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THE. 


erin OF ARCHIMEDES - 


RECENTLY DISCOVERED BY ΡΗΙΒΕΕΌ 


A - SUPPLEMENT. TO THE WORKS 


OF ARCHIMEDES L897. 


EDITED ΒΥ. 


Si THOMAS L. HEATH, 
| K.C.B., Sc.D., F.R.S. 


SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE 


τ Cambridge : 
at the University Press _ 
| | 1912 


Price Two Shillings and Sixpence ek | 


THE 
METHOD OF ARCHIMEDES 


CAMBRIDGE UNIVERSITY PRESS 
Hondon; FETTER LANE, E.C. 
C. F. CLAY, Manacrer 


€vinburgh: 100, PRINCES STREET 
Berlin: A. ASHER AND CO, 
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Bombay and Calcutta: MACMILLAN AND CO., Lrp. 


All rights reserved 


THE 
METHOD OF ARCHIMEDES 


RECENTLY DISCOVERED BY HEIBERG 


OU PPLEMENT TO THE WORKS 
OF ARCHIMEDES 1897 


EDEFED BY 


Sr THOMAS L. HEATH, © 
KC Ben SeD<  HiR:S. 


SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE 


Cambridge : 
at the University Press 


012 


Cambridge : 
PRINTED BY JOHN CLAY, M.A. 
AT THE UNIVERSITY PRESS 


150369 


INTRODUCTORY NOTE 


From the point of view of the student of Greek mathematics 
there has been, in recent years, no event comparable in interest 
with the discovery by Heiberg in 1906 of a Greek MS. containing, 
among other works of Archimedes, substantially the whole of a 
treatise which was formerly thought to be irretrievably lost. 

The full description of the MS. as given in the preface to Vol. 1. 
(1910) of the new edition of Heiberg’s text of Archimedes now in 
course of publication is— 

Codex rescriptus Metochii Constantinopolitani 8. Sepulchri 
monasterli Hierosolymitani 355, 4to. 

Heiberg has told the story of his discovery of this MS. and 
given a full description of it*. His attention having been called 
to a notice in Vol. 1v. (1899) of the ἹἹεροσολυμιτικὴ βιβλιοθήκη of 
Papadopulos Kerameus relating to a palimpsest of mathematical 
content, he at once inferred from a few specimen lines which were 
quoted that the MS. must contain something by Archimedes. As 
the result of inspection, at Constantinople, of the MS. itself, and 
by means of a photograph taken of it, he was able to see what it 
contained and to decipher much of the contents. This was in the 
year 1906, and he inspected the MS. once more in 1908. With 
the exception of the last leaves, 178 to 185, which are of paper 
of the 16th century, the MS. is of parchment and contains writings 
of Archimedes copied in a good hand of the 10th century, in two 
columns. An attempt was made (fortunately with only partial 
success) to wash out the old writing, and then the parchment was 
used again, for the purpose of writing a Euchologion thereon, in the 
12th—13th or 13th—14th centuries. The earlier writing appears 
with more or less clearness’ on most of the 177 leaves; only 29 
leaves are destitute of any trace of such writing; from 9 more 
it was hopelessly washed off; on a few more leaves only a few 
words can be made out; and again some 14 leaves have old writing 


* Hermes xu. 1907, pp. 235 sq. 


6 INTRODUCTORY NOTE 


upon them in a different hand and with no division into columns. 
All the rest is tolerably legible with the aid of a magnifying glass. 
Of the treatises of Archimedes which are found in other MSS., the 
new MS. contains, in great part, the books On the Sphere and 
Cylinder, almost the whole of the work On Spirals, and some parts 
of the Measurement of a Circle and of the books On the Equilibrium 
of Planes. But the important fact is that it contains (1) a con- 
siderable proportion of the work On Floating Bodies which was 
formerly supposed to be lost so far as the Greek text is concerned 
and only to have survived in the translation by Wilhelm von 
Morbeke, and (2), most precious of all, the greater part of the 
book called, according to its own heading, “E¢odos and elsewhere, 
alternatively, “E@odiov or “Edodixov, meaning Method. The portion 
of this latter work contained in the MS. has already been published 
by Heiberg (1) in Greek* and (2) in a German translation with 
commentary by Zeuthen7. The treatise was formerly only known 
by an allusion to it in Suidas, who says that Theodosius wrote a 
commentary upon it; but the Metrica of Heron, newly discovered 
by R. Schone and published in 1903, quotes three propositions from 
101, including the two main propositions enunciated by Archimedes 
at the beginning as theorems novel in character which the method 
furnished a means of investigating. Lastly the MS. contains two 
short propositions, in addition to the preface, of a work called 
Stomachion (as it might be “ Neck-Spiel” or “ Quél-Geist”) which 
treated of a sort of Chinese puzzle known afterwards by the name 
of “loculus Archimedius” ; it thus turns out that this puzzle, which 
Heiberg was formerly disinclined to attribute to Archimedes§, is 
really genuine. 

The Method, so happily recovered, is of the greatest interest for 
the following reason. Nothing is more characteristic of the classical 
works of the great geometers of Greece, or more tantalising, than 
the absence of any indication of the steps by which they worked 
their way to the discovery of their great theorems. As they have 
come down to us, these theorems are finished masterpieces which 
leave no traces of any rough-hewn stage, no hint of the method 
by which they were evolved. We cannot but suppose that the 


* Hermes xui1. 1907, pp. 243—297. 

+ Bibliotheca Mathematica viz, 1906-7, pp. 321—363. 

1 Heronis Alexandrini opera, Vol. 111. 1903, pp. 80, 17; 130, 15; 130, 25. 
§ Vide The Works of Archimedes, p. xxii. 


INTRODUCTORY NOTE 7 


Greeks had some method or methods of analysis hardly less powerful 
than those of modern analysis; yet, in general, they seem to have 
taken pains to clear away all traces of the machinery used and all 
the litter, so to speak, resulting from tentative efforts, before they 
permitted themselves to publish, in sequence carefully thought out, 
and with definitive and rigorously scientific proofs, the results 
obtained. A partial exception is now furnished by the Method; for 
here we have a sort of lifting of the veil, a glimpse of the interior 
of Archimedes’ workshop as it were. He tells us how he discovered 
certain theorems in quadrature and cubature, and he is at the same 
time careful to insist on the difference between (1) the means which 
may be sufficient to suggest the truth of theorems, although not 
furnishing scientific proofs of them, and (2) the rigorous demonstra- 
tions of them by irrefragable geometrical methods which must follow 
before they can be finally accepted as established ; to use Archi- 
medes’ own terms, the former enable theorems to be investigated 
(θεωρεῖν) but not to be proved (ἀποδεικνύναι). The mechanical 
method, then, used in our treatise and shown to be so useful for 
the discovery of theorems is distinctly said to be incapable of 
furnishing proofs of them; and Archimedes promises to add, as 
regards the two main theorems enunciated at the beginning, the 
necessary supplement in the shape of the formal geometrical proof. 
One of the two geometrical proofs is lost, but fragments of the other 
are contained in the MS. which are sufficient to show that the 
method was the orthodox method of exhaustion in the form in 
which Archimedes applies it elsewhere, and to enable the proof to 
be reconstructed. 

The rest of this note will be best understood after the treatise 
itself has been read; but the essential features of the mechanical 
method employed by Archimedes are these. Suppose X to be a 
plane or solid figure, the area or content of which has to be found. 
The method is to weigh infinitesimal elements of X (with or without 
the addition of the corresponding elements of another figure C’) 
against the corresponding elements of a figure δ, B and C being 
such figures that their areas or volumes, and the position of the 
centre of gravity of Bb, are known beforehand. For this purpose 
the figures are first placed in such a position that they have, as 
common diameter or axis, one and the same straight line; if then 
the infinitesimal elements are sections of the figures made by parallel 
planes perpendicular (in general) to the axis and cutting the figures, 


8 INTRODUCTORY NOTE 


the centres of gravity of all the elements lie at one point or other 
on the common diameter or axis. This diameter or axis is produced 
and is imagined to be the bar or lever of a balance. It is sufficient 
to take the simple case where the elements of X alone are weighed 
against the elements of another figure B. The elements which cor- 
respond to one another are the sections of X and B respectively by 
any one plane perpendicular (in general) to the diameter or axis 
and cutting both figures; the elements are spoken of as straight 
lines in the case of plane figures and as plane areas in the case of 
solid figures. Although Archimedes calls the elements straight lines 
and plane areas respectively, they are of course, in the first case, 
indefinitely narrow strips (areas) and, in the second case, indefinitely 
thin plane laminae (solids); but the breadth or thickness (da, as 
we might call it) does not enter into the calculation because it is 
regarded as the same in each of the two corresponding elements 
which are separately weighed against each other, and therefore 
divides out. The number of the elements in each figure is in- 
finite, but Archimedes has no need to say this; he merely says 
that X and B are made up of all the elements in them respectively, 
Le. of the straight lines in the case of areas and of the plane areas 
in the case of solids. 

The object of Archimedes is so to arrange the balancing of the 
elements that the elements of X are all applied at one point of 
the lever, while the elements of 6 operate at different points, 
namely where they actually are in the first instance. He con- 
trives therefore to move the elements of X away from their first 
position and to concentrate them at one point on the lever, while 
the elements of 6 are left where they are, and so operate at their 
respective centres of gravity. Since the centre of gravity of B as 
a whole is known, as well as its area or volume, it may then be 
supposed to act as one mass applied at its centre of gravity; and 
consequently, taking the whole bodies X and & as ultimately placed 
respectively, we know the distances of the two centres of gravity 
from the fulerum or point of suspension of the lever, and also the 
area or volume of δ. Hence the area or volume of X is found. 
The method may be applied, conversely, to the problem of finding 
the centre of gravity of X when its area or volume is known before- 
hand; in this case it is necessary that the elements of Y, and 
therefore X itself, should be weighed in the places where they are, 
and that the figures the elements of which are moved to one single 


τ a 


INTRODUCTORY NOTE 9 


point of the lever, to be weighed there, should be other figures and 
not X. 

