# The Quadrature of the Circle and Hippocrates' Lunes

Author(s):
Daniel E. Otero (Xavier University)

Mathematics as we know it today was born in the Greek world of the sixth century BCE at the hands of Thales of Miletus and his contemporaries, who harnessed the power of logical deduction to discover truths about the mathematical world and to draw forward universal and irreproachable consent for the validity and certainty of these discoveries. Within this context, the first mathematical program was established: the determination, if possible, of methods for determining the areas of plane figures. This paper, adapted from a forthcoming textbook by the author for a course aimed at undergraduate students of the humanities, will investigate some of the work of geometers in the fifth century BCE, especially the advance of Hippocrates of Chios regarding the quadrature of lunes. We do this by commenting through hypertext links on a few important ancient texts that record this work.

# The Quadrature of the Circle and Hippocrates' Lunes - Introduction

Author(s):
Daniel E. Otero (Xavier University)

Mathematics as we know it today was born in the Greek world of the sixth century BCE at the hands of Thales of Miletus and his contemporaries, who harnessed the power of logical deduction to discover truths about the mathematical world and to draw forward universal and irreproachable consent for the validity and certainty of these discoveries. Within this context, the first mathematical program was established: the determination, if possible, of methods for determining the areas of plane figures. This paper, adapted from a forthcoming textbook by the author for a course aimed at undergraduate students of the humanities, will investigate some of the work of geometers in the fifth century BCE, especially the advance of Hippocrates of Chios regarding the quadrature of lunes. We do this by commenting through hypertext links on a few important ancient texts that record this work.

# The Quadrature of the Circle and Hippocrates' Lunes - The Area Problem in the Fifth Century BCE

Author(s):
Daniel E. Otero (Xavier University)

The ancient Babylonians and the Egyptians developed a rich knowledge of basic geometric concepts, like how the areas of simple plane figures - triangles, squares, rectangles, parallelograms, trapezoids, and the like - relate to the lengths of their sides, and how volumes of simple solid bodies - parallelopipeds and pyramids - relate to corresponding appropriate linear dimensions. Indeed, every one of the great human civilizations of the world have geometrized, often quite independently of each other. It seems to be in the nature of civilized human beings to deal in such abstractions.

It is important to note, however, that the geometrical relationships discovered by these early geometers were not expressed algebraically, as we do when, for example we claim that the statement A = [1/2]bh describes the area of a triangle. Algebra, the science of finding unknown numerical values from arithmetical conditions placed on them, especially in its familiar modern symbolic form, did not exist in the ancient world as such. This version of geometry tends to the utilitarian, meant for use by surveyors, builders, and astronomers. The Greeks, however, turned geometry into a theoretical science, a brand new discipline that concerned itself with the properties of a very refined collection of shapes and forms - points, lines, circles, polygons and polyhedra; a science founded on the deductive methods they were systematizing through philosophical dialectics. And for the first time in history, epistemological questions were being studied about mathematical ideas: how do we know that the geometric results we have discovered are true? Can we justify their truth so that others can come to agree to their truth independent of our authority on the matter? Are these ideas structurally interrelated? A tradition of proof through deduction arose to organize these ideas in theoretical, rather than utilitarian, form.

Geometers of the fifth century BCE were more interested in determining what could be asserted, say, about the relationship between the lengths of the pieces into which two intersecting chords in a circle cut each other than in how much land there was in some particular grower's quadrangular pomegranate orchard. Indeed, they understood the power of the logical, deductive method of theory: that once general theorems of geometry were established, they could then be applied in an infinite variety of particular instances to deal with these more prosaic utilitarian issues. And while their objects of study (circles, squares, etc.) were certainly conceived of as representing the underlying forms of physical reality, their theoretical consideration was viewed as applying to highly abstracted versions of this reality.

In any case, by the middle of the fifth century BCE, Greek geometers had worked out a theory for the area of plane figures that successfully handled all rectilinear figures. It is this theory to which we turn our attention.

# The Quadrature of the Circle and Hippocrates' Lunes - More on Area

Author(s):
Daniel E. Otero (Xavier University)

If plane geometry is the study of plane figures, then a natural and fundamental problem in plane geometry is: given a figure, determine its area. That is, how much of the plane does it enclose? And while solutions to this problem were already well known for many types of the simplest plane figures, it is the systematic way in which the problem was being addressed by them theoretically that was new. Let us follow some of this development.

First of all, what is meant by area? The simple answer for us modern readers is that area is a measure of the size of a plane figure. But as soon as one attempts to measure, one needs an appropriate scale to measure against. For the same figure might measure 5 square feet on one scale and 0.4645 square meters on another. Which of these measures is the correct one? Well, both are, of course. This example just illustrates that we measure area, like length and volume, relative to some unit. The value of the measure is dependent on the unit used.

