by Archimedes

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Title: Geometrical Solutions Derived from Mechanics

A Treatise of Archimedes

Author: Archimedes

Release Date: April, 2005 [EBook #7825]

[Yes, we are more than one year ahead of schedule]

[This file was first posted on May 20, 2003]

Edition: 10

Language: English

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*** START OF THE PROJECT GUTENBERG EBOOK GEOMETRICAL SOLUTIONS ***

Gordon Keener

% gbn0305181551: Archimedes, Geometrical Solutions Derived from Mechanics. Gordon Keener

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\usepackage[greek,english]{babel}

\usepackage{wrapfig}

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\title{Geometrical Solutions Derived from Mechanics}

\author{A Treatise of Archimedes}

\date{\vspace{\baselineskip}

{\small Recently discovered and translated from the Greek

by}\\ Dr. J. L. Heiberg\\ {\small Professor of Classical Philology at

the University of Copenhagen}\\

\vspace{\baselineskip} {\small with an introduction by}\\ David Eugene

Smith\\ {\small President of Teachers College, Columbia University, New York}\\

\vspace{\baselineskip}

{\small English version translated from the German by}\\ Lydia

G. Robinson\\ {\small and reprinted from ``The Monist,'' April,

1909}\\

\vspace{\baselineskip}

{\small Project Gutenberg edition:\\

gbn0305181551: Archimedes, Geometrical Solutions Derived from Mechanics.

Gordon Keener $<$gkeener@nc.rr.com$>$. 1909c. 5/19/2003. ok.}

}

\begin{document}

\maketitle

\vfill\pagebreak

\section*{Introduction}

If there ever was a case of appropriateness in discovery, the finding

of this manuscript in the summer of 1906 was one. In the first place

it was appropriate that the discovery should be made in

Constantinople, since it was here that the West received its first

manuscripts of the other extant works, nine in number, of the great

Syracusan. It was furthermore appropriate that the discovery should be

made by Professor Heiberg, \emph{facilis princeps} among all workers

in the field of editing the classics of Greek mathematics, and an

indefatigable searcher of the libraries of Europe for manuscripts to

aid him in perfecting his labors. And finally it was most appropriate

that this work should appear at a time when the affiliation of pure

and applied mathematics is becoming so generally recognized all over

the world. We are sometimes led to feel, in considering isolated

cases, that the great contributors of the past have worked in the

field of pure mathematics alone, and the saying of Plutarch that

Archimedes felt that ``every kind of art connected with daily needs

was ignoble and vulgar''\footnote{Marcellus, 17.} may have

strengthened this feeling. It therefore assists us in properly

orientating ourselves to read another treatise from the greatest

mathematician of antiquity that sets clearly before us his

indebtedness to the mechanical applications of his subject.

Not the least interesting of the passages in the manuscript is the

first line, the greeting to Eratosthenes. It is well known, on the

testimony of Diodoros his countryman, that Archimedes studied in

Alexandria, and the latter frequently makes mention of Konon of Samos

whom he knew there, probably as a teacher, and to whom he was indebted

for the suggestion of the spiral that bears his name. It is also

related, this time by Proclos, that Eratosthenes was a contemporary of

Archimedes, and if the testimony of so late a writer as Tzetzes, who

lived in the twelfth century, may be taken as valid, the former was

eleven years the junior of the great Sicilian. Until now, however, we

have had nothing definite to show that the two were ever

acquainted. The great Alexandrian savant,---poet, geographer,

arithmetician,---affectionately called by the students Pentathlos, the

champion in five sports,\footnote{His nickname of \emph{Beta} is well

known, possibly because his lecture room was number 2.} selected by

Ptolemy Euergetes to succeed his master, Kallimachos the poet, as head

of the great Library,---this man, the most renowned of his time in

Alexandria, could hardly have been a teacher of Archimedes, nor yet

the fellow student of one who was so much his senior. It is more

probable that they were friends in the later days when Archimedes was

received as a savant rather than as a learner, and this is borne out

by the statement at the close of proposition I which refers to one of

his earlier works, showing that this particular treatise was a late

one. This reference being to one of the two works dedicated to

Dositheos of Kolonos,\footnote{We know little of his works, none of

which are extant. Geminos and Ptolemy refer to certain observations

made by him in 200 B. C., twelve years after the death of

Archimedes. Pliny also mentions him.} and one of these (\emph{De

lineis spiralibus}) referring to an earlier treatise sent to

Konon,\footnote{\selectlanguage{greek} T\~wn pot\`i K\'onwna

\'apustal\'entwn jewrhm\'atwn.} we are led to believe that this was

one of the latest works of Archimedes and that Eratosthenes was a

friend of his mature years, although one of long standing. The

statement that the preliminary propositions were sent ``some time

ago'' bears out this idea of a considerable duration of friendship,

and the idea that more or less correspondence had resulted from this

communication may be inferred by the statement that he saw, as he had

previously said, that Eratosthenes was ``a capable scholar and a

prominent teacher of philosophy,'' and also that he understood ``how

to value a mathematical method of investigation when the opportunity

offered.'' We have, then, new light upon the relations between these

two men, the leaders among the learned of their day.

A second feature of much interest in the treatise is the intimate view

that we have into the workings of the mind of the author. It must

always be remembered that Archimedes was primarily a discoverer, and

not primarily a compiler as were Euclid, Apollonios, and Nicomachos.

Therefore to have him follow up his first communication of theorems to

Eratosthenes by a statement of his mental processes in reaching his

conclusions is not merely a contribution to mathematics but one to

education as well. Particularly is this true in the following

statement, which may well be kept in mind in the present day: ``l have

thought it well to analyse and lay down for you in this same book a

peculiar method by means of which it will be possible for you to

derive instruction as to how certain mathematical questions may be

investigated by means of mechanics. And I am convinced that this is

equally profitable in demonstrating a proposition itself; for much

that was made evident to me through the medium of mechanics was later

proved by means of geometry, because the treatment by the former

method had not yet been established by way of a demonstration. For of

course it is easier to establish a proof if one has in this way

previously obtained a conception of the questions, than for him to

seek it without such a preliminary notion. . . . Indeed I assume that

some one among the investigators of to-day or in the future will

discover by the method here set forth still other propositions which

have not yet occurred to us.'' Perhaps in all the history of

mathematics no such prophetic truth was ever put into words. It would

almost seem as if Archimedes must have seen as in a vision the methods

of Galileo, Cavalieri, Pascal, Newton, and many of the other great

makers of the mathematics of the Renaissance and the present time.

The first proposition concerns the quadrature of the parabola, a

subject treated at length in one of his earlier communications to

Dositheos.\footnote{\selectlanguage{greek} Tetragwnismds parabol\~hs.}

He gives a digest of the treatment, but with the warning that the

proof is not complete, as it is in his special work upon the

subject. He has, in fact, summarized propositions VII-XVII of his

communication to Dositheos, omitting the geometric treatment of

propositions XVIII-XXIV. One thing that he does not state, here or in

any of his works, is where the idea of center of

gravity\footnote{\selectlanguage{greek} K\'entra bar\~wn,

\selectlanguage{english} for ``barycentric'' is a very old term.}

started. It was certainly a common notion in his day, for he often

uses it without defining it. It appears in Euclid's\footnote{At any

rate in the anonymous fragment \emph{De levi et ponderoso}, sometimes

attributed to him.} time, but how much earlier we cannot as yet say.

Proposition II states no new fact. Essentially it means that if a

sphere, cylinder, and cone (always circular) have the same radius,

$r$, and the altitude of the cone is $r$ and that of the cylinder

$2r$, then the volumes will be as $4 : 1 : 6$, which is true, since

they are respectively $\frac{4}{3}\pi r^3$, $\frac{1}{3}\pi r^3$, and

$2\pi r^3$. The interesting thing, however, is the method pursued,

the derivation of geometric truths from principles of mechanics. There

is, too, in every sentence, a little suggestion of Cavalieri, an

anticipation by nearly two thousand years of the work of the greatest

immediate precursor of Newton. And the geometric imagination that

Archimedes shows in the last sentence is also noteworthy as one of the

interesting features of this work: ``After I had thus perceived that a

sphere is four times as large as the cone. . . it occurred to me that

the surface of a sphere is four times as great as its largest circle,

in which I proceeded from the idea that just as a circle is equal to a

triangle whose base is the periphery of the circle, and whose altitude

is equal to its radius, so a sphere is equal to a cone whose base is

the same as the surface of the sphere and whose altitude is equal to

the radius of the sphere.'' As a bit of generalization this throws a

good deal of light on the workings of Archimedes's mind.

In proposition III he considers the volume of a spheroid, which he had

already treated more fully in one of his letters to

Dositheos,\footnote{\selectlanguage{greek} Per\`i kwnoeide\~wn kai

sfairoeide\~wn.} and which contains nothing new from a mathematical

standpoint. Indeed it is the method rather than the conclusion that is

interesting in such of the subsequent propositions as relate to

mensuration. Proposition V deals with the center of gravity of a

segment of a conoid, and proposition VI with the center of gravity of

a hemisphere, thus carrying into solid geometry the work of Archimedes

on the equilibrium of planes and on their centers of

gravity.\footnote{\selectlanguage{greek} 'Epip\'edwn \`isorropi\~wn

\^h k\'entra bar\~wn \'epip\'edwn.} The general method is that already

known in the treatise mentioned, and this is followed through

proposition X.