The method will be seen to be, not wnétegration, as certain 
geometrical proofs in the great treatises actually are, but a clever 
device for avoiding the particular integration which would naturally 
be used to find directly the area or volume required, and making 
the solution depend, instead, upon another integration the result of 
which is already known. Archimedes deals with moments about 
the point of suspension of the lever, 1.6. the products of the ele- 
ments of area or volume into the distances between the point of 
suspension of the lever and the centres of gravity of the elements 
respectively ; and, as we said above, while these distances are 
different for all the elements of B, he contrives, by moving the 
elements of X, to make them the same for all the elements of X 
in their final position. He assumes, as known, the fact that the 
sum of the moments of each particle of the figure B acting at 
the point where it is placed is equal to the moment of the whole 
figure applied as one mass at one point, its centre of gravity. 

Suppose now that the element of X is τὺ. αἶα, εὐ being the length 
or area of a section of X by one of a whole series of parallel planes 
cutting the lever at right angles, x being measured along the lever 
(which is the common axis of the two figures) from the point of 
suspension of the lever as origin. This element is then supposed 
to be placed on the lever at a constant distance, say a, from the 
origin and on the opposite side of it from B. If w’.dax is the cor- 
responding element of cut off by the same plane and a its distance 
from the origin, Archimedes’ argument establishes the equation 

k k 

4 nde = | αἰ de. 

h h 
Now the second integral is known because the area or volume of 
the figure B (say a triangle, a pyramid, a prism, a sphere, a cone, 
or a cylinder) is known, and it can be supposed to be applied as 
one mass at its centre of gravity, which is also known; the integral 
is equal to 6U, where ὦ is the distance of the centre of gravity from 
the point of suspension of the lever, and U is the area or content 


of B. Hence 


the area or volume of X = : 


In the case where the elements of X are weighed along with the 
corresponding elements of another figure Οὐ against corresponding 


10 INTRODUCTORY NOTE 


elements of B, we have, if v be the element of C, and JV its area 


or content, 
k k k 
4 ude τα vde= | au'dx 
h h h 


and (area or volume of X + V)a=6bU. 


In the particular problems dealt with in the treatise h is always 
= 0, and & is often, but not always, equal to a. 

Our admiration of the genius of the greatest mathematician of 
antiquity must surely be increased, if that were possible, by a 
perusal of the work before us. Mathematicians will doubtless 
agree that it is astounding that Archimedes, writing (say) about 
250 B.c., should have been able to solve such problems as those of 
finding the volume and the centre of gravity of any segment of a 
sphere, and the centre of gravity of a semicircle, by a method so 
simple, a method too (be it observed) which would be quite rigorous 
enough for us to-day, although it did not satisfy Archimedes himself. 

Apart from the mathematical content of the book, it is in- 
teresting, not only for Archimedes’ explanations of the course which 
his investigations took, but also for the allusion to Democritus as 
the discoverer of the theorem that the volumes of a pyramid and 
a cone are one-third of the volumes of a prism and a cylinder 
respectively which have the same base and equal height. These 
propositions had always been supposed to be due to Eudoxus, and 
indeed Archimedes himself has a statement to this effect*. It 
now appears that, though Eudoxus was the first to prove them 
scientifically, Democritus was the first to assert their truth. I have 
elsewhere Ὁ made a suggestion as to the probable course of Democritus’ 
argument, which, on Archimedes’ view, did not amount to a proof 
of the propositions; but it may be well to re-state it here. Plutarch, 
in a well-known passage{, speaks of Democritus as having raised the 
following question in natural philosophy (φυσικῶς) : “if a cone were 
cut by a plane parallel to the base [by which is clearly meant a 
plane indefinitely near to the base], what must we think of the 
surfaces of the sections? Are they equal or unequal? For, if they 
are unequal, they will make the cone irregular, as having many 
indentations, like steps, and unevennesses; but, if they are equal, 
the sections will be equal, and the cone will appear to have the 
property of the cylinder and to be made up of equal, not unequal, 

* On the Sphere and Cylinder, Preface to Book 1. 


+ The Thirteen Books of Euclid’s Elements, Vol. 111. p. 368. 
+ Plutarch, De Comm. Not. adv. Stoicos xxx1x. 3. 


INTRODUCTORY NOTE 11 


circles, which is very absurd.” The phrase “made up of equal... 
circles” (ἐξ ἴσων συγκείμενος... κύκλων) shows that Democritus already 
had the idea of a solid being the sum of an infinite number of 
parallel planes, or indefinitely thin laminae, indefinitely near to- 
gether: a most important anticipation of the same thought which 
led to such fruitful results in Archimedes. If then we may make 
a conjecture as to Democritus’ argument with regard to a pyramid, 
it seems probable that he would notice that, if two pyramids of the 
same height and with equal triangular bases are respectively cut by 
planes parallel to the base and dividing the heights in the same 
ratio, the corresponding sections of the two pyramids are equal, 
whence he would infer that the pyramids are equal because they 
are the sums of the same infinite numbers of equal plane sections 
or indefinitely thin laminae. (This would be a particular anti- 
cipation of Cavalieri’s proposition that the areal or solid contents 
of two figures are equal if two sections of them taken at the same 
height, whatever the height may be, always give equal straight lines 
or equal surfaces respectively.) And Democritus would of course 
see that the three pyramids into which a prism on the same base 
and of equal height with the original pyramid is divided (as in 
Kucl. x11. 7) satisfy, in pairs, this test of equality, so that the 
pyramid would be one third part of the prism. The extension to 
a pyramid with a polygonal base would be easy. And Democritus 
may have stated the proposition for the cone (of course without an 
absolute proof) as a natural inference from the result of increasing 
indefinitely the number of sides in a regular polygon forming the 
base of a pyramid. 

In accordance with the plan adopted in The Works of Archimedes, 
I have marked by inverted commas the passages which, on account 
of their importance, historically or otherwise, I have translated 
literally from the Greek; the rest of the tract is reproduced in 
modern notation and phraseology. Words and sentences in square 
brackets represent for the most part Heiberg’s conjectural restoration 
(in his German translation) of what may be supposed to have been 
written in the places where the MS. is illegible; in a few cases 
where the gap is considerable a note in brackets indicates what the 
missing passage presumably contained and, so far as necessary, how 
the deficiency may be made good. 

ΤῸ. Ef. 
7 June 1912. 


THE METHOD OF ARCHIMEDES TREATING 
OF MECHANICAL PROBLEMS— 
TO ERATOSTHENES 


“ Archimedes to Eratosthenes greeting. 


I sent you on a former occasion some of the theorems 
discovered by me, merely writing out the enunciations and 
inviting you to discover the proofs, which at the moment 
I did not give. The enunciations of the theorems which I 
sent were as follows. 


1. If in a right prism with a parallelogrammic base a 
cylinder be inscribed which has its bases in the opposite 
parallelograms*, and its sides [1e. four generators] on the 
remaining planes (faces) of the prism, and if through the 
centre of the circle which is the base of the cylinder and 
(through) one side of the square in the plane opposite to 
it a plane be drawn, the plane so drawn will cut off from 
the cylinder a segment which is bounded by two planes 
and the surface of the cylinder, one of the two planes being 
the plane which has been drawn and the other the plane 
in which the base of the cylinder is, and the surface being 
that which is between the said planes; and the segment cut 
off from the cylinder is one sixth part of the whole prism. 


2. If in a cube a cylinder be inscribed which has its 
bases in the opposite parallelograms}+ and touches with its 
surface the remaining four planes (faces), and if there also 
be inscribed in the same cube another cylinder which has 
its bases in other parallelograms and touches with its surface 
the remaining four planes (faces), then the figure bounded 
by the surfaces of the cylinders, which is within both cylinders, 
is two-thirds of the whole cube. 

Now these theorems differ in character from those commu- 
nicated before ; for we compared the figures then in question, 


* The parallelograms are apparently squares. + i.e. squares, 


THE METHOD 13 


conoids and spheroids and segments of them, in respect of size, 
with figures of cones and cylinders: but none of those figures 
have yet been found to be equal to a solid figure bounded by 
planes; whereas each of the present figures bounded by two 
planes and surfaces of cylinders is found to be equal to one of 
the solid figures which are bounded by planes. The proofs then 
of these theorems I have written in this book and now send 
to you. Seeing moreover in you, as I say, an earnest student, 
a man of considerable eminence in philosophy, and an admirer 
[of mathematical inquiry], I thought fit to write out for you 
and explain in detail in the same book the peculiarity of a 
certain method, by which it will be possible for you to get 
a start to enable you to investigate some of the problems in 
mathematics by means of mechanics. This procedure is, I am 
persuaded, no less useful even for the proof of the theorems 
themselves ; for certain things first became clear to me by a 
mechanical method, although they had to be demonstrated by 
geometry afterwards because their investigation by the said 
method did not furnish an actual demonstration. But it is of 
course easier, when we have previously acquired, by the method, 
some knowledge of the questions, to supply the proof than 
it is to find it without any previous knowledge. This is a 
reason why, in the case of the theorems the proof of which 
Kudoxus was the first to discover, namely that the cone is 
a third part of the cylinder, and the pyramid of the prism, 
having the same base and equal height, we should give no 
small share of the credit to Democritus who was the first 
to make the assertion with regard to the said figure* though 
he did not prove it. I am myself in the position of having 
first made the discovery of the theorem now to be published 
[by the method indicated], and I deem it necessary to expound 
the method partly because I have already spoken of 107 and 
I do not want to be thought to have uttered vain words, but 


* περὶ τοῦ εἰρημένου σχήματος, in the singular. Possibly Archimedes may 


have thought of the case of the pyramid as being the more fundamental and as 
really involving that of the cone. Or perhaps “figure” may be intended for 
‘type of figure.” 