Modern geometry is characterized by its highly arithmetized and algebraic form: in treatments of geometry given today, geometric objects are often defined and described through equations or formulas. Another reflection of this perspective more pertinent to our present discussion, however, is the modern need to quantify the chief properties of geometric objects through the measurement of their dimensions. Length, area, angle, volume, and other characteristics of geometric objects are for us numerical measures, relative to some convenient scale (feet, square meters, degrees, cubic centimeters, etc.). This way of doing geometry is quite different from that found in the work of the Greeks.

For them, the area of a figure was not a number computed through some measurement; indeed, area was not conceived of numerically at all. Rather, it was a property attached to the object, the one having to do with its size. For the Greek geometer, the area of a triangle was not a number but a certain extent of two-dimensional space. In fact, in most cases, the area of the object and the object itself were not distinguished each from the other. Instead, to find the area of a figure meant for them to determine the dimensions of a square, the primordial plane figure, having the same two-dimesional content of space as the given figure. As we will see, to the Greek geometer, figures which had the same area were in fact said to be ëqual," illustrating that the figure itself was identified as one and the same with its area. This process was called quadrature (from the Latin word for square, quadratus, literally meaning four-sided), manifesting the importance of reference to the square as the most fundamental of plane figures. Notice also that this form of quadrature of a figure requires no measure, no computation, no number; it is a purely geometric procedure.

We will consider a sequence of propositions from Books I and II of Euclid's Elements, consisting of results that were already well-known by the middle of the fifth century. They systematically solve the quadrature problem for any polygonal figure.

# The Quadrature of the Circle and Hippocrates' Lunes - The Quadrature of Polygons in Euclid's [i]Elements,[/i] Book I

Author(s):
Daniel E. Otero (Xavier University)

Proposition I.36.1] Parallelograms which are on equal bases and in the same parallels equal one another.2

Let ABCD and EFGH be parallelograms which are on the equal bases BC and FG and in the same parallels AH and BG.

Figure 1: Elements, I.36.

I say that the parallelogram ABCD equals EFGH.3

Join BE and CH.4

Since BC equals FG and FG equals EH, therefore BC equals EH.

But they are also parallel, and EB and HC join them. But straight lines joining equal and parallel straight lines in the same directions are equal and parallel, therefore EBCH is a parallelogram.5

And it equals ABCD, for it has the same base BC with it and is in the same parallels BC and AH with it.6

For the same reason also EFGH equals the same EBCH, so that the parallelogram ABCD also equals EFGH.

Therefore parallelograms which are on equal bases and in the same parallels equal one another. QED7

[Proposition I.38.] Triangles which are on equal bases and in the same parallels equal one another.

Let ABC and DEF be triangles on equal bases BC and EF and in the same parallels BF and AD.

Figure 2: Elements, I.38.

I say that the triangle ABC equals the triangle DEF.

Produce AD in both directions to G and H. Draw BG through B parallel to CA, and draw FH through F parallel to DE.

Then each of the figures GBCA and DEFH is a parallelogram, and GBCA equals DEFH, for they are on equal bases BC and EF and in the same parallels BF and GH.

Moreover the triangle ABC is half of the parallelogram GBCA, for the diameter AB bisects it. And the triangle FED is half of the parallelogram DEFH, for the diameter DF bisects it.

Therefore the triangle ABC equals the triangle DEF.

Therefore triangles which are on equal bases and in the same parallels equal one another. qed

[Proposition I.42] To construct a parallelogram equal to a given triangle in a given rectilinear angle.8

Let ABC be the given triangle, and D the given rectilinear angle.

Figure 3: Elements, I.42.

It is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC.

Bisect BC at E, and join AE. Construct the angle CEF on the straight line EC at the point E on it equal to the angle D.9 Draw AG through A parallel to EC, and draw CG through C parallel to EF.

Then FECG is a parallelogram.

Since BE equals EC, therefore the triangle ABE also equals the triangle AEC, for they are on equal bases BE and EC and in the same parallels BC and AG.10 Therefore the triangle ABC is double the triangle AEC.

But the parallelogram FECG is also double the triangle AEC, for it has the same base with it and is in the same parallels with it,11 therefore the parallelogram FECG equals the triangle ABC.

And it has the angle CEF equal to the given angle D.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which equals D. QEF12

Footnotes:

1 The text is Sir Thomas Heath's well-known English translation. We adopt the common convention of referring to propositions from the Elements by book and number, whence I.36 refers to the 36th proposition of Book I.

