Greek text. 

In most of the sections that follow, my commentary is included either in a box or using footnotes.  This will very clearly delineate when I am speaking versus when Eutocius (or another ancient voice) is speaking.  An example of a commentary box is given below:

I suppose you could say that I am thinking inside the box.

Euclid, in Elements VI.13, gives the construction for finding a single mean proportional. That is, given magnitudes Α and Γ, Euclid finds an intermediate magnitude Β so that \begin{equation} \tag{1} \text{A : B :: B : Γ.} \end{equation}

He does this by placing Α and Γ in straight line with each other, and then describing a semicircle with their combined length as its diameter. Extending a perpendicular from the point where they meet to the semicircle gives the length Β.

Above: Euclid's Elements VI.13. Lengths A and Γ can be adjusted by moving the points at the ends of the diameter formed by A and Γ .

Euclid also describes the concept of duplicate and triplicate ratios, in the definitions to Book V. From this definition, and Elements VI.19 cor.^[1] we conclude that if  \begin{equation} \tag{2} \text{ A : B :: B : Γ,} \end{equation}then \begin{equation} \tag{3} \text{A : Γ :: figure on A : figure on B,} \end{equation}where the figures are similar. For example, if we define the figures to be squares then \begin{equation} \tag{4} \text{A : Γ :: sq.(A) : sq.(B).} \end{equation}If we make the ratio Α : Γ the same as 1 : 2, then the square on B will be twice the square on A, and hence we have doubled the square. So if our goal was to double the square, this sequence of propositions gives us the method. But it also gives us more: since we only assumed that Α : Γ is as 1 : 2 at the very end, the problem we have solved is more general. 

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[1] A brief note about the term “corollary” seems appropriate. In Greek, the term Euclid uses is πόρισμα, “porism.”  The sense of the word is what we mean with the word “corollary.”  Since “porism” may seem overly archaic to readers (especially those who have not taken Greek), I will use “corollary,” or the abbreviation “cor.,” for this term.

An immediate generalization to finding one mean proportional comes in Elements, Book XI, on solids. First, we make a definition. Suppose we have magnitudes Α, Β, Γ, and Δ. Then we say that these magnitudes are in continuous proportion if \begin{equation} \tag{5} \text{A : B :: B : Γ :: Γ : Δ.} \end{equation}

In Elements XI.33 cor., Euclid shows that if we have four magnitudes in continuous proportion, then \begin{equation} \tag{6} \text{A : Δ :: parallelepiped solid on A : parallelepiped solid on B.} \end{equation}

If we let Α : Δ be the same as 1 : 2, and take the parallelepipedal solids as cubes, then \begin{equation} \tag{7} \text{1 : 2 :: A : Δ :: cube(A) : cube(B).} \end{equation}This is all well and good, except unlike the single mean proportional, we do not (yet) know how to find Β or Γ given only Α and Δ.

Eratosthenes (next section) details that it was Hippocrates of Chios who realized the connection of the duplication problem with the problem of finding two mean proportionals. Again, the latter problem is more general, since we need not assume that Α : Δ is as 1 : 2. 

(Heiberg 88.4) Eratosthenes to King Ptolemy, greetings!

They say that one of the ancient tragedy writers introduced Minos, who was building a tomb to Glaucos, and upon hearing that the tomb was to be a hundred feet in each dimension, said:

You have described a small part of a royal tomb:
Let it be doubled, but no less fine;
Quickly! Double each side of the tomb!

He seems to have been in error: for when the sides are doubled, the plane figure becomes quadrupled, and the solid figure octupled. This was examined by geometers, too, in what way one could double a given solid figure while keeping the same shape: they called this problem that of duplicating the cube, since assuming a cubical figure, they sought to double it. Having been at a loss for a long time, Hippocrates of Chios conceived that if, between two straight lines, with the larger double the smaller, two mean proportionals were found taken in continuous proportion, then the cube will be doubled. Hence he changed the problem into a different problem, but one no less challenging. 

After a while, they say that the Delians, having been exhorted by an oracle to double some altar, fell upon the exact same problem; they sent messengers to the esteemed geometers who were with Plato in the Academy, asking them to solve their problem. Of those industrious men who devoted themselves to finding two mean proportionals, Archytas of Tarentum is said to have done so by means of half-cylinders, and Eudoxus by so-called kampyles. But as it happened, all of them wrote in a technical sense, and so it was not possible to execute any of them by hand. Except, perhaps, in some small way, that of Menaechmus,^[2] but even still it is difficult. But we have devised a mechanical method, by which we are capable of finding not just two mean proportionals between two given magnitudes, but any number of means that might be demanded! 

Having found this, we will be able to, generally speaking, transform a given solid figure which is contained by parallelograms into a cube, or change the form of one solid into another, both to make it similar and, increasing it, to maintain this similarity, be it for temples or altars. We will also be able to do this both for liquid and dry measures, for instance the μετρητής^[3] or the μέδιμνος;^[4] we will be able to convert a vessel into a cube, and by measuring its sides, determine the capacity in liquid or dry measures.

This idea will be useful to those desiring to augment ballistas or catapults, for it is necessary to increase both the thickness and magnitude of the aperture, the washers, and the sinew.^[5] If the power is desired to be proportionally increased, it is not possible to do so without finding these mean proportionals. I have written to you about both the proof and construction of the aforementioned machines.^[6]

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[2] Netz [2013] takes this differently, translating that Menaechmus did it via the shortness, but there is no clue as to what this term might mean.

[3] A unit of liquid measurement; approximately 9 gallons [Liddell and Scott 1940].

[4] A unit of solid measurement, for grain; approximately 12 gallons [Liddell and Scott 1940].

[5] These terms (aperture, washers, sinew) are technical, and part of a missile-throwing type of catapult. Eratosthenes uses them freely, implying the intended audience (King Ptolemy, if the letter is to be believed) was well-acquainted with the terms. I leave further discussion on the problem and its connection to military texts for future work.

[6] This separate material seems to not be extant, though technical descriptions do appear in other texts. See, for example, [Marsden 1971].

Biographical note: Eutocius (ca 480 CE–ca 530 CE) was born in Ascalon (now Ashqelon, in Israel). His chief writings are commentaries on several of Archimedes’ works (including On the Sphere and the Cylinder, which is the primary source for this article) and an edition with commentary of Apollonius’s Conics. He was likely a student of Ammonius, and a contemporary of Anthemius of Tralles. He dedicates his commentary on Apollonius to Anthemius, who was one of the architects of the Hagia Sophia in Constantinople. Read more about Eutocius at MacTutor.

Eutocius is engaging in two acts in his commentary. The first is mathematical. Since Archimedes’ proposition (SC II.1) depends on finding two mean proportionals, and Archimedes does not provide details as to how to find them, Eutocius is bridging the gap mathematically. While it may have been true that Archimedes and other research mathematicians of his day would have been sufficiently familiar with the result to not need this gap filled in, students or those in Eutocius’ day, who lived in a different era, may have needed it. The second act is historical. Since mathematically it would suffice to include only one method of finding two mean proportionals, Eutocius is clearly doing something different, beyond being mathematically thorough, by including twelve. Fortunately for us, Eutocius tells us exactly what he is doing in his own words:

(Heiberg 54.27) Having assumed the things related to the problem through the analysis of it, namely that it is necessary to find two mean proportionals in continuous proportion, but in the synthesis he says, "let them be found." But how to find them, being not written by him [Archimedes] at all, we have found in the writings on this problem by many famous men. Of these, we have excluded that of Eudoxus of Cnidus, since he says in his preface that he has found them by means of curved lines, but in the proof he does not make use of curved lines; and also because he passes off a discrete proportion as if it were continuous. This is impossible to imagine, I dare say for Eudoxus, but also for anyone even moderately engaged in geometry. And so, in order that the ideas of those men who have come to us might become well-known, the method of solution from each will be also (καί) written here.