+ Cf. Preface to Quadrature of Parabola. 


14 ARCHIMEDES 


equally because I am persuaded that it will be of no little 
service to mathematics; for I apprehend that some, either of 
my contemporaries or of my successors, will, by means of the 
method when once established, be able to discover other 
theorems in addition, which have not yet occurred to me. 
First then I will set out the very first theorem which 
became known to me by means of mechanics, namely that 


Any segment of a section of a right-angled cone (1.e.a parabola) 
is four-thirds of the triangle which has the same base and equal 
height, 
and after this I will give each of the other theorems investi- 
gated by the same method. ‘Then, at the end of the book, 
I will give the geometrical [proofs of the propositions]... 

[I premise the following propositions which I shall use 
in the course of the work.] 

1. If from [one magnitude another magnitude be sub- 
tracted which has not the same centre of gravity, the centre 
of gravity of the remainder is found by] producing [the 
straight line joining the centres of gravity of the whole 
magnitude and of the subtracted part in the direction of 
the centre of gravity of the whole] and cutting off from it 
a length which has to the distance between the said centres 
of gravity the ratio which the weight of the subtracted 
magnitude has to the weight of the remainder. 

[On the Equilibrium of Planes, τ. 8] 

2. If the centres of gravity of any number of magnitudes 
whatever be on the same straight line, the centre of gravity 
of the magnitude made up of all of them will be on the same 


straight line. [Cf. Lbed. τ. 5] 
3. The centre of gravity of any straight line is the point 
of bisection of the straight line. [Cf. Ibid. τ. 4] 


4, The centre of gravity of any triangle is the point in 
which the straight lines drawn from the angular points of 
the triangle to the middle points of the (opposite) sides cut 
one another. [Lbid. τ. 13, 14] 

5. The centre of gravity of any parallelogram is the point 
in which the diagonals meet. [Lbid. 1. 10] 


THE METHOD 15 


6. The centre of gravity of a circle is the point which is 
also the centre [of the circle]. 


7. The centre of gravity of any cylinder is the point of 
bisection of the axis. 


8. The centre of gravity of any cone is [the point which 
divides its axis so that] the portion [adjacent to the vertex is] 
triple [of the portion adjacent to the base]. 


[All these propositions have already been] proved*. [Besides 
these I require also the following proposition, which is easily 
proved : 

If in two series of magnitudes those of the first series are, 
in order, proportional to those of the second series and further] 
the magnitudes [of the first series], either all or some of them, 
are in any ratio whatever [to those of a third series], and if the 
magnitudes of the second series are in the same ratio to the 
corresponding magnitudes [of a fourth series], then the sum 
of the magnitudes of the first series has to the sum of the 
selected magnitudes of the third series the same ratio which 
the sum of the magnitudes of the second series has to the 
sum of the (correspondingly) selected magnitudes of the fourth 
series. [On Conoids and Spheroids, Prop. 1.]” 


Proposition 1. 


Let ABC be a segment of a parabola bounded by the 
straight line AC and the parabola ABC, and let D be the 
middle point of AC. Draw the straight line DBE parallel 
to the axis of the parabola and join AB, BC. 


Then shall the segment ABC be 4 of the triangle ABC. 


From A draw AKF parallel to DEH, and let the tangent 
to the parabola at C meet DBH in HE and AKF in F. Produce 
CB to meet AF in K, and again produce CK to H, making 
KH equal to CK. 


* The problem of finding the centre of gravity of a cone is not solved in 
any extant work of Archimedes. It may have been solved either in a separate 
treatise, such as the περὶ ζυγῶν, which is lost, or perhaps in a larger mechanical 
work of which the extant books On the Equilibrium of Planes formed only a part. 


16 ARCHIMEDES 


Consider CH as the bar of a balance, K being its middle 
point. 

Let MO be any straight line parallel to HD, and let it meet 
CF, CK, AC in M, N, O and the curve in P. 


Now, since CE is a tangent to the parabola and CD the 
seml-ordinate, 
EB=BD; 
“for this is proved in the Elements [of Conics]*.” 
Since ΝΆ, MO are parallel to HD, it follows that 
ΚΞ ΚΑ, ΜΝ -ΝΟ. 
Now, by the property of the parabola, “ proved in a lemma,” 
MO:O0P=CA:AO [Cf. Quadrature of Parabola, Prop. 5] 
== OG KEN, [Ἐπ]. vi. 2] 
== ΗΑ: TEN, 


ΟΞ -ἰ 


Take a straight line 7G equal to OP, and place it with its 
centre of gravity at H, so that TH = HG; then, since WV is the 
centre of gravity of the straight line MO, 
and MO:TG=HK: KN, 

* i.e. the works on conics by Aristaeus and Euclid. Cf. the similar 


expression in On Conoids and Spheroids, Prop. 3, and Quadrature of Parabola, 
Prop. 3. 


THE METHOD iy 


it follows that 7G at H and MO at N will be in equilibrium 
about K. [On the Equilibrium of Planes, τ. 6, 7] 


Similarly, for all other straight lines parallel to DE and 
meeting the arc of the parabola, (1) the portion intercepted 
between FC, AC with its middle point on KC and (2) a 
length equal to the intercept between the curve and AC 
placed with its centre of gravity at H will be in equilibrium 
about Κ΄. 


Therefore K is the centre of gravity of the whole system 
consisting (1) of all the straight lines as /O intercepted between 
FC, AC and placed as they actually are in the figure and (2) of 
all the straight lines placed at H equal to the straight lines 
as PO intercepted between the curve and AC. 


And, since the triangle CF'A is made up of all the parallel 
lines like MO, 


and the segment CBA is made up of all the straight lines like 
PO within the curve, 


it follows that the triangle, placed where it is in the figure, is 
in equilibrium about K with the segment CBA placed with its 
centre of gravity at H. 

Divide KC at W so that CK =3KW; 


then W is the centre of gravity of the triangle ACF; “ for this 
is proved in the books on equilibrium” (ἐν τοῖς ἰσορροπικοῖς). 
[Cf. On the Equilibrium of Planes τ. 15] 


Therefore A ACF: (segment ABC) = HK : KW 


=e dle 
Therefore segment ABC=4A ACF. 
But AAOF=4A ABC. 
Therefore segment ABC = 4A ABC. 


“Now the fact here stated is not actually demonstrated 
by the argument used; but that argument has given a sort 
of indication that the conclusion is true. Seeing then that 
the theorem is not demonstrated, but at the same time 


H, A. 2 


18 ARCHIMEDES 


suspecting that the conclusion is true, we shall have recourse 
to the geometrical demonstration which I myself discovered 


and have already published*.” 


Proposition 2. 


We can investigate by the same method the propositions that 


(1) <Any sphere is (in respect of solid content) four times 
the cone with base equal to a great circle of the sphere and 
height equal to tts radius ; and 


(2) the cylinder with base equal to a great circle of the 
sphere and height equal to the diameter 1s 14 times the sphere. 


(1) Let ABCD be a great circle of a sphere, and AC, BD 


diameters at right angles to one another. 


Let a circle be drawn about BD as diameter and in a plane 
perpendicular to AQ, and on this circle as base let a cone 
be described with A as vertex. Let the surface of this cone 
be produced and then cut by a plane through C parallel to 
its base; the section will be a circle on HF as diameter. On 
this circle as base let a cylinder be erected with height and 
axis AC, and produce CA to H, making AH equal to CA. 


Let CH be regarded as the bar of a balance, A being its 
middle point. 


Draw any straight line MN in the plane of the circle 
ABCD and parallel to BD. Let MN meet the circle in O, P, 
the diameter AC in S, and the straight ines AH, AF in Q, R 


respectively. Join AO. 


* The word governing τὴν γεωμετρουμένην ἀπόδειξιν in the Greek text is 
τάξομεν, a reading which seems to be doubtful and is certainly difficult to 
translate. Heiberg translates as if τάξομεν meant ‘‘ we shall give lower down”’ 
or ‘‘later on,” but I agree with Th. Reinach (Revue générale des sciences pures 
et appliquées, 30 November 1907, p. 918) that it is questionable whether 
Archimedes would really have written out in full once more, as an appendix, 
a proof which, as he says, had already been published (i.e. presumably in the 
Quadrature of a Parabola). τάξομεν, if correct, should apparently mean ‘‘ we 
shall appoint,” ‘‘ prescribe” or ‘‘ assign.” 


THE METHOD 19 


Through MN draw a plane at right angles to AC; 


this plane will cut the cylinder in a circle with diameter MN, 
the sphere in a circle with diameter OP, and the cone in a 
circle with diameter QR. 


Now, since MS= AC, and QS = AS, 
MS.SQ=CA.AS 
eee 
= OS’+ SQ? 