2Notice that it is the areas of the parallelograms that are being discussed here; Euclid makes no distinction between the area of the figure and the figure itself. Further, the statement of the proposition makes it clear that the area of a parallelogram depends only on the length of its base and its corresponding height (the distance between the parallels to which reference is made in the statement of the proposition). This is clear to the modern geometer when she writes that the formula for the area of a parallelogram with base b and height h is A = bh. But Euclid is not a modern geometer, so he sees no need to make reference to the (numerical) height of the parallelogram when the purely geometric form of the pair of parallels enclosing the figure is description enough for him. Euclid does not compute the area of the figure, he simply characterizes it in terms of its "elements."

3It will be obvious from the style of the writing, here and throughout the Elements, that Euclid adopts a very formal style of writing, almost like poetic verse, to lay out geometric propositions. The initial statement of the proposition, given in italics in the text, is called the protasis. It identifies the premises or conditions and the corresponding conclusion of the proposition at hand, often (though not in this instance) in the form "If ... , then ... " The protasis is followed by a second version of the hypotheses of the proposition, the ekthesis, this time with letters referring to a specific, but entirely generic, illustration of the given data. The ekthesis is also accompanied by a diagram to which the letters refer. This is immediately followed by the diorismos, or restatement of the conclusion, often using the words "I say that ... " The remaining parts of the format for Euclidean propositions will be discussed in the next note.

4The fourth part of the propostion is called the kataskeu¯e (it need not appear in all propositions). It calls for any additional lines or curves that should be added to the diagram to help effect the proof. It is followed by the apodeixis, the meat of the proof, which lists the chain of reasons leading from the given premises to the final conclusion.

5Euclid calls here on the fact, which he has proved earlier (in Proposition I.33) that in a quadrilateral, if one pair of opposite sides is parallel and equal, then so must the other pair be.

6Euclid also cites an earlier proposition here to justify this step of the argument. But it is easy enough to see the reason directly from the diagram: since triangles ABE and BCH have equal dimensions (see the figure), then they are clearly equal, and removing each of these triangles in turn from the trapezoid ABCH leaves the parallelograms ABCD and EBCH, which must therefore also be equal.

7The final sentence of the proposition is called the sumperasma; it is simply a recapitulation of the original statement of the proposition, prefaced with a "therefore," and typically followed by the phrase "which was to be demonstrated" (in Latin, quod erat demonstrandum, often abbreviated as QED).

8The form of the statement of this proposition is different from that of the two previous propositions we have displayed here. Whereas Propositions I.36 and I.38 are what are technically called theorems, or assertions of mathematical fact, Proposition I.42 is what the Greeks called a problem, in which the goal is to find some desired point, line, figure, or number. All of Euclid's propositions have one of these two forms.

9Euclid, in I.23, showed how to construct a copy of a given angle along a given line at any particular point on it. It is a straightforward construction procedure.

10You recognize this as precisely the content of Proposition I.38.

11Euclid has established this - that the triangle whose base is the same as a given parallelogram and whose opposite vertex lies on the extensions of the opposite and parallel side of the parallelogram is equal to half the parallelogram - in Proposition I.41. In other words, if a triangle has base b and height h, and a parallelogram has the same dimensions, then the area of the triangle (A = [1/2]bh) is exactly half that of the parallelogram (A = bh).

12Since this proposition is a problem, not a theorem, the sumperasma ends with the phrase "which was to be constructed," in Latin, "quod erat factum."

# The Quadrature of the Circle and Hippocrates' Lunes - The Quadrature of Polygons in Euclid's [i]Elements,[/i] Books I, II

Author(s):
Daniel E. Otero (Xavier University)

[Proposition I.44.] To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle.13

Let AB be the given straight line, D the given rectilinear angle, and C the given triangle.

Figure 4: Elements, I.44.

It is required to apply a parallelogram equal to the given triangle C to the given straight line AB in an angle equal to D.

Construct the parallelogram BEFG equal to the triangle C in the angle EBG which equals D, and let it be placed so that BE is in a straight line with AB.14

Draw FG through to H, and draw AH through A parallel to either BG or EF. Join HB.

Since the straight line HF falls upon the parallels AH and EF, therefore the sum of the angles AHF and HFE equals two right angles.15 Therefore the sum of the angles BHG and GFE is less than two right angles. And straight lines produced indefinitely from angles less than two right angles meet, therefore HB and FE, when produced, will meet.16

Let them be produced and meet at K. Draw KL through the point K parallel to either EA or FH. Produce HA and GB to the points L and M.17

Then HLKF is a parallelogram, HK is its diameter,18 and AG and ME are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF.19

But BF equals the triangle C, therefore LB also equals C.

Since the angle GBE equals the angle ABM, while the angle GBE equals D, therefore the angle ABM also equals the angle D.

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which equals D. QEF

[Proposition I.45.] To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.