Proposition XI is the interesting case of a segment of a right

cylinder cut off by a plane through the center of the lower base and

tangent to the upper one. He shows this to equal one-sixth of the

square prism that circumscribes the cylinder. This is well known to us

through the formula $v = 2r^2h/3$, the volume of the prism being

$4r^2h$, and requires a knowledge of the center of gravity of the

cylindric section in question. Archimedes is, so far as we know, the

first to state this result, and he obtains it by his usual method of

the skilful balancing of sections. There are several lacunae in the

demonstration, but enough of it remains to show the ingenuity of the

general plan. The culminating interest from the mathematical

standpoint lies in proposition XIII, where Archimedes reduces the

whole question to that of the quadrature of the parabola. He shows

that a fourth of the circumscribed prism is to the segment of the

cylinder as the semi-base of the prism is to the parabola inscribed in

the semi-base; that is, that $\frac{1}{4}p : v = \frac{1}{2}b :

(\frac{2}{3} \cdot \frac{1}{2}b)$, whence $v = \frac{1}{6}p$.

Proposition XIV is incomplete, but it is the conclusion of the two

preceding propositions.

In general, therefore, the greatest value of the work lies in the

following:

1. It throws light upon the hitherto only suspected relations of

Archi\-medes and Eratosthenes.

2. It shows the working of the mind of Archimedes in the discovery of

mathematical truths, showing that he often obtained his results by

intuition or even by measurement, rather than by an analytic form of

reasoning, verifying these results later by strict analysis.

3. It expresses definitely the fact that Archimedes was the discoverer

of those properties relating to the sphere and cylinder that have been

attributed to him and that are given in his other works without a

definite statement of their authorship.

4. It shows that Archimedes was the first to state the volume of the

cylinder segment mentioned, and it gives an interesting description of

the mechanical method by which he arrived at his result.

{\hspace*{\fill}\textsc{David Eugene Smith.}}\linebreak

{\hspace*{\fill}\textsc{Teachers~College,~Columbia~University.}}\linebreak

\vfill\pagebreak

\section*{Geometrical Solutions Derived from Mechanics.}

\textsc{Archimedes to Eratosthenes, Greeting:}

Some time ago I sent you some theorems I had discovered, writing down

only the propositions because I wished you to find their

demonstrations which had not been given. The propositions of the

theorems which I sent you were the following:

1. If in a perpendicular prism with a parallelogram\footnote{This must

mean a square.} for base a cylinder is inscribed which has its bases

in the opposite

parallelograms\addtocounter{footnote}{-1}\footnotemark\ and its

surface touching the other planes of the prism, and if a plane is

passed through the center of the circle that is the base of the

cylinder and one side of the square lying in the opposite plane, then

that plane will cut off from the cylinder a section which is bounded

by two planes, the intersecting plane and the one in which the base of

the cylinder lies, and also by as much of the surface of the cylinder

as lies between these same planes; and the detached section of the

cylinder is $\frac{1}{6}$ of the whole prism.

2. If in a cube a cylinder is inscribed whose bases lie in opposite

parallelograms\addtocounter{footnote}{-1}\footnotemark\ and whose

surface touches the other four planes, and if in the same cube a

second cylinder is inscribed whose bases lie in two other

parallelograms\addtocounter{footnote}{-1}\footnotemark\ and whose

surface touches the four other planes, then the body enclosed by the

surface of the cylinder and comprehended within both cylinders will be

equal to $\frac{2}{3}$ of the whole cube.

These propositions differ essentially from those formerly discovered;

for then we compared those bodies (conoids, spheroids and their

segments) with the volume of cones and cylinders but none of them was

found to be equal to a body enclosed by planes. Each of these bodies,

on the other hand, which are enclosed by two planes and cylindrical

surfaces is found to be equal to a body enclosed by planes. The

demonstration of these propositions I am accordingly sending to you in

this book.

Since I see, however, as I have previously said, that you are a

capable scholar and a prominent teacher of philosophy, and also that

you understand how to value a mathematical method of investigation

when the opportunity is offered, I have thought it well to analyze and

lay down for you in this same book a peculiar method by means of which

it will be possible for you to derive instruction as to how certain

mathematical questions may be investigated by means of mechanics. And

I am convinced that this is equally profitable in demonstrating a

proposition itself; for much that was made evident to me through the

medium of mechanics was later proved by means of geometry because the

treatment by the former method had not yet been established by way of

a demonstration. For of course it is easier to establish a proof if

one has in this way previously obtained a conception of the questions,

than for him to seek it without such a preliminary notion. Thus in the

familiar propositions the demonstrations of which Eudoxos was the

first to discover, namely that a cone and a pyramid are one third the

size of that cylinder and prism respectively that have the same base

and altitude, no little credit is due to Democritos who was the first

to make that statement about these bodies without any

demonstration. But we are in a position to have found the present

proposition in the same way as the earlier one; and I have decided to

write down and make known the method partly because we have already

talked about it heretofore and so no one would think that we were

spreading abroad idle talk, and partly in the conviction that by this

means we are obtaining no slight advantage for mathematics, for indeed

I assume that some one among the investigators of to-day or in the

future will discover by the method here set forth still other

propositions which have not yet occurred to us.

In the first place we will now explain what was also first made clear

to us through mechanics, namely that a segment of a parabola is

$\frac{4}{3}$ of the triangle possessing the same base and equal

altitude; following which we will explain in order the particular

propositions discovered by the above mentioned method; and in the last

part of the book we will present the geometrical demonstrations of the

propositions.\footnote{In his ``Commentar,'' Professor Zeuthen calls

attention to the fact that it was aiready known from Heron's recently

discovered \emph{Metrica} that these propositions were contained in

this treatise, and Professor Heiberg made the same comment in

\emph{Hermes}.---Tr.}

1. If one magnitude is taken away from another magnitude and the same

point is the center of gravity both of the whole and of the part

removed, then the same point is the center of gravity of the remaining

portion.

2. If one magnitude is taken away from another magnitude and the

center of gravity of the whole and of the part removed is not the same

point, the center of gravity of the remaining portion may be found by

prolonging the straight line which connects the centers of gravity of

the whole and of the part removed, and setting off upon it another

straight line which bears the same ratio to the straight line between

the aforesaid centers of gravity, as the weight of the magnitude which

has been taken away bears to the weight of the one remaining [\emph{De

plan. aequil.} I, 8].

3. If the centers of gravity of any number of magnitudes lie upon the

same straight line, then will the center of gravity of all the

magnitudes combined lie also upon the same straight line [Cf.

\emph{ibid.} I, 5].

4. The center of gravity of a straight line is the center of that line

[Cf. \emph{ibid.} I, 4].

5. The center of gravity of a triangle is the point in which the

straight lines drawn from the angles of a triangle to the centers of

the opposite sides intersect [\emph{Ibid.} I, 14].

6. The center of gravity of a parallelogram is the point where its

diagonals meet [\emph{Ibid.} I, 10].

7. The center of gravity [of a circle] is the center [of that circle].

8. The center of gravity of a cylinder [is the center of its axis].

9. The center of gravity of a prism is the center of its axis.

10. The center of gravity of a cone so divides its axis that the

section at the vertex is three times as great as the remainder.

11. Moreover together with the exercise here laid down I will make use

of the following proposition:

If any number of magnitudes stand in the same ratio to the same number

of other magnitudes which correspond pair by pair, and if either all

or some of the former magnitudes stand in any ratio whatever to other

magnitudes, and the latter in the same ratio to the corresponding

ones, then the sum of the magnitudes of the first series will bear the

same ratio to the sum of those taken from the third series as the sum

of those of the second series bears to the sum of those taken from the

fourth series [\emph{De Conoid.} I].

\section*{Proposition I}

Let $\alpha\beta\gamma$ [Fig.~1] be the segment of a parabola bounded

by the straight line $\alpha\gamma$ and the parabola

$\alpha\beta\gamma$. Let $\alpha\gamma$ be bisected at $\delta$,

$\delta\beta\epsilon$ being parallel to the diameter, and draw

$\alpha\beta$, and $\beta\gamma$. Then the segrnent

$\alpha\beta\gamma$ will be $\frac{4}{3}$ as great as the triangle

$\alpha\beta\gamma$.

From the points $\alpha$ and $\gamma$ draw $\alpha\zeta \|

\delta\beta\epsilon$, and the tangent $\gamma\zeta$; produce

[$\gamma\beta$ to $\kappa$, and make $\kappa\theta = \gamma\kappa$].

