VERIIFICATION 0F TIEE QUADRATURE 0OF THE CIRCLE.
Let either of the above quadrants be divided into 196 squares, having 14 on each side; then
the line which is the side of the whole inscribed square will divide the quadrant into two
equal parts, each of which contains 98 squares.
And the arc of the circle belonging to the quadrant will divide the outside 98 squares into
two parts, which shall be to each other as 4 is to 3.
For the arc of the quadrant cuts 21 of the squares, which form a complete arch; and there
are 142 squares within those which are eut by the arc and 23 without; then if these 21 squares
be divided in the proportion of 4 to 3, there will be 12 squares that will fall within the circle
and 9 that will fall without it; then 142 + 12 = 154, the number of squares within the arc, and
there are 196 squares in the quadrant.
196   14
Then 154 is to 196 as 11 is to 14, for 54 = 11
Again, of the 154 blocks within the arc, if 98 be deducted, we shall have 56 between the
side of the inseribed square and the arc of the circle; and if to the 33 blocks without the arc 9
be added, we shall have 42 blocks without the arc of the circle; then 56 is to 42 as 4 is to 3.
Again, if the sides of the squares which form the arch of the quadrant, viz: 5 + 2 +- 2+ 2
11 X 2 = 22, be multiplied by 4, we shall have 88 sides for the circumference of the circle,
and as each of the quadrants has 14 squares on each side, we shall have 28 sides for the diameter of the circle.
Then dividing thc given circumference by the given diameter, we have 88 -. 28 = 22 -- 7 7
3.142857, or 31, which is the true ratio of the circumference to the diameter of the given circle.
V~~~~~~~~~~/i




THE
QUADRATURE OF THE CIRCLE,
THE
SQUARE ROOT 0F TWO,
AND THE
RIGHT-ANGLED TRIANGLE,
BY WILLIAM ALEXANDER MYERS,
President of Myers' Commercial College, Louisville, Ky.
SECOND EDITION.
"Where is the wise."-1st Cor., i, 20. "Now the serpent. vas more subtile than any of the
beasts of the earth which the Lord God.had made."-Gen. iii, 1.
CINCINNATI:
WILSTACH, BALDWIN & CO., PRINTERS,
NOS. 141 AND 143 RACE STREET.
1874.




AUTHORS MADE USE OF IN THE PRESENT VOLUME.
Should the student desire more general information upon the subjects treated
of in the present volume, he is referred to the following works which have been
freely used by the author wherever they have been found to be of service to his
cause. They will be found to be among the best of their kind
"Montuclas' History of Mathematics;" "lHutton's Recreations;" "DeMorgan
on the Law of Probabilities;" "Elements of Euclid," by Todhunter; "Elements of Euclid," by Thompson; "Davies' Le Gendre;" "Robinson's Geometry;" "Chauvenet's Geometry;" "Loomis' Geometry and Trigonometry;"
"Bullfinch's Beauties of Mythology;" "Minifee's Draughting and Architecture;" "Home and School Journal;" "Chambers' Encyclopoedia," and
the " Douay Bible."
Entered according to Act of Congress, in the Office of the Librarian at Washingtou, D. C.,
BY WILLIAM ALEXANDER MYERS,
March 31st, 1873.
All Rights Reserved.




TO
THE AMERICAN PEOPLE
WHOSE LOVE FOR LEARNING AND DEVOTION TO THE TRUTH,
ARE ONLY EQUALED
BY THE MAGNIFICENT CONTRIBUTIONS WHIICH THEY HAVE MADE
TO THE CAUSE OF EDUCATION,
TtIIS VOLUME IS RESPECTFULLY INSCRIBED AS A CHEERFUL CONTRIBUTION TO
THE CAUSE WHICEH WE ALL ADVOCATE IN COMMON, AND AS A
SMALL TESTIMONIAL OF THE ESTEEM IN WHICH
THEY ARE HELD
BY THE AUTHOR.








PREFACE.
THE following pages are intended to explain certain mathematical truths, which were discovered by the author while engaged
in a series of investigations made during the hours of rest from
the labors of the college and the counting room. They consist
chiefly of new methods employed in the solution of problems
which have heretofore been regarded by mathematicians as impossible; and, although the author's mind has been employed with
the subject for a number of years, the result of the investigations
are now published for the first time.
If the discoveries should not come up to that standard ofbrilliancy which commands attention, it is hoped that they may be
found worthy of a fair and impartial consideration.
The author scarcely dares to hope, with the many examples of
failure before him, that at the outset the entire mathematical
world will bow in submission to his decree, or submit unconditionally to the power of his reason or the force of his logic; nor
does he desire that the glorious fabric, which the mathematical
genius of the world combined has reared as a monument to the
memory of departed greatness, should crumble into dust by a
single touch. Ah, no! Rather let the ivy of remembrance forever remain green upon their mausoleums, and the vines of gladness encircle their remains. But if Genius, while pursuing her
walks amid these temples of departed greatness, should suddenly
be inspired by 1Wisdom, and conceive Truth, who woulc be so
poor as to refuse a garland with which to crown her brow, where
truth sits enthroned?
The discoveries are as follows:
1. The Quadrature of the Circle.




vi                       PREFACE.
2. A Common Measure of the Side and Diagonal of the Square.
3. An Infinite Series of Right-angled Triangles, with a Rule
for their Solution.
For information concerning the History of the Quadrature of
the Circle, the reader is referred to the Introduction, which begins
on page 9, of this book. But before we proceed too far in our
investigation of the subject, it seems proper to inquire first what
is the circle. If a draughtsman or mechanic take an ordinary
pair of dividers, and with one foot as a center, and the other
starting at a certain point, cause it to describe a curve which is
constantly receding upon itself, this point will return to the point
from whence it started, when it is said to be an inclosed curve;
and the curve, which is described by one point rotating around
the other point within, is said to be the circumference of the
circle, every point of which is equally distant from the point
within; and this point within is called the center of the circle;
and the plane figure which is inclosed by the circumference is
said to be the circle itself. But a nmathemactical circle is more difficult to comprehend. If we say that to make a dot with a pencil that it is a point, the definition is sufficient for mechanical
purposes; but a mathematical point has position only, and no
magnitude, because it has no size. So, also, a mathematical circumference is a curved line constantly receding upon itself; but,
like a mathematical straight line, it has length only, itithout either
breadth or thickness. A mathematical circle, then, is a plane figure, which is inclosed by a curved line so finely defined as to be
invisible, not only to the naked eye but by the means of the most
powerful microscope which it is likely ever will be made, yet its
existence can be as certainly determined, mathematically, as if it.vere drawn mechanically upon wood or paper, and not only its
figure, but its dimensions, and consequently the ratio or pro



PREFACE.                       Vii
portion of its different parts one to another. But it may be objected that it has no real existence, because we can not see it. It
may be answered that it has a mathematical existence, which can
be so plainly demonstrated by sensible figures that the human
mind finds it impossible to doubt it. Just as the Deity has an
eternal existence, for neither can we see Him  " and live," but it
can be shown that He is, for "the invisible things of Ifin,fromn the
creation of the world, are clearly seen, being understood by the things
that are made, His eternal power also, and divinity;" and it is a wonderful truth that space, which is also said to be infinite, can only
even be partially measured by the aid of the power which we obtain from the science of numbers, because, by their aid, we can
reason mathematically and truly far beyond what we can see. It
may be said with truth that the circle, the square, and even the
triangle are emanations from the divine intelligence, as well as the
science of numbers, bv the aid of which they are measured, because they are a contrivance; and if there is contrivance there must
have been design; but there is contrivance, therefore there was design;
and if there was design, there must have been a designer; and it
is very evident this designer was not man, who has expended all
the talent, energy, and ability he ever had in trying to find out
what the circle is, and has ignominiously failed. And after all
the struggle, by simply turning our eyes to Holy Writ, there we
find it, simple and beautiful as truth itself, of which it is a fit emblem, for it has no fault-it is perfect. " Verily the foolishness of
God is wiser than nen." The Designer, then, must have been one
of infinite power and wisdom, "higher than the heavens," " who
dwelleth il light inaccessible; and if so, He must have used such
numbers as in His infinite wisdom woulc best accomplish that result, and these numbers mnst be such as is commonly found in all
His works. Let us " come and see."




Viii                     PREFACE.
By PROPOSITION 2, PART 1, it is proved that, if from the circumference of the given polygon the  lth be deducted, the remaining  - will be exactly equal to the circle itself; and if the
square root of the -4 be extracted, the result will be the number
7, which is the base of the system.
Again, by PROPOSITION 3, PART 1, it is proved that, if from
the sums of the squares of the two sides of any square the -5th
be deducted, the square root of the remaining 4 can be extracted
exactly, which will be 7, the base of the system or the generating
number; and if the square root of the {~th be extracted, it will
give the number 1, which is another generating number.
COINCIDENCE FROM HOLY WRIT.
Levit. xxiii., xxiv., and xxv., see the numbers 7, 49, and 50.
Genesis, Exodus, Leviticus, Numbers, Deuteronomy, and Apocalypse for the number 7.
By Case 2, PART 1, it is shown that the sine No. 1 was divided
into 14 parts, and each part was -,.
COINCIDENCE FROM HOLY WRIT.
See Genesis, Exodus, Leviticus, Numbers, and St. Matthew for
the number 14.
See Genesis and Daniel for the number 70.
By Case 2, the inscribed double triangle for double the number
of sides was proved to be 9-; by Case 3, i92-0; by Case 4,
7 6 8 3 2 8, and so on ad infinitum.
COINCIDENCE FROM, HOLY WRIT.
See Genesis, Exodus, Leviticus, Numbers, and Apocalypse for
the number 2.
See Job, chapters 1 and 2, for the circumference and diameter.
See Genesis for the circle.
See Apocalypse for the square.
"And the stone which the builders rejected was composed of three triangles."




HEIISTOIY
OF THE
QUADRATURE OF THE CIRCLE,
TRANSLATED FROM THE FRENCH OF MONTUCLA,
BY   J. BABIN, A. B., LOUISVILLE, KY.
IT has seemed to us fit to treat here separately this Great Question,
on account of its great celebrity, and we shall not hesitate to give some
of the silly notions to which this problem has given rise in ill-balanced
and enthusiastic minds.
To square the circle is to assign the geometrical dimensions of a
square equal to the circle. The quadrature of the circle has been attempted in several ways, by endeavoring to find a square or any other
rectilineal figure equal to the circle. As it was soon found that the
rectangle of the radius, by half the circumference, is equal to the area
of the circle, the problem was soon reduced to finding the length of the
circumference in terms of the radius. We can not believe that Archimedes was the first who made known this truth, for it is a necessary
consequence of what was already known on the measure of regular
polygons of which the circle is the extreme limit, the last of all.
The circle being after rectilineal figures, the most simple in appearance, geometricians very naturally soon began to seek for its measure.
Thus we find that the philosopher Anaxagoras occupied himself with
this question in his prison. Then Hippocrates of Chios tried the same
problem, and it led him to the discovery of what is called his lune. a sur



10                      INTRODUCTION.
face in the shape of a crescent, bounded by two arcs and exactly equal
to a given square. He also found two unequal lines which were toiether equal to a rectilineal figure, so that if their relation could have
been found the solution of the problem would have been obtained.
But this no one has yet been able to do, nor is it likely ever to be done.
We are also indebted to Simplicius for the history of a disciple of
Pythagoras, named Sextus, who claimed to have solved the problem,
but his reasoning has not been transmitted to us. Finally, this inquiry at that early day became so famous that Aristophanes in ridiculing Meton makes him appear on the stage, in his "Comedy of the
Clouds," as promising to square the circle about 430 years before
Christ. This was all the more amusing from the fact that people generally suppose that to attempt to square the circle is the same thing as
trying to make a circle square, which implies a manifest contradiction.
Yet this was that Meton, so renowned for his discovery of the cycle of
nineteen years, of whom, together with Socrates, the comedian made a
public laughing stock.
Aristotle mentions two of-his contemporaries, Bryson and Antiphon
who worked at the quadrature of the circle. Nothing could have been
more grossly inaccurate than Bryson's pretended quadrature; for he
made the circumference of a circle equal to 3M times the diameter.
But Antiphon stated that having inscribed a square in a circle, if an
isoceles triangle having the chord for the base be inscribed in each of
the remaining segments, and similar triangles in the remaining eight
segments, and so on, the sum of all these rectilineal figureswould be
equal to the circle; nothing could be more true than this, and Aristotle
was undoubtedly wrong in calling Antiphon a paralogist, for one of
Archimedes' two quadratures of the parabola depends upon the same
operation; but this method has not yet succeeded with the circle.
It might be supposed that Archimedes applied himself to the solution
of this problem, and that he gave his approximate measure of the circumference of a circle only for want of the long-sought-for vigorously
exact measure. His discoveries on the spiral, if they have preceded
his book on the dimensions of the circle, might well inspire him with
the hope of finding the length of the circumference. However, that
may be, Archimedes showed, about the year 250 before Christ, that if
the diameter of a circle is 1, its circumference is less than 31-~ or 31,
and more than 31 ~; by taking 3-1 the error is less than the 97 of the
diameter. The calculation of Archimedes is singularly skillful, and




INTRODUCTION.                       11
anticipates the objection made by some of those who reject his account,
for the reason that he could not precisely extract the square roots of
the several numbers used in his calculation. But I have known some
of these individuals, and I have never found a single one who knew
Archimedes otherwise than by name.
* We still further know by the testimony of Simplicius, that Niconedes
and Appolonius had tried to square the circle; the first by means of
the curve, which he calls quadrans or the quadratrix, the discovery of
which, however, is usually ascribed to Dinostratos; and the second, by
means of a certain line which he called the sister of the tortuous line,
or the spiral, and which was nothing else but the quadratrix of Dinostratos. This quadratrix, invented in truth at first to divide the angle
in any way whatever, would give the quadrature if its extreme limit on
the radius could be found. Perhaps Appolonius or Nicomedes discovered this property; be this so or not, Eutocious tells us that Appolonius had carried further than Archimedes the close relation of the
diameter to the circumference, and that another geometrician named
Philo, of Gadares or Gades, had gone still further, so that the error
did not exceed the loo'ooth. The moderns have carried this accuracy
much beyond this point.
Finally, among the ancients there were many of those persons unworthy the name of geometrician, who pretended to have found
in different ways the quadrature of the circle. Jamblicus, cited by
Simplicius, says so expressly. But their false reasoning has not reached
us, and no doubt did not deserve it.
The Arabs, who followed the Greeks in the Culture of the Sciences,
must also have had their quadratures; but all we know about it is that
some of them supposed they had discovered that the diameter being 1,
the circumference is the square root of 10; a very grave error; for it
exceeds 3.162, and the circumference, according to the account of Archimedes, is not quite 3.142857. For the rest we see in the catalogues
of Arabian writings several works entitled de quadrature Circuli;
like several others on the trisection of the angle, the duplication of the
cube, etc.
We pass rapidly over the centuries of ignorance which produced a
few treatises on the quadrature of the circle, manuscripts left remaining
in the dust of libraries, until we reach the period of the revival of let.
ters among ourselves. About this time the famous CARDINAL DE
CUSA distinguished himself by his unfortunate attempts at the solution




12                      INTRODUCTION.
of this problem. Nevertheless he tried an ingenious method; he rolled
a circle on a plane or line, and supposing that its circumference was
applied to it wholly until the point which had first touched it touched
it again; he therefore justly inferred that this line would be equal to
the circumference. He even conceived the outline of the curve, which
the point that first touched the straight line was to describe which
formed the curve, since called the Cycloid. But he supposed, witl
Charles de Bovelle, in the following century, that this curve was itself
an arc of a circle, and from this he claimed to determine it by a geometrical construction which was entirely arbitrary, resting on no real
property of this movement.  He also tried another method, according
to which he gave the following solution of the problem: a circle being
given, add to its radius the side of the inscribed square, and with this
line as diameter describe a circle, in which is inscribed an equilateral triangle, the perinuter of this triangle will, says Cardinal de
Cusa, be equal to the circumference of the first circle.
It was not difficult for Regiomontanus to prove that Cusa was mistaken; this relation of the circumference to the diameter fell outside
the limits demonstrated by Archimedes; that is according to this relation the diameter would be to the circumference as one to a number
greater than 34 already too large. Besides, the Cardinal learned for
his age, though very much addicted to astrology, he presents in the
collection of his works several geometrical tracts which are full of
paralogisms.
We have just spoken of Charles de Bovelle or Carolus Bovillus, distinguished at the time by the title of noble philosopher. He signalized
himself by the strangest ideas. He gave in 1507 a work entitled:
Introductionum Geometricurn, translated into French and republished
in 1552 under the auspices of Oronce Finée, under the title of Geometrie Pratique, Composée par le noble Philosopha, Maitre Charles de
Bovelle, etc. He claims to give there the quadrature of tle circle according to the idea of the Cardinal de Cusa, which, he says, came to
him by seeing a wheel moving on the pavement. But the construction
by which he pretends to give the length of the line to which is applied the circumference of the rolling circle is absolutely arbitrary, and
it would follow that the diameter is to the circumference as 1 is to the
square root of 10, or 3.1618, which is far from the limits of Archimedes.
What is also singular, is that in this same book, and in an appendix
added to the first volume of the preceding works, he speaks of the




INTRODUCTION.                        13
quadrature of the circle made by a poor peasant, according to which
the circle having 8 for diameter is equal to the square having 10 for
diagonal, that is to 50, which is false; for the circle is in this case less
than 502, and more than 501-, and the quadrature of Bovelle does not
agree with that of the peasant, which he considers as true; for the
latter gives the relation of the diameter to the circumference exactly as
1.0000 to 3.1250; the noble philosopher even wanders further fiom
the truth than the peasant does below; and he might have been told
that when one is mistaken he ought not at least to contradict himself.
Bovelle says, falsely, that these relations coincide. Either he had not
performed the calculation himself, or he did not know enough arithmetic to extract by approximation a square root; these works of Bovelle
are pitiable; his nianner of cubing the sphere is preëminently absurd.
We are sorry to find in the same class a royal professor of the 16th
century, named Oronce Finée, who, by his numerous works, acquired a
kind of fame. He gave in his Protomathesis a quadrature of the circle,
a little more ingenious, in truth, than that of Bovelle; but which is,
nevertheless, a paralogism. On the point of dying, in 1555, he urgently
advised his friend, Mizault de Monthuçon, to publish his discoveries, not
only upon this subject, but also on the most famous problems of geometry, such as the trisection of the angle, and the duplication of the cube,
and the inscription in the circle of all regular polygons. Mizault kept his
word, and in 1556 published this assemblage'of paralagisms under the
title of De rebus Mathematicus hactenus desideratis, libri IV. Most of
these problems are solved in various ways by him;. it happens that his
different solutions of the same problem do not agree with one another,
nor with those of Bovelle, and of his rural geometrician which he had
approved by publishing them, it was the height of false reasoning in
geometry; consequently he was easily refuted by the geoinetrician
Buteon, who had been his disciple at the College Royal, by Momus or
Nunez, a Portuguese geometrician, and several others; but still he
died satisfied, fully persuaded that his name would be placed on a level
with those of Archimedes and Apollonius. This scandal was renewed
among the royal professors in 1600, when Monantheuil, one of their
number, published a quadrature of the circle.
One Simon a Quercu (doubtless Duchêne or Van Eck) appeared on
the arena a few years later, in 1585, and proposed a quadrature of the
circle. His pretended discovery was much less wide from the truth
than those of his predecessors and fell within the limits of Archime



14                      INTRODUCTION.
des. So Peter Metius, who undertook to refute him, was obliged to
seek for a closer relation of the diameters to the circumference, and
found that the one was to the other as 113 to 355. The pretended
quadrature of Duchêne could not stand this test, and must be named
only because it led to the curious and elegant discovery of Metius;
for this relation of 113 to 355, reduced to decimals, is the same
as 10000000 to 31415929; which is at the most but  a-oo   of the
diameter in excess. The diameter of the earth being only 6542816
(toises) or 13936912 yards, the error made by this relation of the
circumference of a circle of that size would scarcely be 2 (toises) or
4 yards. If those who connect in their minds the problem of the
quadrature of the circle with that of longitudes, knew what we have
just said they would soon see their mistake.; for, if these problems
were connected with each other, what would be an error of 4 yards
on a track around the earth?  The Spanyard, Sir Jaime Falcon, of
the order of Our Lady (Notre Dame), of Montesa, published in 1587,
at Antwerp, his paralogism on the quadrature of the circle.  His
book is rendered amusing by a dialogue in verse between himself and
the circle, which thanks him very affectionately for squaring it; but
the good and model knight ascribes all the honor thereof to the holy
patroness of his order. The paralogism was apparently so gross that
no one took the trouble to refute it.
But a man much more famous than the foregoing challenged the attention of learned Europe by his pretensions on the quadrature of the
circle; it was the celebrated Joseph Scaliger.. Full of self conceit, he
supposed that he had only to present himself on the field of geometry
anc that nothing that had baffled geometricians until then could resist
a man of letters with his powers. He therefore undertook to find the
quadrature of the circle, and put forward, with much braggadocia, his
discoveries on this subject in a book which appeared in 1592: iNova
Cyclometria; but he had no cause to congratulate himself on having
thus wished to place himself among geometricians. For he was refuted by Clavius, Viete, Adrianus Ronmanus, Christman, etc., who
showed each in his own way that the size which le assigned to the circumference of the circle was only a little less than the inscribed polygon of 192 sides; which being absurd, demonstrated the incorrectness
of Scaliger's reasoning; but he did not surrender; and never did a
man who thought he had discovered the quadrature of the circle, the
trisection of the angle, the duplication of the cube, or perpetual motion,