ΙΗ 


And, since HA = AC, 
ia. as —CA: AS 
=MS:SQ 
= MS?: MS.SQ 
= MS? : (OS?+SQ?), from above, 
= MN? : (OP? + QR’) 
=(circle, diam. MI) : (circle, diam. OP 
+ circle, diam. QF). 
That is, 
HA: AS = (circle in cylinder) : (circle in sphere + circle in cone). 
Therefore the circle in the cylinder, placed where it is, is 
in equilibrium, about A, with the circle in the sphere together 
2---ὦ 


20 ARCHIMEDES 


with the circle in the cone, if both the latter circles are placed 
with their centres of gravity at H. 

Similarly for the three corresponding sections made by a 

plane perpendicular to AC and passing through any other 
straight line in the parallelogram LF parallel to HF. 
‘Tf we deal in the same way with all the sets of three circles 
in which planes perpendicular to AC cut the cylinder, the 
sphere and the cone, and which make up those solids respec- 
tively, it follows that the cylinder, in the place where it is, will 
be in equilibrium about A with the sphere and the cone together, 
when both are placed with their centres of gravity at H. 

Therefore, since A’ is the centre of gravity of the cylinder, 

HA : AK =(cylinder) : (sphere + cone AHF). 

But HA =2AK; 
therefore cylinder = 2 (sphere + cone AEF), 

Now cylinder = ὃ (cone AEF); [Eucl. x1. 10] 
therefore cone AHF =2 (sphere). 

But, since HF'=2BD, 

cone AHF -- (cone ABD); 
therefore sphere = 4 (cone ABD). 


(2) Through B, D draw VBW, XDY parallel to AC; 


and imagine a cylinder which has AC for axis and the circles 
on VX, WY as diameters for bases. 


Then cylinder VY = 2 (cylinder VD) 
= 6 (cone ABD) [πο]. x1. 10] 


= 3 (sphere), from above. 
Q.E.D. 
“From this theorem, to the effect that a sphere is four 
times as great as the cone with a great circle of the sphere as 
base and with height equal to the radius of the sphere, I con- 
ceived the notion that the surface of any sphere is four times as 
great as a great circle in it; for, judging from the fact that any 
circle is equal to a triangle with base equal to the circumference 
and height equal to the radius of the circle, I apprehended 


THE METHOD 21 


that, in like manner, any sphere is equal to a cone with base 
equal to the surface of the sphere and height equal to the 
radius *,” 


Proposition 3S. 


By this method we can also investigate the theorem that 


A cylinder with base equal to the greatest circle in u spheroid 
and height equal to the axis of the spheroid is 14 times the 
spheroid ; 
and, when this is established, it is plain that 


If any spheroid be cut by a plane through the centre and at 
right angles to the axis, the half of the spheroid is double of the 
cone which has the same base and the same aais as the segment 


(9. the half of the spheroid). 


Let a plane through the axis of a spheroid cut its surface in 
the ellipse ABCD, the diameters (i.e. axes) of which are AC, 
BD; and let K be the centre. 


Draw a circle about BD as diameter and in a plane per- 
pendicular to AC; 


imagine a cone with this circle as base and A as vertex 
produced and cut by a plane through C parallel to its base; 
the section will be a circle in a plane at right angles to AC 
and about HF as diameter. 


Imagine a cylinder with the latter circle as base and axis 


AC; produce CA to H, making AH equal to CA. 


Let HC be regarded as the bar of a balance, A being its 
middle point. 


In the parallelogram ZF draw any straight lmne MN 
parallel to HF meeting the ellipse in O, P and AH, AF, AC in 
Q, R, S respectively. 


* That is to say, Archimedes originally solved the problem of finding the 
solid content of a sphere before that of finding its surface, and he inferred the 
result of the latter problem from that of the former. Yet in On the Sphere and 
Cylinder 1. the surface is independently found (Prop. 33) and before the 
volume, which is found in Prop. 34: another illustration of the fact that the 
order of propositions in the treatises of the Greek geometers as finally 
elaborated does not necessarily follow the order of discovery. 


22 ARCHIMEDES : 


If now a plane be drawn through MW at right angles to 
AC, it will cut the cylinder in a circle with diameter MN, the 
spheroid in a circle with diameter OP, and the cone in a circle 
with diameter QR. 


Since HA = AC, 
HA: AS=CA: AS 
SPU 2 ALL 
— ES : 50. 
Therefore HA: AS=MS’*: MS. SQ. 


H 


But, by the property of the ellipse, 
ANS SC: SO-—= A Ke ane 


= 45" : SQ’; 
therefore SQ’: SO?=AS? : AS. SC 

=8Q’ : SQ . QM, 
and accordingly SO’=SQ. QM. 


Add SQ’ to each side, and we have 
SO’? + 8Q’=SQ .SM. 
Therefore, from above, we have 
HA:AS= MS? : (SO? + SQ?) 
= MN’ : (OP? + QR?) 
=(circle, diam. MN ):(circle, diam. OP + circle,diam. QR) 


THE METHOD 23 


That is, 
HA:AS= (circle in cylinder) : (circle in spheroid + circle in cone). 
Therefore the circle in the cylinder, in the place where it is, 
is in equilibrium, about A, with the circle in the spheroid and 
the circle in the cone together, if both the latter circles are 
placed with their centres of gravity at ZH. 


Similarly for the three corresponding sections made by a 
plane perpendicular to AC and passing through any other 
straight line in the parallelogram LF parallel to EF. 


If we deal in the same way with all the sets of three circles 
in which planes perpendicular to AC cut the cylinder, the 
spheroid and the cone, and which make up those figures 
respectively, it follows that the cylinder, in the place where it 
is, will be in equilibrium about A with the spheroid and the 
cone together, when both are placed with their centres of 
gravity at H. 


Therefore, since K is the centre of gravity of the cylinder, 
HA : AK =(cylinder) : (spheroid + cone AFF). 
But HA =2AK; 


therefore cylinder = 2 (spheroid + cone A/F). 
And cylinder =3 (cone ALF); [Eucl. x11. 10] 
therefore cone AHF = 2 (spheroid). 


But, since HF'= 281), 
cone AHF’ =8 (cone ABD); 
therefore spheroid = 4 (cone ABD), 
and half the spheroid = 2 (cone ABD). 


Through B, D draw VBW, XDY parallel to AC; 
and imagine a cylinder which has AC for axis and the circles 
on VX, WY as diameters for bases. 
Then cylinder VY =2 (cylinder VD) 
=6 (cone ABD) 


= 3 (spheroid), from above. 
. Q.E.D. 


24, ARCHIMEDES 


Proposition 4. 


Any segment of a right-angled conoid (1.6. a paraboloid of 
revolution) cut off by a plane at right angles to the ams is 
14 times the cone which has the same base and the same aais 
as the segment. 

This can be investigated by our method, as follows. 

Let a paraboloid of revolution be cut by a plane through 
the axis in the parabola BAC; 
and let it also be cut by another plane at right angles to the 
axis and intersecting the former plane in BC. Produce DA, 
the axis of the segment, to H, making HA equal to AD. 


H 


Imagine that HD is the bar of a balance, A being its 
middle point. 


The base of the segment being the circle on BC as diameter 
and in a plane perpendicular to AD, 


imagine (1) a cone drawn with the latter circle as base and A 
as vertex, and (2) a cylinder with the same circle as base and 
AD as axis. 

In the parallelogram £C let any straight line MN be drawn 
parallel to BC, and through MW let a plane be drawn at right 
angles to AD; this plane will cut the cylinder*in a circle with 
diameter MN and the paraboloid in a circle with diameter OP. 


ὙΦ αμβμ 


THE METHOD 25 


Now, BAC being a parabola and BD, OS ordinates, 
DA AS = BD? = OS", 
ΤῊ ΗΑ: AS = MS? : 80" 
Therefore 
HA : AS=(circle, rad. MS) : (circle, rad. OS) 
=(circle in cylinder) : (circle in paraboloid). 

Therefore the circle in the cylinder, in the place where it is, 
will be in equilibrium, about A, with the circle in the paraboloid, 
if the latter is placed with its centre of gravity at H. 

Similarly for the two corresponding circular sections made 
by a plane perpendicular to AD and passing through any other 
straight line in the parallelogram which is parallel to BC. 

Therefore, as usual, if we take all the circles making up the 
whole cylinder and the whole segment and treat them in the 
same way, we find that the cylinder, in the place where it is, 
is in equilibrium about A with the segment placed with its 
centre of gravity at H. 

If K is the middle point of AD, K is the centre of gravity 
of the cylinder ; 
therefore HA : AK =(cylinder) : (segment). 

Therefore cylinder = 2 (segment). 

And cylinder = 3 (cone ABC); [Eucl. xir. 10] 
therefore segment = 3 (cone ABC). 


Proposition 5. 


The centre of gravity of a segment of a right-angled conoid 
(i.e. a paraboloid of revolution) cut off by a plane at right angles 
to the axis is on the straight line which is the amis of the segment, 
and divides the said straight line in such a way that the portion 
of it adjacent to the vertex 1s double of the remaining portion. 

This can be investigated by the method, as follows. 

Let a paraboloid of revolution be cut by a plane through 
the axis in the parabola BAC; 
and let it also be cut by another plane at right angles to the 
axis and intersecting the former plane in BC. : 


20 ARCHIMEDES 


Produce DA, the axis of the segment, to H, making HA 
equal to AD; and imagine DH to be the bar of a balance, its 
middle point being A. 

The base of the segment being the circle on BC as diameter 
and in a plane perpendicular to AD, 


imagine a cone with this circle as base and A as vertex, so that 
AB, AC are generators of the cone. 