Think of $\gamma\theta$ as a scale-beam with the center at $\kappa$

and let $\mu\xi$ be any straight line whatever $\|

\epsilon\delta$. Now since $\gamma\beta\alpha$ is a parabola,

$\gamma\zeta$ a tangent and $\gamma\delta$ an ordinate, then

$\epsilon\beta = \beta\delta$; for this indeed has been proved in the

Elements [i.e., of conic sections, cf. \emph{Quadr. parab.} 2]. For

this reason and because $\zeta\alpha$ and $\mu\xi \| \epsilon\delta$,

$\mu\nu = \nu\xi$, and $\zeta\kappa = \kappa\alpha$. And because

$\gamma\alpha : \alpha\xi = \mu\xi : \xi o$ (for this is shown in a

corollary, [cf. \emph{Quadr. parab.} 5]), $\gamma\alpha : \alpha\xi =

\gamma\kappa : \kappa\nu$; and $\gamma\kappa = \kappa\theta$,

therefore $\theta\kappa : \kappa\nu = \mu\xi : \xi o$. And because

$\nu$ is the center of gravity of the straight line $\mu\xi$, since

$\mu\nu = \nu\xi$, then if we make $\tau\eta = \xi o$ and $\theta$ as

its center of gravity so that $\tau\theta = \theta\eta$, the straight

line $\tau\theta\eta$ will be in equilibrium with $\mu\xi$ in its

present position because $\theta\nu$ is divided in inverse proportion

to the weights $\tau\eta$ and $\mu\xi$, and $\theta\kappa : \kappa\nu

= \mu\xi : \eta\tau$; therefore $\kappa$ is the center of gravity of

the combined weight of the two. In the

%

\begin{wrapfigure}{r}{0.5\textwidth}

\includegraphics[width=0.5\textwidth]{fig01.png}

\begin{center}

{\small Fig.~1.}

\end{center}

\end{wrapfigure}

%

same way all straight lines drawn in the triangle $\zeta\alpha\gamma

\| \epsilon\delta$ are in their present positions in equilibrium with

their parts cut off by the parabola, when these are transferred to

$\theta$, so that $\kappa$ is the center of gravity of the combined

weight of the two. And because the triangle $\gamma\zeta\alpha$

consists of the straight lines in the triangle $\gamma\zeta\alpha$ and

the segment $\alpha\beta\gamma$ consists of those straight lines

within the segment of the parabola corresponding to the straight line

$\xi o$, therefore the triangle $\zeta\alpha\gamma$ in its present

position will be in equilibrium at the point $\kappa$ with the

parabola-segment when this is transferred to $\theta$ as its center of

gravity, so that $\kappa$ is the center of gravity of the combined

weights of the two. Now let $\gamma\kappa$ be so divided at $\chi$

that $\gamma\kappa = 3\kappa\chi$; then $\chi$ will be the center of

gravity of the triangle $\alpha\zeta\gamma$, for this has been shown

in the Statics [cf. \emph{De plan. aequil.} I, 15, p. 186, 3 with

Eutokios, S. 320, 5ff.]. Now the triangle $\zeta\alpha\gamma$ in its

present position is in equilibrium at the point $\kappa$ with the

segment $\beta\alpha\gamma$ when this is transferred to $\theta$ as

its center of gravity, and the center of gravity of the triangle

$\zeta\alpha\gamma$ is $\chi$; hence triangle $\alpha\zeta\gamma : $

segm. $\alpha\beta\gamma$ when transferred to $\theta$ as its center

of gravity $= \theta\kappa : \kappa\chi$. But $\theta\kappa =

3\kappa\chi$; hence also triangle $\alpha\zeta\gamma = 3$

segm. $\alpha\beta\gamma$. But it is also true that triangle

$\zeta\alpha\gamma = 4\Delta\alpha\beta\gamma$ because $\zeta\kappa =

\kappa\alpha$ and $\alpha\delta = \delta\gamma$; hence

segm. $\alpha\beta\gamma = \frac{4}{3}$ the triangle

$\alpha\beta\gamma$. This is of course clear.

It is true that this is not proved by what we have said here; but it

indicates that the result is correct. And so, as we have just seen

that it has not been proved but rather conjectured that the result is

correct we have devised a geometrical demonstration which we made

known some time ago and will again bring forward farther on.

\section*{Proposition II}

That a sphere is four times as large as a cone whose base is equal to

the largest circle of the sphere and whose altitude is equal to the

radius of the sphere, and that a cylinder whose base is equal to the

largest circle of the sphere and whose altitude is equal to the

diameter of the circle is one and a half times as large as the sphere,

may be seen by the present method in the following way:

\begin{wrapfigure}{r}{0.5\textwidth}

\includegraphics[width=0.5\textwidth]{fig02.png}

\begin{center}

{\small Fig.~2.}

\end{center}

\end{wrapfigure}

Let $\alpha\beta\gamma\delta$ [Fig.~2] be the largest circle of a

sphere and $\alpha\gamma$ and $\beta\delta$ its diameters

perpendicular to each other; let there be in the sphere a circle on

the diameter $\beta\delta$ perpendicular to the circle

$\alpha\beta\gamma\delta$, and on this perpendicular circle let there

be a cone erected with its vertex at $\alpha$; producing the convex

surface of the cone, let it be cut through $\gamma$ by a plane

parallel to its base; the result will be the circle perpendicular to

$\alpha\gamma$ whose diameter will be $\epsilon\zeta$. On this circle

erect a cylinder whose axis $= \alpha\gamma$ and whose vertical

boundaries are $\epsilon\lambda$ and $\zeta\eta$. Produce

$\gamma\alpha$ making $\alpha\theta = \gamma\alpha$ and think of

$\gamma\theta$ as a scale-beam with its center at $\alpha$. Then let

$\mu\nu$ be any straight line whatever drawn $\| \beta\delta$

intersecting the circle $\alpha\beta\gamma\delta$ in $\xi$ and $o$,

the diameter $\alpha\gamma$ in $\sigma$, the straight line

$\alpha\epsilon$ in $\pi$ and $\alpha\zeta$ in $\rho$, and on the

straight line $\mu\nu$ construct a plane perpendicular to

$\alpha\gamma$; it will intersect the cylinder in a circle on the

diameter $\mu\nu$; the sphere $\alpha\beta\gamma\delta$, in a circle

on the diameter $\xi o$; the cone $\alpha\epsilon\zeta$ in a circle on

the diameter $\pi\rho$. Now because $\gamma\alpha \times \alpha\sigma

= \mu\sigma \times \sigma\pi$ ( for $\alpha\gamma = \sigma\mu$,

$\alpha\sigma = \pi\sigma$), and $\gamma\alpha \times \alpha\sigma =

\alpha\xi^2 = \xi\sigma^2 + \alpha\pi^2$ then $\mu\sigma \times

\sigma\pi = \xi\sigma^2 + \sigma\pi^2$. Moreover, because

$\gamma\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$ and

$\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma

= \mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times

\sigma\pi$. But it has been proved that $\xi\sigma^2 + \sigma\pi^2 =

\mu\sigma \times \sigma\pi$; hence $\alpha\theta : \alpha\sigma =

\mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2$. But it is true that

$\mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2 = \mu\nu^2 : \xi\alpha^2 +

\pi\rho^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$

the circle in the cone whose diameter is $\pi\rho$ + the circle in the

sphere whose diameter is $\xi o$; hence $\theta\alpha : \alpha\sigma

=$ the circle in the cylinder $:$ the circle in the sphere $+$ the

circle in the cone. Therefore the circle in the cylinder in its

present position will be in equilibrium at the point $\alpha$ with the

two circles whose diameters are $\xi o$ and $\pi\rho$, if they are so

transferred to $\theta$ that $\theta$ is the center of gravity of

both. In the same way it can be shown that when another straight line

is drawn in the parallelogram $\xi\lambda \| \epsilon\zeta$, and upon

it a plane is erected perpendicular to $\alpha\gamma$, the circle

produced in the cylinder in its present position will be in

equilibrium at the point $\alpha$ with the two circles produced in the

sphere and the cone when they are transferred and so arranged on the

scale-beam at the point $\theta$ that $\theta$ is the center of

gravity of both. Therefore if cylinder, sphere and cone are filled up

with such circles then the cylinder in its present position will be in

equilibrium at the point $\alpha$ with the sphere and the cone

together, if they are transferred and so arranged on the scale-beam at

the point $\theta$ that $\theta$ is the center of gravity of both. Now

since the bodies we have mentioned are in equilibrium, the cylinder

with $\kappa$ as its center of gravity, the sphere and the cone

transferred as we have said so that they have $\theta$ as center of

gravity, then $\theta\alpha : \alpha\kappa =$ cylinder $:$ sphere $+$

cone. But $\theta\alpha = 2\alpha\kappa$, and hence also the cylinder

$= 2 \times$ (sphere $+$ cone). But it is also true that the cylinder

= 3 cones [Euclid, \emph{Elem.} XII, 10], hence 3 cones = 2 cones + 2

spheres. If 2 cones be subtracted from both sides, then the cone whose

axes form the triangle $\alpha\epsilon\zeta =$ 2 spheres. But the cone

whose axes form the triangle $\alpha\epsilon\zeta =$ 8 cones whose

axes form the triangle $\alpha\beta\delta$ because $\epsilon\zeta =

2\beta\delta$, hence the aforesaid 8 cones = 2 spheres. Consequently

the sphere whose greatest circle is $\alpha\beta\gamma\delta$ is four

times as large as the cone with its vertex at $\alpha$, and whose base

is the circle on the diatneter $\beta\delta$ perpendicular to

$\alpha\gamma$.