Let ABCD be the given rectilinear figure and E the given rectilinear angle.

Figure 5: Elements, I.45.

It is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E.

Join DB. Construct the parallelogram FH equal to the triangle ABD in the angle HKF which equals E.20 Apply the parallelogram GM equal to the triangle DBC to the straight line GH in the angle GHM which equals E.21

Since the angle E equals each of the angles HKF and GHM, therefore the angle HKF also equals the angle GHM.

Add the angle KHG to each. Therefore the sum of the angles FKH and KHG equals the sum of the angles KHG and GHM.

But the sum of the angles FKH and KHG equals two right angles, therefore the sum of the angles KHG and GHM also equals two right angles.

Thus, with a straight line GH, and at the point H on it, two straight lines KH and HM not lying on the same side make the adjacent angles together equal to two right angles, therefore KH is in a straight line with HM.22

Since the straight line HG falls upon the parallels KM and FG, therefore the alternate angles MHG and HGF equal one another.

Add the angle HGL to each. Then the sum of the angles MHG and HGL equals the sum of the angles HGF and HGL.

But the sum of the angles MHG and HGL equals two right angles, therefore the sum of the angles HGF and HGL also equals two right angles. Therefore FG is in a straight line with GL.23

Since FK is equal and parallel to HG, and HG equal and parallel to ML also, therefore KF is also equal and parallel to ML, and the straight lines KM and FL join them at their ends. Therefore KM and FL are also equal and parallel. Therefore KFLM is a parallelogram.

Since the triangle ABD equals the parallelogram FH, and DBC equals GM, therefore the whole rectilinear figure ABCD equals the whole parallelogram KFLM.

Therefore the parallelogram KFLM has been constructed equal to the given rectilinear figure ABCD in the angle FKM which equals the given angle E. QEF

[Proposition II.14.] To construct a square equal to a given rectilinear figure.

Let A be the given rectilinear figure.

Figure 6: Elements, II.14.

It is required to construct a square equal to the rectilinear figure A.

Construct the rectangular parallelogram BD equal to the rectilinear figure A.24

Then, if BE equals ED, then that which was proposed is done, for a square BD has been constructed equal to the rectilinear figure A.25

But, if not, one of the straight lines BE or ED is greater. Let BE be greater, and produce it to F. Make EF equal to ED, and bisect BF at G.26

Describe the semicircle BHF with center G and radius one of the straight lines GB or GF. Produce DE to H, and join GH.

Then, since the straight line BF has been cut into equal segments at G and into unequal segments at E, the rectangle BE by EF together with the square on EG equals the square on GF.27

But GF equals GH, therefore the rectangle BE by EF together with the square on GE equals the square on GH.

But the sum of the squares on HE and EG equals the square on GH,28 therefore the rectangle BE by EF together with the square on GE equals the sum of the squares on HE and EG.

Subtract the square on GE from each. Therefore the remaining rectangle BE by EF equals the square on EH.

But the rectangle BE by EF is BD, for EF equals ED, therefore the parallelogram BD equals the square on HE.

And BD equals the rectilinear figure A.29

Therefore the rectilinear figure A also equals the square which can be described on EH.

Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilinear figure A. QEF

Footnotes:

13For Euclid, a "straight line" means a line segment with endpoints, not an infinitely long line. Notice also the special use of the verb "apply," meaning to build with a given segment as side.

14Here Euclid uses the result of I.42.

15In modern geometry, we would say that angles AHF and HFE are supplementary, that is, they add up to 180°. But Euclid never uses the degree as a unit of angular measure. Instead, he measures all angles in reference to the right angle as the basic unit. Here he needs to make use of the fact, which he proves in I.29, that alternate interior angles made by a transversal line across a pair of parallels are supplementary to each other.

16It is clear from the diagram that these lines meet, but Euclid proves this, since he is careful not to rely on the drawing for his argument. After all, the drawing represents a typical, but not universal, situation. Since angle BHG is certainly smaller than AHG, the angles BHG and GFE sum to less than 180°. It follows then (from the famous Parallel Postulate) that the lines HB and FE are not parallel and must meet.

17It is curious that, after going through the trouble to argue why HB and FE must meet at a point K, Euclid does not explain how he knows that HA and GB, when produced, will eventually cut KL in points L and M, respectively. Mathematicians of the nineteenth century studied Euclid's proof-writing techniques carefully and recognized that he wasn't always as rigorous as he could have been. Still, the issue at this point in his proof is not a serious one. One can, if necessary, provide a more complete proof that fills in this small lacuna in the logic.

18A diameter of a parallelogram corresponds to either of the diagonal lines joining opposite corners of the figure.