In the parabola let any double ordinate OP be drawn 
meeting AB, AD, AC in Q, S, R respectively. 


H 


B D Cc 


Now, from the property of the parabola, 
JID OS δά 2 ΑΚ 


= Βὴ 3 OS 
= BD’: BD. QS. 
Therefore OS’? = BD. QS, 
or ΒΒ: Οδξε Ο : οὗ, 
whence BD : QS = OS’: QS’. 
But BD: QS=AD: AS 
SCAR AUS: 


Therefore ΠΑ: TAS 0 Sen) Se 
= (ORs Ne, 


THE METHOD ΟἿ 


If now through OP a plane be drawn at right angles to 
AD, this plane cuts the paraboloid in a circle with diameter 
OP and the cone in a circle with diameter QR. 

We see therefore that 

ΠΑ: AS=(cirele, diam. OP): (circle, diam. QR) 
= (circle in paraboloid) : (circle in cone); 
and the circle in the paraboloid, in the place where it is, is in 
equilibrium about A with the circle in the cone placed with its 
centre of gravity at H. 


Similarly for the two corresponding circular sections made 
by a plane perpendicular to AD and passing through any other 
ordinate of the parabola. 

Dealing therefore in the same way with all the circular 
sections which make up the whole of the segment of the 
paraboloid and the cone respectively, we see that the segment 
of the paraboloid, in the place where it is, is in equilibrium 
about A with the cone placed with its centre of gravity at ἢ]. 

Now, since A is the centre of gravity of the whole system 
as placed, and the centre of gravity of part of it, namely the 
cone, as placed, is at H, the centre of gravity of the rest, 
namely the segment, is at a point K on HA produced such 
that 

HA : AK =(segment) : (cone). 

But segment = 3 (cone). [ Prop. 4] 

Therefore HA = AK; 
that is, K divides AD in such a way that AK =2KD. 


Proposition 6. 


The centre of gravity of any hemisphere [as on the \traight 
line which] is its axis, and divides the said straight line in such 
a way that the portion of τὲ adjacent to the surface of the 
hemisphere has to the remaining portion the ratio which 5 has 
to 3. 

Let a sphere be cut by a plane through its centre in the 
circle ABCD; 


28 ARCHIMEDES 


let AC, BD be perpendicular diameters of this circle, 
and through BD let a plane be drawn at right angles to AC. 
The latter plane will cut the sphere in a circle on BD as 
diameter. 
Imagine a cone with the latter circle as base and A as 
vertex. 
Produce C'A to H, making AH equal to CA, and let HC be 
regarded as the bar of a balance, A being its middle point. 


In the semicircle BAD, let any 
straight line OP be drawn parallel to 
BD and cutting AC in # and the two 
generators AB, AD of the cone in Q, R 
respectively. Join AO. 

Through OP let a plane be drawn 
at right angles to AC; 
this plane will cut the hemisphere in a 
circle with diameter OP and the cone 
in a circle with diameter QR. 

Now 
HA:AK=AC:AE 

= AGEs 25" 

=(OL°+ AE’): AE’ 

=(OL" + QE’): QE? 

= (circle, diam. OP + circle,diam. QR): (circle, diam. QR). 

Therefore the circles with diameters OP, QR, in the places 
where they are, are in equilibrium about A with the circle with 
diameter QR if the latter is placed with its centre of gravity 
at H. 

And, since the centre of gravity of the two circles with 
diameters OP, QF taken together, in the place where they are, 


[There is a lacuna here; but the proof can easily be com- 
pleted on the lines of the corresponding but more difficult case 
in Prop. 8. 

We proceed thus from the point where the circles with 
diameters OP, (Qk, in the place where they are, balance, about A, 


THE METHOD 29 


the circle with diameter QR placed with its centre of gravity 
at ΗΠ. 

A similar relation holds for all the other sets of circular 
sections made by other planes passing through points on AG 
and at right angles to AG. 

Taking then all the circles which fill up the hemisphere 
BAD and the cone ABD respectively, we find that 
the hemisphere BAD and the cone ABD, in the places where 
they are, together balance, about A, a cone equal to ABD placed 
with its centre of gravity at ἢ. 

Let the cylinder + NV be equal to the cone ABD. 

Then, since the cylinder +N placed with its centre of 
gravity at H balances the hemisphere BAD and the cone ABD 
in the places where they are, 
suppose that the portion M of the cylinder, placed with its 
centre of gravity at H, balances the cone ABD (alone) in the 
place where it is; therefore the portion NV of the cylinder placed 
with its centre of gravity at H balances the hemisphere (alone) 
in the place where it 1s. 

Now the centre of gravity of the cone is at a point V such 
that AG=4GV ; 
therefore, since M at ἢ is in equilibrium with the cone, 

M : (cone) =2AG: HA =3AC: AC, 
whence M = (cone). 

But Μ-- Ν =(cone); therefore V-= (cone). 

Now let the centre of gravity of the hemisphere be at W, 
which is somewhere on AG. 

Then, since V at H balances the hemisphere alone, 

(hemisphere): V= HA: AW. 

But the hemisphere BAD = twice the cone ABD; 

[On the Sphere and Cylinder τ. 34 and Prop. 2 above] 
and NV = (cone), from above. 

Therefore De Bit Al eA. W. 

=2AG: AW, 
whence 4 W=8AG, so that W divides AG in such a way that 
AW: WG=5:3.] 


30 ARCHIMEDES 


Proposition 7. 


We can also investigate by the same method the theorem 
that 

[Any segment of a sphere has] to the cone [with the same 
base and height the ratio which the sum of the radius of the 
sphere and the height of the complementary segment has to 
the height of the complementary segment.| 

[There is a lacuna here; but all that is missing is the 
construction, and the construction is easily understood by 


H 


means of the figure. BAD is of course the segment of the 
sphere the volume of which is to be compared with the volume 
of a cone with the same base and height. ] 

The plane drawn through MW and at right angles to AC 
will cut the cylinder in a circle with diameter MJ, the segment 
of the sphere in a circle with diameter OP, and the cone on 
the base HF in a circle with diameter QR. 

In the same way as before [cf. Prop. 2] we can prove that 
the circle with diameter MN, in the place where it is, is in 


THE METHOD 91 


equilibrium about A with the two circles with diameters OP, 
QF if these circles are both moved and placed with their centres 
of gravity at ἢ. 

The same thing can be proved of all sets of three circles 
in which the cylinder, the segment of the sphere, and the 
cone with the common height AG are all cut by any plane 
perpendicular to AC. 

Since then the sets of circles make up the whole cylinder, 
the whole segment of the sphere and the whole cone respec- 
tively, it follows that the cylinder, in the place where it is, 
is in equilibrium about A with the sum of the segment of 
the sphere and the cone if both are placed with their centres 
of gravity at H. 

Divide AG at W, V in such a way that 

AW=WG, AV=38VG. 

Therefore W will be the centre of gravity of the cylinder, 
and V will be the centre of gravity of the cone. 

Since, now, the bodies are in equilibrium as described, 

(cylinder) : (cone AL #’'+ segment BAD of sphere) 
=A AW. 

[The rest of the proof is lost; but it can easily be supplied 
thus. 


We have 
(cone AHF +segmt. BAD) : (cylinder) = AW: AC 
=e We Ae ees 
But (cylinder) : (cone AHF) = AC? : 4#G? 
= AC?: 14." 


Therefore, ea aequali, 
(cone AHF +segmt.BAD):(cone AEF)=AW.AC:14AG? 
=4AC: 1AG, 
whence (segmt. BAD): (cone AFF) =(4AC—1 AG): 44G. 
Again (cone AEF) : (cone ABD) = EG? : DG? 
Ξ- 463: 46. 66 
ΞΟ 
=41AG: 160. 


9.2 ARCHIMEDES 


Therefore, ex aequalt, 
(segment BAD) : (cone ABD) =($AC —4 AG) : τσ 
= (3A4C0-— AG): GC 
= (406 + GC): GC. 
Q.E.D. | 
Proposition 8. 


[The enunciation, the setting-out, and a few words of the 
construction are missing. 

The enunciation however can be supplied from that of 
Prop. 9, with which it must be identical except that it cannot 
refer to “any segment,” and the presumption therefore is that 
the proposition was enunciated with reference to one kind of 
segment only, 1.6. either a segment greater than a hemisphere 
or a segment less than a hemisphere. 

Heiberg’s figure corresponds to the case of a segment 
greater than a hemisphere. The 
segment investigated is of course 
the segment BAD. The setting- 
out and construction are self- 
evident from the figure.] 

Produce AC to H, O, making 
HA equal to AC and CO equal 
to the radius of the sphere ; 


and let HC be regarded as the 
bar of a balance, the middle point 
being A. 

In the plane cutting off the 
segment describe a circle with G 
as centre and radius (GZ) equal 
to AG; and on this circle as 
base, and with A as vertex, let 
a cone be described. AKL, AF 


are generators of this cone. 


Draw KL, through any point 
Q on AG, parallel to HF and cutting the segment in K, L, and 
AE, AF in R, P respectively. Join AK. 


THE METHOD 99 


Now HA»AQ=CA:AQ 
=Ak?: AQ? 
= (KQ + QA?) : QA? 
= (KQ? + PQ?) : PY? 
=(cirele, diam. KL + circle, diam. PR) 
: (circle, diam. PR). 

Imagine a circle equal to the circle with diameter PR 
placed with its centre of gravity at H; 
therefore the circles on diameters KL, PR, in the places where 
they are, are in equilibrium about A with the circle with 
diameter PR placed with its centre of gravity at H. 

