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Chapter 5
Hippocrates of Chios
1
The area problem
The 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 – were related to the lengths
of their sides, and how volumes of simple solid bodies – parallelopipeds and pyramids – related to corresponding appropriate linear dimensions.1 These versions
of geometry tended to be very utilitarian, meant for use by surveyors, builders,
and astronomers. The Greeks however turned geometry into a theoretical science, a new discipline that concerned itself with the properties of a very refined
collection of shapes and forms – points, lines, circles, polygons and polyhedra2 ;
a science founded on the deductive methods they were systematizing through
philosophical dialectics. And for the first time, 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 was arising to organize these ideas in theoretical, rather than utilitarian,
form.
Geometers of the fifth and fourth centuries BCE tended to be 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
1 Indeed,
every one of the great human civilizations of the world have done so, often quite
independently of each other. It seems to be in the nature of human beings to geometrize in
some fashion.
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
A = 12 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 in this form.
2 A polyhedron is a finite solid body whose exterior is made up of plane polygons, like a
cube or a pyramid.
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CHAPTER 5. HIPPOCRATES OF CHIOS
other than in how much land there was in some particular grower’s quadrangular pomegranate orchard. After all, once the general theorems of geometry were
discovered, they could then be applied in an infinite variety of particular instances to deal with these more prosaic utilitarian issues. And while the objects
of study (circles, squares, etc.) were certainly conceived of as representing the
underlying forms of physical reality, their theoretical consideration was viewed
by the classical Greek geometers as applying to highly abstracted and general
versions of this reality.
In any case, by the end 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 now turn our attention.
If plane geometry is the study of plane figures, then a fundamental problem in
plane geometry naturally is: given a figure, determine its area. While solutions
to this problem were already well known to Greek geometers by the fifth century
BCE for many types of plane figures, it is the systematic way in which the
problem was being addressed by them 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 make such a 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.
Now we have already mentioned (see note 9 in Chapter 2) that 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 Greek geometers.
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 from each other. Instead, to find the
area of a figure meant for them to determine the dimensions of a square having
the same two-dimesional content of space as the given figure.3 This process
3 As we will see, to the Greek geometer, figures which had the same area were in fact said
to be “equal,” illustrating that the figure itself was identified as one and the same with its
2. TEXT: THE QUADRATURE OF POLYGONS
63
was called quadrature (from the Latin word for square, quadratus, literally
meaning four-sided ), and it manifests for us the importance of reference to the
square as the most fundamental of plane figures. Notice also that quadrature
of a figure requires no measure, no computation, no number; it is a purely
geometric procedure.
The first set of texts below is a sequence of propositions from Books I and II
of Euclid’s Elements, consisting of results that were already well-known by the
fifth century BCE. They systematically solve the quadrature problem for any
polygonal figure.
2
Text: The quadrature of polygons
From Euclid’s Elements, I.36.
Parallelograms which are on equal bases and in the same parallels equal one
another.4
Let ABCD and EF GH be parallelograms which are on the equal bases BC
and F G and in the same parallels AH and BG.5
Figure 5.1: Elements, I.36
area.
4 Note that it is the areas of the parallelograms that is being discussed here; Euclid makes
no distinction between the area of the figure and the figure itself.
Further, 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 obvious 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 version 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.”
5 It will be obvious from the stylistic form of the writing, here and in all of the subsequent
propositions drawn from 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 above, 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 set-up of the proposition, the ekthesis, this time with letters referring
to a specific, but entirely generic, illustration of the given data. The ekthesis is 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 note 7.
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CHAPTER 5. HIPPOCRATES OF CHIOS
I say that the parallelogram ABCD equals EF GH.6
Join BE and CH.7
Since BC equals F G and F G 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.8
And it equals ABCD, for it has the same base BC with it and is in the
same parallels BC and AH with it.9
For the same reason also EF GH equals the same EBCH, so that the parallelogram ABCD also equals EF GH.