Draw the straight lines $\phi\beta\chi$ and $\psi\delta\omega \|

\alpha\gamma$ through $\beta$ and $\delta$ in the parallelogram

$\lambda\zeta$ and imagine a cylinder whose bases are the circles on

the diameters $\phi\psi$ and $\chi\omega$ and whose axis is

$\alpha\gamma$. Now since the cylinder whose axes form the

parallelogram $\phi\omega$ is twice as large as the cylinder whose

axes form the parallelogram $\phi\delta$ and the latter is three times

as large as the cone the triangle of whose axes is

$\alpha\beta\delta$, as is shown in the Elements [Euclid, \emph{Elem.}

XII, 10], the cylinder whose axes form the parallelogram $\phi\omega$

is six times as large as the cone whose axes form the triangle

$\alpha\beta\delta$. But it was shown that the sphere whose largest

circle is $\alpha\beta\gamma\delta$ is four times as large as the same

cone, consequently the cylinder is one and one half times as large as

the sphere, Q. E. D.

After I had thus perceived that a sphere is four times as large as the

cone whose base is the largest circle of the sphere and whose altitude

is equal to its radius, it occurred to me that the surface of a sphere

is four times as great as its largest circle, in which I proceeded

from the idea that just as a circle is equal to a triangle whose base

is the periphery of the circle and whose altitude is equal to its

radius, so a sphere is equal to a cone whose base is the same as the

surface of the sphere and whose altitude is equal to the radius of the

sphere.

\section*{Proposition III}

By this method it may also be seen that a cylinder whose base is equal

to the largest circle of a spheroid and whose altitude is equal to the

axis of the spheroid, is one and one half times as large as the

spheroid, and when this is recognized it becomes clear that if a

spheroid is cut through its center by a plane perpendicular to its

axis, one-half of the spheroid is twice as great as the cone whose

base is that of the segment and its axis the same.

For let a spheroid be cut by a plane through its axis and let there be

in its surface an ellipse $\alpha\beta\gamma\delta$ [Fig.~3] whose

diameters are $\alpha\gamma$ and $\beta\delta$ and whose center is

$\kappa$ and let there be a circle in the spheroid on the diameter

$\beta\delta$ perpendicular to $\alpha\gamma$; then imagine a cone

whose base is the same circle but whose vertex is at $\alpha$, and

producing its surface, let the cone be cut by a plane through $\gamma$

parallel to the base; the intersection will be a circle perpendicular

to $\alpha\gamma$ with $\epsilon\zeta$ as its diameter. Now imagine a

cylinder whose base is the same circle with the diameter

$\epsilon\zeta$ and whose axis is $\alpha\gamma$; let $\gamma\alpha$

be produced so that $\alpha\theta = \gamma\alpha$; think of

$\theta\gamma$ as a scale-beam with its center at $\alpha$ and in the

parallelogram $\lambda\theta$ draw a straight line $\mu\nu \|

\epsilon\zeta$, and on $\mu\nu$ construct a plane perpendicular to

$\alpha\gamma$; this will intersect the cylinder in a circle whose

diameter is $\mu\nu$, the spheroid in a circle whose diameter is $\xi

o$ and the cone in a circle whose diameter is $\pi\rho$. Because

$\gamma\alpha : \alpha\sigma = \epsilon\alpha : \alpha\pi = \mu\sigma

: \sigma\pi$, and $\gamma\alpha = \alpha\theta$, therefore

$\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$. But $\mu\sigma

: \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$ and $\mu\sigma

\times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$, for $ \alpha\sigma

\times \sigma\gamma : \sigma\xi^2 = \alpha\kappa \times \kappa\gamma :

\kappa\beta^2 = \alpha\kappa^2 : \kappa\beta^2$ (for both ratios are

equal to the ratio between the diameter and the parameter [Apollonius,

\emph{Con.} I, 21]) $ = \alpha\sigma^2 : \sigma\pi^2$ therefore

$\alpha\sigma^2 : \alpha\sigma \times \sigma\gamma = \pi\sigma^2 :

\sigma\xi^2 = \sigma\pi^2 : \sigma\pi \times \pi\mu$, consequently

$\mu\pi \times \pi\sigma = \sigma\xi^2$. If $\pi\sigma^2$ is added to

both sides then $\mu\sigma \times \sigma\pi = \pi\sigma^2 +

\sigma\xi^2$. Therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 :

\pi\sigma^2 + \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2 +

\sigma\pi^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$

the circle with the diameter $\xi o$ + the circle with the diameter

$\pi\rho$; hence the circle whose diameter is $\mu\nu$ will in its

present position be in equilibrium at the point $\alpha$ with the two

circles whose diameters are $\xi o$ and $\pi\rho$ when they are

transferred and so arranged on the scale-beam at the point $\alpha$

that $\theta$ is the center of gravity of both; and $\theta$ is the

center of gravity of the two circles combined whose diameters are $\xi

o$ and $\pi\rho$ when their position is changed,

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{\small Fig.~3.}

\end{center}

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hence $\theta\alpha : \alpha\sigma =$ the circle with the diameter

$\mu\nu :$ the two circles whose diameters are $\xi o$ and

$\pi\rho$. In the same way it can be shown that if another straight

line is drawn in the parallelogram $\lambda\zeta \| \epsilon\zeta$ and

on this line last drawn a plane is constructed perpendicular to

$\alpha\gamma$, then likewise the circle produced in the cylinder will

in its present position be in equilibrium at the point $\alpha$ with

the two circles combined which have been produced in the spheroid and

in the cone respectively when they are so transferred to the point

$\theta$ on the scale-beam that $\theta$ is the center of gravity of

both. Then if cylinder, spheroid and cone are filled with such

circles, the cylinder in its present position will be in equilibrium

at the point $\alpha$ with the spheroid $+$ the cone if they are

transferred and so arranged on the scale-beam at the point $\alpha$

that $\theta$ is the center of gravity of both. Now $\kappa$ is the

center of gravity of the cylinder, but $\theta$, as has been said, is

the center of gravity of the spheroid and cone together. Therefore

$\theta\alpha : \alpha\kappa =$ cylinder $:$ spheroid $+$ cone. But

$\alpha\theta = 2\alpha\kappa$, hence also the cylinder = 2 $\times$

(spheroid $+$ cone) = 2 $\times$ spheroid + 2 $\times$ cone. But the

cylinder = 3 $\times$ cone, hence 3 $\times$ cone = 2 $\times$ cone +

2 $\times$ spheroid. Subtract 2 $\times$ cone from both sides; then a

cone whose axes form the triangle $\alpha\epsilon\zeta$ = 2 $\times$

spheroid. But the same cone = 8 cones whose axes form the

$\Delta\alpha\beta\delta$; hence 8 such cones = 2 $\times$ spheroid, 4

$\times$ cone = spheroid; whence it follows that a spheroid is four

times as great as a cone whose vertex is at $\alpha$, and whose base

is the circle on the diameter $\beta\delta$ perpendicular to

$\lambda\epsilon$, and one-half the spheroid is twice as great as the

same cone.

In the parallelogram $\lambda\zeta$ draw the straight lines $\phi\chi$

and $\psi\omega \| \alpha\gamma$ through the points $\beta$ and

$\delta$ and imagine a cylinder whose bases are the circles on the

diameters $\phi\psi$ and $\chi\omega$, and whose axis is

$\alpha\gamma$. Now since the cylinder whose axes form the

parallelogram $\phi\omega$ is twice as great as the cylinder whose

axes form the parallelogram $\phi\delta$ because their bases are equal

but the axis of the first is twice as great as the axis of the second,

and since the cylinder whose axes form the parallelogram $\phi\delta$

is three times as great as the cone whose vertex is at $\alpha$ and

whose base is the circle on the diameter $\beta\delta$ perpendicular

to $\alpha\gamma$, then the cylinder whose axes form the parallelogram

$\phi\omega$ is six times as great as the aforesaid cone. But it has

been shown that the spheroid is four times as great as the same cone,

hence the cylinder is one and one half times as great as the

spheroid. Q. E. D.