INTRODUCTION.                        15
give in to the plainest reasoning. He will sooner deny the most elementary propositions of geometry, like Moliensis Cana, who found no
less than twenty-seven false propositions in the first book of Euclid.
Scaliger replied with bitterness to the geometricians who had censured
his quadrature; he treated them with contempt, especially Clavius who
had already wounded him by an answer to his attack on the Gregorian
Calender. Unfortunately for the honor of Scaliger abuse is not reasoning, and the established fact remains that Scaliger, an eagle in
literature, was nothing in geometry.
Scarcely had Scaliger disappeared when one Thomas Gephirander
came to take his place. But he had not Scaliger's pride; he acknowledged, even in the title of his book, that his discovery was simply the
result of divine Grace. We shall see many others gifted with this
same spirit of humility. The paralogism of Gephirander nevertheless
was palpable; for it consisted in the pretension that if between two
magnitudes there is any geometrical relation whatever, the same relation will exist if the same quantity is taken from each. Thus, according
to this illuminate, the same relation exists between 2 and 5 as between
3 and 6, since only the same quantity, viz., unity is subtracted'from
each of these numbers. But scarcely any of the follies with which
false reasoning, a false mind, and the conceit of never recanting one's
errors inspire these visionaries have equaled those of Alph. Cano, of
Molina, in a book entitled: Nuevos descubrimientos Geometricos. IHe
remodels the whole of Euclid, and scarcely one of his propositions is
spared by him. Yet who would believe it! he found another fool,
named Janson or Jansen, who translated him into Latin under the
title of Nova reperta Geomnetrica, etc. Moreover Cano admitted that
he had not the least idea of geometry until the Deity, whose delight it is
to humble the proud and enlighten the ignorant, had inspired him.
A similar wiseacre presented at the same time in France his paralogisms upon the quadrature of the circle and the duplication of the
cube. It was a merchant of Rochelle, called De Laleu. This one also
pretended to have received the solution of these problems by divine
revelation, and announced that the union of the Jews, Turks, and Pagans to the Christian religion depended upon the manifestation of this
truth. In fact, according to him, the quadrature of the circle was the
quadrature of the heavenly Temple, and the duplication of the cube
that of the elementary, terrestrial, and aquatic altar, whence was to flow
the conversion of the Jews, idolaters, etc. Accordingly some religion



16                      INTRODUCTION.
ists, overexcited by lueditation, dabbled with the inatter, and the superior of the Jesuits even invited some able geometricians of that time,
like Mydorge, Hardy, etc., to confer with Laleu. The result cau easily
be foreseen-it is impossible to reason with persons who have no principles in common with us. Hardy showed the incorrectness of the
solutions in a manner satisfactory to all geometricians; for, as this Laleu gave several solutions of the same problem, Hardy made it appear
that these did not even agree with each other. But Laleu, backed by
Perjos, his book-keeper, and by a Scotchman named J. de Dunbar,
did not give up the contest until after his death. We pass lightly on
some other quadratures of the circle proposed by an anonymous writer
of his time, and by one Benoit Scotto, whom St. Clair, royal professor,
and Hardy refuted, to come to Longomontanus, who defiled, so to
speak, the last years of his life, by his pretensions on the quadrature
of the circle. That astronomer, formerly a disciple of Tycho Brahé,
and known by a good work on astronomy, imagined, in 1622, that he
had discovered the solution of this celebrated problem, which he published under the title of Cyclomnetria lunulis reciproci demonstrate, etc.
He claimed to have found that the diameter is to the circumference as
1. is to 3.14185. In vain did Snellius, Henry Briggs, Guldin, warn
him with moderation of his mistake, by showing him that the diameter
being 1, the circuniference is more than 3.14159, and less than 3.14160.
But Longomontanus was not at all inclined to give in. He heaped a
thousand bad reasons against the calculations of Vieta, Adrianus, Romanus, Ludolph, Vanceulen, Snellius, who were unanimously opposed
to him. Soon he saw the quadrature of the circle in the mysterious
properties of the numbers 7, 8, 9, and of the proportion sesqui tertiam,
or of 3 to 4; he spent the last years of his life in publishing new vagaries: his different quadratures do not even agree together.  The
geometrician Pell attempted, about 1644, to set him right; he made
him see, by a calculation, without the extraction of any root, that his
relation would make the circumference greater than the circumscribed
polygon of 236: the stubborn and irritable old man died in 1647, persuaded that he alone was right against all.
About the same time a new pretender to the honor of squaring the
circle appeared in France, in the person of one S. Oudart, of Ogen,
author of a work entitled, Supplementum Suplplementi Continens. He
gave a geometrical construction quite ingenious, and which would, in
fact, give a line equal to the circumference, if three points which he




INTRODUCTION.                        17
supposes in a straight line were so in fact; besides, he did not deduce
from it any numerical relation. His reasoning was about the same as
that of Mallemant, of Messange, who,. among many insignificant works,
gave, in 1685, a quadrature of the circle, with a pretty rational history
of the problem. He supposed that the three points were in a circular
line, for which he lad no warrant. They were not deemed worthy of
a refutation; both might have been undeceived by making his construetion only 2 feet in diameter.
The famous Hobbes appeared shortly after, about 1650, with his
claims, not only. to the quadrature: of the circle, but also the rectification ofthe parabola, etc. His pretended solution having been refuted
by Wallis, he took occasion to write. against geometricians and geômetry itself. Almost every year he wrote something new on this subject,
and went always fiom paralogism to paralogism. One of his writings
isentitled: Rosetumrn Geometricium, or the Geometrical Boquet.  He
abused geometricians a great deal, and Wallis in particular, and showed
in several ways that his.pretended discoveries were ridiculous.
Bertrand la Costa published in 1666, and again in 1677, a work entitled: Demonstration of the Quadrature of the Circle. But it was of
no value and was treated with contempt by the French Academy of
Sciences.
Three other semi-mystical visionaries presented to the public some
vagaries on the quadrature of the circle. One John Bachon, of Lyons,
announced in 1657 his discovery by a work entitled: Demonstratio
divini Theorematis Quadraturae Circuli, theologica, philisophica, Geometrica et Mecanicacum ratione quantitatum incommensurabilium.  The
geometrical quadrature would have been enough; and people may judge
of the author from his mixing together all these different methods.
An anonymous writer proclaimed in 1671 that the reign of the
greatest king of the universe was to be rendered illustrious by this
most.brilliant discovery, and undertook to prove it by a pamphlet in
4to, entitled: Demonstration of the Divine Theorem of the Quadrature
of the Circle, and the relation of this theorem with the Visions of
Ezechiel and the Revelation of Stk John. He does not fail, after the
example of his colaborers, to ascribe his discovery to a special favor
of the Divinity, according to this passage of the Scriptures: Revelasti
ea parvulis. In fact, there is found at the end of this work a large
mysterious board, presentingon a common center four decreasing pyramids of circles and angles, which represent the Angelic Hierarchy.
2




18                      INTRODUCTION.
The third fool was named Dethlef Cluver, grandson or nephew of the
celebrated geographer of that name. By ransacking the science of the
infinite, on which he promised a great treatise, he finitely discovered
that this problem, to find the quadrature of the circle, reduces itself to
this one: to construct a world analogous to the divine intelligence:
Construere Mundum divinae Menti Analogum; he promised to give
geometrically and vigorously the solution of the first. Meanwhile, he
(unsquared) dequarrail the parabola, and claimed that all that the
geometricians had found on the curve figures was incorrect. (See Acta
lipsiensia Julii, 1686, October, 1687). Leibnitz propounded doubtless
to amuse himself some doubts on these vagaries. He wanted to pitch
this Cluver against Nieuventit, who at the same time accumulated pitiable objections against the new calculations of the infinite that would
have amused the geometricians. This trick did not succeed.
But among all the discoveries of the quadrature of the circle one is
a kind of phenomenon, the only one who as yet admitted his mistake.
It is Richard Albinus (White), an English Jesuit, author of a work
entitled: Chrysespis sen Quadratura Circuli, in which he gives a false
solution of the problem. But some friends opened his eyes, and he
also acknowledged his error on the rectification of the spiral.
A better knowledge of geometry did not keep the 18th century from
similar follies; there is not even a doubt that succeeding ages will resemble in this respect the past. In 1713, a Mr. G. A. Roerberg undertook
to show. that the circle is equal to the square of the side of an inscribed
equilateral triangle; he did not perceive that it follows that the circumference would be exactly three times the diameter, or equal to the
inscribed hexagon.
The solution of the three problems which have so long puzzled
narrow intellects, the quadrature of the circle, perpetual motion, and
the trisection of the angle was also announced in 1714 with much emphasis. The first discovery was that of one S. Daniel Wayvel, a Dutchman, and it was a palpable paralogism. From it followed that the
diameter being 1. the circumference was, 3.142 exactly, which is altogether too much.
Usually the (quadrateurs) come off with seeing their discoveries
neglected or scoffed at by their contemporaries; but in 1728, Mathulon,
of Lyons, was more unfortunate. HIe announced to the learned world
his signal discoveries of the quadrature of the circle and perpetual motion. He was so confident of his success that he appropriated 1,000




INTRODUCTION.                        19
crowns for whoever would demonstrate to him that he was mistaken on
either of these points. But Nicoli, then very young, demonstrated his
error, and Mathulon admitted it; but he objected to the payment of
this sum which Nicoli had abandoned to the Hôtel Dieu of Lyons.
The matter was brought before the courts of that city, and the 1,000
crowns adjudicated to the poor.
In spite of this ill-success a new aspirant to the honor of squaring
the circle was soon after seen. It was Basselin, professor of the University; his calculations were so complicated and so long that no one
would have been willing to follow and verify them. But there is in
such a case a means of detecting the mistake. Barben du Bourg, who
later devoted himself to medicine, employed it by showing that the results of Basselin fell outside the known limits; for the rest, he was
such a novice in geometry that he did not know that Archimedes had
squared the parabola. Yet I have seen a beautiful Latin poem which
celebrated the glory of Basselin and that of the college which his discovery had made illustrious.
The abbot Falconet, brother of the celebrated academician of that
name, also published, about 1740, a little work in which he claimed to
have discovered the quadrature of the circle. His process was less
awkward than many others, but la Land, who was his friend, vainly
endeavored, a few years later, to undeceive him.
Leistner, an officer in the service of the emperor, made more noise,
and succeeded in having an imperial commission appointed to judge of
the truth, real or assumed, of his discovery. He was, like many others,
persuaded that in the series of numbers there are two which express
the ratio of the diameter to the circumference, and that if the quadrature of the circle was not found it is because no one has been fortunate
enough to put his finger on these numbers. In the first place that is
a very false idea, since it has been demonstrated that there is no two numbers which express exactly the ratio of the side of a square to the diagonal; and it is also demonstrated that there is none which express the
ratio of the diameter to the circumference, but Leistner thought he had
found these highly favored numbers in the following: 1225 and 3844,
for these numbers are two squares, and even prime to each other. They
are derived from 35 and 31, which, according to Leistner, express the
relation of the square to the circumscribed circle, and squaring them
both and quadrupling the last they must express the relation of the
diameter to the circumference. But Marinoni, reporter of the Com



20                      INTRODUCTION.
mission, made it appear that this ratio of 1225 to 3844 is not even as
accurate as that of 1 to 3-, and that it gives a circumference which
falls below the least limits of 37 and 3-, the last of which is less than
the polygon of 196 sides.
There is a quantity of other couples of numbers enjoying the properties deemed so wonderful by Leistner, and which gives a value nearer
the circumference, as was shown by Lambert in his Beytrage or Memoires de Mathematiques, Vol. II, 1770, page 156.  What Leistner
thought he had found by a special favor of God can be found in a thousand ways by an analytical process. The Commission, on the report of
Marinoni, rejected the discovery of Leistner, who, like his compeers,
appealed from the judgment in a work entitled Nodus Gordius. It
afforded Marinoni the occasion of publishing a work, where the process
of determining the ratio of the diameter to the circumference is developed and expounded in a manner to convince any one but a man who
thinks he has found the quadrature of the circle.
In 1751 a pastor or preacher of Kattembourg announced to the public this beautiful discovery, and soon after an inhabitant of Rostock
enrolled himself as a volunteer for the same object. One of the two
claimed that having entertained a Frenchman the latter had seen
some of his papers and taken unfair advantage of the circumstance.
Yet, about 1750, Henry Sullamar announced in England the quadrature of the circle, and found it in the number 666 of the Apocalypse; he
published periodically every two or three years some pamphlet in which
he endeavored to prop his discovery. Paris soon enjoyed a similar
spectacle. In 1753 an officer in the guards, Sir de Causans, who until
then had never had any suspicion of geometry, suddenly found the
quadrature of the circle in having a circular piece of sod cut, and then,
rising from truth to truth, explain by his quadrature original sin and
the Trinity. He pledged himself by a public writing to deposit at a notary to the amount of 300,000 francs to wager with those who might
wish to appear against.him, and actually deposited 10,000 francs for the
benefit of any who might show him his mistake. That was certainly
not difficult; for it followed from his discovery that the square circumscribed about the circle was equal to it and, consequently, the whole to
its part. Some persons undertook to win the 10,000 francs, among
others a young lady sued him; some others, in answer to his challenge,
deposited different sums of money with the notaries. But the king
adjudged that the fortune of a man, who was really innocent, ought not




INTRODUCTION.                         21
to suffer from such eccentricities of mind; for, on every other subject,
Sir de Causans was a very estimable man. The suit was quashed and
the bets declared void. Yet the knight found the means of obtaining
the judgment of the academy which considerately refused to speak, but
which was finally compelled to give its opinion.
We shall pass more rapidly over the names of other discoverers of
the quadrature of the circle. Fondee, of Nangis, found it, not by
measuring curves by straight lines, but straight lines by curves. Liger
filled les Mercurés with similar follies on the quadrature of the circle.
He demonstrated it by the lMéccanisme en pleiu des figures, which
gave him, independently of the quadrature of the circle, the commensurability of the side of the square and its diagonal, by making out
that 288 are equal to 289. Sir de Culant fell on the sanie discovery
some years ago, and would also no doubt have discovered the quadrature of the circle had not death taken him away. La Frenaye, footian to the Duke of Orleans, spent twenty years in wandering fiom
paralogism to paralogism, and in sifting the numbers 7, 8, and 9, which,
according to him, contained the whole mystery of the quadrature.
Clerget saw some contradiction in the relation more or less near of the
diameter to the circumference given decimally, and had besides found
the size of the point of contact of a sphere with a plain. Some years
ago Maure used to weary all those that would listen to him by the recital of the injustice of geometricians and the Academie of Sciences. He
intended to cross to England where he was sure to find more equitable
judges in the Royal Society.
We must not forget one of the modern Quadrateurs who outdid
many others in sanguieneess and absurdity. It is Rohberger de Vausenville.  The challenges he had made to the geometricians of all
nations, even Turk and Arabian, as well as to all academies, the suit
brought forward against the Academy of Sciences to secure for himself
the capital of the prize established by Count de Meslay, his indecent
attacks upon all geometricians who tried to enlighten or teach him,
have made him famous among those who have followed this path.
His final theorem is that the square of the diameter is to that of the
circumference, as 22 times the radius multiplied by the square root of
3 is to 432 times the radius. A more experienced geometrician would
have said as eleven times the square root of three is to 216. That
made the circumference of the circle whose radius is one equal to 3.36,




22                      INTRODUCTION.
which differs from the known proportion even in the 2d figure. It
would follow from the pretended discovery of Vausenville that the circumference of a circle would exceed the circumscribed polygon of 12
sides.
Even at the present time, citizen Tardi, an old engineer, applies to
the institute, the Corps Legislatif and all the world, to show his quadrature. He is having pamphlets printed, but is waiting for the proceeds of subscription. We have also just received a print with the
title: Final Solution of the diameter of the Circle to its Circumference,
or the discovery of the Quadrature of the Circle, by Christian Lowenstein, Architect, Cologne, 1801. His method consists in applying to
a great quarter of a circle a strip of iron and he finds the circumference
to be 3.1426.
These publications come to us more especially in the Spring of the
year, when fits of folly are more frequent, and cit. de la Land, who spent
a year at Berlin, says it was the season when the Academy of Berlin
received most writings of that kind.
We were, perhaps, wrong in dwelling so long upon these follies;
we now pass to a more important article about this subject.
The impossibility of finding the quadrature of the circle was maintained by James Gregory, a Scotch geometrician, in a treatise entitled:
Vera Circuli et hyperbola quadratura, Patav, 1664, in 4to, for he understood by the quadrature that which he obtained by approximation.
One is disposed to think this quadrature impossible to the human
intellect when the useless efforts of geometricians of all times are considered; I do not speak of the pitiable efforts of those we have just
been discussing, but of the efforts of such modern geometricians as St.
Vincent, Wallis, Newton, Leibnitz, Bernoulli Euler, etc., who have
found new methods of determining the area of curves, and who, by
their reasoning, have found that of a quantity of other curves less complicated in appearance than the circle, whereas the latter has always
eluded their efforts.
Besides, a distinction must be made in this respect: there are two
quadratures of the circle, one definite, the other indefinite. The definite quadrature is the one that would give the precise measure of the
entire circle or of a given sector or segment, without giving indefinitely that of any sector or segment whatever. The indefinite quadrature, which would be the most perfect by giving the quadrature of any




INTRODUCTION.                        23
part whatever, would evidently include the other. Scarcely any but
the first is sought by quadrateurs in general.
The conviction is general that there is no demonstration absolutely
convincing that the definite quadrature is impossible.  Yet James
Gregory claimed that he gave an irrefragable demonstration.
It rested upon the progressive course represented by the increase and
decrease of the inscribed and circumscribed polygons whose limit is
the circle itself. But this demonstration did not appear conclusive to
Huygens, and it was the cause of a contest between these two geometricians which occupied the newspapers of the time. It must be admitted that though the reasoning is worthy of a head like that of
Gregory, one of the forerunners of Newton, yet as the last limit of
which he speaks is placed, so as to speak, in the mists of the infinite,
the mind is not struck by an irresistible conviction. Still I would not
put in the same category the assumed demonstration of this impossibility by Hanow. It is only a pitiable reasoning. An anonymous
writer, some years ago, gave a little tract entitled: Demonstration of
the incommensurability, etc. He claimed to have proved the impossibility of the quadrature of the circle. His calculations are exact,
although more complicated than necessary; but it proves neither the
incommensurability of the circumference and diameter, nor the impossibility of measuring the former; for a complication of incommensurable
quantities does not prove demonstratively the incommensurability of
the product or quotient. Two irrational quantities may, when multiplied
together, give a rational quantity. The same is true of a larger number. A quantity may be composed of an infinite number of irrational
quantities. and represent only a rational quantity.  But citizen Legendre, at the end of his Geometry, edition of 1800, page 320, demonstrates that the ratio of the circumference to the diameter and its
square are irrational numbers, and that had been already demonstrated
by Lambert, Mein de Berlin, 1761.
An irrational quantity is susceptible of a geometrical construction.
Thus, supposing the circumference to be irrational or incommensurable
with the diameter, it could, nevertheless, be determined geometrically,
and this would undoubtedly be to find the quadrature of the circle.
As for the indefinite quadrature, Newton seems to have demonstrated
that no enclosed (fermée) curve continually receding upon itself, as the
circle, is capable of it. (Princ. phil. nat. math. lib. I.; Lem. XXVIII,
p. 106.)  This demonstration is connected with the theory of angular




24                      INTRODUCTION,
sections and of equations. I undertook, in 1754, to make it more plain
and develop it more fully in my Histoire des Recherches, etc. I will
have to refer to it and deem it convincing. Besides, although geometry
presents numberless examples of squared curves, I know   of none
among the enclosed curves or curve continually receding (retournant)
upon itself, that can be. Still D.'Alembert, in the fourth volume of
his Opuscules, 1768, says, that he can scarcely assent to Newton's reasoning to prove the impossibility of the quadrature of the circle. I see,
says he, that a similar course of reasoning, applied to the rectification
of the cycloid, would lead to a false conclusion, the only difference, it
seems to me, is tliat the circle is a receding curve and the cycloid is
not. But I see nothing in Newton's reasoning which can be changed
by disparity, more particularly, since the cycloid, if it is not a receding
curve like the circle, is a continued curve whose sides (branches) are
not separated; in a word, the reasoning of Newton rests solely on this
supposition that in the circle an infinite number of areas corresponds
to the same abscissa, whence he infers that the equation between the
arc and the abscissa must be of an infinite degree, and consequently is
not algebraically rectifiable;  now, by applying his reasoning to the
cycloid, I would infer that the equation between the abscissa and the
corresponding arc must also be of an infinite degree, and therefore the
arc is not rectifiable algebraically, which is false. D.'Alembert made
the calculation and concluded by saying, it seems to me that these reflections might deserve the attention of the geomnetricians and induce
them to look-for a more vigorous demonstration of the impossibility of
the quadrature and of the indefinite rectification of oval curves.
We shall now give a brief account of the principal discoveries on the
quadrature of the circle, as most of them are included among the geometrical discoveries already discussed in the former volumes, I will
only give them here without going into details  Archimedes first discovered that the circumference is less than 3-~O or 3-, and more than
31~ times the diameter. Some of the ancients, as Appolonius and Philo,
found nearer relations, but it is not known what they were.
About 1585 Peter Metius, in impugning the false quadrature of Duchêne, gave his near ratio of 113 to 355. It was shown above how near
he was right. About the sanie time Viete and Adrianus Romanus also
published relations expressed decimally which came much nearer to the
truth. Viete carried the approximation to 10 decimal places instead
of 6, and taught besides several somewhat simple constructions which