Similarly for the corresponding circular sections made by 
any other plane perpendicular to AG. 

Therefore, taking all the circular sections which make up 
the segment ABD of the sphere and the cone AHF respec- 
tively, we find that the segment ABD of the sphere and the 
cone AHF, in the places where they are, are in equilibrium 
with the cone AHF assumed to be placed with its centre of 
gravity at ἢ. 

Let the cylinder +N be equal to the cone AHF which 
has A for vertex and the circle on HF as diameter for base. 

Divide AG at V so that AG=4VG; 
therefore V is the centre of gravity of the cone AHF’; “for 
this has been proved before*.” 

Let the cylinder M+ NV be cut by a plane perpendicular to 
the axis in such a way that the cylinder 9 (alone), placed with 
its centre of gravity at H, is in equilibrium with the cone AHF. 

Since +N suspended at ἢ is in equilibrium with the 
segment ABD of the sphere and the cone AZ#F in the places 
where they are, 
while MV, also at H, is in equilibrium with the cone AHF in 
the place where it is, it follows that 

N at Η is in equilibrium with the “segment ABD of the 
sphere in the place where it 1s. 


* Cf, note on p. 15 above, 


34 ARCHIMEDES 


Now (segment ABD of sphere) : (cone ABD) 
= (OG 2 GU: 
“for this is already proved” [Cf On the Sphere and Cylinder 
1. 2 Cor. as well as Prop. 7 ante]. 
And (cone ABD): (cone AEF) 
= (circle, diam. BD) : (circle, diam. HF) 
= BD EB 
= Βα": GE? 
= (WG! (GA) 3 CrP 
= OG 3 GAL 
Therefore, ex aequali, 
(segment ABD of sphere) : (cone AHF’) 
ΞΞΟ ΤΟΣ 
Take a point W on AG such that 
AW: WG=(GA + 4GC) : (GA + 260}. 
We have then, inversely, 
GW: WA=(2GC + GA): (466 + GA), 
and, componendo, 
GA :AW = (6GC + 2GA) : (4GC + GA). 
But GO =4(6GC+2GA), [for GO—-GC=4 (CG+GA)] 
and CV =} (460+ GA); 
therefore GAT AWS 0G Gs 
and, alternately and ‘inversely, 
Οα -Ο4 Ξ- ΑΥ̓͂ : ἹΥΔ4. 
It follows, from above, that 
(segment ABD of sphere) : (cone AE F)=CV: WA. 
Now, since the cylinder M with its centre of gravity at H 
is in equilibrium about A with the cone AHF with its centre 
of gravity at V, 
(cone A/F) : (cylinder M)= HA: AV 
=CA:AV; 
and, since the cone AHF =the cylinder M+WN, we have, 
dividendo and invertendo, 


(cylinder M7) : (cylinder N)= AV: CV, 


THE METHOD 35 


Hence, componendo, — 


ELE AEF) : (cylinder Ν᾽ ye ΟἿ Cie 
= HA: CV. 
But it was proved that 
(segment ABD of sphere): (cone AFF)=CV: WA; 
therefore, ex aequalt, 
(segment ABD of sphere) : (cylinder V)= HA: AW. 
And it was above proved that the cylinder V at Π is 
in equilibrium about A with the segment ABD, in the place 
where it is; 
therefore, since H is the centre of gravity of the cylinder Ν, 
W is the centre of gravity of the segment ABD of the sphere. 


Proposition 9. 

In the same way we can investigate the theorem that 

The centre of gravity of any segment of a sphere is on the 
straight line which is the axis of the segment, and divides this 
straight line in such a way that the part of it adjacent to the 
vertex of the segment has to the remaining part the ratio which 
the sum of the axis of the segment and four times the axis of 
the complementary segment has to the sum of the aais of the 
segment and double the axis of the complementary segment. 

[As this theorem relates to “any segment” but states the 
same result as that proved in the preceding proposition, it 
follows that Prop. 8 must have related to one kind of segment, 
either a segment greater than a semicircle (as in Heiberg’s 
figure of Prop. 8) or a segment less than a semicircle; and 
the present proposition completed the proof for both kinds of 
segments. It would only require a slight change in the figure, 
in any case. | 


Proposition 10. 
By this method too we can investigate the theorem that 
[A segment of an obtuse-angled conoid (1.6. a hyperboloid of 
revolution) has to the cone which has] the same base [as the 


* Archimedes arrives at this result in a very roundabout way, seeing that it 
could have been obtained at once convertendo. Cf. Kuclid x. 14. 


3—2 


90 ARCHIMEDES 


segment and equal height the same ratio as the sum of the axis 
of the segment and three times] the “annex to the axis” (1.6. half 
the transverse axis of the hyperbolic section through the axis of 
the hyperboloid or, in other words, the distance between the 
vertex of the segment and the vertex of the enveloping cone) has to 
the sum of the axis of the segment and double of the “annex” * 
[this is the theorem proved in On Conoids and Spherovds, 
Prop. 25], “and also many other theorems, which, as the method 
has been made clear by means of the foregoing examples, I will 
omit, in order that I may now proceed to compass the proofs 
of the theorems mentioned above.” 


Proposition 11. 


If in a right prism with square bases a cylinder be inscribed 
having its bases in opposite square faces and touching with tts 
surface the remaimng four parallelogrammic faces, and if 
through the centre of the circle which ws the base of the cylinder 
and one side of the opposite square face a plane be drawn, the 
figure cut off by the plane so drawn is one siath part of the 
whole prism. 


“This can be investigated by the method, and, when it 
is set out, I will go back to the Pree! of it by geometrical 
considerations.” 


[The investigation by the mechanical method is contained 
in the two Propositions, 11,12. Prop. 13 gives another solution 
which, although it contains no mechanics, is still of the character 
which Archimedes regards as inconclusive, since it assumes that 
the solid is actually made up of parallel plane sections and that 
an auxiliary parabola is actually made up of parallel straight 
lines in it. Prop. 14 added the conclusive geometrical proof.] 


Let there be a right prism with a cylinder inscribed as 
stated. 


* The text has “triple” (τριπλασίαν) in the last line instead of ‘‘ double.” 
As there is a considerable lacuna before the last few lines, a theorem about the 


centre of gravity of a segment of a hyperboloid of revolution may have fallen 
out. 


THE METHOD 37 


Let the prism be cut through the axis of the prism and 
cylinder by a plane perpendicular 
to the plane which cuts off the 


D B 
portion of the cylinder; let this 
plane make, as section, the paral- 
lelogram AB, and let it cut the 
plane cutting off the portion of / 
the cylinder (which plane is per- F τ. Ε 


pendicular to AB) in the straight 
line BC. 

Let CD be the axis of the 
prism and cylinder, let HF’ bisect 
it at right angles, and through 
EF Ἰοὺ a plane be drawn at right 
angles to CD; this plane will cut the prism in a square and 
the cylinder in a circle. 


Let MN be the square and OPQR the circle, and let the 
circle touch the sides of the square 
in O, P, Q, R [F, F in the first 
figure are identical with O, Q 
respectively]. Let H be the centre 
of the circle. 


Let KL be the intersection of 
the plane through HF’ perpen- 
dicular to the axis of the cylinder 
and the plane cutting off the 
portion of the cylinder; KL is 
bisected by OHQ [and passes 
through the middle point of HQ]. 


Let any chord of the circle, as ST’, be drawn perpendicular 
to HQ, meeting HQ in W; 
and through S87’ let a plane be drawn at right angles to OQ 
and produced on both sides of the plane of the circle OPQR. 


The plane so drawn will cut the half cylinder having the 
semicircle PQR for section and the axis of the prism for height 
in a parallelogram, one side of which is equal to ST and another 
is a generator of the cylinder; and it will also cut the portion 


A Cc Ny: 


38 ARCHIMEDES 


of the cylinder cut off in a parallelogram, one side of which is 
equal to ST and the other is equal and parallel to UV (in the 
first figure). 


UV will be parallel to BY and will cut off, along HG in the 
parallelogram DE, the segment HI equal to QW. 


_ Now, since HC is a parallelogram, and VJ is parallel to GC, 
μα :-61- γα : ΟΥ 
ΞΡ: UN 
ΞΞ ((7 in half cyl.) : ([7] in portion of cyl.). 
And EG = HQ, GI=HW, QH=OH; 
therefore OH: HW=(C in half cyl.) : (© in portion). 


Imagine that the parallelogram in the portion of the 
cylinder is moved and placed at O so that O is its. centre 
of gravity, and that OQ is the bar of a balance, H being its 
middle point. 


Then, since W is the centre of gravity of the parallelogram 
in the half cylinder, it follows from the above that the paral- 
lelogram in the half cylinder, in the place where it is, with its 
centre of gravity at W, is in equilibrium about H with the 
parallelogram in the portion of the cylinder when placed with 
its centre of gravity at O. 

Similarly for the other parallelogrammic sections made by 
any plane perpendicular to OQ and passing through any other 

chord in the semicircle PQR perpendicular to OQ. 

If then we take all the parallelograms making up the half 
cylinder and the portion of the cylinder respectively, it follows 
that the half cylinder, in the place where it is, is in equilibrium 
about H with the portion of the cylinder cut off when the 
latter is placed with its centre of gravity at Ὁ. 


Proposition 12. 


' Let the parallelogram (square) MN perpendicular to the 
axis, with the circle OPQR and its diameters OQ, PR, be 


drawn separately. 