Therefore parallelograms which are on equal bases and in the same parallels
equal one another. Q.E.D.10
From Euclid’s Elements, 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 5.2: 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 F H through F parallel to DE.
6 The
diorismos.
fourth part of the propostion is called the kataskeuē (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.
8 Euclid 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.
9 Euclid 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 (Figure 5.1), 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.
10 The 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” (Q.E.D. See Chapter 2, note 15).
7 The
2. TEXT: THE QUADRATURE OF POLYGONS
65
Then each of the figures GBCA and DEF H is a parallelogram, and GBCA
equals DEF H, 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 F ED is half of the parallelogram
DEF H, 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. Q.E.D.
From Euclid’s Elements, I.42.
To construct a parallelogram equal to a given triangle in a given rectilinear
angle.11
Let ABC be the given triangle, and D the given rectilinear angle.
Figure 5.3: 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.12 Draw AG through A parallel
to EC, and draw CG through C parallel to EF .
Then F ECG 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.13 Therefore the triangle ABC is double the triangle AEC.
But the parallelogram F ECG is also double the triangle AEC, for it has
the same base with it and is in the same parallels with it,14 therefore the parallelogram F ECG equals the triangle ABC.
11 The 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 theorems, or
assertions of mathematical fact, Proposition I.42 is 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.
12 Euclid, 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. See Chapter 7 for more
on Euclidean constructions.
13 You recognize this as precisely the content of Proposition I.38
14 Euclid has established this in Proposition I.41. See also Exercise 7).
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CHAPTER 5. HIPPOCRATES OF CHIOS
And it has the angle CEF equal to the given angle D.
Therefore the parallelogram F ECG has been constructed equal to the given
triangle ABC, in the angle CEF which equals D. Q.E.F.15
From Euclid’s Elements, I.44.
To a given straight line in a given rectilinear angle, to apply a parallelogram
equal to a given triangle.16
Let AB be the given straight line, D the given rectilinear angle, and C the
given triangle.
Figure 5.4: 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 BEF G 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.17
Draw F G 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 HF E equals two right angles.18 Therefore the
sum of the angles BHG and GF E is less than two right angles. And straight
lines produced indefinitely from angles less than two right angles meet, therefore
HB and F E, when produced, will meet.19
Let them be produced and meet at K. Draw KL through the point K
parallel to either EA or F H. Produce HA and GB to the points L and M .20
15 Since this proposition is a problem, not a theorem, the sumperasma ends with the phrase
“which was to be constructed,” in Latin, “quod erat factum.”
16 For Euclid, a “straight line” means a line segment, not an infinitely long line. Notice also
the special use of the verb “apply,” meaning to build with the given conditions.
17 Here Euclid uses the result of I.42.
18 In modern geometry, we would say that angles AHF and HF E 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 right angles. (See Chapter 7.) 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.
19 It is clear from the diagram that these lines meet, but Euclid is careful to prove this fact,
since he does not rely on the drawing for his argument. Since angle BHG is certainly smaller
than AHG, the angles BHG and GF E sum to less than 180◦ . It follows then (from Euclid’s
Parallel Postulate; see Chapter 7, note 7.1) that the lines HB and F E are not parallel and
must meet.
20 It is curious that, after going through the trouble to argue why HB and F E must meet
at a point K, Euclid does not explain how he knows that HA and GB, when produced, will
2. TEXT: THE QUADRATURE OF POLYGONS
67
Then HLKF is a parallelogram, HK is its diameter,21 and AG and M E
are parallelograms, and LB and BF are the so-called complements about HK.
Therefore LB equals BF .22
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. Q.E.F.
From Euclid’s Elements, 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 5.5: 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 F H equal to the triangle ABD in
the angle HKF which equals E.23 Apply the parallelogram GM equal to the
triangle DBC to the straight line GH in the angle GHM which equals E.24
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 F KH and
KHG equals the sum of the angles KHG and GHM .
eventually cut KL in points L and M . 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, fill in a more complete proof that takes fills this small lacuna in the logic.