We perhaps should pay special attention to the adverbial καί here, since in Eutocius’ day, it was clearly the case that these solutions were recorded elsewhere. Not so for us: most of Eutocius’ source texts are no longer extant. Thus, by including all of the solutions he could find, even ones that are nearly the same, he gives us a glimpse into the history of the problem and its solutions. It is worth noting that Eutocius will later engage in this exact sort of philological enterprise, with his commentary and new edition of the Conics.

The methods of solution presented here are likely to seem strange to a modern reader. We have several involving motion, two involving the use of conic sections (which themselves ultimately derive from motion, since they derive from a cone), and several that involve motion in a way that has a mechanical feeling, suggesting their adaptability to practical computation. A reader who has previously studied Greek mathematics may have come across the term neusis to describe a particular type of solution to various problems. In the usual sense, neusis involves finding a line from a given point, so that the segment cut off by two given lines is equal to a given segment. The two lines could be either straight or curved.

For example, in the following figure, we have two curved lines, and a given point A. We seek the particular configuration where the segment between the two curves is equal to some given length. There is a bit of a Goldilocks feel here: some segments, like BC, are too large; others, like DE, are too small; but one, say FG, is just right.

Above: The general setup of a neusis construction.  
The point G is movable.

The reader will note however that the methods below do not precisely fit this mold. For this reason, the term neusis-like (coined by Fried and Unguru, in [2001]), seems an apt choice. Nicomedes’ solution, for example, is built around a curve that is defined in such a way that all the segments between it and a straight line are equal to some given segment. Hence using the curve so generated (which Nicomedes calls a conchoid line) in the usual method of neusis would not be particularly illuminating on its own. But the method of generating it was clearly inspired by the method of neusis; thus, Fried and Unguru’s term neusis-like.

Biographical Note: Plato (ca 428 BCE–ca 348 BCE). Best known as a philosopher, mathematics features prominently in several of his dialogues, e.g. Theaetaetus, Meno, and Republic. It is unlikely that the device and method described here are actually due to Plato, for these reasons: we do not know of Plato actively engaging in mathematical research (though associates of his did), and the mechanical nature of the construction seems at odds with Plato’s philosophy of mathematics. It is somewhat strange to think that Eutocius would not have found the attribution dubious, unless there is another Plato who did mathematics (about whom we know nothing beyond this reference), or the attribution is meant in an ironic fashion. Read more about Plato at MacTutor.

(Heiberg 56.14) Given two straight lines, to find two mean proportionals in continuous proportion.

For let the two given straight lines be ΑΒ, ΒΓ, at right angles to each other, between which it is required to find two mean proportionals. Let them be extended in straight lines to Δ and Ε, and let a right angle ΖΗΘ be constructed; and in one side ΖΗ, let there be a cross-bar ΚΛ, being in a groove in the side ΖΗ, so that when moved, it remains parallel to ΗΘ.

This will be, if also another cross-bar ΘΜ is imagined attached to ΘΗ, parallel to ΖΗ: for having fashioned dove-tail^[7] shaped grooves in the top sides of ΖΗ and ΘΜ, and having attached knobs to ΚΛ in the aforementioned grooves, the movement of ΚΛ will always be parallel to ΗΘ^[8].

So having fashioned these things,^[9] let one side of the angle, say ΗΘ, be placed so that it touches Γ; 

Above: Eutocius’ Diagram for Plato’s Device, First Constraint. The apparatus he describes is shown in red.
The points K and H are movable.

At this point, the apparatus has one constraint in terms of its placement over the figure: The side ΗΘ is placed so that Γ is on it. This still permits a considerable amount of freedom for the apparatus to be moved. It is worth mentioning that the scale of the instrument needs to be reasonable given the scale of the to lines AB and ΒΓ. However, the sides ΑΒ and ΒΓ could easily be scaled up or down to accommodate the use of a particular size tool.

and let both the angle and the side ΚΛ be moved until the point Η is on the side ΒΔ and the side ΗΘ is touching Γ, 

Above: Eutocius’ Diagram for Plato’s Device, Second Constraint.
The points Z, K, and H are movable.

This is the second constraint introduced. No longer can the apparatus rotate freely about Γ. It is limited in the extents of its motion, since now the point Η must be somewhere on the line ΒΔ. At one extreme, when Η is at Β, the entire apparatus is oriented with the bottom edge ΗΘ coincident with ΒΓ, and the left edge ΖΗ coincident with ΑΒ. At the other extreme, it is situated somewhat diagonally, with the bottom edge of the apparatus ΗΘ connecting Δ and Γ.

and the bar ΚΛ touches the side ΒΕ at Κ, 

Above: Eutocius’ Diagram for Plato’s Device, Third Constraint.
The points Z and H are movable.

Now the point Κ must be on the line segment ΒΕ. This links the motion of the second cross-bar ΚΛ with the bottom edge of the apparatus ΗΘ. At one extreme, we have points Η and Κ coinciding at Β, and the second cross-bar ΚΛ is coincident with the bottom of the apparatus ΗΘ and with the line ΒΓ. At the other extreme, K coincides with E, while H coincides with Δ.

and the remaining side at Α, so that, as the diagram shows,^[10] the right angle has position as ΓΔΕ, and the bar ΚΛ has the position which ΕΑ has: 

Above: Eutocius’ Diagram for Plato’s Device, Fourth Constraint.
This is the finished configuration.

The apparatus is now fully constrained. At this point, Δ and E are dynamically redefined. The positions of Δ and Ε were essentially arbitrary: they needed only to be in straight line with ΑΒ and ΒΓ, respectively. For this reason, we can redefine Δ and Ε to be where Η and Κ are, respectively. This sort of game is common enough in Greek mathematics.

for having arranged everything thus, the prescribed thing will be. For since the angles at Δ and Ε are right, 

\begin{equation} \tag{8} \text{ΓB : BΔ :: ΔB : BE :: EB : BA.} \end{equation}

We can view this result as coming from two applications of Elements VI.13. If we imagine two semicircles ΑΕΔ and ΕΔΓ, then in each we can apply VI.13. For ΑΕΔ, this yields  \begin{equation} \tag{9} \text{ ΔB : BE :: EB : BA,} \end{equation} and for ΕΔΓ, it yields \begin{equation} \tag{10} \text{ΓB : BΔ :: ΔB : BE.} \end{equation} Hence the claim.

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[7] Netz takes this word as “axe-shaped”; I’ve chosen “dove-tail” because of the connotation in woodworking.

[8]  Netz takes this differently, as the knobs Κ and Λ being parallel to the ruler HΘ, on the grounds that the manuscripts have a plural article rather than Heiberg’s singular one. On this point, I’m more inclined to agree with Heiberg. 

[9]  i.e. the two rods, the grooves, etc. 

[10]  Eutocius is referring to his diagram here, which is a sub-set of Eutocius' Diagram for Plato's Device, Fourth Constraint (the finished configuration). Eutocius’ figure has only the points Α, Β, Γ, Δ, Ε, and the lines ΕΓ, ΑΔ, ΑΕ, ΕΔ, and ΔΓ. The manuscripts apparently have a second figure showing the apparatus, and a complete study of these diagrams is a project for future work. 

Biographical note: Heron (or Hero) of Alexandria (ca 65 CE–ca 125 CE). Probably most familiar to modern students because of Heron’s formula for the area of a triangle in terms of its sides and semiperimeter. Heron wrote on a wide range of topics, from plane and solid geometry to optics, surveying, and mechanics; not all of his works survive. Read more about Heron at MacTutor.

(Heiberg 58.17) Let there be two given straight lines ΑΒ and ΒΓ, between which it is necessary to find two mean proportionals. Let them be placed, so that they contain a right angle at Β, and let the parallelogram ΒΔ be completed,^[11] and let ΑΓ and ΒΔ be joined.