THE METHOD 39 


Join HG, HM, and through them draw planes at right 
angles to the plane of the circle, 
producing them on both sides of 
that plane. 

This produces a prism with 
triangular section GHM and height 
equal to the axis of the cylinder; 
this prism is 1 of the original 
prism circumscribing the cylinder. 

Let LK, UT be drawn parallel 
to OQ and equidistant from it, 
cutting the circle in K, 7, RP 
in S, F, and GH, HM in W, V respectively. 

Through LK, UT draw planes at right angles to PR, 
producing them on both sides of the plane of the circle; — 


these planes produce as sections in the half cylinder PQR 
and in the prism GHWM four parallelograms in which the 
heights are equal to the axis of the cylinder, and the other 
sides are equal to KS, TF, LW, UV respectively............. 


[The rest of the proof is missing, but, as Zeuthen says*, 
the result obtained and the method of arriving at it are plainly 
indicated by the above. 

Archimedes wishes to prove that the half cylinder PQR, in 
the place where it is, balances the prism GHW, in the place 
where it is, about H as fixed point. 

He has first to prove that the elements (1) the parallelo- 
gram with side = KS and (2) the parallelogram with side = LW, 
in the places where they are, balance about S, or, in other 
words that the straight lines SK, LW, in the places where 
they are, balance about S. 

Now (radius of circle OPQR)? = SK? + SH?, 


or SL? = SK? + SW? 
Therefore LS? —SW?=SK?, 

and accordingly (2S+SW).LW=SK?, 

whence 4(18S+8W):4SK=SK : LW. 


* Zeuthen in Bibliotheca Mathematica vi13, 1906-7, pp. 352-3. 


40 ARCHIMEDES 


And §(ZS+SW) is the distance of the centre of gravity 
of LW from S, 
while 48K is the distance of the centre of gravity of SK from S. 

Therefore Sk and LW, in the places where they are, 
balance about S. 

Similarly for the corresponding parallelograms. 

Taking all the parallelogrammic elements in the half 
cylinder and prism respectively, we find that 
the half cylinder PQA and the prism GHW, in the places 
where they are respectively, balance about ἢ. 

From this result and that of Prop. 11 we can at once deduce 
the volume of the portion cut off from the cylinder. For in 
Prop. 11 the portion of the cylinder, placed with its centre of 
gravity at O, is shown to balance (about H) the half-cylinder 
in the place where it is. By Prop. 12 we may substitute for 
the half-cylinder in the place where it is the prism GHM 
of that proposition turned the opposite way relatively to RP. 
The centre of gravity of the prism as thus placed is at a point 
(say Z) on HQ such that HZ = 2HQ. 

Therefore, assuming the prism to be applied at its centre of 
gravity, we have 

(portion of cylinder) : (prism) = 2HQ : OH 
ΞΞΡ 
therefore (portion of cylinder) = 3 (prism GHW) 
= 1 (original prism). 


Note. This proposition of course solves the problem of 
finding the centre of gravity of a half cylinder or, in other 
words, of a semicircle. 


For the triangle GHWM in the place where it is balances, 
about H, the semicircle PQR in the place where it is. 


If then X is the point on HQ which is the centre of 
gravity of the semicircle, 
2H0.(AGHM) = HX . (semicircle PQR), 
or 2H0O.HO?= HX .40. HO’; 


hal 3 ix = HON 
37 


THE METHOD 41 


Proposition 19. 


Let there be a right prism with square bases, one of which 
is ABCD; 


in the prism let a cylinder be ~ 


inscribed, the base of which is 
the circle HFGH touching the 
sides of the square ABCD in Ε΄ 
E, F, 6, Η. 
Through the centre and 
through the side corresponding 
to CD in the square face opposite 
to ABCD let a plane be drawn; 


this will cut off a prism equal 
to + of the original prism and 
formed by three parallelograms and two triangles, the triangles 
forming opposite faces. 

In the semicircle HFG describe the parabola which has 
FK for axis and passes through £, G; 


draw MN parallel to KF meeting GH in M, the parabola 
in LJ, the semicircle in O and CD in N. 


Then MNT EN Nee 
“for this is clear.” [Cf. Apollonius, Conies τ. 11] 
[The parameter is of course equal to GK or KF] 

Therefore MN: NE =GR aT S” 


Through WN draw a plane at right angles to HG; 


this will produce as sections (1) in the prism cut off from the 
whole prism a right-angled triangle, the base of which is MN, 
while the perpendicular is perpendicular at NV to the plane 
ABCD and equal to the axis of the cylinder, and the hypo- 
tenuse is in the plane cutting the cylinder, and (2) in the 
portion of the cylinder cut off a right-angled triangle the base 
of which is MO, while the perpendicular is the generator of 
the cylinder perpendicular at O to the, plane KW, and the 
My POLCNUSE) 18-2 τ 150. τι τος τοῦτες 


Pere Ξ er | 


42 ARCHIMEDES 


[There is a lacuna here, to be supplied as follows. 
Since ΜΝ: NL=GK?: LS? 
= MN? LS* 
it follows that MN: MZ=MWN?: (MN? —- LS?) 
= MN?: (MN? —- MK") 
= MN?: MO” 
But the triangle (1) in the prism is to the triangle (2) in 
the portion of the cylinder in the ratio of MN? : MO* 
Therefore . 
(A in prism) : (A in portion of cylinder) 
=MN:ML 
= (straight line m rect. DG) : (straight line in parabola). 
We now take all the corresponding elements in the prism, 
the portion of the cylinder, the rectangle DG and the parabola 
EFG respectively ;] 
and it will follow that 
(all the As in prism) : (all the As in portion of cylinder) 
= (all the str. lines in 7 DG) 
: (all the straight lines between parabola and EG). 
But the prism is made up of the triangles in the prism, 
[the portion of the cylinder is made up of the triangles in it], 
the parallelogram DG of the straight lines in it parallel to KF, 
and the parabolic segment of the straight lines parallel to AF 
intercepted between its circumference and EG ; 
therefore (prism) : (portion of cylinder) 
= (7 GD): (parabolic segment HFG). 
But 7 GD = 3 (parabolic segment HFG) ; 
“for this is proved in my earlier treatise.” 
[Quadrature of Parabola] 
Therefore prism = 3 (portion of cylinder). 
If then we denote the portion of the cylinder by 2, the 
prism is 3, and the original prism circumscribing the cylinder 
is 12 (being 4 times the other prism) ; : 


therefore the portion of the cylinder = 4 (original prism). 
Q.E.D. 


THE METHOD 43 


[The above proposition and the next are peculiarly interest- 
ing for the fact that the parabola is an auxiliary curve introduced 
for the sole purpose of analytically reducing the required 
cubature to the known quadrature of the parabola.] 


Proposition 14. 


Let there be a right prism with square bases [and a 
cylinder inscribed therein having its base in the square ABCD 
and touching its sides at E, F, G, H; 
let the cylinder be cut by a plane through HG and the side 
corresponding to CD in the square face opposite to A BCD]. 


This plane cuts off from the prism a prism, and from the 
cylinder a portion of it. 


It can be τρλῦνθα that the portion of the cylinder cut off by 
the plane is 1 of the whole prism. 


But we will first prove that it is possible to inscribe in 
the portion cut off from the cylinder, and to circumscribe about 
it, solid figures made up of prisms which have equal height 
and similar triangular bases, in such a way that the circum- 
scribed figure exceeds the inscribed by less than any assigned 
πα ΡΘΕ το Το δες 

But it was proved that 

(prism cut off by oblique plane) 


< 3 (figure inscribed in portion of cylinder), 
Now 


(prism cut off) : (inscribed figure) 
= (7 DG : (Ls inscribed in parabolic segment) ; 
therefore 47 DG « ὃ ((75 in parabolic segment) : 
which is impossible, since “it has been proved elsewhere” that 
the parallelogram DG is 3 of the parabolic segment. 


Womsequently. ii. c.erecumseets see ts 


Sees essesereese tosses eeeeset ses eeecessseeeeoreseee seeeneses 


44. ARCHIMEDES 


And (all the prisms in prism cut off) 
: (all prisms in circumser. figure) 
=(all Osin O& DG) 


: (all Cys in fig. circumscr. about parabolic segmt.) ; 
therefore 


(prism cut off) : (igure circumscr. about portion of cylinder) 
=( DG): (figure circumscr. about parabolic segment). 


But the prism cut off by the oblique plane is > 2 of 
the solid figure circumscribed about the portion of the 
VAIO ET seein sas ree oe 


i eC ee ee ---- 


[There are large gaps in the exposition of this geometrical 
proof, but the way in which the method of exhaustion was 
applied, and the parallelism between this and other applications 
of it, are clear. The first fragment shows that solid figures 
made up of prisms were circumscribed and inscribed to the 
portion of the cylinder. The parallel triangular faces of these 
prisms were perpendicular to G# in the figure of Prop. 13; 
they divided GE into equal portions of the requisite smallness ; 
each section of the portion of the cylinder by such a plane was 
a triangular face common to an inscribed and a circumscribed 
right prism. The planes also produced prisms in the prism cut 
off by the same oblique plane as cuts off the portion of the 
cylinder and standing on GD as base. 


The number of parts into which the parallel planes divided 
GE was made great enough to secure that the circumscribed 
figure exceeded the inscribed figure by less than a small 
assigned magnitude. 


The second part of the proof began with the assumption 
that the portion of the cylinder is > 2 of the prism cut off; 
and this was proved to be impossible, by means of the use of 
the auxiliary parabola and the proportion 


MN : Mi ΕΝ Oo: 
which are employed in Prop. 13. 