21 A diameter of a parallelogram corresponds to either diagonal line joining opposite corners
of the figure.
22 Parallelograms, rectangles and squares are often labeled by Euclid by referring to one pair
of opposite corners. 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 M E, 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 KF H are equal, as are HAB and BGH, and BM K and
KEB. Removing HAB and BM K from HLK leaves parallelogram LB, and removing BGH
and KEB from KF H leaves parallelogram BF , so these last two are equal as well.
23 Here Euclid uses Proposition I.42.
24 Here he uses Proposition I.44.
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CHAPTER 5. HIPPOCRATES OF CHIOS
But the sum of the angles F KH 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 .25
Since the straight line HG falls upon the parallels KM and F G, therefore
the alternate angles M HG and HGF equal one another.
Add the angle HGL to each. Then the sum of the angles M HG and HGL
equals the sum of the angles HGF and HGL.
But the sum of the angles M HG and HGL equals two right angles, therefore
the sum of the angles HGF and HGL also equals two right angles. Therefore
F G is in a straight line with GL.26
Since F K is equal and parallel to HG, and HG equal and parallel to M L
also, therefore KF is also equal and parallel to M L, and the straight lines KM
and F L join them at their ends. Therefore KM and F L are also equal and
parallel. Therefore KF LM is a parallelogram.
Since the triangle ABD equals the parallelogram F H, and DBC equals GM ,
therefore the whole rectilinear figure ABCD equals the whole parallelogram
KF LM .
Therefore the parallelogram KF LM has been constructed equal to the given
rectilinear figure ABCD in the angle F KM which equals the given angle E.
Q.E.F.
From Euclid’s Elements, II.14.
To construct a square equal to a given rectilinear figure.27
Let A be the given rectilinear figure.
Figure 5.6: Elements, II.14
It is required to construct a square equal to the rectilinear figure A.
25 While 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.
26 This mirrors the argument above to show that angle F GL is a 180◦ angle.
27 Finally, Euclid has accumulated the machinery he needs, in the guise of the propositions listed above, to solve the problem central to our discussion here, the quadrature of any
polygonal figure.
2. TEXT: THE QUADRATURE OF POLYGONS
69
Construct the rectangular parallelogram BD equal to the rectilinear figure
A.28
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.
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.29
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 .30
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,31
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.32
Therefore the rectilinear figure A also equals the square which can be described on EH.
28 He 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.
29 One of the most elementary construction problems, which he solves in Proposition I.10,
is to bisect a line segment.
30 Here 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 = 21 (a + b), and we have that
EG = GF − EF =
1
1
(a + b) − b = (a − b).
2
2
Now it is straightforward to verify algebraically that
1
1
ab + [ (a − b)]2 = [ (a + b)]2 ,
2
2
and when we rewrite this in terms of the lines in this diagram, we get the result that BE ·
EF + EG2 = GF 2 , 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.
31 He is citing the Pythagorean Theorem (Proposition I.47) here, of course.
32 At 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, xs = ys , so resolving the quadrature of the
√
rectangle is equivalent to finding the square with side s = xy. In the diagram (Figure 5.6),
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 = xy is also
known as the geometric mean between x and y.
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CHAPTER 5. HIPPOCRATES OF CHIOS
Therefore a square, namely that which can be described on EH, has been
constructed equal to the given rectilinear figure A. Q.E.F.
3
Squaring the circle?
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 = πr2 , or the less familiar but equivalent formula A = π4 d2 ,
where r represents the radius of the circle and d its diameter, but the simplicity
of the formulas mask the underlying conceptual difficulties. The formula makes
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
A
π
= π (or, alternatively, 2 = ).
2
r
d
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, π.33 But giving the number a
special symbol gives no indication of its value. What is this number π?
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. An Egyptian papyrus, for example,
records that a circle is equal to the square on 89 its diameter. This is equivalent
to saying that π = 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 π = 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).