INTRODUCTION.                         25
gave the value of the circle, or the circumference to within a few millionths. Adrianus Romanus carried the approximation to 17 figures.
But all that is far below what was done by Ludolph Van Ceulen, and
which he published in his book de Circulo et adscriptis, of which
Snellius published a Latin translation, at Leyden, in 1619. Ceulen,
assisted by Petrus Cornelius, found with inconceivable labor a ratio
of 32 decimals; see V. II, p. 6. Snellius found the means of shortening this calculation by some very ingenious theorems, and if he did
not excel Van Ceulen he verified his result, which he put beyond attack. His discoveries on this subject are found in the book entitled
Willebrordi Snelli Cyclometricus de Circuli dimensione, etc. Descartes
also found a geometrical construction which, carried to infinity, would
give the circular circumference, and from which he could easily deduce an expression in the form of a series. (See his Opera posthuma.)
Gregpoire de Saint-Vincent is one of those who are most distinguished in this field; true, he claimed incorrectly to have found the
quadrature of the circle and of the hyperbola, but the failure in this
respect was preceded by so great a number of beautiful geometrical
discoveries, deduced with much elegance according to the method of
the ancients, that it would hâve been unjust to have placed him among
the paralogists we have mentioned. He announced, in 1647, his discoveries in a book entitled,: Opus Geometricumr quadratzirae Circuli et
Sectionern Coni libris, X, Comprehensum. All the beautiful things conitained in this book are admired; only the conclusion is impugned.
Gregoire de Saint-Vincent lost himself in the maze of his proofs which
he calls prolortionalities, and which he introduces in his speculations.
It was the subject of quite a lively quarrel between his disciples on the
one hand, and his adversaries on the other, Huygens, Mersenne, and
Leotand, from 1652 to 1664.
If that skillful geometrician had not been mistaken, it would only
have followed from his investigations that the quadrature of thc circle
depends upon logarithms, and consequently on that of the hyperbola.
That would still be a handsome discovery, but it did not even have
that advantage. This furnished Huygens the occasion of divers investigations on this subject. He demonstrated several new and curious
theorems on the quadrature of the circle: Theoremata de quadratura
hyper, ellipsis et Circuli, 1651; De Circuli MIagnitude inventa, 1654.
He gave several methods of approaching his quadrature much shorter




26                     INTRODUCTION.
than the usual way. He demonstrated a theorem which Snellius had
taken for granted. There are also many very simple geometrical constructions which give lines singularly near any given area. If, for
example, the arc is 60~ the error is scarcely -gloth.
James Gregory distinguished himself in this controversy, and whatever may be our opinion of his demonstration of the impossibility of
the definite quadrature of the circle, he can not be denied the authorship of many curious theorems on the relation of the circle to the inscribed and circumscribed polygons, and their relation to each other.
By means of these theorems he gives with infinitely less trouble than
by the usual calculations, and even those of Snellius the measure of
the circle and of the hyperbola (and consequently the construction
of the logarithms) to more than twenty decimal places. Following
the example of Huygens, he also gave constructions of straight lines
equal to arcs of the circle, and whose error is still less. For example,
let the chord of the arc of a circle be a, the sum of the two equal inscribed chords equal to b, then make this proportion: A B: B:: B:
O; if you take the following quantity, c+8 B-A it does not exceed the
3- 50-th for a semicircle, and for 120~ it would be less than  0o oth;
finally the error for a quarter of a circle will not be * Uio th.
The discoveries of Wallace, found in his Arithmetica infinituz,
published in 1655, lead him to a singular expression of the relation
of the circle to the square of its diameter; it is a fraction in this
form                S i3X35X5X7X7X9X9X1X  et11
2X>14XtX 66XsX8XXio0X10X12Î
This fraction, carried to infinity, expresses exactly the above relation, Arithmet. infinit., prop. 191; but if we confine ourselves, as is
necessary, to a finite number of terms we have alternately a relation
greater or less than the true one according as we take an odd or an
even number of terms of the numerator and denominator. Thus 3
gives too great a relation, and 3X3 give too small a relation. The fraction 32-45.5I. 7; is too small, and 3.3:.55.;:9 too great. But to bring
the one near to the other, Wallis directs to multiply this product by
the square root of a fraction formed by the unity plus unity, divided
by the last figure which ends the series. Then the product, although
much nearer, will be too large if the figure is the last of the numerator, and too small if the last of the denominator. The values of.83.5_.i_î1_   3..5-5 i.- 11;.3.3.5.5.7.7 /1_ +  3-3.5.5 15/l - { i7
2.4.4.6   5.. 4.4... r   V"+{ 2.4.6.6.46+7, 2.464,.6.8  V-.




INTRODUCTION.                        27
etc., alternately too small or too great, will fall within the known
limiits.
Here is another expression of the relation of the circle to the square
of the diameter, found by Lord Brunker about the same time.
The circle being one, the square is expressed by the following fraction carried to infinity:
2+25
2$-25
2+49
2+etc.
It will be seen that this fraction is such that the denominator is an
integer plus a fraction, whose denominator is 2 plus the square of one
of the odd numbers 1, 3, 5, 7, etc.; when brought to an end the limits
obtained are alternately in excess or too small.
Such was the knowledge of geometricians on this famous problem when
Newton and Leibnitz appeared on the arena. In 1682, Leibnitz gave
out in his Actes de Leipsig what he had discovered as early as 1673,
namely, that the square of the diameter being one, the area of the circle
is expressed by the infinite series 1 —+  + 1- 4-   etc. It follows from
his discovery about the same time that the radius of the circle being unity
and the tangent of 'an arc t, this arc itself is t —  t3-+ — t5-  t7, etc.
If then the arc is 45~, the tangent t is equal to the radius or one.
Thus the arc of 45~ is 1- 1-+-l --  etc.; multiplying by 4 we shall
have the semi-circumference, which multiplied by the radius will give
the area of the circle equal 4 4-+-4   etc. the square of the diam4
eter being 4. Thus the square of the diameter being made unity, the
area of the circle will be 1-_+_-+-   etc. to infinity.
The area can also be expressed by +    +     +T     etc., viz.: by
adding together the two first terms, and the next two by two, or else
in this way, 1-22, —        etc., where it is easy to see that the denominators are successively in the first the squares of 2, 6, 10, etc.,
diminished by unity, and in the second the squares of 4, 8, 12, etc.,
similarly reduced. But it must be conceded that these different series
do not converge rapidly enough to derive from them a value sufficientlyaccurate without the addition of a prodigious number of terms;
but Euler found a remedy.
The discoveries made by Newton, even before Leibnitz, had also
placed him in possession of various methods of expressing the circumference and the area of the circle, as also of segments by infinite
series.




28                      INTRODUCTIO-N.
Nothing is better known to-day by all those who have any knowledge, even elementary, of the new calculations; but among these series,
those that have been used most successively for that purpose are the
following 
FIGURE 1.
F
D/    \                  \
S      Q     B  a
Let C, A, F, be a quarter of a circle, whose radius A Q is unity and
Q B an abscissa,  x, we find the area of the segment Q, B, A, F, represented by this series, x — 13.-  3-_.~.  1 x —...T X7, etc.
Now, supposing B Q or x equal to ~, this series is reduced to the
following: -     4 ~ ----  00-       -      etc. Calculating therefore in decimal fractions each of these terms, and taking the sum of the
negatives from the-first, which is positive, we obtain the value of the
segment Q, B, A, F, from which it will be necessary to take the value
of the triangle A, B, Q, calculated in decimals, and there will remain
the sector F, A, Q, which is the third of the quarter of the circle, or
the 12th part of the entire circle. Thus, multiplying this value by 12,
we shall obtain that of the circle for a diameter equal to 2; and finally
dividing this last one by 4, we shall have that of the circle to the
diameter.  The first ten terms above given, calculating in this manner, give the near ratio of the diameter to the circumference of 1 to
3.141.
The tangent of a small arc as that of 30~ might also be employed
for the same purpose; for this tangent is /3, or V/; besides, if the
radius, unity, and the tangent t, the arc is, as already shown, t-' t3+
5 —t 7+t 9   1 1. Thus t being made =,/, this series is reduced to
I/    1                          1 /  1  +. +I3 1/+  T2
/V — 3.-t33+-. —./3+ etcrwhose progression can easily be seen. — 52
+ T-~.51(; —r s.1 -s-3 5.l'3, etc., whose progression can easily be seen.




INS'TOD UCTION.                      29
This series, by transferring the radical to the numerator, becomes j/X
— '+ +i/- i-}/, etc., which can be expressed thus: 1/i (1-V+-. —.
— T -; thus -/* must be taken in as many decimals as are to be used
in the approximation or a little more, to be more certain of the last
figures; then this value must be divided successively by 3.3 or 9 by
5.9 or 45, by 7.27 or 189, by 9.81 or 7.29, etc.; this will give the approximate value of each terni in as many decimals as there are in the
value of. Add all the positive terms together, and from the sum
subtract that of the negatives.  This will give very nearly the arc of
30~, which being multiplied by 12, will be the value of the circumference with the diameter 2, consequently the half will be that of the
circumference with the diameter 1.
It is by this means and others similar that Samuel Sharp extended
to 75 decimals the approximate ratio of Ludolph, which had only 35.
Machine, towards the beginning of the present century, carried it as far
as 100. Lagny, in 1719, carried it to 128, and another to 155. Thus,
the diameter of the circle being 1, followed by 128 zeros, the
circumference is according to Lagny, greater than 3.14159, 26535,
89793, 23846, 26433, 83279, 50  288, 41971, 69399, 37510, 58209,
74944, 59230, 78164, 06286, 20899, 86280, 34825, 34211, 70679, 82 -148, 08651, 32723, 06647, 09384, 46, and less than the same number
increased by unity (added to the last figure). I have separated by a
dash the 32 decimals of Ceulen. The error for a circle with a diameter
100 millions times greater than that of the sphere of the fixed stars.
Supposing the parallax of the terrestrial orb to be only one second,
would still be several billions of billions times less than the breadth
of a hair. The. 114th figure of the seven which is underlined, ought
to be 8; Mr. Vega ascertained this as appears from his large tables of
Logarithms, page 633, where he gives the values of the series. Baron
de Zach saw, in a manuscript of the library of Ratcliffe at Oxford, the
calculation carried still further, and as far as 155 figures; after 446
add 0955058, 22317, 25359, 40812, 84802.
The expedient found by Euler for using the series which the arc by
the tangent gives, will bring us nearer the truth and with less trouble.
The expedient deserves to be inserted here.
It consists in the remark made by that great geometrician that every
arc is rational or commensurable with the radius (the arc of 450 for
example, whose tangent is 1), can be divided into 2 arcs whose tangents




30                      INTRODUCTION.
much less, shall also be commensurable to it. It is a consequence of
the theorem which gives the tangent of the sum and difference of two
arcs whose tangents are given; for there being no extraction of square
roots in this formula, if the arcs have their tangents rational, the
tangent of the sum will be so also; and vice versa, an arc with a rational tangent can be divided into two arcs, whose tangents much
smaller will be rational. Thus the arc 45~ can be divided into two
(incommensurable really to each other), the tangent of one of which
will be 2, that of the other W. We shall therefore find by the series of
the arc by the tangent each of these arcs, and their sum (though irrational to each other and to the radius) will, nevertheless, be to the
arc of 45~, which, multiplied by 4, will give the ratio of the relation
of the semi-circumference to the radius, or of the circumference to the
i         1     i
diameter; for the first of these series will be 2  323   -27.2Ï 11.    etc., or      +160    896      822       etc.; and the
9.2911.1'   2c o       24+1-6-896+4608        22528! 1       1I                          1   1
second will be -_33, etc., that is to say, --
3              7373Y)5 7î37
1      1       1         1          1
1225  15309+177147     17537553 +165526791e'
Now, in both these series, the terms dimiinish rapidly enough to attain very nearly their real values; for, in the second, by taking only
seven terms the error would already be less than T-,G-oa,0lUth.
Euler shows that it is possible to obtain this result even more rapidly;
for he remarks that the arc whose tangent is 1 can be divided into two
whose tangents will be ~ and  -, which gives the arc of 45 equal to twice
the second of the above series, plus this one 7 —31 7+575 etc., or --
1      1
i129so89035
Finally, we shall observe that the arc whose tangent is ~ can be
divided into two whose tangents shall be  - and ï, which affords the
means of obtaining two series still more converging than the one which
gives the alc answering to    the tangent ~. It would be easier to find
by these 3 or 4 series the circumference of a circle to 200 decimals
than it was for Viete or Romanus to calculate by their methods to 10
or 15 decimals.:-,We pass in silence over the other artifices of calculations presented by Euler in the same treatise, and in another, Vol.




INTRODUCTION.                       31
XI, for 1739, by which it would require only 80 hours of work to find
128 figures of Lagny; there are also some in Stirling, Summatione Serierum, in Simpson; The Doctrine of Fluxions, 1750. Kraft, in the
13th volume of Petersbourg for 1741, page 121, gave some mechanical
constructions very simple and very near.
All these methods which, undoubtedly, mutually confirm each other,
leave no doubt as to the numerical expression"of the approximate ratio
of the diameter to its circumference; this ratio is the true touchstone
to try the pretended quadratures of the circle without entering into the
maze of the pitiable and often obscure reasoning of those who claim to
be the happy explainers of this enigma. Nothing better can be done
than by leaving them to their pleasing allusion; for experience has
shown attempts to open their eyes would be fruitless; that is what induced the Academy of Sciences to give notice that it would examine
no more quadratures of the circle any more than trisections of the
angle, or duplications of the cube, or perpetual motions. Histoire de
l'Acadernie, 1775, page 64. This is the way in which the Secretary of
the Academy, Condorset, who was himself a very great geolmetrician
expresses himself.
This solution may be considered in two lights. In fact, we may
look for the quadrature of the entire circle or the quadrature of any of
its sectors whose chord is assumed as known. The last of these problems is considered as having no solution. Gregory and Newton, whose
authority is so great, even in a science where authority goes for so little, have given different demonstrations of the impossibility of the
indefinite equation. John Bernoulli has proved that the required sector was expressed by a real logarithmetical function, but which in form
contains imaginary quantities. It follows that no real functions, either
algebraical or logarithmetical and real in form, can represent the value
of the sector of an indefinite circle; that the equation between the sector and the chord can not be constructed by the intersection of tie lines
of real or curve surfaces, and we may infer from this consideration the
absolute impossibility of the indefinite quadrature.
Geometricians are not as well agreed as to the impossibility of the
first problem, for it often happens that for particular values quantities
are found whose expression is in general impossible; but an experience
of more than 70 years taught the Academy that none of those who used
to send in solutions of these problems understood either their nature




32                      INTRODUCTION.
or their difficulties, that none of the methods which they used would
have led them to the solution. This long experience sufficed to satisfy
the Academy of the little utility that would accrue to the sciences by
examining all these pretended solutions. Other considerations have
also decided the Academy. There is a popular report that governments
have promiised considerable rewards to the person who might succeed
in solving the problem of the quadrature of the circle, and that this
problem is the subject of the investigations of the most celebrated geometricians. On the strength of these reasons a crowd of men, much
more considerable than is supposed, have given up useful occupations
to devote themselves to the discovery of this problem, often without
understanding it, and always without the information necessary to try
its solution with success. Nothing was better calculated to undeceive
them than the declaration the Academy thought proper to make. Many
had the misfortune to believe they had succeeded, and would not yield
to the reasoning with which geometricians attacked their solutions.
Often they could not comprehend them, and ended by accusing them
of envy or bad faith. Sometimes their obstinacy degenerated into madness; but it is not considered as such if the opinion which forms this
folly does not clash with the received ideas of men, if it has not influence upon the conduct of life, and if it does not disturb the order of
society. The madness of the quadrateurs would therefore have no other
inconvenience for them but loss of time often useful to their families;
but unfortunately madness is seldom confined to a single object, and
the habit of reasoning falsely is acquired and extends like that of reasoning accurately. That happened more than once to the quadrateurs.
Besides, unable not to realize how singular it would be if they should
discover, without study, truths which the most celebrated men have
sought in vain, they almost all become persuaded that it is by a special
protection of Providence that they have succeeded, and there is only
one step from this idea and believing all the strange combinations of
ideas which occur to them are so many inspirations. Humanity, therefore, required that the Academy, persuaded of the uselessness of the
examination it might have made of these solutions of the quadrature
of the circle, should seek to dispel, by a public declaration, popular
opinions which have been fatal to many families. Considerations so
wise could not excite the animadversion of a writer like Linguet in his
Political Annals.  Ie had also found that it is not true that the pictures of exterior objects are taken inverted (upside down) in the retina,




INTRODUCTION.                       33
and the tide of the Amazon does not ascend up to Pauxis, where Condamine noticed it. Nothing surprises me more than to see persons ot'
understanding have so little good sense as to persist in things they do
not understand, with as much assurance and warmth as if they had
occupied their whole lives in studying them, and had acquired a real
superiority in the same; but mankind is subject to these inconsistencies.
3




34                       INTRODUCTION.
PRACTICAL REMARKS AND EXAMPLES ILLUSTRATING
THE QUADRATURE OF THE CIRCLE.
THE Author, who proposes for investigation a new theory upon any
subject in the present advanced state of science, when nearly every
truth concerning the same is believed to have been discovered, fully
demonstrated and applied, must be happy if he succeeds in establishing
the truth of what he advances; for, if he does not have to disprove all
former theories he of necessity calls them in question and creates a
doubt of their correctness before he can hope to succeed in establishing
his own. This is especially.so with regard to the science of mathematics; for, when a fact has been once established by actual mathematical demonstration which is universally admitted to be true, it is adhered to as firmly as one's own faith, and it is next to impossible to doubt
its correctness. Any discovery, therefore, made after such fact has
been so demonstrated and received, which goes to prove even a slight
error in the result obtained is received with great caution; but if it
should seriously call in question the truth of former theories or directly
contradict a result already established, it must be brilliant indeed to
secure the attention and merit the consideration of the learned. It
would also have to be profitable, for in this age the money tree is pruned
before all others; even imperfect theories are often preferred after they
have been proved erroneous, because they are better understood, and
there are so few who can bear the mental exertion necessary to examine newer and truer ones.
There are two classes of propositions in mathematics which deserve
to be considered with attention, the first of which needs only to be referred to here, as the truths upon which they are based have been put
beyond any question of doubt; such for example as
THEOREM 1. In any right-angled triangle the square described on
the side subtending the right angle is equal to the sum of the squares described on the sides which contain the right angle.




INTRODUCTION.                           35
TIIEORENI 2.   If the one-fiftieth of the sum of the squares of the
two sides of any square be deducted therefrom, the square root of the
remaining ' 4  can be extracted exactly.
THEORErNI 3.   In any right-angle triangle if a perpendicular be
drawn from the right angle to the base the triangles on each side are simr
ilar to the whole triangle and to one another, and the perpendicular is a
mean proportional between the segments of the base, and each of the sides
of the triangle is a mEean proportional between the whole base and the
segment adjacent to that side.
THEORESM 4.    If three straight lines are proportionals the rectangle
contained by the extremes is equal to the square on the mean, and if the
rectangle contained by the extremes is equal to the square on the mean the
three straight lines are proportionals.
THEOREM    5.  Ahiy regular polygon inscribed in a circle is a mean
proportional between the inscribed and circumscribed polygons of half
the Lnumber of sides.
THEOREMI 6. If two straight lines cut one another the opposite or
vertical angles are equal.
THIEOREM 7.    Triangles which have the same altitude are to each
other as tkeir bases, and their areas are to each other as the squares described on those bases.
Or, by trigonometry,
The cosine is to the sine as the radius is to the tangent, consequently
the secant is to the tangent as the radius is to the sine, therefore the square
of the secant is to the square of the tangent as the square of the radius
is to the square of the sine.
The second class embraces those problems which either have not
been fully demonstrated, or have been given up as impossible.  Respecting these Mr. Todhunter says:
" There are three famous problems which are now admitted to be beyond the power
of geometry, namely: To find a straight line equal in length to the circumference
of a given circle, to trisect any given angle, and to find two mean proportionals
between two given straight lines. The grounds upon which the geometrical
solution of these problems is admitted to be impossible can not be explained
without a knowledge of the higher parts of mathematics; the student of the
elements may however be content with the fact that innumerable attempts have
been made to obtain solutions, and that these attempts have been made in vain.




36                          INTRODUCTION.
The first of these problems is usually referred to as the quadrature of the circle. For
the history of it the student should consult the article in the English Cyclopedia
under that head, and also a series of papers in the Atheneum for 1863 and subsequent years, entitled A Budget of Paradoxes, by Professor DeMorgan.  'or the
approximate solutions of the problem we may refer to Davies' edition of Hutton's Course of 2Mathematics, Art. I, page 400; the Lady's and Gentleman's Diary
for 1855, page 86, and the Philosophical  [agazine for April, 1862. The third of
the three problems is often referred to as tle duplication of the cube. See note on
VI, 13, in Lardner's Euclid and a dissertation by C. H. Biering, entitled Ilistoria
Problematis Cubi Duplicandi-Haunio, 1844.
Under the head of " Geometrical Analysis " Mr. Todhunter says:
"Geometrical analysis has sometimes been described in language which
might lead to the expectation that directions could be given which would enable a student to proceed to tle demonstration of any proposed theorem, or the
solution of any proposed problem with confidence of success; but no such directions can be given.  We will state thé: exact extent of these directions: Suppose that
a new theorem is proposed for investigation, or a new problem  for trial, assume the truth of the theorem, or the solution of the problem, and deduce consequences from. this assunrption combined with results which have already been
established. If a consequence can be deduced which contradicts some result
already established, this amounts to a demonstration that our assumption is
inadmissible; that is the theorem is not true, or the problem can not be solved.
If a consequence can be deduced which coincides with some result already established, we can not say that the assumption is not inadmissible; and it may happen that by starting fiomn the consequence which we deduced, and retracing our
steps, we can succeed in giving a synthetical demonstration of the theorem or
a solution of the problem. These directions however are very vague, because
ino certain rule can be prescribed by which we are to combine our assumption with resuits already established; and, moreover, no test exists by which we can ascertain
whether a valid consequence which we have drawn from an assumption wiill enable us
to establish the assumption itself. That a proposition may be false and yet furnish
consequences which are true, can be seen from a simple example. Suppose a
theorem were proposed for investigation in the following words: one angle of a
triangle is to another as the side opposite to the first angle is to the side opposite to the
other. If this be assumed to be true we can immediately deduce Euclid's result
in I, 19; but from Euclid's result in I, 19, we can not retrace our steps and establish the proposed tleorem, and in fact the proposed theorem.is false.
Thus the only definite statement in the directions respecting geometrical analysis is,
that if a consequence can be deduced from an assumed proposition which contradicts
a result already established, that assumed proposition must be false. We may mention,
in particular, that a consequence would contradict results already established if we
could show that it would lead to the solution of a problem already given up as impossible."