[It is clear that they (ΑΓ and ΒΔ) are equal and they bisect each other (at E): for the circle being drawn around one of them will also pass through the extremities of the other, since the parallelogram is right-angled.]^[12]

Let ΔΓ and ΔΑ be projected [to Ζ, Η], and let a straightedge^[13] be conceived, as ΖΒΗ, being moved around some knob that remains fixed at Β, and moving, until it cuts the two segments from E equally, that is, ΕΗ and ΕΖ.

And let it (the straightedge) be conceived, having as position ΖΒΗ, and having cut equally, as was said before, the segments ΕΗ and ΕΖ. 

Above: Heron’s Diagram. I have added the circles: by moving the point Η, the radii of the circles are adjusted. When they coincide, the lines ΗΕ and ΕΖ are equal. It should be noted that this isn’t exactly what Heron is doing. The text refers to moving the ruler ΖΒΗ around Β, and hence the position of Η depends on the position of the ruler. In the diagram, I have made the ruler’s position depend on the position of Η instead. This makes the figure a bit easier to manipulate, and also seemed easier to implement in GeoGebra.

 

Let ΕΘ be drawn perpendicular to ΓΔ from Ε: it is clear that it bisects ΓΔ. So since it bisects ΓΔ at Θ, and ΓΖ is added, we have, by Elements II.6,  \begin{equation} \tag{11} \text{rect.(ΔZΓ) + sq.(ΓΘ) = sq.(ΘZ),} \end{equation}

where the notation \(\text{rect.(ΔZΓ)}\) denotes the rectangle with sides \(\text{ΔZ}\) and \(\text{ZΓ}.\)

Let the square on ΕΘ be added in common: therefore \begin{equation} \tag{12} \text{rect.(ΔZΓ) + sq.(ΓΘ) + sq.(ΘE) = sq.(ZΘ) + sq.(ΘE).} \end{equation}

Therefore by the Pythagorean Theorem [Elements I.47] \begin{equation} \tag{13} \text{rect.(ΔZΓ) + sq.(ΓE)  = sq.(EZ).} \end{equation}

Similarly it will be shown that \begin{equation} \tag{14} \text{rect.(ΔHA) + sq.(AE)  = sq.(EH).  } \end{equation}

And \begin{equation} \tag{15} \text{AE = EΓ, and HE = EZ,} \end{equation}

therefore \begin{equation} \tag{16} \text{rect.(ΔZΓ) = rect.(ΔHA). } \end{equation}

[But if the rectangle contained by the extremes is equal to the rectangle contained by the means, then the four straight lines are in proportion:] therefore, by Elements VI.16 \begin{equation} \tag{17} \text{ZΔ : ΔH :: AH : ΓZ} \end{equation}

The bracketed passage is a reference (and abbreviated quotation) of Elements VI.16. The Greek in Eutocius’ text is ἐὰν δὲ τὸ ὑπὸ τῶν ἄκρων ἴσον ᾖ τῷ ὑπὸ τῶν μέσων, αἱ τέσσαρες εὐθεῖαι ἀνάλογόν εἰσιν whereas the Greek text from Euclid is κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται.

But \begin{equation} \tag{18} \text{ZΔ : ΔH :: ZΓ : ΓB :: BA : AH.} \end{equation}

[Using Elements VI.2, for ΓΒ has been drawn parallel to one side of the triangle ΖΔΗ, namely ΔΗ, and ΑΒ has been drawn parallel to ΔΖ.] Therefore \begin{equation} \tag{19} \text{BA : AH :: AH : ΓZ :: ΓZ : ΓB.} \end{equation}

Therefore AH and ΓΖ are mean proportionals between ΑΒ and ΒΓ [which was required to find]. 

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[11] Really, this is a rectangle. 

[12] This bracketed section is an insertion. In the Greek, ΑΓ, ΒΔ, and Ε are not explicitly referred to (only with pronouns). For clarity, I have added them here in parentheses. 

[13] I’ve chosen to translate τὸ κανόνιον here as straightedge, rather than cross-bar like I did with Plato’s construction, since Hero is not defining a sort of mechanical instrument so much as he is putting an ordinary line into a certain position. 

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. 

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[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. 

Biographical note: Apollonius of Perga (ca 262 BCE–ca 190 BCE) is perhaps most well known for his treatise on conic sections, the Conics, wherein the terms parabola, hyperbola, and ellipse were used for the first time (the curves themselves were studied earlier, but by different names). The work significantly extends previous knowledge of conic sections, and contains many original results. His other achievements include results in mathematical astronomy and a method for approximating \(\pi.\) Some of his works do not survive, including the last book (Book VIII) of the Conics. Read more about Apollonius at MacTutor.

(Heiberg 64.16) Let there be two given straight lines ΒΑ and ΑΓ, between which it is necessary to find two mean proportionals, ΒΑΓ containing a right angle at Α. And with center Β and radius ΑΓ, let the perimeter of a circle ΚΘΛ be drawn. Again, with center Γ and radius ΑΒ, let the perimeter of a circle ΜΘΝ be drawn, and let it cut ΚΘΛ at Θ, and let ΘΑ, ΘΒ, and ΘΓ be joined.

Above: Apollonius’ Diagram. Netz notes that the manuscripts have arcs for the circles instead of full circles (ΝΘΜ and ΚΛΘ). I have added the red circle. It has center Ξ and variable radius, which here is controlled by the slider bar. It is desired that the blue line ΔΕ pass through Θ: this will occur by adjusting the slider bar to an appropriate value.

 

Therefore ΒΓ is a parallelogram, and ΘΑ is its diameter. Let ΘΑ be bisected at Ξ. With center Ξ, let a circle be drawn, cutting the segments ΑΒ and ΑΓ (having been extended) at Δ and Ε, so that ΔΕ is in straight line with Θ. This will happen^[16] when a ruler is moved around Θ, cutting ΑΔ and ΑΕ, and being guided until the segments from Ξ to Δ and Ε become equal.

For when this has occurred, the thing being sought will be: for the construction is the same as those given by Heron and Philon, and it is clear that the same proof will suffice. 

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[16] Aorist optative + ἄν: Smyth 1828. 

Biographical note: Diocles (ca 240 BCE–ca 180 BCE). In addition to his cube duplication method using the cissoid curve, Diocles also was the first to prove the focal property of a parabola. This result is of immense practical value in the modern world, for everything from flashlights to satellite dishes. On Burning Mirrors, other than the loose excerpt provided here by Eutocius, survives only in Arabic. Read more about Diocles at MacTutor.

(Heiberg 66.9) In a circle, let two diameters ΑΒ and ΓΔ be drawn at right angles, and let two equal arcs ΕΒ and ΒΖ be cut off on each side of B, and through Ζ, let ΖΗ be drawn parallel to ΑΒ, and let ΔΕ be joined. I say that ΖΗ and ΗΔ are two mean proportionals between ΓΗ and ΗΘ. 

Above: This diagram is not present in the manuscripts, but I include it for clarity. This is a sub-diagram of the sketch below.