33 It may be surprising to learn that, despite the fact that we use a Greek letter to represent
this number, this notation does not come to us from the Greeks. The first use of the symbol
π for this number (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;
see Chapter 8). dates to the early eighteenth century in Britain, and was popularized by the
noted Swiss mathematician Leonhard Euler (see Chapter ???).
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71
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
!
"
1
1
1
7
+
−
−
,
8 8 × 29
8 × 29 × 6 8 × 29 × 6 × 8
which is equivalent to having π = 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.
We will return to a discussion of the quadrature of the circle later on (see
Chapter 6), for work on this problem helped to fuel the development of ideas that
are central to calculus. But for now, we will look at the quadrature of another
figure, called a lune, that it was thought might provide a fruitful inroad towards
resolution of the problem of squaring the circle. A lune is the shape formed by
two arcs of intersecting circles, the region that is interior to one of the circles
but exterior to the other (Figure 5.7).
Figure 5.7: A lune.
Toward the end of the fifth century BCE, Hippocrates of Chios34 , 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, Hippocrates may have been the source for much
of what eventually was included by Euclid in his Elements 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 after Hippocrates’
life; none of his own writings survive today.
4
Text: The quadrature of a lune
From Joannes Philoponus, In Aristotelis Physica.35
34 Do not confuse him with Hippocrates of Cos, a comtemporary, but renowned as a physician. It is this other 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.
35 Philoponus was a sixth century CE philosopher and Christian theologian who studied the
works of Greek philosophers, especially those of Aristotle, and wrote extensive commentaries
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CHAPTER 5. HIPPOCRATES OF CHIOS
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.36
From Simplicius, In Aristotelis Physica.37
Eudemus38 , however, in his History of Geometry says that Hippocrates did
not demonstrate the quadrature of the lune on the side of a square39 but generally, as one might say.40 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.41 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
on them. This excerpt comes from a commentary on Aristotle’s Physics. A native of Alexandria in Egypt (a city about which much will be said later), he lives at the end of the period
of Greek progress in the sciences.
36 We 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 resolution
to the quadrature of the circle. Alas, this was not to be.
37 Simplicius 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 at Athens at Plato’s
Academy.
38 Eudemus 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.
39 By 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.
40 Simplicius 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.
41 A 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!
4. TEXT: THE QUADRATURE OF A LUNE
73
his proofs in abridged form in conformity with the ancient practice. He writes
thus in the second book of the History of Geometry:
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.42 And this he
proved by showing that the squares on the diameters have the same
ratios as the circles.43
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
42 A segment of a circle is any portion of the circle cut of 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 Figure 5.8).
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
b2
= 2.
S
B
43 This result is the key idea behind this quadrature. We will carefully study a proof of it
in the next chapter. For now, however, we will take it as given. The result says that if one
circle c has diameter d and the other circle C has diameter D, then
c
d2
= .
D2
C
Hippocrates uses this to derive the result about segments of circles stated above. While the
details of this proof are unknown to us, here is one 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. (See Figure 5.8.) 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
S
= ,
c
C
or alternately,
s
c
= .
S
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
d
= .
B
D
Combining these last three proportions yields
s
c
d2
b2
=
= 2 = 2,
S
C
D
B
hence the desired result.
74
CHAPTER 5. HIPPOCRATES OF CHIOS
cut off by the sides.44 Since the segment about the base is equal to
44 The figure described here is displayed in Figure 5.9. The right angle is at B. The semicircle
circumscribed around it 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 P R of the triangle
similar to the two segments between the semicircle and the other two sides P Q and QR of
the triangle. Each of these three segments cut off a 90o arc from its circle.
Simplicius described this lune as one ”on the side of a square”. To see that he is talking
about the same lune as the one described by Eudemus, consider the entire circle that forms
the inner circumference of this lune (Figure 5.10). 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.
Figure 5.8: Similar segments.
Figure 5.9: Hippocrates’ lune.