INTRODUCTION.                               37
A brief review of the above quotations will be attempted, which, for
the sake of convenience, I shall classify as follows:
lst. '" No directions can be given in language which would enable a student
to proceed to the demonstration of any proposed theorem  or the solution of
any proposed probleri with confidence of success."
The above quotation will apply with more force to the demonstration of a new theorem    or the solution of a new problem; for it would
be necessary to know     something of the nature of a theorem and the
construction of a problem before directions could be given for the
demonstration of the one or the solution of the other; but if a theorem
is proposed for demonstration, or a problem for solution, which have certain parts in common with similar theorems or problems before know
and demonstrated, the demonstration of such theorems and the solution of such problems would be rendered easy in the same proportion as
they contained those parts in common. The mathematician, therefore,
who should venture to give explicit directions for the demonstration of
any theorem   or the solution of any problem      whatever-such for instance as the quadrature of the circle, the trisection of the angle, or the
duplication of the cube-would certainly pretend to a superior knowledge concerning those things in which the ablest and wisest mathematicians have failed; for the     discovery of a new method for the
demonstration of the above      theorems or the solution of the above
problems might involve new laws which no one knows anything of except the author who makes the discovery.
2d. " Assume the truth of the problem, and deduce consequences from this
assumption combined with results already established.?  If a consequence can
be deduced which contradicts some result already established, tlis amounts to a
demonstration tliat our assumption is inadmissible; tliat is, tlie theorem is not
true or the problem can not be solved."
NOTE.'-A proposition consists of various parts; we have lirst the general enuinciation of the
problem or theorem, as for example: To describe an equilateral triangle on a given. finite sllaight
line, or any two angles of a triangle are together less than two right angles. After the general enunciation follows the discussion of the proposition. First, the enunciation is repeated and applied to the particular figure which is to be considered, as for example: Let A B be the given
straight line; it is required to describe an equilateral triangle on A B. The construction then
usually follows straight lines and circles, which must be drawn in order to constitute the solltion of the problem, or to furnish assistance in the demonstrations of the theorem. Lastly, we
have the demonstration itself, which shows that the problem has been solved or that the theorermis trué.
Sometimes however, no construction is required, and sometimes tlie construction and deionstration are combined. The demonstration is a process of reasoning, in which we draw infer



38                         INTRODUCTION.
With reference to point 2nd, it must be confessed that so far as
those problems are concerned, for which solutions have been already
obtained and which are universally received as final, it may be true that
an assunied proposition would be false which would contradict a result so well established and so universally admitted to be correct; but
it does not follow as a necessary consequence that an assumed proposition would be false which would give a result different fiom the one
already established with respect to those problems which are admitted
to be impossible —such, for example, as the quadrature of the circle,
etc.; for as long as the final solution of this problem  is not obtained
it is difficult to determine which solution is true and which false as
long as they are confined within certain known limits. Thus it is
mathematically certain that the ratio of the circumference to the diameter is either 31 exactly or very near it, and any ratio which claims to
be either much above or below it can not be true.   For example, suppose a circle is described with a radius equal to the square root of two,
and cosine of the given arc is -, which is not quite though very near
the true radius, and the sine of the same arc is 5, which is not quite
though very near the 1 of the true circumference, now, the double of
the cosine i= -, is not quite the true diameter though very near it,
and the double of the sine 1-=_     is not quite the 21 part of the true
circumference though very near it, being the chord of double the arc,
whose sine is -. If then we assume 14 to be the true diameter, and 44
to be the true circumference, and they are very near it, then dividing
44                  14,                          44
the circumference -   by the diameter -   we have by cancellation 5
5                   5
1444 4J       22
=     X-     7  which is equal to 3.i428574285,  2857, or 31 to in44                                  14
finity, so that if  - were the true circumference and   the true diameter, the true ratio of the circumference to the diameter, would be
3.142857 or 3~.
3rd. "These directions are very vague because no certain rule can be prescribed by which we are to combine our assumption with results already estabences froin results already obtained. (These results consist partly of truths established in
former propositions, or are admitted as obvious in commencing the subject; andpartly of truthS
which folloui fromn the construction that has been made, or which are given in the supposition of the
proposition itself. The word hypothesis is used in the same sense as supposition.)




INTRODUCTION.                         39
lished, and no test exists by which we can ascertain whether a valid consequence,
which we have drawn from an assumption, will enable us to establish the
assumption itself."
A fheorem is a truth which becomes evident by a train of reasoning
called a demonstration; the demonstration proceeds from the premises
by a regular deduction; a deduction is the logical consequences which
follow an assumption; an assumption is based upon certain truths
which are found in the construction of a figure, or which result from a
combination of other truths.  Every combination of truths and every
construction of a figure necessarily contain within themselves the
logical reasons for their demonstration; and if a supposition is made in
conformity with the truths developed in the construction of a figure,
or an assumption be made which is based upon truths which are the
necessary consequence of a combination of other truths, this assumption
will lead to a valid consequence which may result ini the demonstration
of the truth involved. But if, from a misconception of those truths, a
false assumption is made, a valid consequence can not be drawn and the
assumption will necessarily lead to an absurdity.  It is also next to
impossible to prove the truth of one method by another, or of combining our results with those already established; it is only when the
two methods both contain certain principles in common that they can
be compared; that is when they contain elements that may be reduced
to the same denomination, to the same dimensions, the same weight,
or the same measure.
In reference to point 4th, it is admitted that so far as those problems are concerned which have received a final solution that is universally admitted to be correct, it may be true that an assumed proposition
would be false if it should contradict a result already established; but
with respect to those problems which are admitted to be impossible it
does not follow that an assumed proposition would be false which contradicts a result already established; for it is impossible to know which
is true, or which is falsè, until a final solution is obtained. Consequently point 5th may very readily be acquiesced in, for the author
says:
" We may mention in particular, that a consequence would contradict a result
already established if we could show that it would lead to the solution of a
problem already given up as impossible."
It is asserted by apologists and advocates of the present quadrature




40                      INTRODUCTION.
of the circle, that it is an established fact and is therefore incontrovertible; they contend that the colnmonly received solution is the nearest approximation that the ablest and wisest mathematicians of every
age and country with their vast learning and great genius have ever
been able to make towards the truth, and then we are told how many
decimal places each particular author has been able to carry it to, and
the greatest mathematician is the one who las carried the result to the
greatest nuniber of decimal places, and the only reason why the final
solution of the problem has not been obtained is because no one has
ever been able to extract the square roots to infinity. It is only necessary to return to Euclid to find that he labored very much under
the same difficulty, and it looks very much like the fact that two thousand years' experience which has witnessed improvements in nearly
every other branch of mathematical science has not made much progress respecting this problem, for the same ancient method is used
that occupied Appolonious while in prison.
To establish a fact does not of necessity prove it to be a fact; though
it may be urged in reply that a fact must be proved before it can be
established. If a majority of the wisest and most intelligent of mankind were convinced of a fact which has been thoroughly tested and
commonly received, it is said to be an established fact. And a fact,
when we speak mathematically, may more properly be styled a truth.
The establishing of a mathematical truth is usually done in the sale
manner as establishing an ordinary fact; two or three credible witnesses
are sufficient to establish a fact in law, but a mathematical fact is first
investigated by a rigid course of mathematical analysis, and after it
has been thoroughly tested by the ablest mathematicians, and the result of all these various calculations agree, it is said to be an established mathematical truth. But to make such a truth binding if it
depend upon the solution of a problem or the demonstration of a
theorem, the truth of the theorem or the problem can not be said to be
established unless there can be found a final demonstration of the
theorem or a final solution of the problem which shall be mathematically demonstrated beyond a reasonable doubt; and if a problem which
has been established has not been finally solved, any other solution of
the same problem, which shall be a complete and final solution of the
given problem, may be said to disestablish an established truth, or to
successfully contradict a result already established and seriously call
in question its correctness.




INTRODUCTION.                           41
Numerous instances can be cited where facts (truths) have been established and believed by the great majority of the wisest and ablest
of mankind which were afterwards proved to be erroneous, and in some
cases they have been shown to be supremely ridiculous.
For example:
"The Greeks, and through them the Romans and other nations, believed the
earth to be flat and circular, their own country occupying the miiddle of it, the
central point being Mount Olympus, the abode of the gods or Delphi, so famous
for its oracle. The circular disc of the earth was crossed from west to east,
and divided into two parts by the sea as they called the Mediterranean, and its
continuation the Euxina, the only seas with which they were acquainted.
Around the earth flowed the _River Ocean, its course being from south to north
on the western side of the earth, and in a contrary direction on the eastern side.
It flowed in a steady, equable current unvexed by storm or tempest. The sea
and all the rivers on the earth received their waters frorn it.
The northern portion of the earth was supposed to be inhabited by a happy
race named the Hyperboreans, dwelling in everlasting bliss and spring beyond
the lofty mountains whose caverns were supposed to send forth the north wind
which chilled the people of Hellas (Greece). Their country was inaccessible by
land or sea. They lived exempt from disease or old age; from toils and warfare. Moore lhas given us the 'Song of the Hyperborean,' beginning:
'I come from a land in the sun-briglit deep,
Where the golden gardens glow;
Where the winds of the north becalmed in sleep
Their conch shells never blow.'
On the south side of the earth, close to the stream of ocean, dwelt a people
happy and virtuous as the Hyperboreans. They were named the ZEthiopians.
The gods favored them so highly that they were wont to leave at times their
Olympian abodes and go share their sacrifices and banquets.
On the western margin of the earth, by the stream of ocean, lay a happy place
named the 'Elysian Plain,' whither inortals favored by the gods were transported, without tasting of death, to enjoy an immortality of bliss. This happy
region was also called the 'Fortunate Fields' and the 'Isles of the Blessed.^
We thus see that the Greeks of the early ages knew little of any real people
except those to the east and south of their own country, or near the coast of the
Mediterranean. Their imagination meantime peopled the western portion of
this sea with giants, monsters, and enchantresses, while tliey placed around the
disc of the earth, which they probably regarded as of no great width, nations
enjoying the peculiar favor of the gods and blessed with happiness and longevity. The dawn, the sun, and the moon were supposed to rise out of the ocean
on the eastern side, and to drive through the air, giving light to gods and men.
The stars also, except those forming the Wain or Bear and others near them,
rose out of and sank into the stream of ocean. There the Sun-god embarked in




42                        INTRODUCTION.
a winged-boat, which conveyed him round by the northern part of the earth
back to his place of rising in the east. Milton alludes to this ill his Comrus:
'Now the gilded car of day
His golden axle doth allay
In the steep Atlantic stream,
And the slope sun his upward beam
Shoots against the dusky pole,
Racing toward the other goal,
Of his chamber il the east.'
The abode of the gods was on the summit of Mount Olympus, in Thessaly. A
gate of clouds kept by the goddess named the Seasons, opened to permit tile
passage of the celestials to earth and to receive them on their return. The
gods had their separate dwellings; but all, when summoned, repaired to the
palace of Jupiter, as did also those deities, whose usual abode was the earth,
the waters, or the underworld.
It was also in the great hall of the palace of the Olympian King that the gods
feasted each day on ambrosia and nectar, tleir food and drink; the latter being
handed round by the lovely goddess Hebe. Here tley conversed of the affairs
of heaven and earth; and as they quaffed tleir nectar Appollo, the God of
Music, deliglgted them with the tones of his lyre, to which the Muses sang in
responsive strains. When the sun was set tihe gods retired to sleep in their respective dwellings. The following lines from the Odyssey will show how
Homer conceived of Olympus:
'So saying Minerva, goddess azure-eyed,
Rose to Olympus, the reputed seat,
Eternal of the gods, which never storms
Disturb, rains drench, or snow invades; but calm
The expanse, and cloudless shines-with purest day;
There the inhabitants, divine, rejoice
Forever.'"
The above fable though very beautiful, was one of those silly notions which the wisest of all the ancients believed as an established
fact, and from it may be seen how few, even among those who profess
to be wise, really think for themselves-and how the vast majority
of mankind allow others to do their thinking for them; furthermore
by far the greater number of mathematical truths which are now
valued so highly, and referred to with so much confidence, were discovered and first demonstrated by these very Greeks and their contemporaries; and the most plausible method known up to this present
time for the solution of the quadrature of the circle was discovered, it
is supposed, during the highest cultivation of the arts and sciences, by




INTRODUCTION.                          43
these people, whose genius for invention probably exceeded that of any
nation which followed them till we come to the American. Mr. Pope,
in eulogizing the genius of Homer, says:
"Homer is universally allowed to have had the greatest invention of any
writer whatever. Tie praise of judgment Virgil has justly contested with him,
and others may have their pretensions as to particular excellencies, but his invention remains yet unrivalled. Nor is it a wonder if lie has ever been acknowledged the greatest of poets who most excelled in that which is the very
foundation of poetry.
It is the invention that in different ages distinguishes all great geniuses. The
utmost stretch of human study, learning, and industry, which masters everything besides, can never attain to this. It furnishes art with all her materials,
and without it judgment itself can but steal wisely; for art is only like a prudent steward that lives on managing the riches of nature. Whatever praises
may be given to works of judgment, there is not even a single beauty in them
to which the invention must not contribute; as in the most regular gardens
art can only reduce the beauties of nature to more regularity, and such a figure
which the common eye may better take in, and is therefore more entertained
with. And the reason why common critics are inclined to prefer a judicious
and methodical genius to a great and fruitful one, is because they find it easier
for themselves to pursue their observations through an uniform and bounded
walk of art, than to comprehend the vast extent of nature."
When Copernicus maintained, in opposition to the theory of the ancients, that the earth moved on its own axis, and that it moved around
the sun instead of the sun moving around the earth, history teaches us
with what difficulty he maintained his position, though in the end he
succeeded in establishing his theory in opposition to all.
When Columbus, following in the footsteps of Copernicus, maintained
that the earth being round another continent was necessary on the opposite side of the earth to maintain its equilibrium, he was treated as a
visionary and madman, until through the eharity of "Her Most Catholic
Majesty," " Isabella, then Queen of Castile and Arragon," he was provided with means to prosecute to a successful issue his discoveries; and
in return for the ingratitude of nations he gave them a " New World."
In reference to those cases where it is necessary to give instructions
for the demonstration of a theorem or the solution of a problem, or of
combining our assumption with results already established, it is to be
observed, supposing the problem to be the solution of the quadrature
of the circle,
Ist. That the csine must always intercept the sine at right angles,
and the radius must always intercept the tangent at right angles.




44                      INTRODUCTION.
2d. The tangent is always greater than the arc to which it belongs,
and the sine is always less than the arc to which it belongs, for the
tangent lies wholly without the circle while the sine lies wholly within
the circle, and the arc to which they belong must lie between them;
therefore, as far as the sine and the tangent agree with one another,
that is as far as they can be expressed by the same quantity, either
one of them may be taken for the arc of the circle to which they belong.
3d. The inscribed polygon must not at any time extend outside the
given circle nor the circumscribed polygon come within it; and any
method which may be adopted for the solution of the quadrature of the
circle by the means of regular inscribed and circumscribed polygons,
which are made to approach the circle (one from within and the other
from without) by doubling the number of sides, will not give a correct
solution as long as there is danger of forcing a limit between the two
polygons; neither does it follow as a necessary consequence that the
circle is the limit of the two polygons.
4th. If by any means the inscribed polygon could be made to approach the circle from within, without regard to the circumscribed polygon, by doubling the number of sides, and this result could be reduced
to a commensurable quantity, it is very evident that such a method
would give the true quadrature as far as it could be carried; for in that
case the inscribed polygon could never be forced to extend beyond
or become greater than the circle, and if it could be continued till it
would reach the circle, that is to infinity, which is impossible, it would
give the true quadrature.
5th. A common measure of two straight lines, as for example the
sine and tangent, can not be regarded as a measure of the circle, as no
part of it, however small, is straight.
6th. Neither is it necessary for the solution of the quadrature of the
circle that the polygon should be regular, for it was demonstrated by
Gauss, a mathematician of Gottingen, in his disquisitiones, Arithmeticae,
published in 1801, that polygons of 17 sides, 257 sides, and in general
any number of sides expressed by 2n-1, can be inscribed in a circle
when 2n+1 is a prime number.
7th. If the square root of two be taken for the radius of the given
circle, and a radius be assumed in terms of the inscribed square (the side
of which will be 2 and the area 4), and this radius as a variable be
made to approach as its constant the true radius, so near that the difference between it and the true radius shall be less than any assignable




INTRODUCTION.                            45
quantity, so that it may be taken for the true radius, and at the same
time if a circumference be assumed in terms of the inscribed square and
of the radius, and this circumference as a variable be made to approach
as its constant the true circumference, so near that the difference between it and the true circumference shall be made less than any assignable quantity, so that it may be taken for the true circumference, provided
that this radius and circumference are selected in accordance with
theorem second, then will the ratio between the assumed diameter and
the assumed circumference be the same as the ratio between the true
diameter and the true circumference, and this ratio will be the repeating
decimal 3.142857 or 31 to infinity; and, consequently, when the assumed
radius becomes the true radius and the assumed circumference becomes
the true circumference, both of which happen at the same moment, then
the circle is infinite, that is, no part, however small, is straight.
The following selection consists of the most approved methods of
solving the quadrature of the circle; they are inserted here, justas
they are to be found in the works of the most approved authors recently published, for the purpose of giving the student a more general
knowledge of the subject.
The first method is taken from " Robinson's Elements of Geometry,"
which gives the approximate ratio of the circumference to the diameter, by means of inscribed and circumscribed polygons to 6144 sides,
commencing with the hexagon; it is as follows:
PROPOSITION III.-THEOREM.
When the radius of a circle is unity, its area and semi-circumference are numericallly equal.
Let R represent the radius of any circle, and the Greek letter Ir, the half circumference of a circle whose radius is unity. Since circumferences are to each
other as their radii, when the radius is R, the semi-circumference will be expressed by rR.
Let m denote the area of the circle of which R is the radius; then, by Theoren
1, we shall have, for the area of this circle, rR2 = m, whicli, when R = 1, reduces to mT = m.
This equation is to be interpreLed as meaning that the semi-circumference
contains its unit, the radius, as many times as the area of the circle contains its
unit, the square of the radius.
REMARK.-The celebrated problem of squaring the circle has for its object to find a line, the
square on which will be equivalent to the area of a circle of a given diameter; or, in other words
it proposes to find the ratio between the area of a circle and the square of its radius.




46                         INTRODUCTION.
An approximate solution only of this problem has been as yet discovered, but the approximation is so close that the exact solution is no longer a question of any practical importance.
PROPOSITION IV.-PROBLEM.
Given, the radius of a circle unity, to find the areas of regular inscribed and circumscribed hexagons.
Conceive a circle described with the radius CA, and in this circle inscribe a
regular polygon of six sides (Prob. 28, B. IV),
and each side will be equal to the radius CA;      D      A     B
hence, the whole perimeter of this polygon must
be six times the radius of the circle, or three
times the diameter. The chord bd is bisected
by CA. Produce Cb and Cd, and through the
point A, draw BD parallel to bd; BD will then             C         c E
be a side of a regular polygon of six sides, circumscribed about the circle, and we can compute the length of this line, BD, as
follows: The two triangles, Cbd and CBD, are equiangular, by construction;
therefore,
Ca: bd:: CA: BD.
Now, let us assume CA = Cd = the radius of the circle, equal unity; then
bd = 1, and the preceding proportion becomes
Ca    1::: 1: BD     (1)
In the right-angled triangle Cad, we have,
(Ca)2+(ad)2=(Cd)2,      (Th. 39, B. I).
That is,        (Ca)2+=-1, because Cd=l, a   and ad=.
Whence, Ca -= 1/3. This value of Ca, substituted in proportion (1), gives
1  -                                2
213: 1: 1: BD; hence, BD =1/.
But the area of the triangle Cbd is equal to bd (-= 1,) multiplied by ~Ca = —
1/3; and the area of the triangle CBD is equal to BD multiplied by ~CA.
Whence,       area, Cbd =-  1/3.
and          area, CBD =  /
But the area of the inscribed polygon is six times that of the triangle Cbd, and
the area of the circumscribed polygon is six times that of the triangle CBD.
Let the area of the inscribed polygon be represented by p, and that of the
circumscribed polygon by P.
Then p-     /3, and P =        2X3 -      21/3
2                     1/3
3              3
Whencep: P::      1/3: 21/3:: 2:: 3: 4:: 9  12
p=    -1/3  2.59807621. P= 21/3      3.46410161.