19Parallelograms, rectangles and squares are often labeled by Euclid by referring to a pair of opposite corners of the figure. In the diagram accompanying this proposition, the large parallelogram HLKF has been divided into four smaller parallelograms, two straddling the diameter HK, namely AG and ME, and two others, LB and BF, called complements by Euclid.

Since the diameter of a parallelogram cuts the figure in half, into two congruent triangles, it follows that triangles HLK and KFH are equal, as are HAB and BGH, and BMK and KEB. Removing HAB and BMK from HLK leaves parallelogram LB, and removing BGH and KEB from KFH leaves parallelogram BF, so these paralleograms are equal as well.

20Here Euclid uses Proposition I.42.

21Here he uses Proposition I.44.

22While it may seem apparent from the diagram that HM is a simple extension of the line KH, it was not added to the diagram as an extension of this line, so Euclid must prove that it is so, which he does by showing that angle KHM is a 180°angle.

23This mirrors the argument above to show that angle FGL is a 180°angle.

24Finally, Euclid has accumulated the machinery he needs, in the guise of the propositions we have worked through above, to solve the problem central to our discussion here, the quadrature of any polygonal figure.

25He uses here Proposition I.45. Notice that he chooses the angle that determines the parallelogram to be a right angle, so that the parallelogram is in fact a rectangle. This is important to the rest of the proof.

26One of the most elementary construction problems, which he solves in Proposition I.10, is to bisect a line segment.

27Here Euclid is citing Proposition II.5, which we will interpret in algebraic terms here to make faster work of his argument, despite the glaring anachronism. Let a = BE and b = EF. Then since BF = a + b, GF = [1/2](a+b), and we have that

 EG = GF - EF = 1 2 (a+b) - b = 1 2 (a-b).

Now it is straightforward to verify algebraically that

 ab + [ 1 2 (a-b)]2 = [ 1 2 (a+b)]2,

and when we rewrite this in terms of the lines in this diagram, we get the result that BE ·EF + EG2 = GF2, which is the algebraic form of the statement that Euclid is asserting here, "the rectangle BE by EF" being given by multiplying its length by its width, and the squares EG and GF being obtained by squaring their sides.

28Of course, he is citing the Pythagorean Theorem (Proposition I.47) here!

29At this point in the argument, Euclid has shown how to square a rectangle. The algebraic form of this result is a significant one: given an x ×y rectangle, the square with side s of equal area satisfies the relation s2 = xy, or equivalently, [(x)/(s)] = [(s)/(y)], so resolving the quadrature of the rectangle is equivalent to finding the square with side $$s=\sqrt{xy}.$$ In the diagram (Figure 6), x = EF, y = BE, and s = HE, and since HE is necessarily intermediate in length between EF and BE, it follows that s is between x and y in value. For this reason $$s=\sqrt{xy}$$ is also known as the geometric mean between x and y.

# The Quadrature of the Circle and Hippocrates' Lunes - The Quadrature of Curved Figures

Author(s):
Daniel E. Otero (Xavier University)

The most interesting problems with quadrature deal with figures having curved sides, the quintessential example being the quadrature of the circle. Today, we express our knowledge about the quadrature of the circle through the familiar area formula A = $$\pi$$r2 where r represents the radius of the circle. We can also use a less familiar but equivalent (and useful) formula A = [$$\pi$$ /4]d2, where d is the circle's diameter. But the simplicity of the formulas masks the underlying conceptual difficulties. The formulas make clear that the area of a circle is a certain multiple of the area of the square whose side is the radius (or, respectively, the diameter) of the circle, so an immediate connection to quadrature is made. This relationship can be made even clearer by expressing the area formula as

 A r2 = $$\pi$$ (or, alternatively, A d2 = $$\pi$$ 4 ).

This brings forward the fact that the ratio between the area of the circle and the area of the square on its radius (or its diameter) is a constant value, independent of how large or small the circle is. This constant ratio between area of the circle and the area of the square on its radius we have recognized to be so important a constant that we have given it a special name, $$\pi$$.30 But giving the number a special symbol gives no indication of its value. What is this number $$\pi$$?

Ancient pre-Greek mathematicians dealt with the area problem for the circle in various ways, but nearly all of these presented solutions that were at best good approximations to an exact quadrature. For instance, the Rhind Papyrus from Egypt (a 17th century BCE copy of a 19th century original) records that a circle is equal to the square on 8/9 its diameter. This is equivalent to saying that $$\pi$$ = 256/81 = 3.16049..., a bit larger than it should be. In the Hebrew Scriptures, in the first Book of Kings, we read about the construction of the ceremonial bowl that was erected at Solomon's Temple in Jerusalem (1 Kgs 7:23) that it was to be 10 cubits across and 30 cubits around, giving a value of $$\pi$$ = 3, which is too small. This latter was a very commonly used approximation, no doubt for its simplicity, and was used in Babylonia as well. In the Indian Sulvasutras, the main source for ancient Indian mathematics prior to the seventh century BCE, we find the following verse:

If you wish to turn a circle into a square, divide the diameter into 8 parts, and again one of these eight parts into 29 parts; of these 29 parts, remove 28, and moreover the sixth part (of the one part left) less the eighth part (of the sixth part).