Figure 5.10: Hippocrates’ lune, expanded view.
75
5. SCHOLIUM: THE DEFINITE INTEGRAL
the sum of those about the sides45 , 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.46 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.
5
Scholium: The definite integral
At the other end of the timeline of the trajectory of ideas that began with the
area problem in ancient Greece is the modern calculus concept of the definite
integral. Loosely speaking, the integral can be viewed geometrically as an area
calculator. Given a line or curve in thexy-plane with equation y = f (x) (here,
f (x) represents some formula in x), there is a region bounded by this curve
and the x-axis. If we assume for the moment that the curve lies above the axis
(that is, that all values of y = f (x) are positive), then the region has this curve
as its top edge and the x-axis as bottom edge (Figure 5.11). If we specify two
numbers a < b on the x-axis and erect the vertical lines perpendicular to the
axis at these points, then these lines can serve as left and right edges of the
region (these lines have equations x = a and x = b).
We therefore have a region bounded on three sides by straight lines and on
#b
the fourth by a curve whose equation we know. We use the notation a f (x) dx
to represent the area of this region.
This is read “the integral of f (x) between a
#
and b.” The integral symbol was first used by Leibniz in 1686. (We will catch
up to this in Chapter ???.) The values placed as subscript and superscript next
to the integral symbol are respectively called the lower limit and upper limit of
integration; the expression that follows is called the integrand ; and the symbol
dx to the right of the integrand is called a differential. The notation gives
Thus, the lune can be seen to rest on neighboring vertices of a square inscribed in the circle
that forms its inner circumference.
45 This is determined by combining the result on areas of similar segments with the
Pythagorean theorem. Referring to Figure 5.9, the Pythagorean theorem says that (P Q)2 +
(QR)2 = (P R)2 . But b = P Q = QR is the common length of the base of the two small
segments, and B = P R is the length of the base of the larger segment, so we can express this
relation as b2 + b2 = B 2 , or in terms of ratios,
b2
b2
+ 2 = 1.
2
B
B
But by the result on similar segments, we can transform this into a relationship between the
corresponding segments themselves:
s
s
+
= 1,
S
S
or more simply,
s + s = S.
This shows that “the segment about the base is equal to the sum of those about the sides.”
46 The lune in Figure 5.9 is comprised of both of the two small circular segments and the
isosceles right triangle. 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 P QR,
Hippocrates has therefore shown that the lune equals the triangle.
76
CHAPTER 5. HIPPOCRATES OF CHIOS
Figure 5.11: The definite integral as area calculator.
all the necessary information for describing the region whose area we wish to
determine: the integral symbol indicates that an area above the x-axis is being
calculated, the limits of integration indicate the left- and right-side bounds of
the region, the integrand determines the upper boundary of the region, and the
differential points to the x as the underlying input variable.
For example, the integral
$ 5
2x + 1 dx
2
represents the area of the trapezoid bounded by the x-axis, the vertical lines
x = 2 and x = 5, and the line y = 2x + 1. The reader is urged to sketch
a graph of this region (perhaps with the aid of a graphics calculator) to see
which trapezoid we are considering here. Notice that the very same region is
represented by the integral
$
5
2t + 1 dt
2
as well, since the only thing that has changed is the name of the input variable.47
Since there is a large class of plane figures whose areas we can calculate from
basic geometry, there is then a large class of integrals that we can compute, even
without the benefit of sophisticated machinery from calculus.
For instance,
$
2
2 dx = 8,
−2
since the integral asks for the area of a rectangle (the graph of y = 2 is a
horizontal line) 4 units wide by 2 units tall. This generalizes to produce areas
of any rectangle that sits on the x-axis: if k is any constant,
$ b
k dx = (b − a)k.
a
We also have
$
8
3x dx = 96
0
47 You can imagine the same geometric picture as being displayed on a ty-plane instead of
an xy-plane.