INTRODUCTION.                             47
Now, it is obvious that the area of the circle must be included between the
areas of these two polygons, and not far from, but somewhat greater than, their
half sum, which is 3.03 +; and this may be regarded as the first approximate
value of the area of the circle to the radius unity.
PROPOSITION V. —PROBLEM.
Given, the areas of two regular polygons of the same number of sides, the one inscribed in and the other circumscribed about, the same circle, to find the areas of regular inscribed and circumscribed polygons of double the number of sides.
Let p represent the area of the given inscribed polygon, and P that of the
circumscribed polygon of the same number of sides. Also denote by p' the area
of the inscribed polygon of double the number of sides, and by P' that of the
corresponding circumscribed polygon. Now, if the arc KAL be some exact
part, as one-fourth, one-fifth, etc., of the circumference of the circle, of which C
is the center and CA the radius, then will KL be the side of a regular inscribed
polygon, and the triangle KCL will be the same part of the whole polygon that
the arc KAL is of the whole circumference, and the triangle CDB will be a like
part of the circumscribed polygon. Draw CA to the point of tangency, and
bisect the angles ACB and ACD, by the lines CG and CE, and draw KA.
It is plain that the triangle ACKis an exact
part of the inscribed polygon of double the   D    E    A    G     B
number of sides, and that the V ECG is a like
part of the circumscribed polygon of double    KMt
the number of sides. Represent the area of the 
LCK by a, and the area of the A BCD by
b, that of the A ACKby x, and that of the A 
ECG by y, and suppose the A's, KCL andDBC, to be each the nth part of their respective
polygons.
Tlien,     na —p, nb     P, 2nx = p'
and,      2ny =P-;
But, by (Th. 33, B. I), we have
CM. MK= a        (1)
CA. AD = b       (2)
CA. MK-= 2x      (3)
Multiplying equations (1) and (2), member by member, we have
(CM. AD)      ( CA. MK) =  ab    (4)
From the similar ~'s CMK and CAD, we have
CM: MK:: CA: AD
whence                 CM. AD       CA. MK
But from equation (3) we see that each member of this last equation is equal
to 2x; hence equation 4 becomes
22x  2x = ab




48                         INTRODUCTION.
If we multiply both members of this by n2 == n. n, we shall have
4n2x2 = na.nb = p.P
or, taking the square root of both members,
2nx =      V/p.P
That is, the area of the inscribed polygon of double the number of sides is a meant
proportional between the areas of the given inscribed and circumscribed polygons
p and P.
Again, since CE bisects the angle ACD, we have, by, (Th. 24, B. II),
AE: ED: CA: CD:: CM: CK:: CM: CA
hence,           AE: AE + ED:: iCM: CM+ CA.
Multiplying the first couplet of this proportion by CA, and the second by MK,
observing that AE + ED -= AD, we shall have
AE.CA: AD.CA:: CIM.MK: (CM + CA) MK.
But AE. CA measures the area of the A CEG, which we have called y, AD. CA
==  CBD = b, CI.MK = A CKL == a, and ( CM - CA) MK         A CKL +
2 A CAK- a + 2x, as is seen from equations (1) and (3). Therefore, the
above proportion becomes
y: b:: a: a + 2.
Multiplying the first couplet by 2n, and the second by n, we shall have
2ny: 2nb:: a: na + 2nx
That is,              P/:  2P::p: p + P
whence,                           2Pp
P +P/
and as the value of p' has been previously found equal to 1/Xp, the value of P'
is known from this last equation, and the problem is completely solved.
PROPOSITION     VI.-PROBLEM.
To determine the approximate numerical value of the area of a circle, when the radius is unity.
We have now found (Prob. 4) the areas of regular inscribed and circumscribed hexagons, when the radius of the circle is taken as the unit; and Prob. 5
gives us formulae for computing from these the areas of regular inscribed and
circumscribed polygons of twelve sides, and from these last we may pass to polygons of twenty-four sides, and so on, without limit. Now, it is evident that, as
the number of sides of the inscribed polygon is increased, the polygon itself will
increase, gradually approaching the circle, which it can never surpass. And it
is equally evident that, as the number of sides of the circumscribed polygon is
increased, the polygon itself will decrease, gradually approaching the circle, less
than which it can never become.




INTRODUCTION.                             49
The circle being included between any two corresponding inscribed and circumscribed polygons, it will differ from either less than they differ from each
other; and the area of either polygon may then be taken as the area of the circle, from which it will differ by an amount less than the difference between the
polygons.
It is also plain that, as the areas of the polygons approach equality, their
perimeters will approach coincidence with each other, and with the circumference of the circle.
Assuming the areas already found for the inscribed and circumscribed hexagons, and applying the formula of Prob. 5 to them and to the successive results
obtained, we may construct the following table:
NUMBER OF SIDES.     INSCRIBED POLYGONS.       CIRCUMSCRIBED POLYGONS
3 /- 
6.................  3  2.59807621....................21/3  3.46410161
12
12...................3 - 3.0000000.................. - 3.2153904
2+1/3
6
24..             - 3.1058286.............................3.1596602
~1/2+4-ri
48........................... 3.1326287..............................  3.1460863
96........................... 3.1393554...................   3.1427106
192....................... 3.1410328............................ 3.1418712
384.........................  3.1414519.............................  3.1416616
768................... 3.1415568................................  3.1416092
1536................... 3.1415829................................. 3.1415963
3072........................... 3.1415895................................. 3.1415929
6144..........................  3.1415912...........................  3.1415927
Thus we have found, that when the radius of a circle is 1, the semi-circumference must be more than 3.1415912, and less than 3.1415927; and this is as
accurate as can be determined with the small number of decimals here used. To
be more accurate we must have more decimal places, and go through a very
tedious mechanical operation; but this is not necessary, for the result is well
known, and is 3.1415926535897, plus other decimal places to the 100th, without
termination. This result was discovered through the aid of an infinite series in
the Differential and Integral Calculus.
The number, 3.1416, is the one generally used in practice, as it is much more
convenient than a greater number of decimals, and it is sufficiently accurate for
all ordinary purposes.
In analytical expressions it has become a general custom with mathematicians to represent this number by the Greek letter TT, and, therefore, when any
diameter of a circle is represented by D, the circumference of the same circle
must be rD. If the radius of a circle is: represented hy B, the circumference
must be represented by 27rR.
4




50                          INTRODUCTION.
SCHOLIUM.-The side of a regular inscribed hexagon subtends an arc of 60~, and the side of
a regular polygon of twelve sides subtends an arc of 30~; and so on, the length of the arc subtended by the sides of the polygons, varying inversely with the number of sides.
Angles are measured by the arcs of circles included between their sides; they may also be
measured by the chords of these arcs, or rather by the half chords called sines in Trigonometry. For this purpose it becomes necessary to know the length of the chord of every possible
arc of a circle.
The second method is taken from "Davies' Legendre," and gives
the approximate ratio of the circumference to the diameter by means
of inscribed and circumscribed polygons to 16384 sides, commencing
with the square.
PROPOSITION XI.-PROBLEM.
The area of a regular inscribed polygon, and that of a similar circumscribed polygon being given, to find the areas of the regular inscribed and circumscribed polygons
having double the number of sides.
Let AB be the side of the given inscribed, and EF that of the given circumscribed polygon. Let C be their common center, AMB a portion of the circumference of the circle, and M the middle point of the arc AMB.
Draw the chord AM, and at A and B draw
'  F   P     M     0      F
the tangents AP and BQ; then will AMbe             P           Q 
the side of the inscribed polygon, and PQ the 
side of the circumscribed polygon of double    \              /    /B
the number of sides (P. VII.) Draw CE, CP,
CM, and CF.
Denote the area of the given inscribed polygon by p, the area of the given circumscribed
polygon by P, and the areas of the inscribed
and circumscribed polygons having double
the number of sides, respectfully by p' and P/.          c
1~. The triangles CAD, CAM, and the CEM, are like parts of the polygons
to which they belong: hence, they are proportional to the polygons themselves.
But CAM is a mean proportional between CAD and CEM (B. IV., P. XXIV.,
C. 2); consequently p' is a mean proportional between p and P: hence,
/ -l/pX P.... (1.)
2~. Because the triangles CPM and CPE have the common altitude CM,
they are to each other as their bases: hence,
CPM: CPE:: PM: PE;
and because CP bisects the angle ACM, we have (B. IV., P. XVII.),




INTRODUCTION.                              51
PMl: PE::  CM: CE:  CD: CA;
hence (B. II., P. II.),
CPM: CPE:: CD:CA or CM.
But, the triangles CAD and CAM     have the common altitude AD; they are
therefore, to each other as their bases: hence,
CAD: CAM:: CD: CM;
or, because CAD and CAM are to each other as the polygons to which they
belong,
p: p'::    CD: CM;
hence (B. II., P. IV.), we have,
CPM: CPE:: p: p',
and, by composition,
CPM: CPM+CPE or CME::  p: p+p';
llence (B. II., P. VII.),
2CPM or CMPA: CME:: 2p: p+p'.
But, CMPA and CME are like parts of P' and P, hence,
p': P::   2p: p+p';
or,
2p X    P2.)
P + P
Scholium. By means of Equation (1), we can find p', and then, by means of
Equation (2), we can find P'.
PROPOSITION XII.-PROBLEM.
Tofind the approximate area of a circle whose radius is 1.
The area of an inscribed square is equal to twice the square of the radius, or 2
(P. III., S.), and the area of a circumscribed square is 4. Making p equal to 2,
and P equal to 4, we have, from Equations (1) and (2) of Proposition XI.,
p  =    1/8      2.8284271... inscribed octagon;
16
P/'           - - 3.3137085... circumscribed octagon.
2 +1/8
Making p equal to 2.8284271, and P equal to 3.3137085, we have from the same
equations,
p' = 3.0614674... inscribed polygon of 16 sides.
P' - 3.1825979... circumscribed polygon of 16 sides.




52                         INTRODUCTION.
By a continued application of these equations, we find the areas indicated in
the following
TABLE.
NUMBER OF SIDES.     INSCRIBED POLYGONS.    CIRCUMSCRIBED POLYGONS.
4                2.0000000               4.0000000
8                2.8284271               3.3137085
16                3.0614674                3.1825979
32                3.1214451               3.1517249
64                3.1365485                3.1441184
128                3.1403311               3.1422236
256                3.1412772                3.1417504
512                3.1415138                3.1416321
1024                3.1415729               3.1416025
2048                3.1415877               3.1415951
4096                3.1415914                3.1415933
8192                3.1415923                3.1415928
16384                3.1415925                3.1415927
Now, the areas of the last two polygons differ from each other by less than
the millionth part of a unit, but the area of the circle differs from either by less
than they differ from each other; hence, the value of the area of either will differ
from that of the circle by less than a millionth part of a unit. Taking the fig.
ures as far as they agree, and denoting the number of units in the required area
by r, we have, approximately,
w = 3.141592;
that is, the area of a circle whose radius is 1, is 3 141592.
Scholium. For practical computation, the value of wr is taken equal to 3.1416.
The third method is taken from " Chauvenet's Elementary Geometry," and gives the approximate ratio of the circumference to the diameter, by two' different methods, to 8192 sides each. The first is
called the Method of Perimeters, and the second the Method of Isoperimeters. The above approximate ratio is followed by a chapter on
the Doctrine of Limits taken from the same author.




INTRODUCTION.                             53
PROPOSITION      X.-PROBLEM.
25. Given the perimeters of a regular inscribed and a similar circumscribed polygon, to compute the perimeters of the regular inscribed and circumscribed polygons of
double the number of sides.
Let AB   be a side of the given inscribed       F     E     G       D
polygon, CD a side of the similar circum-,,         J
scribed polygon, tangent to the arc AB at its /          H          \
middle point E. Join AE, and at A and B         \ 
draw the tangents AFand BG; then AE is a
side of thé regular inscribed polygon of double     " \ i 
the number of sides, and FG is a side of              ",\  /
the circumscribed polygon   of double the 
number of sides (4).
Denote the perimeters of the given inscribed and circumscribed polygons by
p and P respectively; and the perimeters of the required inscribed and circumscribed polygons of double the number of sides by p' and P/ respectively.
Since OC is the radius of the circle circumscribed about the polygon whose
perimeter is P, we have (10),
P    OC    OC
p     -OA o OE;
and since OFbisects the angle COE, we have (III. 21),
OC CF
OE    FE'
therefore,
P CF
p    FE
whence, by composition,
P- p_ CF+ FE CE
2p        2FE       FG'
Now FG is a side of the polygon whose perimeter is P', and is contained as
many times in P' as CE is contained in P, hence (III. 9),
CE     P
PG      '
and therefore,
P+p P
2p     P'
whence
p_    [2pP1]
P'=. ----p                             D]J




54                         INTRODUCTION.
Again, the right triangles AEH and EFLV are similar, since their acute angles
EAH and FEN are equal, and give
AH     EN                   0            E     G      D
-AE.C                        FGF 
Since AH and AE are contained the same     A"=                    ' B
number of times in p and p/, respectively, we /                / '
have                                                   'H     /  "
AH_ p
AE     p"                            \^, /
and since EN and EF are contained the same,
number of times in p' and P', respectively,            o
we have
EN_ p'.
therefore, we have
p - p'
P/   P/'
whence
'- =  /pXP.                            [2]
Therefore, from the given perimeters p and P, we compute P/ by the equation [1], and then with p and P/ we compute p/ by tle equation [2].
26. Definition. Two polygons are isoperimetric when their perimeters are
equal.
PROPOSITION     XI.-PROBLEM.
27. Given the radius and apothem of a regular polygon, to compute the radius and
apothem of the isoperimetrie polygon of double the number of sides.
Let AB be a side of the given regular polygon, O the center of this polygon,
OA its radius, OD its apothem. Produce DO to
meet the circumference of the circumscribed circle  A/            B
in 0'; join O'A, O'B; let fall OA' perpendicular to 
O'A, and through A' draw A'B' parallel to AB. 
Since the new polygon is to have twice as many
sides as the given polygon, the angle at its center    A    o  B'
must be one-half the angle AOB; therefore the an- 
gle AO'B, which is equal to one-half of AOB (II. 
57), is equal to the angle at the center of the new
polygon.                                                   0'
Since the perimeter of the new polygon is to be equal to that of the given
polygon, but is to be divided into twice as many sides, each of its sides must be
equal to one-half of AB; therefore A'B', which is equal to one-half of AB
(I. 121), is a side of the new polygon; O'A' is its radius, and O'D/ its apothem.
If, then, we denote the given radius OA by R, and the given apothem OD by




INTRODUCTION.                          55
r, the required radius O'A' by R', and the apothem O'D' by r', we have
O'D    00   - + OD
2   ~     2
or
R --     9[1]
In the right triangle OA'O', we have (III. 44),
0'A'2 = 00' X OD',
or
R'= 1R X     ';                       [2]
therefore, -' is an arithmetic mean between R and r, and RB is a geometric mean
between R and r'.
MEASUREMENT       OF THE     CIRCLE.
The principle which we employed in the comparison of incommensurable
ratios (II. 49) is fundamentally the same as that whicl we are about to apply
to the meastrementof the circle, but we shall now state it in a much more general form, better adapted for subsequent application.
28. Definitions. I. A variable quantity, or simply, a variable, is a quantity which
has different successive values.
II. When the successive values of a variable, under the conditions imposed upon it, approach more and more nearly to the value of some fixed or
constant quantity, so that the difference between the variable and the constant
may become less than any assigned quantity, without becoming zero, the
variable is said to approach indefinitely to the constant; and the constant is
called the limit of the variable.
Or, more briefly, the limit of a variable is a constant quantity to which the
variable, under the conditions imposed upon it, approaches indefinitely.
As an example, illustrating these definitions, let a point be required to move
from A to B under the following conditions: it
shall first move over one-half of AB, that is to C;   C     C' C"
then over one-half of CB, to C(; then over one- A                 B
half of C0B, to C; and so on indefinitely; then
the distance of the point from A is a variable, and this variable approaches indefinitely to the constant AB, as its limit, without ever reaching it.
As a second example, let A denote the angle of any regular polygon, and n
the number of sides of the polygon; tlen, a right angle being taken as the
unit, we have (8),
2    4
A - 2 —.
n
The value of A is a variable depending upon n; and since n may be taken so
great that - shall be less than any assigned quantity however small, the value
n




56                         INTRODUCTION.
of A approaches to two right angles as its limit, but evidently never reaches
that limit.
29. PRINCIPLE OF LIMITS. Theorem. If tio variable quantities are always
equal to each other and each approaches to a limit, the two limits are necessarily equal.
For, two variables always equal to each other present in fact but one value,
and it is evidently impossible that one variable value shall at the same time
approach indefinitely to two unequal limits.
30. Theorem. The limit of the product of two variables is the product of their
limits. Thus, if x approaches indefinitely to the limit a, and y approaches indefinitely to the limit b, the product xy;nust approach indefinitely to the product ab; that is, the limit of the product xy is the product ab of the limits of x
and y.
31. Theorem. If tuIo variables are in a constant ratio and each approaches to a
limit, these limits are in the same constant ratio.
Let x and y be two variables in the constant ratio m, that is, let x = my; and
let their limits be a and b respectively. Since y approaches indefinitely to b,
my approaches indefinitely to mb; therefore we have x and my, two variables,
always equal to each other, whose limits are a and mnb, respectively, whence, by
(29), a = mb; that is, a and b are in the constant ratio m.
PROPOSITION XVI.-THEOREM.
42. The area of a circle is equal to half the product of its circumference by its
radius.
Let the. area of any regular polygon circumscribed     A    D    B
about the circle be denoted by A, its perimeter by P, and
its apothem which is equal to the radius of the circle b  /  \  /yb
R; then (22),
A                                  o
A -- P X R or       -     R. 
Let the number of the sides of the polygon be continually doubled, then A
approaches the area S of the circle as its limit, and P approaches the circumference 0 as its limit; but A and Pare in the constant ratio j R; therefore
their limits are in the same ratio (31), and we have
C- i R,orS =- CZ    X.                   [I
43. Corallary I. The area of a circle is equal to the square of its radius multiplied
by the constant number ir. For, substituting for C its value 2-mR in [1], we have
S = 7R2.
44. Corollary II. The area of a sector is equal to half the product of its arc by the
radius. For, denote the arc ab of the sector a O b by c, and the area of the sector by s; then, since c and s are like parts of C and S, we have (III. 9),
s _c _-cX R
s   c        ~cX R'
But S= ~ CX R; therefore s = ~c X R.




INTRODUCTION.                            57
45. Scholium. A circle may be regarded as a regular polygon of an infinite number of sides. In proving that the circle is the limit towards which the inscribed
regular polygon approaches when the number of its sides is increased indefinitely, it was tacitly assumed that the number of sides is always`finite. It was
shown that the difference between the polygon and the circle nmay be made less
than any assigned quantity by making the number of sides sufficiently great;
but an assigned difference being necessarily a finite quantity, there is also some
finite number of sides sufficiently great to satisfy the imposed condition. Conversely, so long as the number of sides is finite, there is some finite difference
between the polygon and the circle. But if we make the hypothesis that the
number of sides of the inscribed regular polygon is greater than any finite number,
that is, infinite, then it must follow that the difference between the polygon and
the circle is less than any finite quantity, that is zero; and consequently, the circle
is identical with the inscribed polygon of an infinite number of sides.
This conclusion, it will be observed, is little else than an abridged statement
of the theory of limits as applied to the circle; the abridgment being effected
by the hypothetical introduction of the infinite into the statement.
PROPOSITION XVII.-PROBLEM.
46. To compute the ratio of the circumference of a circle to its diameter approximately.
FIRST METHOD, called the METHOD OF PERTMETERS. In this method, we
take the diameter of the circle as given and compute the perimeters of some
inscribed and a similar circumscribed regular polygon. We then compute the
perimeters of inscribed and circumscribed regular polygons of double the number of sides, by Proposition X. Taking the last-found perimeters as given, we
compute the perimeters of polygons of double the number of sides by the same
method; and so on. As the number of sides increases, the lengths of the perimeters approach to that of the circumference (36); hence, their successively
computed values will be successive nearer and nearer approximations to the
value of the circumference.
Taking, then, the diameter of the circle as given = 1, let us begin by inscrib.
ing and circumscribing a square. The perimeter of the inscribed square = 4 X
s X 1/2 = 2i/2 (13); that of the circumscribed square = 4; therefore, putting
P -4.
p = 2/2 = 2.8284271,
we find, by Proposition X., for the perimeters of the circumscribed and inscribed regular octagons,
P/    2  X P    3.3137085,
P.+ p
p/= t/p     p  = 3.0614675.
Then taking these as given quantities, we put
P = 3.3137085, p = 3.0614675,




58                         INTRODUCTION.
and find by the same formule for the polygons of 16 sides
P/ = 3.1825979, p' = 3.1214452.
Continuing this process, the results will be found as in the following
TABLE.E
NUMBER OF SIDES.   PERIMETER OF CIRCUMSCRIBED  PERIMETER OF INSCRIBED
POLYGON.                 POLYGON.
4                 4.0000000                2.8284271
8                 3.3137085                3.0614675
16                 3.1825979                3.1214452
32                 3.1517249                3.1365485
64                 3.1441184                3.1403312
128                 3.1422236                3.1412773
256                 3.1417504                3.1415138
512                 3.1416321                3.1415729
1024                 3.1416025                3.1415877
2048                 3.1415951                3.1415914
4096                 3.1415933                3.1415923
8192                 3.1415928                3.1415926
From the last two numbers of this table, we learn that the circumference of
the circle whose diameter is unity is less than 3.1415928 and greater than
3.1415926; and since, when the diameter  1, we have C= ir (40), it follows
that
w = 3.1415927
within a unit of the seventh decimal place.
SECOND METgOD, called the METHOD OF ISOPERIMETERS. This method is
based upon Proposition XI. Instead of taking the diameter as given and computing its circumference, we take the circumference as given and compute the
diameter; or we take the semi-circumference as given and compute the radius.
Suppose we assume the semi-circumference  C =- 1; then since C = 27rR, we
have
i0_ 1
R     R'
that is, the value of 7r is the reciprocal of the value of the radius of the circle
whose semi-circumference is unity.
Let ABCD be a square whose semi-perimeter = 1; then    A     E     B
each of its sides = ~. Denote its radius OA by R, and its
apothem OE by r; then we have
r = -  -= 0.2500000,
R =- V/2 = 0.3535534.
D )G
* The computations have been carried out with ten decimal places in order to ensure the accuracy of the seventh place as given in the table.