 

For let ΕΚ be drawn through Ε parallel to ΑΒ: therefore ΕΚ is equal to ΖΗ, and ΚΓ is equal to ΗΔ. For this will be clear, once the straight lines from Λ to Ε and Ζ are drawn: for the angles ΓΛΕ and ΖΛΔ are made equal and the angles at Κ and Η are right. Therefore all the sides and angles are equal^[17] since ΛΕ is equal to ΛΖ: and therefore the remaining segment ΓΚ is equal to the remaining segment ΗΔ. So since [by Elements VI.2], \begin{equation} \tag{25} \text{ΔK : KE = ΔH : HΘ} \end{equation} and \begin{equation} \tag {26} \text{ΔK : KE = EK : KΓ} \end{equation}

[by Elements VI.8 since angle ΓΕΔ is right], therefore ΕΚ is a mean proportional between ΔΚ and ΚΓ. Therefore \begin{equation} \tag{27} \text{ΔK : KE = EK : KΓ = ΔH : HΘ.} \end{equation}Also, ΔΚ is equal to ΓΗ, ΚΕ is equal to ΖΗ, and ΚΓ is equal to ΗΔ: therefore \begin{equation} \tag{28} \text{ΓH : HZ = ZH : HΔ = ΔH : HΘ.} \end{equation}If two [different] equal arcs ΜΒ and ΒΝ are chosen on each side of Β, and through Ν parallel to ΑΒ the line ΝΞ is drawn, and ΔΜ is joined, then again, ΝΞ and ΞΔ will be two mean proportionals between ΓΞ and ΞΟ.

So similarly, producing successive parallels between Β and Δ, and cutting arcs from Β in the direction of Γ, these arcs equal to the ones cut off by the parallels, and joining lines from Δ to the points so created, such as ΔΕ and ΔΜ, the parallels drawn between Β and Δ will be cut [by the lines such as ΔΕ and ΔΜ] at some points, which in the diagram are the points O and Θ. Through these points, placing a ruler joining them, we will then have a certain curve drawn in the circle, on which, if some point be chosen at random, and through this point a line be drawn parallel to ΛΒ, the line joined (from the point on the circle cut off and the diameter) and the one it cuts off of the diameter (on the Δ side) will be [two^[18]] mean proportionals between the cut off part of the diameter (on the Γ side) and the and the part of the [parallel] from the point on the curve to the diameter ΓΔ.^[19]

Above: Diocles’ First Diagram. Diocles has defined a series of points, such as Θ and Ο. The locus described is shown in red, and is a segment of the famous Cissoid of Diocese. Point E generates the locus.

Note: Eutocius changes to a new diagram here, which is labelled differently than the previous one.

Having prepared these things in advance, let there be two straight lines Α and Β, between which it is necessary to find two mean proportionals. And let there be a circle, in which two diameters ΓΔ and ΕΖ are drawn at right angles. And in the circle, let the curve ΔΘΖ be drawn through the successive points, as has been said before. Let it be made [using Elements VI.12] that \begin{equation} \tag{29} \text{A : B = ΓH : HK,} \end{equation}and having joined and extended ΓΚ, let it cut the curved line at Θ. And through Θ, let ΛΜ be drawn parallel to ΕΖ. 

Therefore ΜΛ and ΛΔ are two mean proportionals between ΓΛ and ΛΘ [by above]. And since [by Elements VI.2] \begin{equation} \tag{30} \text{ΓΛ : ΛΘ = ΓH : HK} \end{equation}and \begin{equation} \tag{31} \text{ΓH : HK = A : B,} \end{equation}if we insert means Ν and Ξ between Α and B in the same ratio as ΓΛ, ΛΜ, ΛΔ, and ΛΘ, then Ν and Ξ will have been taken as two mean proportionals: the very thing it was necessary to find. 

Diocles’ Second Diagram. The intersection point Θ was difficult to define in GeoGebra: since GeoGebra cannot find the intersection of a locus and a line, I had to do a conic approximation of the cissoid. The intersection of this conic and the line ΛΜ is defined as Θ, which is sufficiently close to the intersection point of ΛΜ and the cissoid for visual and computational purposes. The ratios referenced are numerically computed at the side, to demonstrate approximate accuracy (rounded to two decimal places, for cleanliness).

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[17] In Greek, πάντα ἄρα πᾶσιν, lit. “each to each.” Equality is not explicitly mentioned, but is implied by the previous sentences. Compare the language of Euclid. 

[18] The number two is not explicitly stated here, but it is clear from context. 

[19] This passage was a bit difficult to render in a readable way that also was faithful to the Greek. What Eutocius is trying to say is that, as in the previous paragraph, we can continue to generate a series of points like Ο and Ξ. The ruler is used to connect these successive points, producing, as it were, a piecewise linear approximation of a curve. The full curve would, of course, require an infinite number of applications of this procedure. The property of certain lines being mean proportionals between others is true regardless of the point chosen on the curve, which has been demonstrated by the previous paragraph for O and Ξ. In some sense this is backwards of the usual arrangement of a Greek proof: normally the general claim is stated first, as the protasis, and the specific claim (with reference to a specific diagram) is referenced second. 

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. 

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[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.”

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 \begin{equation} \tag{48} \text{ΚΘ : ΘΗ :: ΜΒ : ΒΛ,} \end{equation}

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

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

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

Biographical note: Menaechmus (ca 380 BCE–ca 320 BCE) is responsible for discovering the conic sections during his investigation of how to solve the cube duplication problem. The 10th century Greek encyclopedia, the Suda, also has a reference apparently to him, suggesting he also worked in philosophy. Read more about Menaechmus at MacTutor.

The Greek view of conic sections is considerably different than our own, and so I think it wise to provide a brief introduction. I personally am always surprised when some of my students have no idea why the conic sections are called “conic sections”: in their view, parabolas are synonymous with equations like \begin{equation}y = x^{2}. \end{equation} But we use the term we do precisely because of the Greek mathematical heritage; namely, the conic sections are the results of cutting a cone with a plane. 

Greek mathematics actually had two ways of defining the conic sections, corresponding to two different ways to define a cone. The first, likely due to Menaechmus himself, and still in use by Euclid and Archimedes, was that the three types of cones came from rotating the different kinds of right triangles about one of their legs. The three types of triangles are:

1. A triangle where the adjacent angle, \(\angle ACB,\) is half a right angle (so the triangle is also isosceles):

2. A triangle where the adjacent angle, \(\angle ACB,\) is greater than half a right angle:

3. A triangle where the adjacent angle, \(\angle ACB,\) is less than half a right angle:

Hence the three conic sections came from cutting these three different kinds of cones. In each situation the cutting plane was aligned at right angles to the “side” of the cone, where here “side” refers to the hypotenuse of the generating triangle. The three kinds of sections are:

1. The section of a right-angled cone (our “parabola”). Here the cutting plane is perpendicular to one side of the cone, and parallel to the other. Move point D to change the location of the cut.

2. The section of an obtuse-angled cone (half of our “hyperbola”). The obtuse angle here is the angle at the vertex. Apollonius did consider the "other half" of the cone, but as a different object. For our purposes, it is sufficient to consider only one "half" of the cone, as presented here, giving us only one of the two branches in a “modern” hyperbola. Move point D to change the location of the cut:

3. The section of an acute-angled cone (our “ellipse”). Move point D to change the location of the cut:

Under this view the circle would not have been considered a conic section in the same sense, because the only way to get one from a cone was to have the cutting plane parallel to the base of the cone, not perpendicular to an edge. 

The second view, due to Apollonius and attested to in the Conics, redefines cones in such a way that only one definition is necessary. Apollonius defines a cone by first considering a circle (the “base” and a point not in the plane of that circle (the “vertex”). For each point on the circle, draw the line from that point to the vertex. Hence Apollonius’ cones can be “lopsided,” whereas the Euclidean view does not permit that. Apollonius calls the “lopsided” variety oblique. When the line drawn perpendicular to the plane of the circle, through the circle’s center, passes through the vertex, he calls the cone right (but note: right to Apollonius is not the same thing as right-angled to Euclid). In this way, Apollonius can form all three conic sections (which he dubs parabola, hyperbola, and ellipse) by cutting a single cone in three different ways. 