77
6. SUMMARY
since the region we are considering here is a triangle with base 8 units long and
height 3 × 8 = 24 units tall, whence the area of the triangle is 21 (8 × 24) = 96.
Other simple figures can be represented by not-so-simple looking integrals.
For instance,
$ 4
[8 − 2|x − 1| − 2|x − 3|] dx
0
asks for the area of a trapezoid, and
$
4
−4
%
16 − x2 dx
the area of a semicircle. (Check these out with a graphics device.)
As we consider the development of more tools for handling the area problem,
we will be able to extend the class of integrals we can compute even further.
The exercises at the end of this chapter include some more explorations of these
ideas.
6
Summary
• By the fifth century BCE, Greek geometry had become a theoretical
science whose truths were tested through deductive logic applied to abstracted shape and form.
• A natural and important geometric problem was the area problem: given
a figure, is it possible to find (that is, construct a diagram of) the square
with the same area?
• Propositions from Books I and II of Euclid’s Elements, culminating in
Proposition II.14, show how the area problem had been resolved for polygonal figures.
• Euclid organized his mathematics into theorems, assertions of some mathematical truth, and problems, procedures for drawing a geometrical object.
Both types of proposition were laid out in a formal style consisting of
protasis (initial statement of the entire proposition), ekthesis (clear enunciation of the givens or hypotheses), diorismos (declaration of the conclusion), kataskeuē (additional machinery needed to execute the proof),
apodeixis (the main argument), and sumperasma (recapitulation). Propositions proved earlier in the text can be used in the course of proving later
propositions.
• Babylonian, Egyptian and Greek geometers all knew that the area of a
circle had a fixed ratio with the square on its diameter (or radius), or
numerically, with the square of its diameter (or radius). The use of the
symbol π to denote this ratio, however, dates only to the early eighteenth
century CE. The value of this ratio was approximated to varying degrees
78
CHAPTER 5. HIPPOCRATES OF CHIOS
of accuracy in the ancient world, but could not be determined exactly.
This computational problem is related to the geometric problem of the
quadrature of the circle, a problem that remained unresolved in Greek
mathematics.
• Hippocrates of Chios, an accomplished geometer of the mid-fifth century,
discovered how to work the quadrature of a certain type of lune. His discovery was based on knowing that circles are to each other as the squares
on their diameters. This caused a certain amount of excitement because it
was thought that knowing how to square the lune could lead to a method
for squaring the circle.
#b
• The definite integral, denoted a f (x) dx, is used in modern calculus to
represent the area of the region bounded by the x-axis, the vertical lines
x = a, x = b, and the graph of the equation y = f (x).
Know how to...
• find the geometric mean
√
xy between two given values x and y
• describe the quadrature of Hippocrates’ lune
• interpret and calculate definite integrals that represent the areas of familiar figures from plane geometry
7
Exercises
1. A trapezoid is a four-sided figure with just one pair of parallel opposite
sides, called its bases. (In a parallelogram, both pairs of opposite sides are
parallel.) We typically orient a trapezoid so that its bases are horizontal
with the longer base below the shorter one, whence the height of the
trapezoid is the distance between the bases. Draw a trapezoid, label its
longer base B, its shorter base b, and its height h. By dissecting the
trapezoid into smaller figures, derive the area formula A = 12 (B + b)h.
2. (a) In note 30 (page 69), the algebraic identity
ab +
!
a−b
2
"2
=
!
a+b
2
"2
appears. By multiplying through by 4 and expanding both sides of
the equation, show that the identity is correct.
(b) Draw a square and mark off a point on the top edge somewhere to
the right of the midpoint. Label the piece of this edge to the left
of your point a, and the remaining piece to the right b. (So a > b.)
Now working clockwise around the square, divide the other three
7. EXERCISES
79
sides into pieces of length a and b so that line segments of length a
and b alternate around the perimeter of the square. In each corner
of the square, these line segments determine four a × b rectangles;
insert the necessary lines to display these four rectangles. Why must
the remaining quadrilateral in the center of the diagram be a square,
and how big is this square?