INTRODUCTION.                            59
Now, by Proposition XI., we compute the apothem r' and the radius R' of
the regular polygon of 8 sides having the same perimeter as this square; we find
rt - R    2 _ 0.3017767,
R/ = -/R X r = 0.3266407.
Again, taking these as given, we put
v = 0.3017767, R  0.3266407,
and find by the same formulae, for the apothem and radius of the isoperimetric
regular polygon of 16 sides, the values
' =- 0.3142087, R' - 0.3203644.
Continuing this process, the results are found as in the following
TABLE.
NUMBER OF SIDES.          APOTHEMI.                RADIUS.
4                0.2500000               0.3535534
8                0.3017767                0.3266407
16                0.3142087               0.3203644
32                0.3172866               0.3188218
64                0.3180541               0.3184376
128                0.3182460               0.3183418
256                0.3182939               0.3183179
512                0.3183059               0.3183119
1024                0.3183089               0.3183104
2048                0.3183096               0.3183100
4096                0.3183098               0.3183099
8192                0.3183099               0.3183099
Now, a circumference described with the radius r is inscribed in the polygon,
and a circumference described with a radius R is circumscribed about the polygon; and the first circumference is less, while the second is greater, than the
perimeter of the polygon. Therefore the circumference which is equal to the
perimeter of the polygon has a radius greater than r and less than R; and this
is true for each of the successive isoperimetric polygons. But the r and R of
the polygon of 8192 sides do not differ by so much as.0000001; therefore the
radius of the circumference which is equal to the perimeter of the polygons,
t-hat is, to 2, is 0.3183099 within less than.0000001; and we have
1
r'0.318309 —3.141593
within a unit of the sixth decimal place.
47 Scholium I. Observing that in this second method the value of r=-, for




60                         INTRODUCTION.
the square, is the arithmetic mean of 0 and ~, and that R=4-/2 is the geomet-'
ric mean between ~ and 4, we arrive at the following proposition:
The value of 1 is the limit approached by the successive numbers obtained by starting
from the numbers 0 and ~ and taking alternately the arithmetic mean and the geometric mean between the two which precede.
48. Scholium. II. ARCHIMEDES (born 287 B. C.) was the first to assign an
approximate value of r. By a method similar to the above " first method," he
proved that its value is between 31 and 31~, or, in decimals, between 3.1428 and
3.1408; he therefore assigned its value correctly within a unit of the third decimal place. The number 31, or Y2, usually cited as Archimedes' value of ir
althoughl it is but one of the two limits assigned by him), is often used as a
sufficient approximation in rough computations.
METIUS (A. D. 1640) found the much more accurate value -15v, which correctly represents even the sixth decimal place.  It is easily remembered by
observing that the denominator and numerator written consecutively, thus
1131355, present the first three odd numbers each written twice.
More recently the value has been found to a very great number of decimals
by the aid of series demonstrated by the Differential Calculus. CLAUSEN and
DASE, of Germany), about A. D. 1846), computing independently of each
other, carried out the value to 200 decimal places, and their results agreed to
the last figure. The mutual verification thus obtained stamps their results as
thus far the best established value to the 200th place. (See SCHUMACHER'S
Astronomische Nachrichten, No. 589.) Other computers have carried the value
to over 500 places, but it does not appear that their results have been verified.
The value to fifteen decimal places is
-T - 3.141592653589793.
For the greater number of practical applications, the value r- =- 3.1416 is sufficiently accurate.
I shall close the introduction to the present volume by the following
account of the quadrature of the circle, which I have been permitted to
copy from "Chambers Encyclopedia," through the courtesy of the
Superintendent of the Young Men's Mercantile Library of Cincinnati,
Ohio, from the copy of the same in their possession:
QUADRATURE OF THE CIRCLE.
This is one of the grand problems of antiquity, which unsolved and probably
unsolveable, continue to occupy even in the present day, the minds of many curious speculators. The trisection of the angle, the:duplication of the cube, and
the perpetual motion have found, in every age of the world since geometry and
physics were thought of, their hosts of patient devotees. The physical question




INTRODUCTION.                             61
involved in the perpetual motion (q v) is treated of under that head; and we
shall now take the opportunity of noticing the mathematical questions involved
in the other problems above mentioned; but more especially that of the quadrature of the circle, in which the difficulty is of a different nature from that involved in the other two geometrical ones. A few words about them, however,
will help as an introduction to the subject. According to the postulates of ordinary geometry, all constructions must be made by the help of the circle and
straight line. Straight lines intersect each other in but one point; and a straiglit
line and circle, or two circles, intersect in two points only. From the analytical
point of view we may express these facts by saying that the determination of the
intersection of two straight lines involves an equation of the first degree only;
while that of the intersection of a straight line and a circle, or of two circles, is
reducible to an equation of the second degree.
But the trisection of an angle, or the duplication of the cube, requires for its
accomplishment, the solution of an equation of the third degree; or, geometrically, requires the intersections of a straight line and a curve of the third
degree, or of two conics, etc., all of which are excluded by the postulates of the science.
If it were allowed lihat a parabola or ellipse could be described with a given
fucus and directrix, as it is allowed that a circle can be described with a given
radius about a given center, the trisection of an angle and the duplication of the
cube would be at once brought under the category of questions resolvable by
pure geometry; so that the difficulty in these cases is one of mere restriction
of the postulates of what is to be called geometry.
It is very different in the case of the quadrature of the circle, which (the
reader of the preceding article will see at once) means the determination of the
area of a circle of given radius, literally, the assigning of the side of a square
whose area shall be equal to that of the given circle.
The common herd of "squarers of the circle," which grows more numerous
every day, and which includes many men of undoubted sanity, and even of the
highest business talents, rarely have any idea of the nature of the problem they
attempt to solve. It will, therefore, be our best course to show, first of all, what
has been done towards the solution of the problem; we shall then venture a few remarks as to what may yet be done, and in what direction philosophic "squarers of
the circle" inust look for real advance  In the first place, then, we observe that
mechanical processes are utterly inadmissible. A fair approximation mny, no doubt,
be got by measuring the diameter of a circular disc of uniform material, and
comparing the weight of the disc with that of a square portion of the same material of a given side. But it is almost impossible to execute any measurement to
more than six places of significant figures; hence, as will soon be shown, this
process is at best but a rude approximation. The same is to be said of such obvious processes as wrapping a string round a cylinder post of known diameter
and comparing its length with the diameter of the cylinder; only a rude approximation to the ratio of the circumference of a circle to its diameter can thus
be obtained. Before entering on the history of the problem, it must be remarked
that the Greek geometers knew that the area of a circle is half the rectangle un



62                         INTRODUCTION.
der its radius and circumference (see CIRCLE), so that the determination of the
length of the circumference of a circle of given radius is precisely the same
problem as that of the quadrature of the circle.
Confining ourselves strictly to the best ascertained steps in the history of the
question, we remark that Archimedes proved that the ratio of the diameter to
the circumference is greater than 1 to 31~ or 31, and less than 1 to 31g; the
difference between these two extreme limits is less than the 1000th of the whole
ratio. Archimedes' process depends upon the obvious truth, that the circumference of an inscribed polygon is less, while that of a circumscribed polygon
is greater, than that of the circle. His calculations were extended to regular
polygons of 96 sides.
Little more seems to have been done by mathematicians till the end of the
16th century, when P. Métius gave expression for the ratio of the circumference
to the diameter as the fraction 3 —, which, in decimals, is true to the seventh
significant figure inclusive; curiously enough it happens that this is one of the
fractions which express in the lowest possible terms the best approximation to
the required number. Métius seers to have employed, with the aid of far superior arithmetical notation, a process similar to that of Archimedes. Viete
shortly afterwards gave the ratio in a form true to the tenth decimal place, and
was the first to give, though of course in infinite terms, an exact formula. Designating, as is usual in mathematical works, the ratio of the circumference to
the diameter wr, Viete's formula is
Ir       M2/'\V2v2V//\V2i          1      /       etc.! — i/i'X 1/ +     1/t"  X 1/'  +  1/~ +  v' x etc.
Shortly afterwards, Adrianus Romanus, by calculating the length of the side
of an equilateral inscribed polygon of 1073741824 sides, determined the value of
wT to 16 significant figures; and Ludolph Van Ceulen, his contemporary, by calculating that of the polygon of 36.893488147419103232 sides, arrived (correctly)
at 36 significant figures. It is scarcely possible to give, in the present day, an
idea of the enormous labor which this mode of procedure entails even when only
8 or 10 figures are sought; and when we consider that Ludolph was ignorant of
logarithms, we wonder that a life time sufficed for the attainment of such a result
by the method he employed.
The value of r was thus determined to   10  of its amount, a fraction of
which, after Montucla, we shall attempt to give an idea, thus: suppose a circle
whose radius is the distance of the nearest fixed star (250,000 times the earth's
distance from the sun), the error in calculating its circumference by Ludolph's
result would be so excessively small a fraction of the diameter of a human hair
as to be utterly invisible, not merely under the most powerful microscope yet
made, but under any which future generations may be able to construct.
These results were, as we have pointed out, all derived by common arithmetical operations, based on the obvious truth that the circumference of a circle
is greater than that of any inscribed, and less than any circumscribed polygon.




INTRODUCTION.                           63
They involve none of those more subtle ideas connected with limits, infinitesimals, or differentials, which seem to render more recent results suspected by modern "squarers."  If one of that unhappy body would only consider this simple
fact, he could( hardly have the presumption to publish his 3.1250 or whatever it
may be, as the accurate value of a quantity which, by common arithmetical processes, founded on an obvious geometrical truth, was several centuries ago
shown to be the greater than 314159265358979323846264338327950288 and less
than 314159265358979323846264338327950289.
We now know by far simpler processes its exact value to more than 600 places
of decimals; but the above result of Van Ceulen is much more than sufficient
for any possible practical application even in the most delicate calculations in
astronomy.
Snellius, Huyghens, Gregory de St. Vincent, and others, suggested simplifications of the polygon process, which are in reality some of the approximate
expressions derived from modern trigonometry. In 1668 the celebrated James
Gregory gave a demonstration of the impossibility of effecting exactly the quadrature of the circle, which, although objected to by Huyghens, is now received as
quite satisfactory. We may merely revert to the speculations of Fermat, Roberval, Cavalleri, Wallis, Newton, and others, as to quadrature in general; their
most valuable result was the invention of the Differential and Integral Calculus,
by Newton, under tlie name of Fluxions or Fluents. Wallis, however, by an
2.4.4.6.6.8.8.10.10, etc., w
ingenious process of interpolation, showed that 4  3 2 5.5.747.6 91.1   etc, which
4= 3.3.5.5.7.7.9.9.11, etc.,
is interesting as being the first recorded example of the determination, in a
finite form, of value of the ratio of two infinite products.
Lord Bruncker, being consulted by Wallis as to the value of the expression,
put it in the form of an infinite continued fraction, thus:
7r 1
4-1+1
2+9
2+25
2+49
2+, etc.
in which 2 and the squares of the odd numbers appear. This formula has
been employed to show that not only rT, but its square, is incommensurable.
Perhaps the neatest of all the formulas which have been given for the quadrature of the circle, is that of James Gregory, for the arc in terms of its tangentnamely, 0 = tan. 0 -3 tan. s3 + - tan. 50-etc.
This was appropriated by Liebnitz, and formed, perhaps, the first of those
audacious series of peculations from English mathematicians which have forever dishonored the name of a man of real genius.
If we notice that, by ordinary trigonometry, the arc, whose tangent is unity,
(the arc of 45~ or). falls short of four times the arc whose tangent is ~ by an




64                         INTRODUCTION.
angle whose tangent is 1~, we may easily calculate - to any required number
of decimal places by calculating from Gregory's formula the values of the
arcs corresponding to ~ and 3~ as tangents. And it is, in fact, by a slight modification of this process (which was originally devised by Machin) that ir has
been obtained by independent calculators to 600 decirnal places. It is not yet
proved, and it may not be true, that the area or circumference of a circle can
not be expressed in finite terms; if it can be, these must, of course, contain
irrational quantities. The Integral Calculus gives, among hosts of others, the
following very simple expression in terms of a definite integral:
T   r  dx
0
Now it very often happens that the value of a definite integral can be assigned when that of the general integral can not; and it is not impossible, so far
as is yet known, that the above integral niay be expressed in such form as
l/X + Vy
where!/x and l/'y are irrational numbers. Such an expression, if discovered, would undoubtedly be hailed as a solution of the grand problem. But
this, we need hardly say, is not the species of solution attempted by "squarers."
We could easily, from our own experience alone, give numerous instances of
their helpless absurdities, but we spare the reader and refer him, for further
information on this painful yet ridiculous subject, to a recent series of papers,
by Professor De Morgan, in the Athenoeurm, and to the very interesting work
oi Montucla, Histoire des Recherches sur la Quadrature du Cercle.




PART FIRST,
CONTAINING THE
GEOMETRICAL AND FINAL SOLUTION
OF THE
QUADRATURE OF THE CIRCLE,
BY AN ENTIRELY NEW METHOD, TOGETHER WITH AMPLE PROOFS OF
THE SAME.
5                                         (65)




QUADRATURE OF THE CIRCLE.
DEFINITIONS.
Of Lines, Angles, etc., see Plate 1.
1. Science is knowledge systematized.
2. Art is the skill with which the principles of a science are practically applied.
3. Quantity is anything which can be increased, diminished, or
measured.
4. Mathematics is the science of quantity.
5. Geometry is that branch of mathematics which treats of the properties of extension and figure.
LINES.
6. A line has only one dimension, namely, length; without either
breadth or thickness; lines are either straight or curved.
7. The extremities of a line are points.
8. A point has no dimensions, and therefore it has no size; and is
usually represented to the eye by a dot, as at AA.
9. A straight line is the shortest distance between any two given
points, as at B.
10. A curved line does not lie evenly between its extreme points;
but constantly changes its course as at C.
11. A waved line is composed of curved lines as at E.
12. Parallel straight lines are in the same plane, but have no inclination towards one another; and being produced ever so far, both ways
can not meet, as at D.
13. Parallel curved lines if produced would form two concentric
circles; that is, having a common center, as at F.
14. When a straight line standing on another straight line makes
the adjacent angles equal to one another, each of the angles is a right
angle, and the straight line which stands on the other is called a perpendicular to it; IKL are perpendiculars to the line GH.
(66)




PLATE 1.
DEFINITIONAS OF LINES AND ANGLES
A             A
C
B
E
F
S.   R
-1-7r~x 
Y~C
w~~~~~~~L








DEFINITIONS.                         67
15. Inclined lines, as I and I if produced, would meet in a point as
K forming an anle of which the point K is the vertex, and the lines
IH and I the legs of the angle.
16. When two straight lines cut each other so as to make the adjacent angles equal to one another, each of the angles is a right angle;
and the straight lines which so eut each other are perpendicular to
one another; thus QR is perpendicular to ST and XYto ZV; but
perpendicular lines are not always vertical and horizontal,; XY is a
horizontal line, and ZVis a vertical line.
ANGLES.
17. If there is only one angle at a point it may be denoted by a
letter placed at the vertex, as at K. But if several angles are at one
point, any one of them is expressed by three letters, of which the
middle one is at the vertex; thus the angle which is contained by the
straight lines UWYis called the angle UWYor YWU.
REMIARK.-Angles like other quantities may be added, subtracted, multiplied, and divided;
thus the angle UIWYis the sum of tle angles UWV, VIVX, XIVWY and the angle UWV is the
difference between the two angles, YWU and YWV
18. An acute angle is less than a right angle; thus the angle at K
is an acute angle.
19. An obtuse angle is greater than a right angle; thus the angle at
P is an obtuse angle.
20. An oblique angle is formed by one straight line meeting another
straight line on one side of it, so as to make the adjacent angles equal
to two right angles.
21. A plane angle is the inclination of two lines to one another in
a plane which meet together, but are not in the same direction.
22. A plane rectilineal angle is the inclination of two straight lines'
to one another which meet together, but are not in the same straight
line.
23. A THEOREM is a truth requiring demonstration.
24. An AXIOi is a self-evident truth.
25. A PROBLEM is a question requiring a solution.
26. A POSTULATE is a self-evident problem.
Theorems, Axioms, Problems, and Postulates, are all called Prolositions.




68                             DEFINITIONS.
27. A LEMA is an auxiliary proposition.
28. A COROLLARY is an obvious consequence of one or more propositions.
29. A SCHOLIUM is a remark made upon one or more propositions,
with reference to their connection, their use, their extent, or their
limitation.
30. An HYPOTHESIS is a supposition made, either in the statement
of a proposition, or in the course of a demonstration.
31. Magnitudes are equal to each other, when each contains the same
unit an equal number of times.
32. Magnitudes are equal in all their parts, when they may be so
placed as to coincide throughout their whole extent.
Of Plane and Rectilineal Surfaces.
1. A superfices or surface has two dimensions; length and breadth
REMARK.-The extremities of superfices are lines. A term or boundary is the extremity of
anything. A figure is enclosed by one or more boundaries.
NoTE.-Rectilineal figures are contained by straight lines. Trilateral figures or triangles, by
three straight lines; quadrilateral figures, by four straight lines; multilateral figures or polygons,
by more than four straight lines.
2. An equilateral triangle has three equal sides, as A.
3. An isoceles triangle has two equal sides, as B.
4. A right-angled triangle has one right angle, as C.
REMARK.-The side opposite the right angle is called the hypothenuse.
5. A scalene triangle has three unequal sides, as D.
6. An obtuse-angled triangle has one obtuse angle as 0.
7. An acute-angled triangle has three acute angles, as A.
8. A square has all its sides equal, and all its angles right angles,
as E.
9. A rhombus has all its sides equal, but its angles are not all right
angles, as H.
10. A rhomboid has its opposite sides equal to one another, but all
its sides are not equal, nor its angles right angles, as FG.
11. A trapezoid has only two of its sides parallel, as IK.
12. All other four-sided figures beside these are called trapeziums.
NOTE.-The terms oblong and rhomboid are not often used; practically the following definitions are used: Any four-sided figure is called a quadrilateral. A line joining two opposite
angles of a quadrilateral is called a diagonal. A quadrilateral which has its opposite sides parallel is called a parallelogram. The words square and rhombus are used in the sanle as defined by
Euclid, and the word rectangle is used instead of the word oblong.
Some writers propose to restrict the word trapeziumn to a quadrilateral which has two of its
sides parallel; and it would certainl y be convenient if this restriction were universally adoiterl.




PLATE     11.
IDEFLVITIONS OFPLANTEAND RE CTZiNEAL SURFACES
A             B      C    D
C
E                F                7








DEFINITIONS OF THIE CIRCLE.                   69
Of the Circle, see Plate 3.
1. A circle is a plain figure bounded by a curved line called the circumference, every point of which is equally distant from a point within*
And this point within is called the center A, F, C, B, I3, D, G is the
circumference and E the center of the circle.  Fig. 2 Plate 3.
2. A radius is a straight line drawn from the center to any point of
the circumference AE, CE, BE, DE. Fig. 2 Plate 3.
3. A diameter is a straight line drawn through the center and terminating in the circumference, as AB, CD. Fig. 2 Plate 3
REMARK.-All radii of the same circle are equal. All diameters are eqial, and each is double
of the radius.
4. An arc is any part of the circumference, as AF. Fsg. 1 Plate 3.
5. A chord is a straight line joining the extremities of an arc, as
FH. Fig. 1 Plate 3.
6. A segment is that part of a circle between an arc and its chord
AG, HB, and GHD; Fig. 2 Plate 3 are segments of the circle ABC.D.
7. A sector is that part of a circle included within an arc and the
radii drawn to its extremities FEC; Fig. 2 Plate 3 is a sector of the
circle ABCD, also AEF, CEB.
REMARK.-AEF is also called a sextant because it is one-sixth part of the circle, and CEB is
called a quadrant, becai.se it is one-fourth part of the circle.
TRIGONOMETRY.
Trigonometry is the science which teaches how to determine the several
parts of a triangle from having certain parts given.
Plane trigonometry treats of plane triangles; spherical trigonometry
treats of spherical triangles.
The circumference of a circle is supposed to be divided into 360
equal parts, called degrees; each degree into 60 minutes, and each minute into 60 seconds.  Degrees, minutes, and seconds are designated by
the characters, ', ".  Thus 23~ 14' 35" is read 23 degrees, 14
minutes, and 35 seconds.
Since an angle at the center of a circle is measured by the arc intercepted by its sides, a right angle is measured by 90~, two right angles
by 180~, and four right angles are measured by 360~.
The complement of an arc is what remains after subtracting the arc
from 90~. Thus the arc DF, Fig. 1 Plate 3, is the complement of
AF. The complement of 25~ 15' is 64~ 45'.




70               DEFINITIONS OF THE CIRCLE.
In general, if we represent any arc by A, its complement is 90~-A.
Hence, if an arc is greater than 90~, its complement must be negative.
Thus the complement of 100~ 15' is-10~ 15'. Since the two acute
angles of a right-angled triangle are together equal to a right angle
each of them must be the complement of the other.
The supplement of an arc is what remains after subtracting the arc
from 180~. Thus the arc BDF is the supplement of the arc AF.
The supplement of 25~ 15' is 154~ 45'. In general, if we represent
any arc by A, its supplement is 180~-A.  Hence, if an arc is greater
than 180~, its supplement must be negative. Thus the supplement
of 200~ is-20~. Since in every triangle the sum of the three angles
is 180~, either angle is the supplement of the sum of the other two.
The sine of an arc is the perpendicular let fall from one extremity of
the arc on the radius passing through the other extremity. Thus FG
is the sine of the arc AF, or of the angle A CF.
Every sine is half the chord of double the arc. Thus the sine FG is
the half of FH, which is the chord of the arc FAH, double of FA.
The chord which subtènds the sixth part of the circumference, or the
chord of 60~, is equal to the radius (Loomis' Geom., Prop. IV., Book
VI.); hence the sine of 30~ is equal to half of the radius.
T/7e versed sine of an arc is that part of the diameter intercepted between the sine and the arc. Thus GA is the versed sine of the arc
AF.
The tangent of an arc is the ine which touches it at one extremity, and
is terminated by a lne drawn from the center through the other extremity. Thus AI is the tangent of the arc AF, or the angle ACF.
The secant of an arc is the line drawn from the center of the circle
through one extremity of the arc, and is limited by the tangent drawn
through the other extremity. Thus CI is the secant of the arc AF, or
of the angle A CF.
The cosine of an arc is the sine of the complement of that arc.  Thus
the arc DF, being the complement of AF, FK is the sine of the arc
DF, or the cosine of the arc AF.
The cotangent of an arc is the tangent of the complement of that arc.
Thus, DL is the tangent of the arc DF, or the cotangent of the arc
AF.
The cosecant of an arc is the secant of the complement of that arc.
Thus CL is the secant'of the arc DF, or the cosecant of the arc AF.