\section*{Proposition IV}

That a segment of a right conoid cut by a plane perpendicular to its

axis is one and one half times as great as the cone having the same

base and axis as the segment, can be proved by the same method in the

following way:

Let a right conoid be cut through its axis by a plane intersecting the

surface in a parabola $\alpha\beta\gamma$ [Fig.~4]; let it be also cut

by another plane perpendicular to the axis, and let their common line

of intersection be $\beta\gamma$. Let the axis of the segment be

$\delta\alpha$ and let it be produced to $\theta$ so that

$\theta\alpha = \alpha\delta$. Now imagine $\delta\theta$ to be a

scale-beam with its center at $\alpha$; let the base of the segment be

the circle on the diameter $\beta\gamma$ perpendicular to

$\alpha\delta$; imagine a cone whose base is the circle on the

diameter $\beta\gamma$, and whose vertex is at $\alpha$. Imagine also

a cylinder whose base is the circle on the diameter $\beta\gamma$ and

its axis $\alpha\delta$, and in the parallelogram let a straight line

$\mu\nu$ be drawn $\| \beta\gamma$ and on $\mu\nu$ construct a plane

perpendicular to $\alpha\delta$; it will intersect the cylinder in a

circle whose diameter is $\mu\nu$, and the segment of the right conoid

in a circle whose diameter is $\xi o$. Now since $\beta\alpha\gamma$

is a parabola, $\alpha\delta$ its diameter and $\xi\sigma$ and

$\beta\delta$ its ordinates, then [\emph{Quadr. parab.} 3]

$\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$. But

$\delta\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma

= \mu\sigma^2 : \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2$ = the

circle in the cylinder whose diameter is $\mu\nu$ : the circle in the

segment of the right conoid whose diameter is $\xi o$, hence

$\theta\alpha : \alpha\sigma$ = the circle with the diameter $\mu\nu$

: the circle with the diameter $\xi o$; therefore the circle in the

cylinder whose

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\includegraphics[width=0.5\textwidth]{fig04.png}

\begin{center}

{\small Fig.~4.}

\end{center}

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diameter is $\mu\nu$ is in its present position, in equilibrium at the

point $\alpha$ with the circle whose diameter is $\xi o$ if this be

transferred and so arranged on the scale-beam at $\theta$ that

$\theta$ is its center of gravity. And the center of gravity of the

circle whose diameter is $\mu\nu$ is at $\sigma$, that of the circle

whose diameter is $\xi o$ when its position is changed, is $\theta$,

and we have the inverse proportion, $\theta\alpha : \alpha\sigma$ =

the circle with the diameter $\mu\nu$ : the circle with the diameter

$\xi o$. In the same way it can be shown that if another straight line

be drawn in the parallelogram $\epsilon\gamma \| \beta\gamma$ the

circle formed in the cylinder, will in its present position be in

equilibrium at the point $\alpha$ with that formed in the segment of

the right conoid if the latter is so transferred to $\theta$ on the

scale-beam that $\theta$ is its center of gravity. Therefore if the

cylinder and the segment of the right conoid are filled up then the

cylinder in its present position will be in equilibrium at the point

$\alpha$ with the segment of the right conoid if the latter is

transferred and so arranged on the scale-beam at $\theta$ that

$\theta$ is its center of gravity. And since these magnitudes are in

equilibrium at $\alpha$, and $\kappa$ is the center of gravity of the

cylinder, if $\alpha\delta$ is bisected at $\kappa$ and $\theta$ is

the center of gravity of the segment transferred to that point, then

we have the inverse proportion $\theta\alpha : \alpha\kappa$ =

cylinder : segment. But $\theta\alpha = 2\alpha\kappa$ and also the

cylinder = 2 $\times$ segment. But the same cylinder is 3 times as

great as the cone whose base is the circle on the diameter

$\beta\gamma$ and whose vertex is at $\alpha$; therefore it is clear

that the segment is one and one half times as great as the same cone.

\section*{Proposition V}

That the center of gravity of a segment of a right conoid which is cut

off by a plane perpendicular to the axis, lies on the straight line

which is the axis of the segment divided in such a way that the

portion at the vertex is twice as great as the remainder, may be

perceived by our method in the following way:

Let a segment of a right conoid cut off by a plane perpendicular to

the axis be cut by another plane through the axis, and let the

intersection in its surface be the parabola $\alpha\beta\gamma$

[Fig.~5] and let the common line of intersection of the plane which

cut off the segment and of the intersecting plane be $\beta\gamma$;

let the axis of the segment and the diameter of the parabola

$\alpha\beta\gamma$ be $\alpha\delta$; produce $\delta\alpha$ so that

$\alpha\theta = \alpha\delta$ and imagine $\delta\theta$ to be a

scale-beam with its center at $\alpha$;

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\begin{wrapfigure}{r}{0.5\textwidth}

\includegraphics[width=0.5\textwidth]{fig05.png}

\begin{center}

{\small Fig.~5.}

\end{center}

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%

then inscribe a cone in the segment with the lateral boundaries

$\beta\alpha$ and $\alpha\gamma$ and in the parabola draw a straight

line $\xi o \| \beta\gamma$ and let it cut the parabola in $\xi$ and

$o$ and the lateral boundaries of the cone in $\pi$ and $\rho$. Now

because $\xi\sigma$ and $\beta\delta$ are drawn perpendicular to the

diameter of the parabola, $\delta\alpha : \alpha\sigma = \beta\delta^2

: \xi\sigma^2$ [\emph{Quadr. parab.} 3]. But $\delta\alpha :

\alpha\sigma = \beta\delta : \pi\sigma = \beta\delta^2 : \beta\delta

\times \pi\sigma$, therefore also $\beta\delta^2 : \xi\sigma^2 =

\beta\delta^2 : \beta\delta \times \pi\sigma$. Consequently

$\xi\sigma^2 = \beta\delta \times \pi\sigma$ and $\beta\delta :

\xi\sigma = \xi\sigma : \pi\sigma$, therefore $\beta\delta : \pi\sigma

= \xi\sigma^2 : \sigma\pi^2$. But $\beta\delta : \pi\sigma =

\delta\alpha : \alpha\sigma = \theta\alpha : \alpha\sigma$, therefore

also $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$. On

$\xi o$ construct a plane perpendicular to $\alpha\delta$; this will

intersect the segment of the right conoid in a circle whose diameter

is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Now

because $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$ and

$\xi\sigma^2 : \sigma\pi^2$ = the circle with the diameter $\xi o$ :

the circle with the diameter $\pi\rho$, therefore $\theta\alpha :

\alpha\sigma$ = the circle whose diameter is $\xi o$ : the circle

whose diameter is $\pi\rho$. Therefore the circle whose diameter is

$\xi o$ will in its present position be in equilibrium at the point

$\alpha$ with the circle whose diameter is $\pi\rho$ when this is so

transferred to $\theta$ on the scale-beam that $\theta$ is its center

of gravity. Now since $\sigma$ is the center of gravity of the circle

whose diameter is $\xi o$ in its present position, and $\theta$ is the

center of gravity of the circle whose diameter is $\pi\rho$ if its

position is changed as we have said, and inversely $\theta\alpha :

\alpha\sigma$ = the circle with the diameter $\xi o$ : the circle with

the diameter $\pi\rho$, then the circles are in equilibrium at the

point $\alpha$. In the same way it can be shown that if another

straight line is drawn in the parabola $\| \beta\gamma$ and on this

line last drawn a plane is constructed perpendicular to

$\alpha\delta$, the circle formed in the segment of the right conoid

will in its present position be in equilibrium at the point $\alpha$

with the circle formed in the cone, if the latter is transferred and

so arranged on the scale-beam at $\theta$ that $\theta$ is its center

of gravity. Therefore if the segment and the cone are filled up with

circles, all circles in the segment will be in their present positions

in equilibrium at the point $\alpha$ with all circles of the cone if

the latter are transferred and so arranged on the scale-beam at the

point $\theta$ that $\theta$ is their center of gravity. Therefore

also the segment of the right conoid in its present position will be

in equilibrium at the point $\alpha$ with the cone if it is

transferred and so arranged on the scale-beam at $\theta$ that

$\theta$ is its center of gravity. Now because the center of gravity

of both magnitudes taken together is $\alpha$, but that of the cone

alone when its position is changed is $\theta$, then the center of

gravity of the remaining magnitude lies on $\alpha\theta$ extended

towards $\alpha$ if $\alpha\kappa$ is cut off in such a way that

$\alpha\theta : \alpha\kappa$ = segment : cone. But the segment is one

and one half the size of the cone, consequently $\alpha\theta =

\frac{3}{2}\alpha\kappa$ and $\kappa$, the center of gravity of the

right conoid, so divides $\alpha\delta$ that the portion at the vertex

of the segment is twice as large as the remainder.

\section*{Proposition VI}

[The center of gravity of a hemisphere is so divided on its axis] that

the portion near the surface of the hemisphere is in the ratio of $5 :

3$ to the remaining portion.