The Brahmin author means to say that the area of the circle is equal to the area of a square whose side is a certain fraction of the diameter of the circle, that fraction being

 7 8 + 1 8 × 29 - 1 8 × 29 × 6 - 1 8 × 29 × 6 × 8 ,

which is equivalent to having $$\pi$$ = 3.0883...

The problem of the quadrature of the circle is an old and thorny problem, so much so that, to this day, we use the phrase "squaring the circle" to refer to any intractable problem.

Figure 7: A lune.

Toward the end of the fifth century BCE, Hippocrates of Chios,31 a geometer of considerable prowess, lived and taught in Athens. As for many other figures in the ancient world, we know very little about his life. Over 800 years later, the historian Proclus reported that Hippocrates was the first to write a work on elements of geometry. Indeed, there is evidence to suggest that Hippocrates may have been the source for much of what eventually was included by Euclid in his Elements when it was written some 100 years or so later. Our main interest in Hippocrates here is his discovery of the quadrature of certain lunes. We will learn how he did this by consulting two short excerpts from other historians of geometry who wrote some centuries later; none of his own writings survive today.

Footnotes:

30It may be surprising to learn that, despite the fact that we use a Greek letter to represent this number, the use of the letter $$\pi$$ for this number does not come to us from the Greeks. The first use of the symbol (the first letter of the Greek word periphereia, meaning circumference, from the related fact that it is also the ratio of the circumference to the diameter of any circle) dates to the early eighteenth century in Britain, and was popularized by the noted Swiss mathematician Leonhard Euler .

31Do not confuse him with Hippocrates of Cos, a comtemporary, but renowned as a physician. It is this second Hippocrates for whom the famous oath of physicians, to above all do no harm to their patients, is named. Both Cos and Chios are islands in the Aegean Sea.

# The Quadrature of the Circle and Hippocrates' Lunes - Hippocrates' Lunes

Author(s):
Daniel E. Otero (Xavier University)

[From Joannes Philoponus, In Aristotelis Physica.]32

Hippocrates of Chios was a merchant who fell in with a pirate ship and lost all his possessions. He came to Athens to prosecute the pirates and, staying a long time in Athens by reason of the indictment, consorted with philosophers, and reached such proficiency in geometry that he tried to affect the quadrature of the circle. He did not discover this, but having squared the lune he falsely thought from this that he could square the circle also. For he thought that from the quadrature of the lune the quadrature of the circle could also be calculated.33

[From Simplicius, In Aristotelis Physica.]34

Eudemus,35 however, in his History of Geometry says that Hippocrates did not demonstrate the quadrature of the lune on the side of a square36 but generally, as one might say.37 For every lune has an outer circumference equal to a semicircle or greater or less, and if Hippocrates squared the lune having an outer circumference equal to a semicircle and greater and less, the quadrature would appear to be proved generally.38 I shall set out what Eudemus wrote word for word, adding only for the sake of clearness a few things taken from Euclid's Elements on account of the summary style of Eudemus, who set out his proofs in abridged form in conformity with the ancient practice. He writes thus in the second book of the History of Geometry:

Figure 8: Similar circular segments.
The quadratures of lunes, which seemed to belong to an uncommon class of propositions by reason of the close relationship to the circle, were first investigated by Hippocrates, and seemed to be set out in correct form; therefore we shall deal with them at length and go through them. He made his starting-point, and set out as the first of the theorems useful to this purpose, that similar segments of circles have the same ratios as the squares on their bases.39 And this he proved by showing that the squares on the diameters have the same ratios as the circles.40 Having first shown this he described in what way it was possible to square a lune whose outer circumference was a semicircle. He did this by circumscribing about a right-angled isosceles triangle a semicircle and about the base a segment of a circle similar to those cut off by the sides.41 Since the segment about the base is equal to the sum of those about the sides,42 it follows that when the part of the triangle above the segment about the base is added to both, the lune will be equal to the triangle.43

Figure 9: Hippocrates' lune.

Therefore the lune, having been proved equal to the triangle, can be squared. In this way, taking a semicircle as the outer circumference of the lune, Hippocrates readily squared the lune.