Since each conic section is formed by the intersection of a cone and a plane, we can restrict our attention to this cutting plane only, and the conic section within it. This will more easily allow us to state the details of the basic properties, or symptoms of each section. This is of course different than proving these basic properties: for this, Apollonius needs to keep the sections and the cone in view. Conics I.11 and I.12 contain the details of Apollonius’ proofs for the parabola and hyperbola, respectively. The case of the ellipse is not necessary for what follows, and so I omit it.

For the parabola, we start with the diameter, which in many cases resembles what the modern reader might call the axis. A segment of the diameter is called an abscissa, from Latin abscindere, to cut or separate. A line drawn from the diameter to the section is called an ordinate, and for our purposes these ordinates are perpendicular to the diameter.^[27] We then consider a segment, called the parameter, so that the square on the ordinate is equal to the rectangle contained by the parameter and the abscissa. 

Above: In this figure, the diameter is AB, the abscissa is AC, and the ordinate is CD.
The parameter is AE, and satisfies sq.(CD) = rect.(AC, AE). Point C is movable.

 

The case of the hyperbola, as far as Menaechmus’ solutions are concerned, is a bit different than what Apollonius first defines in Conics I.12. Apollonius does not deal with the concept of asymptotes until Book II of the Conics, and hence his symptom of the hyperbola has to be defined in a different manner (see Stoudt’s article [2015] for more details on Apollonius’ approach in I.12). Menaechmus uses the property that if a hyperbola has perpendicular asymptotes, then the rectangle contained by perpendiculars drawn from a point on the hyperbola to each of the asymptotes has a constant area. Apollonius gives the details in Conics II.12; but it should be noted that for Apollonius, this is not the symptom of a hyperbola, but rather a property that follows from it.^[28]

Above: In this figure, the lines ACF and AEB are the asymptotes. The two constant lengths are AF and AE.
For a given point D, the lines CD and BD are such that rect.(BD, CD) = rect.(AF, AE). Point D is movable.

---

[27] In Apollonius, this is not necessarily the case, but for Menaechmus and what follows, we can assume it is. As usual, see the Conics for full details! 

[28] In his commentary on the Conics, Eutocius uses makes the distinction between the principal properties, τὰ ἀρχικὰ συμπτώματα, and the resulting properties, τὰ παρακολουθήματα συμπτώματα. For more details on the distinction, see my dissertation [2010]. 

(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.

In another way. 

Analysis. Let ΑΒ and ΒΓ be two given straight lines, at right angles to each other, and let the means ΔΒ and BE between them supposed as being found, so that \begin{equation} \tag{84} \text{ΓΒ : ΒΔ = ΒΔ : ΒΕ = ΒΕ : ΒΑ, } \end{equation}  and let ΔΖ and ΕΖ be drawn at right angles. 

Since \begin{equation} \tag{85} \text{ ΓΒ : ΒΔ = ΔΒ : ΒΕ, } \end{equation} therefore \begin{equation} \tag{86} \text{ rect.(ΓΒΕ) = rect.(a given segment, ΒΕ) = sq.(ΒΔ) = sq.(ΕΖ).} \end{equation}

Since the rectangle contained by the given line and ΒΕ is equal to the square on ΕΖ, therefore the point Ζ is on a parabola whose axis is the line ΒΕ.

Again, since \begin{equation} \tag{87} \text{ ΑΒ : ΒΕ = ΒΕ : ΒΔ,} \end{equation}therefore \begin{equation} \tag{88} \text{ rect.(ΑΒΔ) = rect.(a given segment, ΒΔ) = sq.(ΕΒ) = sq.(ΔΖ),} \end{equation}therefore Ζ is on a parabola whose axis is the line ΒΔ. But it is also on the other parabola whose axis is ΒE: therefore Ζ is given. So too then are ΖΔ and ΖΕ: therefore Δ and Ε are given. 

Above: Menaechmus’ Second Diagram. The construction of this diagram was similar to that for Menaechmus’ first solution. I show here only the completed one. Unlike the previous diagram, the two given magnitudes are placed on the main figure (rather than to the side). Moving the points Α and Γ will adjust the ratio of ΑΒ to ΒΓ, which by default is 2:1.

 

Synthesis. This will be synthesized thusly. For let two given straight lines ΑΒ and ΒΓ be at right angles to one another, and let these straight lines be produced indefinitely from Β. With axis ΒΕ, let a parabola be drawn, so that the ordinates to ΒΕ are equal in square to the areas applied to ΒΓ.

Again, let a parabola be drawn with axis ΔΒ, so that the ordinates are equal in square to the areas applied to ΑΒ. Therefore the parabolas will intersect one another. Let them intersect at Ζ, and from Ζ, let ΖΔ and ΖΕ be drawn perpendicularly. 

So since ΖΕ (that is, ΔΒ) has been drawn ordinatewise to a parabola, therefore, by Conics I.11, for the blue parabola,  \begin{equation} \tag{89} \text{ rect.(ΓΒΕ) = sq.(ΒΔ).  } \end{equation} Therefore \begin{equation} \tag{90} \text{ ΓΒ : ΒΔ = ΔΒ : ΒΕ.} \end{equation} Again, since ΖΔ (that is, ΕΒ) has been drawn ordinatewise to a parabola, therefore, by Conics I.11, for the red parabola, \begin{equation} \tag{91} \text{rect.(ΔΒΑ) = sq.(ΕΒ). } \end{equation} Therefore \begin{equation} \tag{92} \text{ ΔΒ : ΒΕ = ΒΕ : ΒΑ, } \end{equation} and so \begin{equation} \tag{93} \text{ ΓΒ : ΒΔ = ΒΔ : ΒΕ = ΕΒ : ΒΑ,} \end{equation} the very thing it was required to find. 

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. 

Biographical note: Eratosthenes of Cyrene (ca 276 BCE–ca 194 BCE). Eratosthenes is known for a variety of mathematical topics, including the so-called sieve (related to prime number theory) and a relatively accurate estimate for the circumference of the Earth. He was a polymath but not “best” in any field, and supposedly was nicknamed “Beta” for this reason. Read more about Eratosthenes at MacTutor.

(Heiberg 84.13) Let there be two unequal straight lines, ΑΕ and ΔΘ, between which it is required to find two means in continuous proportion. Let ΑΕ be placed at right angles to some straight line ΕΘ, and on ΕΘ, let three parallelograms be set up, ΑΖ, ΖΙ, and ΙΘ, and let the diameters ΑΖ, ΛΗ, and ΙΘ be drawn in each: hence these diameters will be parallel.

Above: Figure 13.1: Eratosthenes’ First Diagram. This diagram is static. The point F is unlabeled in the manuscripts.
In the next diagram, it is movable, and I label it here for continuity.

 

When the middle parallelogram ΖΙ remains fixed, let the parallelogram ΑΖ be slid along the top surface of the middle one, and ΙΘ along the bottom surface of the middle one, as in the second diagram, until the points Α, Β, Γ, and Δ are in a straight line. Let the line through the points Α, Β, Γ, and Δ be drawn, and let it intersect ΕΘ (having been extended) at Κ.

Above: Eratosthenes’ Second Diagram. The objective with this diagram is to have the points Β and Γ on the red line ΑΔ. This is accomplished by moving the doors (by moving points Α and F). The point Δ is also moveable; adjusting it changes the ratio ΑΕ : ΔΘ. By default, ΑΕ is twice ΔΘ. I have added a few labels to this diagram. First, the point F, which is unlabelled in the manuscripts. Second, the manuscripts only label the points Λ and Ι (and their counterparts, Λ2 and Ι2) in the first diagram. For completeness, I include them here; when the doors are moved to their starting positions, the Λ points coincide and the Ι points coincide. Similarly for the points Ζ and Η: while they are also labeled in the second diagram, their counterparts Z₁ and Η₁ are not. (Note that Eratosthenes’ First Diagram corresponds to a static version of this diagram.)