Let a sphere be cut by a plane through its center intersecting the

surface in the circle $\alpha\beta\gamma\delta$ [Fig.6],

$\alpha\gamma$ and $\beta\delta$ being two diameters of the circle

perpendicular to each other. Let a plane be constructed on

$\beta\delta$ perpendicular to $\alpha\gamma$. Then imagine a cone

whose base is the circle with the diameter $\beta\delta$, whose vertex

is at $\alpha$ and its lateral boundaries are $\beta\alpha$ and

$\alpha\delta$; let $\gamma\alpha$ be produced so that $\alpha\theta =

\gamma\alpha$, imagine the straight line $\theta\gamma$ to be a

scale-beam with its center at $\alpha$ and in the semi-circle

$\beta\alpha\delta$ draw a straight line $\xi o \| \beta\delta$; let

it cut the circumference of the semicircle in $\xi$ and $o$, the

lateral boundaries of the cone in $\pi$ and $\rho$, and $\alpha\gamma$

in $\epsilon$. On $\xi o$ construct a plane perpendicular to

$\alpha\epsilon$; it will intersect the hemisphere in a circle with

the diameter $\xi o$, and the cone in a circle with the diameter

$\pi\rho$. Now because $\alpha\gamma : \alpha\epsilon = \xi\alpha^2 :

\alpha\epsilon^2$ and $\xi\alpha^2 = \alpha\epsilon^2 + \epsilon\xi^2$

and $\alpha\epsilon = \epsilon\pi$, therefore $\alpha\gamma :

\alpha\epsilon = \xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$. But

$\xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$ = the circle with the

diameter $\xi o$ + the circle with the diameter $\pi\rho$ : the circle

with the diameter $\pi\rho$, and $\gamma\alpha = \alpha\theta$, hence

$\theta\alpha : \alpha\epsilon$ = the circle with the diameter $\xi o$

+ the circle with

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\includegraphics[width=0.33\textwidth]{fig06.png}

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{\small Fig.~6.}

\end{center}

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the diameter $\pi\rho$ : circle with the diameter $\pi\rho$. Therefore

the two circles whose diameters are $\xi o$ and $\pi\rho$ in their

present position are in equilibrium at the point $\alpha$ with the

circle whose diameter is $\pi\rho$ if it is transferred and so

arranged at $\theta$ that $\theta$ is its center of gravity. Now

since the center of gravity of the two circles whose diameters are

$\xi o$ and $\pi\rho$ in their present position [is the point

$\epsilon$, but of the circle whose diameter is $\pi\rho$ when its

position is changed is the point $\theta$, then $\theta\alpha :

\alpha\epsilon$ = the circles whose diameters are] $\xi o$ [,$\pi\rho$

: the circle whose diameter is $\pi\rho$. In the same way if another

straight line in the] hemisphere $\beta\alpha\delta$ [is drawn $\|

\beta\delta$ and a plane is constructed] perpendicular to

[$\alpha\gamma$ the] two [circles produced in the cone and in the

hemisphere are in their position] in equilibrium at $\alpha$ [with the

circle which is produced in the cone] if it is transferred and

arranged on the scale at $\theta$. [Now if] the hemisphere and the

cone [are filled up with circles then all circles in the] hemisphere

and those [in the cone] will in their present position be in

equilibrium [with all circles] in the cone, if these are transferred

and so arranged on the scale-beam at $\theta$ that $\theta$ is their

center of gravity; [therefore the hemisphere and cone also] are in

their position [in equilibrium at the point $\alpha$] with the cone if

it is transferred and so arranged [on the scale-beam at $\theta$] that

$\theta$ is its center of gravity.

\section*{Proposition VII}

By [this method] it may also be perceived that [any segment whatever]

of a sphere bears the same ratio to a cone having the same [base] and

axis [that the radius of the sphere + the axis of the opposite segment

: the axis of the opposite segment] \dotfill and [Fig.~7] on $\mu\nu$

construct a plane perpendicular to $\alpha\gamma$; it will intersect

the cylinder in a circle whose diameter is $\mu\nu$, the segment of

the sphere in a circle whose diameter

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\includegraphics[width=0.6\textwidth]{fig07.png}

\begin{center}

{\small Fig.~7.}

\end{center}

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%

is $\xi o$ and the cone whose base is the circle on the diameter

$\epsilon\zeta$ and whose vertex is at $\alpha$ in a circle whose

diameter is $\pi\rho$. In the same way as before it may be shown that

a circle whose diameter is $\mu\nu$ is in its present position in

equilibrium at $\alpha$ with the two circles [whose diameters are $\xi

o$ and $\pi\rho$ if they are so arranged on the scale-beam that

$\theta$ is their center of gravity. [And the same can be proved of

all corresponding circles.] Now since cylinder, cone, and spherical

segment are filled up with such circles, the cylinder in its present

position [will be in equilibrium at $\alpha$] with the cone + the

spherical segment if they are transferred and attached to the

scale-beam at $\theta$. Divide $\alpha\eta$ at $\phi$ and $\chi$ so

that $\alpha\chi = \chi\eta$ and $\eta\phi = \frac{1}{3}\alpha\phi$;

then $\chi$ will be the center of gravity of the cylinder because it

is the center of the axis $\alpha\eta$. Now because the above

mentioned bodies are in equilibrium at $\alpha$, cylinder : cone with

the diameter of its base $\epsilon\zeta$ + the spherical segment

$\beta\alpha\delta = \theta\alpha : \alpha\chi$. And because

$\eta\alpha = 3\eta\phi$ then [$\gamma\eta \times \eta\phi$] =

$\frac{1}{3}\alpha\eta \times \eta\gamma$. Therefore also $\gamma\eta

\times \eta\phi = \linebreak\frac{1}{3}\beta\eta^2$. \dotfill

\section*{Proposition VIIa}

In the same way it may be perceived that any segment of an ellipsoid

cut off by a perpendicular plane, bears the same ratio to a cone

having the same base and the same axis, as half of the axis of the

ellipsoid + the axis of the opposite segment bears to the axis of the

opposite segment. \dotfill

\section*{Proposition VIII}

\dotfill\linebreak produce $\alpha\gamma$ [Fig.~8] making

$\alpha\theta = \alpha\gamma$ and $\gamma\xi$ = the radius of the

sphere; imagine $\gamma\theta$ to be a scale-beam with a center at

$\alpha$, and in the plane cutting

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\begin{wrapfigure}{r}{0.4\textwidth}

\includegraphics[width=0.4\textwidth]{fig08.png}

\begin{center}

{\small Fig.~8.}

\end{center}

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off the segment inscribe a circle with its center at $\eta$ and its

radius = $\alpha\eta$; on this circle construct a cone with its vertex

at $\alpha$ and its lateral boundaries $\alpha\epsilon$ and

$\alpha\zeta$. Then draw a straight line $\kappa\lambda \|

\epsilon\zeta$; let it cut the circumference of the segment at

$\kappa$ and $\lambda$, the lateral boundaries of the cone

$\alpha\epsilon\zeta$ at $\rho$ and $o$ and $\alpha\gamma$ at $\pi$.

Now because $\alpha\gamma : \alpha\pi = \alpha\kappa^2 : \alpha\pi^2$

and $\kappa\alpha^2 = \alpha\pi^2 + \pi\kappa^2$ and $\alpha\pi^2 =

\pi o^2$ (since also $\alpha\eta^2 = \epsilon\eta^2$), then

$\gamma\alpha : \alpha\pi = \kappa\pi^2 + \pi o^2 : o\pi^2$. But

$\kappa\pi^2 + \pi o^2 : \pi o^2$ = the circle with the diameter

$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle

with the diameter $o\rho$ and $\gamma\alpha = \alpha\theta$; therefore

$\theta\alpha : \alpha\pi$ = the circle with the diameter

$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle

with the diameter $o\rho$. Now since the circle with the diameter

$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle

with the diameter $o\rho$ = $\alpha\theta : \pi\alpha$, let the circle

with the diameter $o\rho$ be transferred and so arranged on the

scale-beam at $\theta$ that $\theta$ is its center of gravity; then

$\theta\alpha : \alpha\pi$ = the circle with the diameter

$\kappa\lambda$ + the circle with the diameter $o\rho$ in their

present positions : the circle with the diameter $o\rho$ if it is

transferred and so arranged on the scale-beam at $\theta$ that

$\theta$ is its center of gravity. Therefore the circles in the

segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ are

in equilibrium at $\alpha$ with that in the cone

$\alpha\epsilon\zeta$. And in the same way all circles in the segment

$\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ in their

present positions are in equilibrium at the point $\alpha$ with all

circles in the cone $\alpha\epsilon\zeta$ if they are transferred and

so arranged on the scate-beam at $\theta$ that $\theta$ is their

center of gravity; then also the spherical segment $\alpha\beta\delta$

and the cone $\alpha\epsilon\zeta$ in their present positions are in

equilibrium at the point $\alpha$ with the cone $\epsilon\alpha\zeta$

if it is transferred and so arranged on the scale-beam at $\theta$

that $\theta$ is its center of gravity. Let the cyIinder $\mu\nu$

equal the cone whose base is the circle with the diameter

$\epsilon\zeta$ and whose vertex is at $\alpha$ and let $\alpha\eta$

be so divided at $\phi$ that $\alpha\eta = 4\phi\eta$; then $\phi$ is

the center of gravity of the cone $\epsilon\alpha\zeta$ as has been

previously proved. Moreover let the cylinder $\mu\nu$ be so cut by a

perpendicularly intersecting plane that the cylinder $\mu$ is in

equilibrium with the cone $\epsilon\alpha\zeta$. Now since the segment

$\alpha\beta\delta$ + the cone $\epsilon\alpha\zeta$ in their present

positions are in equilibrium at $\alpha$ with the cone

$\epsilon\alpha\zeta$ if it is transferred and so arranged on the

scale-beam at $\theta$ that $\theta$ is its center of gravity, and

cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$ and the two cylinders

$\mu + \nu$ are moved to $\theta$ and $\mu\nu$ is in equilibrium with

both bodies, then will also the cylinder $\nu$ be in equilibrium with

the segment of the sphere at the point $\alpha$. And since the

spherical segment $\beta\alpha\delta$ : the cone whose base is the

circle with the diameter $\beta\delta$, and whose vertex is at $\alpha

= \xi\eta : \eta\gamma$ (for this has previously been proved [\emph{De

sph. et cyl.} II, 2 Coroll.]) and cone $\beta\alpha\delta$ : cone

$\epsilon\alpha\zeta$ = the circle with the diameter $\beta\delta$ :

the circle with the diameter $\epsilon\zeta = \beta\eta^2 :