(c) Show that the diagram in part (b) provides a geometric verification
of the algebraic identity described in part (a).
3. What is the area of an equilateral triangle whose side length is one unit?
(Hint: Drawing in the altitude of the triangle divides the equilateral triangle into two congruent right triangles; since you know the length of
the short side and the hypotenuse of one of these right triangles, the
Pythagorean theorem will determine the length of the other side.)
4. What is the area of a regular hexagon whose side length is one unit? By
regular, we mean that all sides are equal and all interior angles are equal.
5. What is the arithmetic mean of the numbers 24 and 54? What is their
geometric mean?
6. Find two pairs of whole numbers (like 24 and 54) whose arithmetic mean
and geometric mean are both whole numbers.
7. Evaluate each of these integrals by sketching graphs of the corresponding
regions and using area formulas from plane geometry:
#9
(a) −1 5 dx
'
#7&
(b) 0 2 − 72 x dx
#5
(c) −3 (4 − |x − 1|) dx
#4
(d) 0 (8 − 2|x − 1| − 2|x − 3|) dx
#4√
(e) 1 −x2 + 8x − 7 dx
8. The “lune on the side of the equilateral triangle” has a quadrature which
is resolved by a procedure similar to the one Hippocrates gives for the
“lune on the side of the square.” Inscribe an equilateral triangle ABC in
a circle with center O and build semicircles on the sides of the triangles.
This creates three equal lunes, numbered 1, 2, 3 in the diagram below.
Also, let D be the point opposite B on the diameter of the large circle.
(a) What are the measures of the angles in triangle OAD? How can you
tell?
(b) What are the measures of the angles in triangle BAD? How can you
tell?
(c) Show that AD = 21 BD.
80
CHAPTER 5. HIPPOCRATES OF CHIOS
Figure 5.12: The lune on the side of an equilateral triangle.
(d) Let x = AB be the length of the side of the equilateral triangle ABC,
and let d = BD be the diameter of the large circle. Then AD = d2 .
We can apply the Pythagorean
Theorem to the right triangle BAD.
√
Show from this that x = 23 d.
(e) It follows that the ratio of the areas of the semicircle on diameter
AB to the semicircle on diameter BD is 34 . Why?
(f) Use the diagram to argue that, in terms of area,
3(semicircle onAB)+(triangleABC) = 3(lune)+2(semicircle onBD),
and from this show that
lune =
1
1
(triangleABC) + (semicircleBD).
3
12
The quadrature of this lune is dependent on finding the quadrature of
the circle that forms its inner edge. Therefore, unlike the quadrature
of Hippocrates’ lune, the quadrature of this lune is unresolved.
9. Hippocrates is said to have attempted the quadrature of another lune, the
“lune on the side of the hexagon,” which we repeat here. Inscribe a regular
hexagon (regular means that all the sides and interior angles are equal) in
a circle centered at C. Drawing in the diagonals that join opposite vertices
of the hexagon dissects the hexagon into six triangles. On each side of the
hexagon build a semicircle. This constructs six lunes, marked 1 through
6 in the diagram (Figure 5.13).
(a) In triangle ABC, like the other five triangles into which the hexagon
is dissected, is equilateral. Why?
(b) The semicircles containing lunes 1, 2, 3, and 4 have area equal to the
upper half of the large circle. Why?
7. EXERCISES
81
Figure 5.13: The lune on the side of a regular hexagon.
(c) The three lunes 1, 2, and 3, together with the semicircle containing
lune 4 have total area equal to the upper half of the hexagon (which
is a trapezoid). Why?
(d) It follows that the area of one of these lunes equals one-third the
difference between the area of the trapezoid and the area of the
semicircle that contains the lune. Why? Because this quadrature
is dependent on that of the circle that forms the inner edge of the
lune (just as in the previous exercise), the quadrature of this lune is
unresolved.
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CHAPTER 5. HIPPOCRATES OF CHIOS