THE METHOD 45 


We may supply the missing proof as follows*. 
In the accompanying figure are represented (1) the first 


C’ N’ D’ 


element-prism circumscribed to the portion of the cylinder, 
(2) two element-prisms adjacent to the ordinate OM, of which 
that on the left is circumscribed and ¢ 
that on the right (equal to the other) 
inscribed, (3) the corresponding element- 
prisms forming part of the prism cut 
off (CC’G EDD’) which is 3 of the original 
prism. 

In the second figure are shown , 
element-rectangles circumscribed and 
inscribed to the auxiliary parabola, 
which rectangles correspond exactly to 
the circumscribed and inscribed element- 
prisms represented in the first figure 
(the length of GM is the same in both 
figures, and the breadths of the element- 
rectangles are the same as the heights of the element-prisms) ; 


E 1D) 


* It is right to mention that this has already been done by Th. Reinach in 
his version of the treatise (‘Un Traité de Géométrie inédit d’Archiméde” in 
Revue générale des sciences pures et appliquées, 30 Nov. and 15 Dec. 1907) ; 
but I prefer my own statement of the proof. 


46 ARCHIMEDES 


the corresponding element-rectangles forming part of the 
rectangle GD are similarly shown. 

For convenience we suppose that GH is divided into an 
even number of equal parts, so that GK contains an integral 
number of these parts. 

For the sake of brevity we will call each of the two element- 
prisms of which OM is an edge “el. prism (O)” and each of 
the element-prisms of which MN’ is a common face “el. 
prism (V).” Similarly we will use the corresponding abbrevia- 
tions “el. rect. (Z)” and “el. rect. (V)” for the corresponding 
elements in relation to the auxiliary parabola as shown in 
the second figure. 

Now it is easy to see that the figure made up of all the 
inscribed prisms is less than the figure made up of the circum- 
scribed prisms by twice the final circumscribed prism adjacent 
to F'K, 1.6. by twice “el. prism (V)”; and, as the height of this 
prism may be made as small as we please by dividing GK into 
sufficiently small parts, it follows that inscribed and circum- . 
seribed solid figures made up of element-prisms can be drawn 
differing by less than any assigned solid figure. 

(1) Suppose, if possible, that 

(portion of cylinder) > 2 (prism cut off), 
or (prism cut off) < 8 (portion of cylinder). 

Let (prism cut off) = 3 (portion of cylinder — X), say. 

Construct circumscribed and inscribed figures made up of 
element-prisms, such that 

(circumscr. fig.) — (inser. fig.) « X. 

Therefore (inser. fig.) > (circumser. fig. — X), 
and a fortiort > (portion of cyl. — X). 

It follows that 

(prism cut off) < 8 (inscribed figure). 
Considering now the element-prisms in the prism cut off 
and those in the inscribed figure respectively, we have 
el. prism (JV) : el. prism (O) = MN? : MO? 
= MN: ML [88 in Prop. 13] 
= el. rect. (1) : el. rect. (LZ). 


THE METHOD 47 


It follows that 
Σ {el. prism (V)} : Σ {el. prism (0)} 
= > {el. rect. (V)} : Σ {el. rect. (Z)}. 

(There are really two more prisms and rectangles in the 
first and third than there are in the second and fourth terms 
respectively; but this makes no difference because the first 
and third terms may be multiplied by a common factor as 
n/(n—2) without affecting the truth of the proportion. Cf. 
the proposition from On Conoids and Spheroids quoted on p. 15 
above.) 

Therefore 

(prism cut off) : (figure inser. in portion of cyl.) 
=(rect. GD) : (fig. inser. in parabola). 
But it was proved above that 
(prism cut off) < 3 (fig. inser. in portion of cyl.) ; 
therefore (rect. GD) < 3 (fig. inser. in parabola), 
and, a fortiore, 
(rect. GD) < 3 (parabolic segmt.): 
which is impossible, since 
(rect. GD) = 3 (parabolic segmt.). 
Therefore 
(portion of cyl.) is not greater than 2 (prism cut off). 


(2) In the second lacuna must have come the beginning of 
the next reductio ad absurdum demolishing the other possible 
assumption that the portion of the cylinder is < 2 of the prism 
cut off. 

In this case our assumption is that 

(prism cut off) > 3 (portion of cylinder) ; 
and we circumscribe and inscribe figures made up of element- 
prisms, such that 

(prism cut off) > 3 (fig. circumser. about portion of cyl.). 


48 ARCHIMEDES 


We now consider the element-prisms in the prism cut off 
and in the circumscribed figure respectively, and the same 
argument as above gives 

(prism cut off) : (fig. circumser. about portion of cyl.) 
= (rect. GD) : (fig. cireumser. about parabola), 
whence it follows that 
(rect. GD) > 3 (fig. circumscribed about parabola), 
and, a fortiori, 
(rect. GD) > 3 (parabolic segment) : 
which is impossible, since 
(rect. GD) = 3 (parabolic segmt.). 
Therefore 
(portion of cyl.) is not less than 2 (prism cut off). 

But it was also proved that neither is it greater; 
therefore (portion of cyl.) = 2 (prism cut off) 

= 1 (original prism).] 


[Proposition 15.]| 


[This proposition, which is lost, would be the mechanical 
investigation of the second of the two special problems 
mentioned in the preface to the treatise, namely that of the 
cubature of the figure included between two cylinders, each 
of which is inscribed in one and the same cube so that its 
opposite bases are in two opposite faces of the cube and its 
surface touches the other four faces. 

Zeuthen has shown how the mechanical method can be 
applied to this case*. 

In the accompanying figure VWYX is a section of the 
cube by a plane (that of the paper) passing through the axis 
BD of one of the cylinders inscribed in the cube and parallel 
to two opposite faces. 

The same plane gives the circle ABCD as the section of 
the other inscribed cylinder with axis perpendicular to the 


* Zeuthen in Bibliotheca Mathematica vitz, 1906-7, pp. 356-7. 


THE METHOD 49 


plane of the paper and extending on each side of the plane 
to a distance equal to the radius of the circle or half the side 
of the cube. 

AC is the diameter of the circle which is perpendicular 
to BD. 

Join AB, AD and produce them to meet the tangent at C 
to the circle in Α΄, F. 

Then £C = CF=CA. 


Let LG be the tangent at A, and complete the rectangle 
EFGL. 


H 


Draw straight lines from A to the four corners of the 
section in which the plane through BD perpendicular to AK 
cuts the cube. These straight lines, if produced, will meet the 
plane of the face of the cube opposite to A in four points 
forming the four corners of a square in that plane with sides 
equal to HF or double of the side of the cube, and we thus 
have a pyramid with A for vertex and the latter square for 
base. 

Complete the prism (parallelepiped) with the same base 
and height as the pyramid. 

Draw in the parallelogram LF any straight line MN parallel 
to HF, and through MN draw a plane at right angles to AC. 


δ0 ARCHIMEDES 


This plane cuts— 


(1) the solid included by the two cylinders in a square with 
side equal to OP, 


(2) the prism in a square with side equal to MN, and 
(3) the pyramid in a square with side equal to QR. 


Produce CA to H, making HA equal to AC, and imagine 
HC to be the bar of a balance. 


Now, as in Prop. 2, since MS = AC, QS = AS, 
MIS, SIO = GAL AUS 


=A 
= OS? + SQ? 
Also 
ΠΑ: "AiS = CA: AS 
= MS: SQ 
= MS?: MS. SQ 


= MS? : (OS? + SQ?), from above, 
= MN? : (OP? + QR?) 
= (square, side MV): (sq., side OP +sq., side QR). 

Therefore the square with side equal to MN, in the place 
where it is, is in equilibrium about A with the squares with 
sides equal to OP, QR respectively placed with their centres of 
gravity at ἢ. 

Proceeding in the same way with the square sections 
produced by other planes perpendicular to AC, we finally 
prove that the prism, in the place where it is, is in equilibrium 
about A with the solid included by the two cylinders and the 
pyramid, both placed with their centres of gravity at H. 

Now the centre of gravity of the prism is at K. 

Therefore HA : AK =(prism) : (solid + pyramid) 
or 2 : 1 =(prism) : (solid + 4 prism). 

Therefore 2 (solid) +.2 (prism) = (prism). 

It follows that 

(solid included by cylinders) = 4 (prism) - 
=2 (cube). QED. 


THE METHOD 51 


There is no doubt that Archimedes proceeded to, and 
completed, the rigorous geometrical proof by the method of 
exhaustion. 


As observed by Prof. C. Juel (Zeuthen l.c.), the solid in the 
present proposition is made up of 8 pieces of cylinders of the 
type of that treated in the preceding proposition. As however 
the two propositions are separately stated, there is no doubt 
that Archimedes’ proofs of them were. distinct. 


In this case AC would be divided into a very large number 
of equal parts and planes would be drawn through the points 
of division perpendicular to AC. These planes cut the solid, 
and also the cube VY, in square sections.. Thus we can inscribe 
and circumscribe to the solid the requisite solid figures made 
up of element-prisms and differing by less than any assigned 
solid magnitude; the prisms have square bases and their 
heights are the small segments of AC. The element-prism 
in the inscribed and circumscribed figures which has the square 
equal-to OP? for base corresponds to an element-prism in the 
cube which has for base a square with side equal to that of 
the cube; and as the ratio of the element-prisms is the ratio 
OS? : BK?, we can use the same auxiliary parabola, and work 
out the proof in exactly the same way, as in Prop. 14.] 


CAMBRIDGE: PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS 


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