\eta\epsilon^2$, and $\beta\eta^2 = \gamma\eta \times \eta\alpha$,

$\eta\epsilon^2 = \eta\alpha^2$, and $\gamma\eta \times \eta\alpha :

\eta\alpha^2 = \gamma\eta : \eta\alpha$, therefore cone

$\beta\alpha\delta$ : cone $\epsilon\alpha\zeta = \gamma\eta :

\eta\alpha$. But we have shown that cone $\beta\alpha\delta$ :

segment $\beta\alpha\delta$ = $\gamma\eta : \eta\xi$, hence

{\selectlanguage{greek}di' \~isou} segment $\beta\alpha\delta$ : cone

$\epsilon\alpha\zeta$ = $\xi\eta : \eta\alpha$. And because

$\alpha\chi : \chi\eta = \eta\alpha + 4\eta\gamma : \alpha\eta +

2\eta\gamma$ so inversely $\eta\chi : \chi\alpha = 2\gamma\eta +

\eta\alpha : 4\gamma\eta + \eta\alpha$ and by addition $\eta\alpha :

\alpha\chi = 6\gamma\eta + 2\eta\alpha : \eta\alpha +

4\eta\gamma$. But $\eta\xi = \frac{1}{4} (6\eta\gamma + 2\eta\alpha)$

and $\gamma\phi = \frac{1}{4} (4\eta\gamma + \eta\alpha)$; for that is

evident. Hence $\eta\alpha : \alpha\chi = \xi\eta : \gamma\phi$,

consequently also $\xi\eta : \eta\alpha = \gamma\phi : \chi\alpha$.

But it was also demonstrated that $\xi\eta : \eta\alpha$ = the segment

whose vertex is at $\alpha$ and whose base is the circle with the

diameter $\beta\delta$ : the cone whose vertex is at $\alpha$ and

whose base is the circle with the diameter $\epsilon\zeta$; hence

segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi

: \chi\alpha$. And since the cylinder $\mu$ is in equilibrium with the

cone $\epsilon\alpha\zeta$ at $\alpha$, and $\theta$ is the center of

gravity of the cylinder while $\phi$ is that of the cone

$\epsilon\alpha\zeta$, then cone $\epsilon\alpha\zeta$ : cylinder

$\mu$ = $\theta\alpha : \alpha\phi = \gamma\alpha : \alpha\phi$. But

cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence by subtraction,

cylinder $\mu$ : cylinder $\nu$ = $\alpha\phi : \gamma\phi$. And

cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence cone

$\epsilon\alpha\zeta$ : cylinder $\nu$ = $\gamma\alpha : \gamma\phi =

\theta\alpha : \gamma\phi$. But it was also demonstrated that segment

$\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi :

\chi\alpha$; hence {\selectlanguage{greek}di' \~isou} segment

$\beta\alpha\delta$ : cylinder $\nu$ = $\zeta\alpha : \alpha\chi$.

And it was demonstrated that segment $\beta\alpha\delta$ is in

equilibrium at $\alpha$ with the cylinder $\nu$ and $\theta$ is the

center of gravity of the cylinder $\nu$, consequently the point $\chi$

is also the center of gravity of the segment $\beta\alpha\delta$.

\section*{Proposition IX}

In a similar way it can also be perceived that the center of gravity

of any segment of an ellipsoid lies on the straight line which is the

axis of the segment so divided that the portion at the vertex of the

segment bears the same ratio to the remaining portion as the axis of

the segment + 4 times the axis of the opposite segment bears to the

axis of the segment + twice the axis of the opposite segment.

\section*{Proposition X}

It can also be seen by this method that [a segment of a hyperboloid]

bears the same ratio to a cone having the same base and axis as the

segment, that the axis of the segment + 3 times the addition to the

axis bears to the axis of the segment of the hyperboloid + twice its

addition [\emph{De Conoid.} 25]; and that the center of gravity of the

hyperboloid so divides the axis that the part at the vertex bears the

same ratio to the rest that three times the axis + eight times the

addition to the axis bears to the axis of the hyperboloid + 4 times

the addition to the axis, and many other points which I will leave

aside since the method has been made clear by the examples already

given and only the demonstrations of the above given theorems remain

to be stated.

\section*{Proposition XI}

When in a perpendicular prism with square bases a cylinder is

inscribed whose bases lie in opposite squares and whose curved surface

touches the four other parallelograms, and when a plane is passed

through the center of the circle which is the base of the cylinder and

one side of the opposite square, then the body which is cut off by

this plane [from the cylinder] will be $\frac{1}{6}$ of the entire

prism. This can be perceived through the present method and when it is

so warranted we will pass over to the geometrical proof of it.

\begin{figure}[t]

\begin{minipage}[5]{0.45\textwidth}

\includegraphics[width=\textwidth]{fig09.png}

\begin{center}

{\small Fig.~9.}

\end{center}

\end{minipage}

\hfill

\begin{minipage}[5]{0.45\textwidth}

\includegraphics[width=\textwidth]{fig10.png}

\begin{center}

{\small Fig.~10.}

\end{center}

\end{minipage}

\end{figure}

Imagine a perpendicular prism with square bases and a cylinder

inscribed in the prism in the way we have described. Let the prism be

cut through the axis by a plane perpendicular to the plane which cuts

off the section of the cylinder; this will intersect the prism

containing the cylinder in the parallelogram $\alpha\beta$ [Fig.~9]

and the common intersecting line of the plane which cuts off the

section of the cylinder and the plane lying through the axis

perpendicular to the one cutting off the section of the cylinder will

be $\beta\gamma$; let the axis of the cylinder and the prism be

$\gamma\delta$ which is bisected at right angles by $\epsilon\zeta$

and on $\epsilon\zeta$ let a plane be constructed perpendicular to

$\gamma\delta$. This will intersect the prism in a square and the

cylinder in a circle.

Now let the intersection of the prism be the square $\mu\nu$

[Fig.~10], that of the cylinder, the circle $\xi o\pi\rho$ and let the

circle touch the sides of the square at the points $\xi$, $o$, $\pi$

and $\rho$; let the common line of intersection of the plane cutting

off the cylinder-section and that passing through $\epsilon\zeta$

perpendicular to the axis of the cylinder, be $\kappa\lambda$; this

line is bisected by $\pi\theta\xi$. In the semicircle $o\pi\rho$ draw

a straight line $\sigma\tau$ perpendicular to $\pi\chi$, on

$\sigma\tau$ construct a plane perpendicular to $\xi\pi$ and produce

it to both sides of the plane enclosing the circle $\xi o\pi\rho$;

this will intersect the half-cylinder whose base is the semicircle

$o\pi\rho$ and whose altitude is the axis of the prism, in a

parallelogram one side of which = $\sigma\tau$ and the other = the

vertical boundary of the cylinder, and it will intersect the

cylinder-section likewise in a parallelogram of which one side is

$\sigma\tau$ and the other $\mu\nu$ [Fig.~9]; and accordingly $\mu\nu$

will be drawn in the parallelogram $\delta\epsilon \| \beta\omega$ and

will cut off $\epsilon\iota = \pi\chi$. Now because $\epsilon\gamma$

is a parallelogram and $\nu\iota \| \theta\gamma$, and

$\epsilon\theta$ and $\beta\gamma$ cut the parallels, therefore

$\epsilon\theta : \theta\iota = \omega\gamma : \gamma\nu = \beta\omega

: \upsilon\nu$. But $\beta\omega : \upsilon\nu$ = parallelogram in the

half-cylinder : parallelogram in the cylinder-section, therefore both

parallelograms have the same side $\sigma\tau$; and $\epsilon\theta =

\theta\pi$, $\iota\theta = \chi\theta$; and since $\pi\theta =

\theta\xi$ therefore $\theta\xi : \theta\chi$ = parallelogram in

half-cylinder : parallelogram in the cylinder-section. lmagine the

parallelogram in the cylinder-section transferred and so brought to

$\xi$ that $\xi$ is its center of gravity, and further imagine

$\pi\xi$ to be a scale-beam with its center at $\theta$; then the

parallelogram in the half-cylinder in its present position is in

equilibrium at the point $\theta$ with the parallelogram in the

cylinder-section when it is transferred and so arranged on the

scale-beam at $\xi$ that $\xi$ is its center of gravity. And since

$\chi$ is the center of gravity in the parallelogram in the

half-cylinder, and $\xi$ that of the parallelogram in the

cylinder-section when its position is changed, and $\xi\theta :