DEFINITIONS OF THE CIRCLE
Fig. 1              Cota
B   ________ C  i6s.  Vers.  A
A3                      1A
~     EA
M
C
Fig 2 
/  \        "%\








DEFINITIONS OF THE      CIRCLE.               71
In general, if we represent any angle by A,
Cos. A - sine (90~-A).
Cot. A = tang. (90~-A).
Cosec. A - sec. (90 —A).
Since, in a right-angled triangle either of the acute angles is the
complement of the other, the sine, tangent, and secant of one of these
angles is the cosine, cotangent, and cosecant of the other.
The sine, tangent, and secant of an arc are equal to the sine, tangent, and secant of its supplement. Thus FG is the sine of the arc
AF, or its supplement, BDF. Also, AI, the tangent of the arc AF,
is equal to BM, the tangent of the arc BDF; and CI, the secant of
the are AF, is equal to CM, the secant of the arc BDF.
The versed sine of an acute angle, A CF, is equal to the radius minus
the cosine CG. The versed sine of an obtuse angle BCF, is equal to
radius plus the cosine CG; that is to BG.
The relations of the sine, cosine, etc., to each other, may be derived
from the proportions of the sides of similar triangles.  Thus the triangles CGF, CAI, CDL, being similar, we have,
1. CG:: CF: CA: AI; that is, representing the arc by A,
and the radius of the circle by R, cos. A: sin. A: R: tang. A;
Whence tang. A     R s-n. A
cos. A4
2. CG::   CF:  CA      CI; that is, cos. A: R::   R:
sec. A;
R2
Whence sec. A     --
cos. A'
3. GF: CG::  CD: DL; that is, sin. A: cos. A:: R
cot. A.
R cos. A
Whence cot. A == s. A
sin. A
4. GF: CF:   CD: 'L; that is, sin. A: R::   R: cosec A;
R2
Whence cosec. A     -.
sin. A'
5. AI: AC::  CD: DL; that is, tang.A: R:  R
cot. A.
R2
Whence tan. A     ---
cot. A
Also in the right-angled triangle CGF, we find GG2 +    GF2 =




72                       POSTULATES.
CF2; that is, sin. 2A + cos. 2A = R 2; or, the square of the sine of an
arc, together with the square of its cosine, is equal to the square of the
radius;
Hence sin. A  -+~   R 2 - cos. 2A,
And cos. A    4    ~  /R 2 _ sin. 2A.
POSTULATES.
Let it be granted:
1. That a straight line may be drawn from any one point to any other
point.
2. That a terminated straight line may be produced to any length in
a straight line.
3. And that a circle may be described from any center, and with
any radius.
4. That no part of the circumference of a circle is straight.
5. That it is possible to find a straight line equal in length to the
circumference of a given circle.
6. That it is possible to find a common measure between the side
and diagonal of a square in integers to infinity.
7. That it is possible to inscribe in any circle a polygon with any
given number of sides expressed by 2n + 1, provided, 2n + 1 is a
prime number.
AXIOMS.
1. Things which are equal to the same thing are equal to one another.
2. If equals be added to equals the wholes will be equal.
3. If equals be taken from equals the remainders are equal.
4. If equals be added to unequals the wholes will be unequal.
5. If equals be taken from unequals the remainders will be unequal.
6. Things which are double of the same thing are equal to one another.
7. Things which are halves of the same thing are equal to one
another.
8. Magnitudes which coincide with one another, that is, which exactly fill the same space are equal to one another.
9. The whole is greater than its part.
10. Two straight lines can not inclose a space.
11. All right angles are equal to one another.




ARTICLES.                         73
12. If a straight line meet two straight lines, so as to make the two
interior angles on the same side of it taken together less than two
right angles, these straight lines being continually produced, shall at
length meet on that side on which are the angles which are less than
two right angles.
ARTICLES.
1. The double of the cosine or secant is the constantly assumed
diameter.
2. The sum of the sines or tangents is the constantly assumed circumference.
3. The sine is a mean proportion between the double of the cosine
and the second tangent.
4. The second tangent is a mean proportional between the double of
the second secant and the third tangent, etc., etc., etc.
5. The rectangle contained by the sine and cosine is equal to the
area of the inscribed double triangle.
6. The rectangle contained by the radius and the tangent is equal
to the area of the circumscribed double triangle.
7. The rectangle contained by the radius and the sine is a mean proportional between the inscribed and circumscribed double triangles
of half the number of sides.
8. The rectangle contained by the inscribed double triangle and the
number of sides of the entire polygon is equal to the area of the inscribed polygon.
9. The rectangle contained by the circumscribed double triangle and
the number of sides of the entire polygon is equal to the area of the
circumscribed polygon.
10. The rectangle contained by the rectangle of the.radius and the
sine and the number of sides contained in the given polygon, is equal
to the area of the entire inscribed polygon of double the number of
sides.
SIGNS.
The following are the principal signs employed:
The Sign of Addition, -, called plus:
Thus, A - B, indicates that B is to be added to A.
The Sign af Subtraction, -, called minus:
Thus, A - B, indicates that B is to be substracted from A.




74                          SIGNS.
The Sign of Multiplication, X:
Thus, A X B, indicates that A is to be multiplied by B.
The Sign of Division, -:
A
Thus, A - B, or, B, indicates that A is to be divided by B.
The Exponential Sign:
Thus, A3, indicates that A is to be taken three times as a factor, or
raised to the third power.
The Radical Sign, /-:
Thus, 1/A, ieB, indicate that the square root of A, and the cube
root of B, are to be taken.
When a compound quantity is to be operated upon as a single quantity, its parts are connected by a vinculum or by a parenthesis:
Thus, A + B X C, indicates that the sum of A and B is to be multiplied by C; and (A + B). C, indicates that the sum of A and B
is to be divided by C.
A number written before a quantity, shows how many times it is to
be taken.
Thus, 3(A + B), indicates that the sum of A and B is to be taken
three times.
The Sign of Equality:
Thus, 4A    B — C, indicates that A is equal to the sum of B and C.
The expression, A = B  - C, is called an equation. The part on
the left of the sign of equality, is called the.first member; that on the
right, the second member.
The Sign of Inequality, <:
Thus, VA <    -YB, indicates that the square root of A is less than
the cube root of B. The opening of the sign is towards the greater
quantity.
The sign,.'. is used as an abbreviation of the word hence, or consequently.
The general truths of Geometry are deduced by a course of logical reasoning, the premises being definitions and principles previously
established.  The course of reasoning employed in establishing any
truth or principle, is called a demonstration.








PLATE 1V.
PRROPOSITION 1 TflEOREf
Gr
FH
B      _




PROPOSITION 1. THEOREM.                     75
PROPOSITION 1. THEOREM.
In any right-angled triangle, the square which is described on the side
subtending the right angle is equal to the sum of the squares described on
the side which contains the right angle.
Let ABC be a right-angled triangle, having the right angle BA C;
the square described on the side BC shall be equal to the sum of the
squares described on BA, AC. See Plate 4 Fig. 1.
On BC describe the square BDEC, and on BA, A C describe the
squares GB, HC;                            Todhunter's Euclid, [1.46,]
through A draw AL parallel to BD or CE;                    [1.31,]
and join AD, EC.
Then, because the angle BAC is a right angle,       [Hypothesis,]
and that the angle BAG is also a right angle,      [Definition 30,]
the two straight lines A C, A G, on the opposite sides of A.B, make with
it at the point A the adjacent angles equal to two right angles; therefore CA is in the saine straight line with AG.             [1.14.]
For the same reason, AB and AH are in the same straight line.
Now the angle DBC is equal to the angle FBA, for each of them is
a right angle.                                         [Axiom 11.]
Add to each the angle ABC. Therefore the whole angle DBA is
equal to the whole angle FBC.                           [Axiom 2.]
And because the two sides AB, BD are equal to the two sides FB, BC,
each to each;                                       [Definition 30,]
and the angle DBA is equal to the angle FBC; therefore the triangle
ABD is equal to the triangle FBC.                           [1.4.]
Now the parallelogram BL is double of the triangle ABD, because
they are on the same base BD, and between the same parallels
BD, AL.                                                    [1.41.]
And the square GB is double of the triangle FBC, because they are
on the same base FB, and between the same parallels FB, GC, [1.41.]
But the doubles of equals are equal to one another;     [Axiom 6.]
Therefore the parallelogram BL is equal to the square GB.
In the same manner, by joining AE, BK, it can be shown that the
parallelogram CL is equal to the square CH.
Therefore the whole square BDEC is equal to the two squares
GB, HC.                                                [Axiom 2.]
And the square BDECis described on BC, and the squares GB, HC




76                PROPOSITION 2. THEOREM.
on BA, AC. Therefore the square described on the side BCis equal
to the squares described on the sides BA, AC.
CORALLARY 1. Let ABCD be a square, and A C its diagonal; the triangle ABC being right-angled and isoceles, we have A C2=AB2+B C2
=2AB2.
Therefore, the square described on the diagonal of a square, is double
of the square described on the side.
If we extract the square root of each member of this equation, we
shall have
AB/2-=AC; or AB: A C:: 1:       2.
See Plate 4 Fig. 2.
PROPOSITION 2. THEOREM.
ARGUMENT 1. If ten (10) right-angled triangles, the base, perpendicular, and hypothenuse of which are respectively three (3), four
(4), and five (5), be placed so as to form a polygon of five sides and a
little more, that is to say, with all their bases outwards, and the perpendicular of one triangle joined to the perpendicular of another, so
that they shall touch one another in those points which contain the
right angles, and also in those points which form the verticle angles,
so that any two of the bases when taken together shall be equal to six;
and the hypothenuse of each triangle joined to the hypothenuse of another triangle, so that the vertical angles, that is the angles opposite
the bases shall form a common center, then there will be a lap equal to the
one-fiftieth of the circumference of the whole polygon. See Plate 5 Fig. 1.
For, because the straight lines A, CD, are tangents to the radius BC,
therefore they are both in the same straight line, and the radius BC
is a perpendicular to them. For the same reason the straight lines
DE, EF are in the same straight line, and the radius BE is a perpendicular to them, also FG, G1, and BG. HI, IJ, and BI. JK, KZ, and
BK.                                                  [Hypothesis.]
Again, because the radius BC is perpendicular to the straight line
AD, therefore, the angles BCD and BCA are both right angles; for
the same reason BED and BEF are right angles, as also BGF and
BGH, BIH, and BIJ, BKJand BKZ.                       [Hypothesis.]
Again, because the angles ADF, DFH, FHJ, HJZ are inscribed in
an arc which is less than a semicircle, therefore they are greater than
a right angle.                                       [Hypothesis.]








z vH
z I
(I  *A SAIid\;Y




PROPOSITION 2. THEOREM.                    77
Now let the right-angled triangles BCA and BCD, BED, and BEF,
BGF, and BGH, etc., etc., be the given polygon, and let two circles
be drawn around the center B; one with a radius of four or BC, and
the other with a radius of five or BD; so that the first circle shall pass
through the points which form the right angles, namely, CEGIK, and
the second through the points at the other extremity of the bases,
namely, ADFHJZ; then if a straight line starting at the point Z
shall be made to pass around the polygon, changing its direction at
every point where it intercepts the circle whose radius is five, so that
it shall not at any time pass beyond it, nor at any time come within
the circle whose radius is four, this straight line, if continued in this
manner till it is applied exactly forty-nine times, will return to the point
A. from whence it first started; thus forming a lap equal to the one.
fiftieth of the whole polygon, and the circle so formed by the intersection of these lines, which will have a radius of four, will be exactly
forty-nine fiftieths of the entire polygon, or exactly one-fiftieth less than
the polygon contained within the given circle. See Plate 5 Fig. 2.
Therefore AZ = 1Tth of entire polygon =  9 of the entire circle. Consequently,
If from the number of sides of the above polygon -ltth be deducted, the
square root of'the remaining 49 can be extracted exactly; thus 50 - 1
= 49, and the i/ 49  7; and if from the sum of the squares of the
two sides of any square expressed- in integers the one-fiftieth be deducted, the square root of the remaining forty-nine fiftieths can be exgracted exactly.
Demonstration.
Let 5 be the side of the given square 5 X 5 = 25, which multiplied
by two gives 50, then 50 - 1 = 49 and the / 49=- 7;
Again, let the side of the given square be 1, then 1 X 1 = 1, which
doubled, gives 2; the -wth of 2 - -2 or.04 and 2.00-.04 =  1.96,
and the 1/1.96 = 1.4 or 15 = ] QED.
CORALLARY 1. If the u-8U-th part of the sums of the squares of
the two sides of any square be added thereto, the square root of the surn
can be extracted exactly.
Demonstration.
Let 1 be the side of the given square, then 1 X 1 = 1, which mul



78               PROPOSITION 3. THEOREM.
2  e'9800    then 2 90 0~ —
tiplied by 2 = 2, the -'  of 2 = -- or    -; then 2 + 
9801 and thle -r/o -     QED.
PROPOSITION 3. THEOREM.
ARGUMENT 1. If a circle be described with the square root of two
for a radius, and the one-fiftieth of the square described upon the radius
be deducted therefrom, the square root of the remaining forty-nine
fiftieths can be extracted exactly.
2. The square root of the -~ so deducted will be the sine of the
given arc.
3. The square root of the remaining 4-9 will be the cosine of the
given arc;
For, let the straight line fh, fm, Plate 6, Fig. 1 and 2 be the given
radius, and ADBC the given circle, whose arc is A, and let the straight
lines gh, ge, lm, in, be the sine of the given arc, then will the straight
line fg,fl, be the cosine of the given arc A;
Put R - radius of circle, whose arc is A,
Put sin. - sine A,
Put cos. -- cosine A.
Then by trigonometry:
R2    sin. A2= cos. A2 and R2 - cos. A 2  sin. A2;
Consequently, cos. A2 + sin. A 2 _  2.
Substituting the numbers, we have (1/2)2= 2, and '- of 2 =5  =
= -.04. Then 2.00-.04         1.96, and the V1/.96 = 1.4 = 1- or -7 -which equals the cosine of the given arc, and 2.00 -1.96 -.04, and
the 1/.04 -.2 or -, which equals the sine of the given arc.
ARGUMENT 4. Now, if the cosine of the given arc 7 be multiplied
by two, it will be 1-, which is very near though not quite the true
diameter; and if the sine of the given arc 1 be multiplied by two it
will be, which is very near though not quite the -2 part of true circumference being the chord of twice the arc, whose sine is very near
though not quite the - part of the true circumference. If then we
assume j-4 to be the true diameter and -A4 to be the true circumference
of the given circle, and they are very near it, then dividing the cir44   14
cumference 44 by the diameter 14-, we have, by cancellation, -- -
~         5 '.q~~0 




PLATE VI....E  rst
Fig
22
15   








PROPOSITION 3. THEOREI.                    79
44.,    22
=  -    X  -= = -, which is equal to 3.142857 or 31; which is the true
ratio of the circumference to the diameter of any given circle.
Consequently,
If the double of the cosine of an arc of any given circle, the sine of
which is obtained by extracting the square root of the one-fiftieth of
the square described upon the radius of the saine circle, and the cosine
by extracting the square root of the remaining forty nine fiftieths, be
divided into the sum of the sines, in which case the cos.: sin.::
7: 1, then the ratio will be the same as the ratio between the true
circumference and the true diameter of the given circle, viz.: 3.142857
or 3.
Demonstration.
Letf,fh7n, be the radius of the given circle, and gh, ge, or lm, ln, the
sine of the given arc A, then will fg, fl, be the cosine of the same arc.
See Fig. 1 and 2 Plate 6.
Put R = radius of the given circle
and sin. = sine of the given arc,
and cos. = cosine of the given arc,
then R 2 - si. 2 - cos. 2. Consequently the angles fgh, fimn, fge, fln,
are both right angles, therefore the straight lines hg, ge; ml, ln, are
both in the same straight line.
Again, the line gh, lm, is equal to the line ge, ln, each being the sine
of the given arc A, and one-half the chord of double the arc A, therefore the straight line he is double the sine of the arc A, for the same
reason the chord, which is the base of the double triangle fl, is double
the sine of the arc A, so also are the chords which are the bases of
the double triangles f2, f3, etc., etc.,f 22 double the sine of the arc
A, and each of the chords is assumed to be the -2 part of the whole
circumference; also the cosine f g is equal to the cosine f 1, f 2,
etc., etc., f 22 and each of them is equal to the one-half of the assumed
diameter, therefore the chord 2 X 22 = 44 - the assumed circumference and the cosine 7 X 2 =A4=  the assumed diameter. Then, dividing the assumed circumference 4%! by the assumed diameter 1-4, we
44    14   4,a.      9 22
have by cancellation-.        - X     -= 72  3.142857 or 3- the
te ra5 o5                       7 
true ratio of the circumference to the diameter of any given circle.




80                PROPOSITION 4. THEOREM.
Consequently, Cor. 1, if the double of the secant be assumed for the true
diameter, instead of the double of the cosine, and the sum of the tangents be assumed for the true circumference, instead of the sum of the
sines, the ratio will be the same as the ratio between the double of the
cosine and the sum of the sines, that is it will be the same as the ratio
between the true circumference and true diameter, that is 3.142857
or 3.
PROPOSITION 4. THEOREM.
To Find a Mean Proportional Between Two Given Straight Lines.
Let AB, BC be two given straight lines; it is required to find a
mean proportional between them. Place AB, BC in a straight line,
and on AC describe the semicircle ADC; from the point B draw BD
at right angles to A C. See Plate 7 Fig. 1.               [1.11.]
BD, shall be a mean proportional between AB and BC.
Join AD, DC. Then the angle ADCbeing in a semicircle is a right
angle;                                                  [I. 31,]
and because in the right-angled triangle ADC, DB is drawn from the
right angle perpendicular to the base, therefore DB is a mean proportional between AB, BC, and the segments of the base. [VI. 8, Corollary.]
Wherefore between the two given straight lines AB, BC, a mean proportional DB is found.
PROPOSITION 5. THEOREM.
In a right-angled triangle, if a perpendicular be drawn from the right
angle to the base, the triangles on each side of it are similar to the whole
triangle and to cach other. See Plate 7 Fig. 1.
Let DA C be a right-angled triangle, having the right angle ADC,
and from the point D let DB be drawn perpendicular to the base AC,
the triangles BAD, DBC, shall be similar to the whole triangle DA C
and to one another.
For, the angle ADCis equal to the angle ABD, each of them being
a right angle,                                        [Axiom 11.]
and the angle at A is common to the two triangles DAC, BAD, therefore the remaining angle DUCA is equal to the remaining angle BDA;
therefore the triangle DA Cis equiangular to thé triangle BAD, and
the sides about their equal angles proportionals; therefore the triangles are similar. VI. 4.                          [Definition 1.]