 

Indeed, in the parallels ΑΕ and ΖΒ, we have by Elements VI.4, \begin{equation} \tag{97} \text{ ΑΚ : ΚΒ = ΕΚ : ΚΖ;} \end{equation}and in the parallels ΑΖ and ΒΗ,  \begin{equation} \tag{98} \text{ΑΚ : ΚΒ = ΖΚ : ΚΗ. } \end{equation}Therefore [by Elements V.11]  \begin{equation} \tag{99} \text{ ΑΚ : ΚΒ = ΕΚ : ΚΖ = ΚΖ : ΚΗ.} \end{equation}Again, in the parallels ΒΖ and ΓΗ, \begin{equation} \tag{100} \text{ ΒΚ : ΚΓ = ZK : ΚΗ;} \end{equation}and in the parallels ΒΗ and ΓΘ, \begin{equation} \tag{101} \text{ ΒΚ : ΚΓ = ΗΚ : ΚΘ.} \end{equation}Therefore \begin{equation} \tag{102} \text{ΒΚ : ΚΓ = ΖΚ : ΚΗ = ΗΚ : ΚΘ. } \end{equation}But \begin{equation} \tag{103} \text{ΖΚ : ΚΗ = ΕΚ : ΚΖ; } \end{equation}and therefore \begin{equation} \tag{104} \text{ ΕΚ : ΚΖ = ΖΚ : ΗΚ = ΗΚ : ΚΘ.} \end{equation}

But \begin{equation} \tag{105} \text{ΕΚ : ΚΖ = ΑΕ : ΒΖ,} \end{equation}and \begin{equation} \tag{106} \text{ ΖΚ : ΚΗ = ΒΖ : ΓΗ,} \end{equation}and \begin{equation} \tag{107} \text{ ΗΚ : ΚΘ = ΓΗ : ΔΘ;} \end{equation}therefore \begin{equation} \tag{108} \text{ ΑΕ : ΒΖ = ΒΖ : ΓΗ = ΓΗ : ΔΘ.} \end{equation}Therefore both ΒΖ and ΓΗ have been found as two mean proportionals between ΑΕ and ΔΘ. 

Biographical note: Nicomedes (ca 280 BCE–ca 210 BCE). We know very little about Nicomedes, except that he used the conchoid curve to solve both the problem of trisecting an angle and of duplicating a cube. He apparently also worked on the problem of squaring the circle. Read more about Nicomedes at MacTutor.

(Heiberg 98.2) Nicomedes, in his work On Conchoid Lines, writes about the construction of a mechanism satisfying the same requirement [of finding two mean proportionals]. From the book, he seems to have been especially proud of himself while ridiculing Eratosthenes’ solutions as awkward and lacking any geometrical skill. So as to not exclude anyone who dealt with this problem, and to compare with Eratosthenes, we add it to those already mentioned. He writes essentially^[29] the following. 

It is necessary to imagine two rulers joined at right angles to each other, so that they remain in one surface, just as ΑΒ and ΓΔ are. And in the ruler ΑΒ, imagine a dove-tail shaped groove, through which a peg is able to slide; and in the ruler ΓΔ, near the Δ side, and in the line through the middle of it, a small cylinder attached to the ruler and extending above it a tad. And in another ruler EZ, after some small distance from the endpoint Ζ, imagine it having a cut, as ΗΘ, through which the cylinder at Δ is able to move; and in a small distance from the endpoint Ε, a round hole, through which a small bolt is passed and attached to the peg in the dove-tailed shaped groove ΑΒ. And so fitting the ruler EZ to the cut ΗΘ with the cylinder Δ, and at the hole Ε, fitting it to the hole Ε with the bolt attached to the sliding knob, if one takes hold of the the extremity Κ of this ruler and moves it, first towards Α, and then towards Β, the point Ε will always remain on the ruler ΑΒ, and the cut ΗΘ will be moved about the cylinder near Δ, with the imaginary mid-line of ΗΘ always passing through the axis of the cylinder near Δ, and the excess ΕΚ remaining constant. 

Above: Nicomedes’ First Diagram. I have simplified this diagram slightly,
removing the physical parts described, such as the groove ΗΘ.

 

If we imagine a pencil at Κ, touching the base-plane, it will draw a certain curve, such as ΜΛΝ, which Nicomedes calls the first conchoid line, the length of ΕΚ its interval, and Δ its pole. With respect to this curve, he shows the result that it converges to the ruler ΑΒ, and if some straight line is drawn between the curve and the ruler ΑΒ, this straight line always cuts the curve. The first of these properties is easily understood by means of a different diagram. For imagining a ruler ΑΒ, pole Γ, interval ΔΕ, and conchoid line ΖΕΗ, let the lines from Γ, such as ΓΘ and ΓΖ, be drawn to meet the conchoid line [at Θ and Ζ]: with clearly the resulting lines ΚΘ and ΛΖ being equal.

I say that the perpendicular ΖΜ is less than the perpendicular ΘΝ.^[30] For since the angle ΜΛΓ is larger than the angle ΝΚΓ, the remaining angle, ΜΛΖ, being less than two right angles, is less than the remaining angle ΝΚΘ. Because of this, the angle at Ζ will be larger than the angle at Θ, with the angles at Μ and Ν being right. And if we make the angle ΜΖΞ equal to the angle at Θ, then ΚΘ, that is ΛΖ, will have to ΘΝ the same ratio which ΞΖ has to ΖM: \begin{equation} \tag{109} \text{ΚΘ : ΘΝ :: KΘ : ΛΖ :: ΞΖ : ΖΜ, } \end{equation}so that \begin{equation} \tag{110} \text{ΖΛ : ΘΝ < ΖΛ : ΖΜ, } \end{equation}and because of this, ΘΝ is larger than ΖΜ. 

This could use a bit more explanation. We know that ΖΛ is larger than ΖΞ. Hence \begin{equation} \tag{111} \text{ΖΞ : ΖΜ < ΖΛ : ΖΜ. } \end{equation}Since by construction, triangle ΘΝΚ is similar to triangle ΖΜΞ, we have that \begin{equation} \tag{112} \text{ΖΞ : ΖΜ :: ΘΚ : ΘΝ.  } \end{equation}Hence \begin{equation} \tag{113} \text{ΘΚ : ΘΝ < ΖΛ : ΖΜ.  } \end{equation}But by definition of the conchoid line, ΘΚ and ΖΛ are equal; hence \begin{equation} \tag{114} \text{ΖΛ : ΘΝ < ΖΛ : ΖΜ.} \end{equation}

The second property was that any straight line drawn between the ruler ΑΒ and the conchoid line cuts the conchoid line. And this becomes familiar:^[31] for the drawn line is either parallel to ΑΒ or not.

Let it first be parallel, as ΖΗΘ, and let it be supposed, that \begin{equation} \tag{115} \text{ΔΗ : ΗΓ :: ΔΕ : some other straight line Κ.} \end{equation}With center Γ and radius Κ, let the perimeter of a circle being drawn cut ΖΗ at Η, and let ΓΖ be joined.

Above: Nicomedes’ Second Diagram. Netz notes that some of the manuscripts have Δ and Ν coinciding. Not stated is that Θ should be closer to the “center” of the conchoid line, by which I mean the vertical axis of symmetry, than Ζ: for otherwise, the role of Θ and Ζ would reverse.

 

Therefore, by Elements VI.2 we have \begin{equation} \tag{116} \text{ΔΗ : ΗΓ = ΛΖ : ΖΓ. } \end{equation}But \begin{equation} \tag{117} \text{ΔΗ : ΗΓ :: ΔΕ : Κ :: ΔΕ : ΓΖ [by definition of K],} \end{equation}therefore ΔΕ is equal to ΛΖ. But this is impossible, for it is necessary that Ζ be on the conchoid line. 