\theta\chi$ = the parallelogram whose center of gravity is $\chi$ :

the parallelogram whose center of gravity is $\xi$, then the

parallelogram whose center of gravity is $\chi$ will be in equilibrium

at $\theta$ with the parallelogram whose center of gravity is

$\xi$. In this way it can be proved that if another straight line is

drawn in the semicircle $o\pi\rho$ perpendicular to $\pi\theta$ and on

this straight line a plane is constructed perpendicular to $\pi\theta$

and is produced towards both sides of the plane in which the circle

$\xi o\pi\rho$ lies, then the parallelogram formed in the

half-cylinder in its present position will be in equilibrium at the

point $\theta$ with the parallelogram formed in the cylinder-section

if this is transferred and so arranged on the scale-beam at $\xi$ that

$\xi$ is its center of-gravity; therefore also all parallelograms in

the half-cylinder in their present positions will be in equilibrium at

the point $\theta$ with all parallelograms of the cylinder-section if

they are transferred and attached to the scale-beam at the point

$\xi$; consequently also the half-cylinder in its present position

will be in equilibrium at the point $\theta$ with the cylinder-section

if it is transferred and so arranged on the scale-beam at $\xi$ that

$\xi$ is its center of gravity.

\section*{Proposition XII}

Let the parallelogram $\mu\nu$ be perpendicular to the axis [of the

circle] $\xi o$ [$\pi\rho$] [Fig.~11]. Draw $\theta\mu$ and

$\theta\eta$ and erect upon them two planes perpendicular to the plane

in which the semicircle $o\pi\rho$ lies and extend these planes on

both sides. The result is a prism whose base is a triangle similar to

$\theta\mu\eta$ and whose altitude is equal to the axis of the

cylinder, and this prism is $\frac{1}{4}$ of the entire prism which

contains the cylinder. In the semicircle $o\pi\rho$ and in the square

$\mu\nu$ draw two straight lines $\kappa\lambda$ and $\tau\upsilon$ at

equal distances from $\pi\xi$; these will cut

%

\begin{wrapfigure}{r}{0.5\textwidth}

\includegraphics[width=0.5\textwidth]{fig11.png}

\begin{center}

{\small Fig.~11.}

\end{center}

\end{wrapfigure}

%

the circumference of the semicircle $o\pi\rho$ at the points $\kappa$

and $\tau$, the diameter $o\rho$ at $\sigma$ and $\zeta$ and the

straight lines $\theta\eta$ and $\theta\mu$ at $\phi$ and $\chi$. Upon

$\kappa\lambda$ and $\tau\upsilon$ construct two planes perpendicular

to $o\rho$ and extend them towards both sides of the plane in which

lies the circle $\xi o\pi\rho$; they will intersect the half-cylinder

whose base is the semicircle $o\pi\rho$ and whose altitude is that of

the cylinder, in a parallelogram one side of which = $\kappa\sigma$

and the other = the axis of the cylinder; and they will intersect the

prism $\theta\eta\mu$ likewise in a parallelogram one side of which is

equal to $\lambda\chi$ and the other equal to the axis, and in the

same way the half-cylinder in a parallelogram one side of which =

$\tau\zeta$ and the other = the axis of the cylinder, and the prism in

a parallelogram one side of which = $\nu\phi$ and the other = the axis

of the cylinder.\dotfill

\section*{Proposition XIII}

Let the square $\alpha\beta\gamma\delta$ [Fig.~12] be the base of a

perpendicular prism with square bases and let a cylinder be inscribed

in the prism whose base is the circle $\epsilon\zeta\eta\theta$ which

touches the sides of the parallelogram $\alpha\beta\gamma\delta$ at

$\epsilon$, $\zeta$, $\eta$, and $\theta$. Pass a plane through its

center and the side in the square opposite the square

$\alpha\beta\gamma\delta$ corresponding to the side $\gamma\delta$;

this will cut off from the whole prism a second prism which is

$\frac{1}{4}$ the size of the whole prism and which will be bounded by

three parallelograms and two opposite triangles. In the semicircle

$\epsilon\zeta\eta$ describe a parabola whose origin is $\eta\epsilon$

and whose axis is $\zeta\kappa$, and in the parallelogram $\delta\eta$

draw $\mu\nu \| \kappa\zeta$; this will cut the circumference of the

semicircle at $\xi$, the parabola at $\lambda$, and $\mu\nu \times

\nu\lambda = \nu\zeta^2$ (for this is evident [Apollonios, \emph{Con.}

I, 11]). Therefore $\mu\nu : \nu\lambda = \kappa\eta^2 :

\lambda\sigma^2$. Upon $\mu\nu$ construct a plane parallel to

$\epsilon\eta$; this will intersect the prism cut off from the whole

prism in a right-angled triangle one side of which is $\mu\nu$ and the

other a straight line in the plane upon $\gamma\delta$ perpendicular

to $\gamma\delta$ at $\nu$ and equal to the axis of the cylinder, but

whose hypotenuse is in the intersecting plane. It will intersect the

portion which is cut off from the cylinder by the

%

\begin{wrapfigure}{r}{0.5\textwidth}

\includegraphics[width=0.5\textwidth]{fig12.png}

\begin{center}

{\small Fig.~12.}

\end{center}

\end{wrapfigure}

%

plane passed through $\epsilon\eta$ and the side of the square

opposite the side $\gamma\delta$ in a right-angled triangle one side

of which is $\mu\xi$ and the other a straight line drawn in the

surface of the cylinder perpendicular to the plane $\kappa\nu$,

\linebreak and the hypotenuse\dotfill\linebreak and all the triangles

in the prism : all the triangles in the cylinder-section = all the

straight lines in the parallelogram $\delta\eta$ : all the straight

lines between the parabola and the straight line $\epsilon\eta$. And

the prism consists of the triangles in the prism, the cylinder-section

of those in the cylinder-section, the parallelogram $\delta\eta$ of

the straight lines in the parallelogram $\delta\eta \| \kappa\zeta$

and the segment of the parabola of the straight lines cut off by the

parabola and the straight line $\epsilon\eta$; hence prism :

cylinder-section = parallelogram $\eta\delta$ : segment

$\epsilon\zeta\eta$ that is bounded by the parabola and the straight

line $\epsilon\eta$. But the parallelogram $\delta\eta = \frac{3}{2}$

the segment bounded by the parabola and the straight line

$\epsilon\eta$ as indeed has been shown in the previously published

work, hence also the prism is equal to one and one half times the

cylinder-section. Therefore when the cylinder-section = 2, the prism =

3 and the whole prism containing the cylinder equals 12, because it is

four times the size of the other prism; hence the cylinder-section is

equal to $\frac{1}{6}$ of the prism, Q. E. D.

\section*{Proposition XIV}

[Inscribe a cylinder in] a perpendicular prism with square bases [and

let it be cut by a plane passed through the center of the base of the

cylinder and one side of the opposite square.] Then this plane will

cut off a prism from the whole prism and a portion of the cylinder

from the cylinder. It may be proved that the portion cut off from the

cylinder by the plane is one-sixth of the whole prism. But first we

will prove that it is possible to inscribe a solid figure in the

cylinder-section and to circumscribe another composed of prisms of

equal altitude and with similar triangles as bases, so that the

circumscribed figure exceeds the inscribed less than any given

magni-\-\linebreak tude.\dotfill

But it has been shown that the prism cut off by the inclined plane

$<\frac{3}{2}$ the body inscribed in the cylinder-section. Now the

prism cut off by the inclined plane : the body inscribed in the

cylinder-section = parallelogram $\delta\eta$ : the parallelograms

which are inscribed in the segment bounded by the parabola and the

straight line $\epsilon\eta$. Hence the parallelogram $\delta\eta

<\frac{3}{2}$ the parallelograms in the segment bounded by the

parabola and the straight line $\epsilon\eta$. But this is impossible

because we have shown elsewhere that the parallelogram $\delta\eta$ is

one and one half times the segment bounded by the parabola and the

straight line $\epsilon\eta$, consequently is \dotfill not greater

\dotfill

And all prisms in the prism cut off by the inclined plane : all prisms

in the figure described around the cylinder-section = all

parallelograms in the parallelogram $\delta\eta$ : all parallelograms

in the figure which is described around the segment bounded by the

parabola and the straight line $\epsilon\eta$, i. e., the prism cut

off by the inclined plane : the figure described around the

cylinder-section = parallelogram $\delta\eta$ : the figure bounded by

the parabola and the straight line $\epsilon\eta$. But the prism cut

off by the inclined plane is greater than one and one half times the

solid figure circumscribed around the

cylinder-section\dotfill\linebreak.\dotfill\linebreak

\vfill

\end{document}

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