Footnotes:

32Philoponus was a sixth century CE philosopher and Christian theologian who studied the works of Greek philosophers, especially those of Aristotle, and wrote extensive commentaries on them. This excerpt comes from a commentary on Aristotle's Physics. A native of Egyptian Alexandria, he lives at the end of the period of Greek progress in the sciences.

33We will analyze in detail Hippocrates' quadrature of one type of lune in the next text below. It is enough for now to understand the point Philoponus makes here concerning the quadrature of the circle. For Hippocrates' work represented the first time that someone had determined the area of a figure with curved sides - in fact, with circular sides - and it was thought that the techniques he used for the quadrature of the lune might lead to a positive resolution to the quadrature of the circle. Alas, as we shall see below, this was never to be!

34Simplicius was another sixth century commentator on early Greek texts, notably on the works of Aristotle and Euclid. He was born in Cilicia, a Roman province which is today part of Turkey. He studied in Alexandria under a pupil of Proclus, and also at Athens at Plato's Academy.

35Eudemus of Rhodes lived in the late fourth century BCE and was the first historian of mathematics. He was a fellow student with Aristotle in Athens and authored histories of geometry, arithmetic and astronomy, none of which survive today. What we know of his work is based on references and quotes in other works like this commentary on Aristotle's Physics by Simplicius. Here Simplicius quotes from Eudemus' history of geometry. Because Eudemus is writing so much closer in time to Hippocrates and the other Greek geometers, he is considered a valuable historical source.

36By this, Simplicius means the quadrature of the lune constructed as follows: given a circle and an inscribed square, build the circle whose diameter is one of the sides of the square. The lune being described here is the one inside the second smaller circle and outside the larger.

37Simplicius seems to believe that Hippocrates had claimed to have successfully squared the circle as a result of his work on lunes, but the argument he gives does not accomplish this, so Simplicius is not to be trusted on this account.

38A lune has two boundary arcs; the line segment that joins the endpoints of these arcs forms a chord in each of the two circles. Simplicius is here describing the situation in which the arc of the outer circle is a semicircle, or is less than a semicircle, or more than a semicircle. If Hippocrates had successfully worked out the quadrature of the lune in all these cases, then by arranging that the outer arc of the lune increases to its greatest possible extent and the inner arc shrinks to a point, then the lune becomes a full circle and the quadrature of the circle would be resolved!

39A segment of a circle is any portion of the circle cut off by a single line. Segments are similar if the arcs that bound them come from equal central angles in their respective circles (the shaded areas labeled s and S in the figure below).

Thus, to say that "similar segments of circles have the same ratios as the squares on their bases" means that the ratio of the areas of the segments is the same as the ratio of the squares of the lengths of their straight line bases. In symbols, if one segment s has base b and the other segment S has base B, then

 s S = b2 B2 .

40This result (Elements, XII.2) is the key idea behind this quadrature. It says that if one circle c has diameter d and another circle C has diameter D, then

 d2 D2 = c C .
Hippocrates uses it to derive the result about segments of circles stated above. While the details of this proof are unknown to us, here is a plausible reconstruction.

Draw the diameter d through one of the endpoints of b, then complete the triangle two of whose sides are b and d. (Refer to Figure 7.) Similarly, in circle C, draw diameter D through one of the endpoints of B, then complete the triangle two of whose sides are B and D. Then, on the one hand, similarity of the segments means that segment s is the same fraction of circle c that segment S is of circle C, or

 s c = S C ,     or alternately, s S = c C .
On the other hand, since the triangles in the two circles and have equal corresponding angles, they are similar, so their corresponding sides are proportional:
 b B = d D .
Combining these last three proportions yields
 s S = c C = d2 D2 = b2 B2 ,
hence the desired result.

41The semicircle circumscribed around the right triangle in Figure 9 forms the outer circumference of the lune. The inner circumference is formed by the arc of a second circle that makes a segment with the base PR of the triangle similar to the two segments between the semicircle and the other two sides PQ and QR of the triangle. Each of these three segments cuts off a 90 degree arc from its circle.

Simplicius described this lune as one ön the side of a square." To see that he is talking about the same lune as the one described by Eudemus, consider from the diagram below the entire circle that forms the inner circumference of this lune. The base of the semicircle that forms the outer circumference of the lune is one side of a square that can be inscribed in this inner circle. Thus, the lune can be seen to rest on neighboring vertices of a square inscribed in the circle that forms its inner circumference.

Figure 10

42This is determined by combining the result on ratios of similar segments with the Pythagorean theorem. Referring to this figure, the Pythagorean theorem says that (PQ)2 + (QR)2 = (PR)2. But b = PQ = QR is the common length of the base of the two small segments, and B = PR is the length of the base of the larger segment, so we can express this relation as b2 + b2 = B2, or in terms of ratios,

 b2 B2 + b2 B2 = 1.
But by the result on similar segments, we can transform this into a relationship between the corresponding segments themselves:
 s S + s S = 1,     or more simply,     s + s = S.
This shows that "the segment about the base is equal to the sum of those about the sides."