Above: Nicomedes’ Third Diagram. Netz notes that the manuscripts only have an arc of a circle, representing the conchoid line. The actual circle (drawn here as a blue arc) is omitted. By definition of Ζ, the arc passes through Ζ. What is not obvious, at least not yet, is that Ζ must therefore also be on the conchoid line. The point Θ is not necessarily on the conchoid line: it’s merely a third point along the line ΖΗ. However, if we define Θ as the other point where the circle cuts the line ΖΗΘ, then it plays a symmetric role to Ζ. Also for Ν: it is not necessarily on either the conchoid line or the circle, but the argument that Nicomedes (through Eutocius) gives shows that the line ΜΗΝ would, if extended, meet both.

I wonder if γάρ (“for”) has been corrupted from ἄρα. Netz notes that Heiberg disliked the proof, because nowhere did Eutocius suppose that Ζ wasn’t on the conchoid line. I suppose to some extent we should look past this, since the flow of contradiction proofs such as this is, as I said above, was familiar. The essential argument here is that: ΛΖ is equal to ΔΕ.  ΔΕ is an excess by which the moving ruler extends past the horizontal fixed ruler.  Since the curve is formed by the motion of the moving ruler, and the construction of the mechanism forces these excesses to remain equal, the segment ΛΖ is itself one of these excesses. Hence:  The point Ζ is in fact on the conchoid line.

Hence let the drawn line be not parallel, as ΜΗΝ. Let ΖΗ be drawn through Η parallel to ΑΒ. Therefore ΖΗ will intersect the curve: so clearly ΜΝ will too.  

Probably an appeal to the continuity of the conchoid line and the symmetry of the figure. Since Γ is the pole of this conchoid line, the line ΓΕ is the axis of symmetry for the conchoid line. So the mirror image of Ζ also passes through both the circle and the conchoid line, say, at Θ’. Thus we have a continuous line from Ε to Θ’, and straight lines ΕΗ and ΗΘ’. Since ΗΘ extends to ΗΘ’, we have “trapped” ΗΝ inside a sort of quasi-triangle. Hence ΗΝ, if extended, also must intersect the conchoid line (and, if extended further, would also intersect the circle). It would have perhaps been easier had Eutocius (or Nicomedes) put Ν on the other side of the axis: then we could directly appeal to ΖΗ and ΗΕ.

With these results following from the mechanism, its usefulness to the given problem is shown thus. Given an angle Α and a point Γ outside, to draw ΓΗ and make ΚΗ equal to a given line. Let the perpendicular ΓΘ be drawn from the point Γ to ΑΒ, and let it be extended. Let ΔΘ be equal to the given line. With pole Γ, radius ΔΘ, and ruler ΑΒ, let the first conchoid line ΕΔΖ be drawn. Therefore it intersects ΑΗ by the previous result. Let it intersect at Η, and let ΓΗ be joined: therefore ΚΗ is equal to the given line. 

Above: Nicomedes’ Fourth Diagram. The angle is adjustable by moving the point Η along the conchoid line.

 

Having proved this, let two segments ΓΛ and ΛΑ have been given at right angles to one another. It is necessary to find two mean proportionals in succession between them. Let the parallelogram ΑΒΓΛ be completed, and let ΑΒ and ΒΓ be bisected at the points Δ and Ε (respectively). Let ΔΛ, having been joined, be extended and intersect ΓΒ (having also been joined) at Η. Let ΕΖ, having been joined, be extended and intersect ΒΓ at right angles. Let ΓΖ be extended, being equal to ΑΔ. Let ΖΗ and its parallel ΓΘ be joined. And the angle ΚΓΘ being equal to an angle, let ΖΘΚ be drawn from the given point Ζ, making ΘΚ equal to either ΑΔ or ΓΖ: for this is possible, as was shown with the conchoid line. Let ΚΛ, having been joined, be produced and intersect ΑΒ (itself having been produced) at Μ. 

Above: Nicomedes’ Fifth Diagram. Not shown in the manuscripts is the conchoid line: I have added it here in red, and the ruler for it in blue. The given angle ΚΓΘ is first drawn, with Θ otherwise undetermined. A perpendicular from Ζ to ΓΘ is drawn, serving as the length from the pole Ζ to the ruler ΓΘ. The given length is here ΑΔ (or ΓΖ, which is equal to it). In the particular configuration shown here, ΓΖ is itself that perpendicular, but this is because of the specific ratio of ΛΓ to ΛΑ (here, 2:1). Since GeoGebra is not capable of finding the intersection of the conchoid line and the like ΓΚ, I was forced to make this diagram static. The trick I used for other diagrams of using a conic to interpolate the locus is rather unsuitable here, since the conchoid line is both nearly straight and appears to have an inflection point near Κ.

 

I say that \begin{equation} \tag{118} \text{ΓΛ : ΚΓ :: ΚΓ : ΜΑ :: ΜΑ : ΑΛ.} \end{equation}Since ΒΓ has been bisected at Ε, and ΚΓ has been added to it, therefore [by Elements II.6] \begin{equation} \tag{119} \text{rect.(ΒΚΓ) + sq.(ΓΕ) = sq.(ΕΚ).} \end{equation}Let the square on ΕΖ be added in common: therefore [by several applications of Elements I.47]  \begin{align}  \tag{120} \text{rect.(ΒΚΓ) + sq.(ΓΕ) + sq.(EΖ)} & = \text{rect.(ΒΚΓ) + sq.(ΓΖ)} \\\\ & = \tag{121} \text{sq.(ΚΕ) + sq.(ΕΖ)} \\\\ & =  \tag{122} \text{sq.(KZ).} \end{align}And since [by two uses of Elements VI.2] \begin{equation} \tag{123} \text{ΜΑ : ΑΒ :: ΜΛ : ΛΚ,} \end{equation}and \begin{equation} \tag{124} \text{ΜΛ : ΛΚ :: ΒΓ : ΓΚ,} \end{equation}therefore \begin{equation} \tag{125} \text{ΜΑ : ΑΒ :: Β Γ: ΓΚ. } \end{equation}And since ΑΔ is half of ΑΒ, and ΓΗ is double ΒΓ (since ΛΓ is also double ΔΒ), therefore \begin{equation} \tag{126} \text{ΜΑ : ΑΔ :: ΗΓ : ΚΓ.} \end{equation}But [using Elements VI.2 yet again] \begin{equation} \tag{127} \text{ΗΓ : ΓΚ :: ΖΘ : ΘΚ} \end{equation}since ΗΖ and ΓΘ are parallel. And therefore componendo [Elements V def. 14] \begin{equation} \tag{128} \text{ΜΔ : ΔΑ :: ΖΚ : ΚΘ.} \end{equation}But ΑΔ is established as equal to ΘΚ, since ΑΔ is also equal to ΓΖ: therefore ΜΔ is equal to ΖΚ. Therefore \begin{equation} \tag{129} \text{sq.(ΜΔ) = sq.(ΖΚ). } \end{equation}And since [by Elements II.6] \begin{equation} \tag{130} \text{sq.(ΜΔ) = rect.(ΒΜΑ) + sq.(ΔΑ),} \end{equation}and it was shown that \begin{equation} \tag{131} \text{rect.(ΒΚΓ) + sq.(ΓΖ) = sq.(ΖΚ) [by (122)],} \end{equation}and \begin{equation} \tag{132} \text{sq.(ΑΔ) = sq.(ΓΖ),} \end{equation}since ΑΔ was established as equal to ΓΖ. Therefore by Elements VI.16 \begin{equation} \tag{134} \text{ΜΒ : ΒΚ :: ΚΓ : ΑΜ.} \end{equation}But using Elements VI.2 we have \begin{equation} \tag{135} \text{ΒΜ : ΒΚ :: ΓΛ : ΓΚ,} \end{equation}and therefore \begin{equation} \tag{136} \text{ΛΓ : ΓΚ :: ΓΚ : ΑΜ.  } \end{equation}But [Elements VI.2 again] \begin{equation} \tag{137} \text{ΛΓ : ΓΚ :: ΜΑ : ΑΛ, ΒΜ : ΒΚ :: ΓΛ : ΓΚ, } \end{equation}therefore \begin{equation} \tag{138} \text{ΛΓ : ΓΚ :: ΓΚ : ΑΜ :: ΑΜ : ΑΛ.} \end{equation} 

---

[29] δυνάμει here functioning adverbially. 

[30] This is of course only true if the perpendiculars are on the same “side” of Γ. 

[31] Netz has “and this is made understood as follows”. I took γνώριμον as more “familiar”, as in the argument for the second property is a familiar one. 

We have now seen, in as interactive a manner as possible, the wealth of solutions offered by Greek mathematicians to solve the problem of finding two mean proportionals, and hence to solve the problem of duplicating a cube. The variety of approaches has always fascinated me, and I continue to look in awe at Archytas’ construction.

My plans for future work on this topic include looking at the duplication techniques recorded by other authors. Pappus, in Book III of the Collection, gives four methods. The methods presented there are a subset of those given by Eutocius: Pappus gives the constructions of Eratosthenes, Nicomedes, and Heron. For Nicomedes, though, he omits Eutocius’ preparatory materials on the construction and properties of the conchoid line. Pappus also gives his own method. As for Heron and Philon, each gives a construction in their artillery manuals, and describe the context of how finding mean proportionals is related to the construction of artillery. I also plan to look in more detail at the methods Dürer gives in his Painter’s Manual.

I also plan to direct attention to the mechanical methods of solving the problem. As I’ve presented them here, their “mechanical” nature is somewhat subdued, and they are really more abstract mechanical thought-experiments than actual machines. But it seems clear to me from Eutocius’ assessment of the utility of Philon’s construction that he at least imagined actually using a ruler, rather than the mathematical abstraction of one. Similarly, his description of Plato’s construction and Nicomedes’ conchoid lines are full of clearly mechanical terms: dove-tail grooves, knobs that slide in grooves, etc. I suspect that at least some of the mathematicians involved in the story had constructed (or commissioned) these machines. Eratosthenes himself went so far as to dedicate an altar and included on it his mesolabe; however, there is no archaeological record of it or of any of the other devices. That said, reconstruction of these machines should prove to further enrich our understanding and appreciation of the ancient techniques. 

Readers interested in reading more about the problem, its history, and the textual issues are invited to consult the following works.

• Knorr's two books, The Ancient Tradition of Geometric Problems [1993 reprint] and the later Textual Studies in Ancient and Medieval Geometry [1989]. The latter includes very thorough textual comparison between the methods given by Eutocius and those preserved in other authors. This includes Arabic traditions, such as Diocles’ work, which is not extant in Greek. 
• Saito’s paper, Doubling the Cube: A new interpretation of its significance in early Greek geometry [1995]. The paper deals with technical details of justifying Hippocrates’ reduction of the cube duplication problem to finding two mean proportionals.  
• More philosophically-minded readers will no doubt be interested in the connection of Plato to the story. While the solution Eutocius attributes to Plato is almost certainly not genuinely due to the Plato, mathematical themes certainly play prominently in the Platonic corpus. Kouremenos’ paper, “The tradition of the Delian problem and its origins in the Platonic corpus” [2011], discusses the lack of historicity of the “legend” of the Delians needing to actually duplicate a cubical temple, despite its repetition in various forms in authors from Plato to Eutocius to Dürer.  

References

Apollonius of Perga. 2013. Conics: Books I-IV. Translated by R. Catesby Taliaferro, Michael N. Fried, and William H. Donahue. Santa Fe, NM: Green Lion Press.

Archimedes of Syracuse. 2004. On the Sphere and the Cylinder. In The Works of Archimedes: The Two Books On the Sphere and the Cylinder, edited by Reviel Netz, vol. 1. Cambridge: Cambridge University Press.

Aristophanes. 2000. Birds. Edited and translated by Jeffrey Henderson. Loeb Classical Library 179. Cambridge, MA: Harvard University Press.

Cawthorne, Stephanie, and Judy Green. 2014. Cubes, Conic Sections, and Crocket Johnson. Convergence 11.

Dilworth v. Dudley, et al. 1996. 75 F.3d 307 (7th Cir.).

Dudley, Underwood. 1994. The Trisectors. Spectrum, vol. 16. Mathematical Association of America.

Dürer, Albrecht. 1977. The Painter’s Manual: A Manual of Measurement of Lines, Areas, and Solids by Means of Compass and Ruler Assembled by Albrecht Dürer for the Use of All Lovers of Art with Appropriate Illustrations Arranged to be Printed in the Year MDXXV. New York: Abaris Books.

Euclid of Alexandria. 2003. Elements. Translated by Sir Thomas Little Heath and edited by Dana Densmore. Santa Fe, NM: Green Lion Press.

Eutocius of Ascalon. 1891. Commentary on Archimedes’ On The Sphere and the Cylinder. In Archimedis Opera omnia, cum commentariis Eutocii. E codice Florentino recensuit, edited by Johan Ludvig Heiberg, vol. 1. Leipzig: B. G. Tuebneri.

Eutocius of Escalon. 2004. Commentary on Archimedes’ On The Sphere and the Cylinder.  In The Works of Archimedes: The Two Books On the Sphere and the Cylinder, edited by Reviel Netz, vol. 1. Cambridge: Cambridge University Press.

Eutocius of Ascalon. 2013. Commentary on Archimedes’ On The Sphere and the Cylinder. In Archimedis Opera omnia, cum commentariis Eutocii. E codice Florentino recensuit, edited by Johan Ludvig Heiberg, vol. 1. Leipzig: B. G. Tuebneri. Reprint; Cambridge: Cambridge University Press.

Fried, Michael N., and Sabetai Unguru. 2001. Apollonius of Perga’s Conica: Text, Context, Subtext. Leiden: Brill.

Ishizu, Hideko (石津秀子). 2009. Another solution to the polyhedron in Dürer’s Melencolia: A visual demonstration of the Delian problem. Aesthetics (13):179–194. 

Joyce, David E., ed. 1996–1998. Euclid’s Elements. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html.

Knorr, Wilbur Richard. 1989. Textual Studies in Ancient and Medieval Geometry. Boston: Birkhaüser.

Knorr, Wilbur Richard. 1993. The Ancient Tradition of Geometric Problems. New York: Dover.

Kouremenos, Theokritos. 2011. The tradition of the Delian problem and its origins in the Platonic corpus. Trends in Classics 3:341–364.

Liddell, Henry George, and Robert Scott. 1940. Greek-English Lexicon. Oxford: Clarendon Press.

Marsden, E. W. 1971. Greek and Roman Artillery: Technical Treatises. Oxford: Clarendon Press.

McKinney, Colin B. P. 2010. Conjugate Diameters: Apollonius of Perga and Eutocius of Escalon.  PhD diss., University of Iowa.

Pantelia, Maria, dir. 2014. Thesaurus Linguae Graecae.

Saito, Ken (斎藤憲). 1995. Doubling the Cube: A New Interpretation of Its Significance for Early Greek Geometry. Historia Mathematica 22:119–137.

Stoudt, Gary S. 2015. Can You Really Derive Conic Formulae from a Cone? Convergence 12.

Thomas, Ivor, ed. 1939a. Greek Mathematical Works, Volume 1: Thales to Euclid. Loeb Classical Library, no. 335. Cambridge, MA: Harvard University Press.

Thomas, Ivor, ed. 1939b. Greek Mathematical Works, Volume 2: Aristarchus to Pappus. Loeb Classical Library, no. 362. Cambridge, MA: Harvard University Press.