43Hippocrates' lune is comprised of both of the two small circular segments and the isosceles right triangle (see Figure 9). Since the sum of the two small segments is equal to the one larger one, and replacing the two smaller areas with the larger one produces the triangle PQR, Hippocrates has therefore shown that the lune equals the triangle!

# The Quadrature of the Circle and Hippocrates' Lunes - Tantalizing Prospects

Author(s):
Daniel E. Otero (Xavier University)

Having found the means to effect the quadrature of any polygonal figure (as we saw in Elements, II.14), the inability to square the circle by similarly direct means stood as an inviting challenge to geometers of the fifth and fourth centuries. So when Hippocrates' lune was found to be equal to a triangle, a tantalizing hope was raised that some similar analysis might effect the quadrature of the full circle. Thus Hippocrates' result stands as one of the more important advances in geometry of that time. But the problem of the quadrature of the circle remained unresolved as no way was found to replicate Hippocrates' success in application to the full circle. The best that could be done was to achieve approximate quadratures.

Antiphon, a contemporary of Hippocrates who taught in Athens, attempted to resolve the problem by inscribing a polygon within the circle and effecting the quadrature of the polygon by finding lower bound approximations for the area of the circle. He performs a similar process to find an upper bound approximation for the area of the circle by circumscribing a polygon about its circumference. Then, by successively doubling the number of sides of these polygons, one inside and the other outside the circle, the error of approximation can be reduced. Effectively, the area of the circle is ëxhausted" by taking more and more sides for these approximating polygons. Because the sources we have are quite fragmentary, we cannot be quite sure whether Antiphon believed that this procedure would actually result in finding the exact quadrature of the circle, or whether he knew that the best this would accomplish is the determination of a pair of values which are close but never equal to the true area.

Figure 11: "Exhausting" the area of the circle with inscribed and circumscribed polygons.

Another contemporary, Bryson of Heraclea, is said to have asserted that the area of the circle was at once greater than the area of any and all possible polygons that could be inscribed within it and smaller than the area of any and all possible polygons that could be circumscribed about it. This principle would come to fruition in the work of Eudoxus of Cnidos, who working at the start of the fourth century BCE would give the first proof of Elements, XII.2, the theorem that circles are to each other as the squares on their diameters. This method of exhaustion would be one that later geometers would return to again and again over many centuries to apply to quadratures of a variety of curved shapes. Archimedes (in the third century BCE) used it often, to demonstrate results he had discovered regarding the quadrature of regions like the parabolic segment44 and the cubature45 of the sphere. Much later, when Greek geometry was studied once again in the Europe of the sixteenth and seventeenth centuries CE, a resurgence of interest in these methods took place. Geometers like Gregoire a Saint-Vincent (1584-1667) and his student Alfonso Antonio de Sarasa (1618-1667), both Jesuit scientists, applied the method to show that the area between a hyperbola and its asymptote behaves like a logarithm.

As for the original problem of the quadrature of the circle, final resolution would not come until the nineteenth century, when it was shown using the tools of modern algebra (not geometry!) that although the Greek construction tools of straightedge and compass are capable of producing an infinite number of line segments, these segments all have lengths that belong to a restricted subset of real numbers, and p is not in this set, making the quadrature, in its strictest form, impossible.

The work of the fifth century geometers set a course for the pursuit of quadratures that led to the generation of much new and important mathematics. As geometers came across new types of curves, they considered new sorts of plane regions and solids and asked questions about their quadrature and cubature. This work came to a high point in the seventeenth and eighteenth centuries CE with the development of integral calculus. The fifth century area problem was the first mathematical "program," an enterprise representing the efforts of many individuals over a long period of time, all contributing to the understanding of a single type of problem through the development of new and more powerful mathematical methods. Even early on, as we have seen from direct reference to some of the texts written at the time, it involved surprising discoveries, patient systematization, and the realization that to adequately resolve the problem would require not just cleverness, but willingness to change the way in which the problem was being considered. These features have characterized progress in mathematical programs throughout history.

Footnotes:

44Like a circular segment, a parabolic segment is the closed region bounded on one side by an arc of a parabola and on the other by a line that cuts the parabola in two points.

45Naturally, cubature is for three-dimensional solids what quadrature is to two-dimensional regions. The cubature of a solid is obtained by constructing a line segment equal to the side of a cube which is equal in volume to the given solid.

# The Quadrature of the Circle and Hippocrates' Lunes - References

Author(s):
Daniel E. Otero (Xavier University)

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