PLATE Vil.
Fi. I  _.D
F~~Fi. 2 B:1.~)
Fig. 4
Fig.3 3
C
A          1)
|B  i
Dû  -D








PROPOSITION 6. THEOREM.                    81
In the same manner it may be shown that the triangle BAD to the
triangle DA C are similar to each other.
Wherefore in a right-angled triangle, etc., QED.
Corallary from this it is manifest that the perpendicular drawn
from the right-angle of a right-angled triangle to the base, is a mean
proportional between the segments of the base, and also that each of
the sides is a mean proportional between the base and the segment of
the base adjacent to that side.
For in the triangles BAD, BDC, AB is to BD       as BD is to
BC;                                                      [V. 4.]
And in the triangles DAC, BAD, AC is to AD as to AD is to
AB;                                                      [VI. 4.]
and in the   triangles DA, BDC, A     is to CD   as CD is to
CB.                                                     [VI. 4.]
PROPOSITION 6. THEOREM.
If three straight lines be proportionals, the rectangle contained by the
extremes is equal to the square on the mean; and if the rectangle contained by the extremes is equal to the square on the mean, the three
straight Unes are proportionals.
See Plate 7 Fig. 2.
Let the three straight lines ABCbe proportionals, namely: let A
be to B as B is to C; the rectangle contained by A and C shall be
equal to the square on B. Take D equal to B. Then, because A is
to B as B is to C,                                   [Hypothesis.]
and that B is equal to D; therefore A is to B as D is to C.  [V. 7.]
But if four straight lines be proportionals, the rectangle contained by
the extremes is equal to the rectangle contained by the means; [VI. 16.]
therefore the rectangle contained by A and Cis equal to the rectangle
contained by B and D. See Plate 7 Fig. 3 and 4. But the rectangle by B  and D   is the square on B, because B     is equal
to D;                                               [Construction.]
therefore the rectangle contained by A and C is equal to the square
on B. Next, let the rectangle contained by A and C be equal to the
square on B; A shall be to B as B is to C. For, let the same construction be made, then, because the rectangle contained by A and
Cis equal to the square on B,                        [Hypothesis.]
and that the square on B is equal to the rectangle contained by B and
D, because B is equal to D;
6




82              QUADRATURE OF THE CIRCLE.
therefore the rectangle contained by A and C is equal to the rectangle contained by B and D; but if the rectangle contained by the
extremes be equal to the rectangle contained by the means, the four
straight lines are proportionals;                        [VI. 16.]
therefore A is to B as D is to C.
But B is equal to D;                                 [Construction.]
Therefore A is to B as B is to C.                          [. 7.]
Wherefore if three straight lines, etc., QED.
QUADRATURE 0F THE CIRCLE.
CASE 1.
In which the Inscribed and Circumscribed Polygons are Carried to
22 Sides.
It is required to find the quadrature of the circle, that is when the
radius is known it is required to find a straight line (in terms of the
given radius) which shall be equal in length to the circumference of
the given circle.                                     [Postulate 5.]
For this purpose let the square root of two be taken for the given radius fh, fm, and with this radius describe the given circle ADBC.
[Postulate 3.]
See Plate 6 Fig. 1 and 2. Then let the square ABCD be inscribed in
the given circle, the side of the inscribed square will be two and the
area four.                                            [IHypothesis.]
For if each of the sides of the inscribed square ABCD be bisected the
whole square will be divided into four smaller squares, each of the sides
of which as well as their respective areas will be equal to one, consequently they will be equal to one another.              [Axiom 1.]
Then, by (PROPOSITION 1, COR.), the diagonals of each of the smaller
squares will be equal to the square root of two. But the radius of the
given circle is equal to the square root of two, consequently the diagonals of each of the smaller inscribed squares is equal to the radius
of the given circle.                                    [Axiom 1.]
Again, all the radii are equal and each is half the diameter; [Definition 3.]
therefore the diameter of the given circle is twice the square root of
two, or the square root of eight, and it is equal to the diagonal of the
square whose side is equal to the side of the inscribed square. For
the half of the diameter of the given circle is the V/z, and the half of




QUADRATURE OF THE CIRCLE.                    83
the side of the inscribed square is 1, consequently, by (Theorem 1,
Cor.), the side of any square is to its diagonal as 1: 1/2  [Axiom 7.]
therefore the one half of the side of the inscribed square is the side
of a square whose diagonal is the square root of two, which is equal to
the radius of the circle or half the diameter.
Now, by (PROPOSITION 2), the sine of the given arc is the square
root of the 5-'-th of the square described upon the radius, and the cosine of the given arc is the square root of the remaining 49. For,
by trigonometry, R2- sin.2 -cos.2, and R2- cos.2 = sin.2  consequently cos.2+ sin.2 = R2.   Substituting  the  numbers we have
(1/2)2=2.00 and 1 of 2 - =- =.04  and the -'.04- 2 or
-    =sine of the given arc; and 2.00-.04 = 1.96 and the 1/1.96 = 1.4
= 1-2 or 7-= cosine of the given arc.
Again, if the side of the small inscribed square, which is equal to
1, be divided into five equal parts each part is equal to, which is
equal to the sine of the given arc; and seven of the same parts is
equal to -7, which is equal to the cosine of the given arc; consequently
(})2-'5 anrd (.)_2 49 tlhet en  +  -       whichh is equal to the
square described upon the radius; and the 1/  -/ 2- which is equal to
the radius. Now, by Theorenm 3, the sine - is not quite though very
near the Tth part of the entire circumference; and the cosine 7 is not
quite though very near ~ the entire diameter; therefore - X 44 =-4-4 —
the assumed circumference; and 7 X 2 = 14-  the assumed diameter;
then, dividing the assumed circumference by the assumed diameter, we
44    14   4,         22
have by cancellation     5   -   X       2    3.142857 or 3 -the true ratio between the assumed circumference and the assumed
diameter of the given circle.
Again, the cosine of the given arc is —, and the sine of the given arc
is -, then, by trigonometry, cos.: sine:: R: tang. Substituting the
numbers we have cos. -: sin.:: radius /2:: tan. T/ 2, for 1 X
1I                                           7   1 71
/2= 5V2; then, by division and cancellation, 5    1/2  5  /2 X
7     7 1/2 = tangent of the given arc.
Again, by trigonometry, R2 + tan.2 - sec.2. Substituting the numbers
we have rad. (1/)2= 2 and tang. (71/2)2 =-_ then 2+       14oo
2   -2-;ith-9          4 9-,




84              QUADRATURE OF THE CIRCLE.
and the 1/1-f =      1.4142857, or = 1.4 = secant of the given arc.
-= 1.4-= cosine of the given arc.   =.2 =sine of the given arc.
41/2 =.2020305089 - tangent of the given arc.
Again, the side of the inscribed square being divided into five parts,
by applying the sine to it as a common measure each part is expressed
by 5; then 5 X - =-   = the size of the square which is to be the first
unit of superficial measure for the circle.
For, by ARTICLE 6, the area of the circumscribed double trianglefPE,
f PE is equal to the product of the radius and the tangent; thus, radius 1-/2 X tangent 4-1/2 =; and, by ARTICLE 9, the area of the entire
circumscribed polygon is equal to the product of the circumscribed
double triangle, and the number of sides contained in the given polygon; thus - X 22 - 4- = 62 the area required.
Now, 44 2                5 X 5     00  1571; therefore the area of
7   -25     7 X     -   7
157~    1100   44
the entire circumscribed polygon is equal to 25-7  17-7     6;
that is, there is 1574 blocks each of which is equal in area to -5, QED.
CASE 2.
In which the Inscribed and Circumscribed Polygons are Carried to 311 -Sides.
By Case 1, the sine is - and the double of the cosine is -4, and, by
ARTICLE 3, the sine is a mean proportional between the double of the
cosine and the tangent No. 2.  For, by (PROPOSITIONS 4 and 5),
sin. (5)2 =, and cos. 7 X 2  1, then, by division and cancellation,
i    14    1          1 
25    5 —     X 1-=-tangent No. 2.         (By PROPOSITION 6),
the rectangle contained by the double of the cosine and the tangent
No. 2 is equal to the rectangle contained by the square described upon the sine of the given arc.
For put double cosine =  A
and put sine         =   B
and put tangent No. 2 =  C
Then, by proportion, we have A: B   B: C; then A X C= B
X B = (A X C= B2), consequently, B2 - A = C. Substituting the




QUADRATURE OF THE CIRCLE.                    85
1a4   1    1
numbers we have by division and cancellation  X   = -   and (~)2
o        -y)   25 j
1                i    14     1        i
=25' consequently 2         -   X e=n=tangent No. 2.
By trigonometry, R 2 - tang. 2 = sec. 2. Substituting the numI          I          9801
berswehave R(j/)2=2; T(~))2 =490. Then,2+ 490          49
/9801         99
secant sequare; extracting the 49  we have - = secant No. 2.
7   98
By Case 1, the cosine =5- 70; and, by Case 2, the secant No. 2 =
99      99    98    1            1
70; then 7 - 7=     0; therefore 7 has been added to the cosine,
2
the assumed radius; and   to the assumed diameter.  Now, by hypothesis, the circumference of a circle is to the diameter as 3q is to 1;
therefore 31 times   must be added to the assumed circumference;
thus,   X 3x   6-.
By Case 1, the sine of the given arc is 5; and, by Case 2, the tangent of the given arc is 7; then, dividing the sine by the tangent, we
i    1    1   ~0
have by cancellation. 7-    X e - 14. Therefore the sine lias
been divided into 14 equal parts, each of which is equal to the tangent
of 7. Now, by Case 1, there were 22 sides to the given polygon, each
2         44   616
of which contained two sines, 2 X 22 =-   7 —. Now, by hypothesis, 7 must be added to the circumference to complete the polygon;
then 616         622 = the circumference of the polygon No 2.
70 i   -0   70
By ARTICLE 1, the double of the secant is the constantly assumed
diameter, and, by ARTICLE 2, the sum of the sines or tangents is the
constantly assumed circumference.




86              QUADRATURE OF THE CIRCLE.
99       99       198
By Case 2, the secant is 5-, then  0 X 2    -- =- the assumed
diameter.
Then, dividing the assumed circumference by the assumed diameter,
622     198   62       0    6222    4356
we have by cancellation 70              -622X 198 _-       4356
70     70      0  X198     198     1386
22
- -     3.142857, or 3-, the true ratio between circumference and diameter of the circle.
Then, by trigonometry, sec.2: tang.2:  R2: sin.2.
By Case 2, the secant is 9;9 the tangent v1y; and the radius i/i.
Then, ()2; and (-)2 -O     and (V))2- 2; and by pro-hen   70?-  4(l2-:i'Y9r0
portion 98      o      2      2
1           2
For,      X 2    4900 then, by division and cancellation, we have
4c/UO      -4900u
2     9801     2     4900     2
4900    4900 -90 X       9801  98-01  extracting the square root we
have   9801 9=1/-2 = sine No. 2; again, by trigonomefry, R2 -sin.2  cos..2
19602             2
Let 98     R2 and9       - sin.2.
9801       '    9801
Then, by subt action, -1962- 9-  -   98oo; extracting the square
root l/-'-0 we have 140 - cosine No. 2.
Now, by ARTICLE 5, the rectangle contained by the sine and the cosine,
is elqal to area of the inscribed double triangle:
140    1.      140
Thus 99 X 991/2    980l1/2.
And, by ARTICLE 6, the rectangle contained by the radius and the tangent, is equal to the area of the circumscribed double triangle;
I    1
Thus /2 X 70 - 70/
Then, by ARTICLE 7, the rectangle contained by the radius and the
sine, is a mean proportional between the inscribed and circumscribed
double triangles of half the number of sides;
s40        1                           4
Thus 980 1     /2 X  - ]/2  by cancellation, =  -801; extracting the




QUADRATURE OF THE CIRCLE.                   87
1 4           2
square root 98    we have 9, the area of the inscribed double triangle for double the number of sides.
And, again, /2 X   1/2 - - 9   area of inscribed double triangle.
BY ARTICLE 10, the rectangle contained by the rectangle of the radius and the sine, and the number of sides contained in the given polygon;
is equal to the area of the entire inscribed polygon of double the number
of sides.
For, by Case 2, the number of sides contained by the given polygon
2        2
is 311f, and the rectangle of the radius and the sine is 99; then 9X
311   6221   4356    44                       44        22
311      99   693      7   and (1/2     2; then- 7 
3.142857, or 3-, which is the true ratio of the circumference to the diameter of the given circle.
Again, by ARTICLE 8, the rectangle contained by the inscribed double
triangle, and the number of sides is equal to the area of the entire in140   -      1   43560
scribed polygon. For -98  1/  X311 7    9801  area of the inscribed
polygon.
And, by ARTICLE 9, the rectangle contained by the double circumscribed triangle and the number of sides, is equal to the area of the entire
1          1   3111        2178
circumscribed polygon.  For  0 1/2 X3117-   70    2 =  4901/2.
2178       21346578 
Now the area of circumscribed polygon 490  2-40249 1/2;
43560       21344400
Again, the area of inscribed polygon 9801 1/2 =  802490 1/2;
21346578   _   21344400         2178      1
4802490  / 2-  4802490 v/2,~4802490- -22051/
Which is the difference between the areas of the circumscribed and
inscribed polygons of the given circle whose radius is the square root
of 2.




88              QUADRATURE OF THE CIRCLE.
SUMMARY, CASE 2.
Tangent of the given arc       =.0142857 +.
Sine of the given arc    -2 =.0142849 +.
Cosine of the given arc 1940 /' = — 1.4141414 +.
Radius of the given arc   V/'    1.4142135 -+-.
Secant of the given arc       = — 1.4142857+.
By Case 2, the tangent is 70; consequently, the square which forms
the unit of comparison is (qo)2  4 900
Again, by Case 2, the rectangle contained by the radius and the
2                                        2            6222
sine is —; and the number of sides 311; therefore  ~ X 3111 --
4356    44                                        44     1
=-  6   -7 — =area of the inscribed polygon. Now,     4 
-   7                                        7   ' 4900
44   4900   215600
7X -1-             =-30800; therefore there are exactly 30800 squares
~~~~~~7  7~~~~1
in the given polygon, each of which is expressed by; then
~1         30800   44
4  X 30800 =        =9-  = 6 -= area of the inscribed polygon.
CASE 3.
In which the Inscribed and Circumscribed Polygons are Carried to
616064 - 31 - 616031 Sides.
1                  99
By Case 1, the tangent is; and the secant is7   Then, by ARTICLE 3 and (PROPOSITIONS 4 and 5), tangent (7)2 4o  and sec. (.-9)
198                                      1     198     1
X 2 =70.     Then, by division and cancellation, 4900 
7~0     1
X 198   -13860   tangent No. 3.
By (PROPOSITION 6) and ARTICLE 4, the rectangle contained by the
double of the secant No. 2 and tangent No. 3, is equal to the rectangle
contained by the square described on tangent No. 2.
For, put double of secant - A;
and tangent No. 2       = B;
and tangent No. 3       -.




QUADRATURE OF THE CIRCLE.                     89
Then, by proportion, we have
A: B:: B: C; then A X C= B XB          (A X C    B2). Substituting the numbers we have, by division and cancellation, -  X  ~
I                1
= 4 00" alud (T1T)2 X 4C —00.
4900 4and(  9)2400
By trigonometry, R2 + tan. 2 = sec. 2, substituting the numbers we
have R(-/î)2    2; T(-/u    )2   1 ^^'; then, 2 +
hiave R (i-/Z)2-;  T(1-386) 192099600;   then, 2   192099600
384199201                                                   19601
192099600 extracting the square root, we have  192099600    13860
192099600'
secant No. 3.
99   1-9602
By Case 2, secant No. 2   70 - 13860; and by Case 3, secant No. 3,
19601        19602    19601      1                1
13860        13860    1386013860       therefo  13860 has been
2
subtracted from secant No. 2, the assumed radius and      from the
1386O
assumed diameter.  Now, by hypothesis, the circumference of a circle
2       62
is to the diameter as 34 is to 1; therefore 31 times  2   or 
-y 13860  13860
must be subtracted from the circumference.
By Case 2, the tangent of the given arc is -, and, by Case 3, the
tangent of the given arc is 3860; then dividing tangent No. 2 by tan_ I              i
gent No. 3 we have, by cancellation, 70 - 13860      0 X 
70   13860            i   -
198. Therefore the tangent No. 2 has been divided into 198 equal
1
parts, each of which is equal to the'tangent No. 3, or 13860
By Case 2, the given polygon was carried to 311- sides, each of
1        2            6222
which contained two tangents each of    then    X 311= - 
10   70    7     70
1232124                      62
136, and, by hypothesis, 1-86 must be subtracted from the cir13860                     '13860
1232124      62
cumference to complete the given polygon. Then 13         1
13860     13860
1232062
123206 7 - the circumference of the polygon No. 3.
13860 




90              QUADRATURE OF THE CIRCLE.
By ARTICLE 1, the double of the secant is the constantly assumed diameter; and, by ARTICLE 2, the sumn of the tangents is the constantly assumed circumference.
19601   h   19601        39202
By Case 3, the secant No. 3 is; then       X 2
13860;       13860 X     13860
the assumed diameter.
Then, dividing the assumed circumference by the assumed diameter,
1232062-   39202   1232062    ~$0
we have by cancellation    13860     13860=X39202
13860     13860 - XO$b0      39202
1232062    862444   22
123206    862444   -22   3.142857, or 31 the true ratio between the
39202     274414    7
circumference and diameter of the given circle.
Now, by trigonometry, sec.2 tang.2:  R: sin.2, by Case 3, the: 19601              1 
secant is 13860' the tangent 3860and the radius 
_ 19601 2 384199201        1    2      1 an  
For (1386      192099600     13860)   192099600 a   (1/2)
38419920         1               2
By proportion we have 192099600   192099600    2   384199201
2       384199201
For, by division and cancellation, we have 192099600  192099600
2         000990000       2
=  9209900 X 384199201 - 384199201 extracting the square root,
wehave\       2          1   1li = sine No. 3.
we ave  384199201  19601 V    sie No 3 -Again, by trigonometry, R2 -sin.2 =cos.2.
768398402 -     2 and      2
Let 384199201 -  and 384199201 = sin..  768398402     2        768398400
Then, by subtracton, 384199201 - 384199201 = 384199201; ex
768398400         27720
tracting the square root, \ 38499209 we have 120  cosine No. 3.
'384199201'        19601
Now, by ARTICLE 5, the rectangle contained by the sine and the cosine,
27720
is equal to the area of the inscribed double triangle; thus 1-960  X
1     _ _   27720
iÎ9601 2    3841992011 2/




QUADRATURE OF THE CIRCLE.                    91
And, by ARTICLE 6, the rectangle contained by the radius and the
tangent, is equal to the area of the circumscribed double triangle; thus
I        1
J/2 x 13860 -13860V2'
Then, by ARTICLE 7, the rectangle contained by the radius and the
sine, is a mean proportional between the inscribed and circumscribed
double triangles of half the number of sides; thus, by cancellation
3841992011          0/' we have 384199201; extracting the square
4               2
root,        we have
384199201 we have 19601'
1           2
And, again, /2 X 101    /     19601  =z area of inscribed double
triangle for half the number of sides.
By ARTICLE 10, the rectangle contained by the rectangle of the radius
and the sine, and the number of sides contained in the given polygon; is
equal to the area of the entire inscribed polygon of double the number of
sides.
For, by Case 3, the number of sides contained by the given polygon
is 616031, and the rectangle of the radius and sine is 9601  then
2               1232062-  862444    44                     44
9601 X 61603 -     19601  - 137207 =-7; and (1/V)2    2; then 7
19601x            19601      137207   7'
22.- 2  - -   3.i42857, or 31, which is the true ratio of the circumference to the diameter of the given circle.
Again, by ARTICLE 8, the rectangle contained by the inscribed double
triangle and the number of sides, is equal to the area oJ the entire in27720     -            1707639120
scribedpoygon; then 3841992012 X 61603 =      3841992011/
And, by ARTICLE 9, the rectangle contained by the circumscribed
double triangle and the number of sides, is equal to the area of the entire
1                  431222
circumscribed polygon; thus 138602/  X 61603  9-00 1/2.
13860'/X    1    97020 
431222
Now, the area of the circumscribed polygon is 97020 /
165675147853622
37275006481020 




92              QUADRATURE OF THE CIRCLE.
1707639120
Again, the area of the inscribed polygon is 3841992011/2 
165675147422400
37275006481020 
165675147853622   _    165675147422400   _
37275006481020 -V2     37275006481020/2
431222               1
4372750024820 /  - 860       1/2, which is the difference between
37275006481020       86440410    '
the areas of the circumscribed and inscribed polygons of the given
circle, whose radius is the square root of two.
SUMMARY OF CASE 3.
Tangent No. 3 = 1      -.00007215007215.
13860
Sine No. 3= — 1    /2=.00007215007205.
19601
27720
Cosine No. 3  19601      1.41421356237309504,
Radius No. 3    /2    = 1.41421356237309504.
19601
Secant No. 3   13860     1.41421356237309504.
1
By Case 3, the tangent is 13860 consequently the square which
forms the unit of comparison is (138o)   192099600
Again, by Case 3, the rectangle contained by the radius and the
2                                              2
sine is 1960and the number of sides 616031; therefore 160 X
11232062   862444   44
61603-              137207   T -----  area of the inscribed polygon.
44        1       44   192099600   8452382400 
Now 7   192099600-7           1            7        1207
483200; therefore there are exactly 1207483200 squares in the given
1               1
polygon, each of which is expressed by 19; then 19
192099600'      192099600
1207483200   44     _
X 1207483200 = 19209600        -    6 =- area of inscribed polygon
for double the number of sides.




QUADRATURE OF THE CIRCLE.                      93
CASE 4.
In which the Inscribed and Circumscribed Polygons are Carried to
2414966406- - 31= 24149664031 Sides.
19601
By Case 3, the tangent is     and the secant is 3. And by
ARTICLE 3 and (PROPOSITIONS 4 and 5), tang. (1-38) — 192099600
19601        39202
and sec. 1      6 X 2= 13 0; then, by division and cancellation, we
13860        138 60
1        39202         1 iM          0 i      1
ave  92099600   13860    19209900 X    39202 ~ 543339720
tangent No. 4.
By (PROPOSITION 6) and ARTICLE 4, the rectangle contained by the
double of the secant No. 3 and tangent No. 4 is equal to the rectangle
contained by the square described in tangent No. 3.
For, put double of secant =A;
and tangent No. 3       = B;
and tangent No. 4       - C;
Then, by proportion, we have A: B:  B: C:; then A X C= B
X B=(A X C=B2).
Substituting the numbers, we have, by division and cancellation,
__W          1            1 _ _/       1   \( 1 2    1
13860 X            =4 0u 192099600 an   13860)  192099600'
By trigononmtry, R2 +- tang.2 =-sec.2, substituting the numbers, we
have rad. (/)2 =- 2.tang. (543339720) - 295218051329678400'
1             590436102659356801
then 2 +   295218051329678400 - 295218051329678400      extracting
/590436102659356801      768398401
the square root, we' have   295218051329678400 =   543339720 =
secant No. 4.
19601    768398402
By Case 3, sec. No. 3  161         39  0; and, by Case 4, sec
13860    543339720'
768398401         768398402    768398401        1
No. 4,    - 543339720;  t   543339720    543339720      543339720;




94               QUADRATURE OF THE CIRCLE.
therefore          has been subtracted from the assumed radius, and
543  from the assumed diameter.
543339720
Now, by hypothesis, the circumference of a circle is to the diameter
2            62
as 3- is to 1; therefore 31 times 5433397 or 54333      must be
54333920'    54333972
subtracted from the circumference.
By Case 3, the tangent of the given arc is 3860' and, by Case 4, the
tangent of the given arc is 54333; then, dividing the tangent No.
543339720'
1          i
3 by the tangent No. 4, we have by cancellation 13860. 54339720
=   i     $   $ X  $90 - 39202; therefore the tangent No. 3 has
been divided into 39202 equal parts, each of which is equal to the
tangent No. 4, or 543339720'
By Case 3, the given polygon was carried to 616034- sides, each of
which contained two tangents, and each tangent was 13860; then
2                 1232062   48299328124 -X   616031          7-             Y; and, by hypothesis,
13860        0       13860-     543339720         by hypo
62
fy   must be subtracted from the circumference to complete the
543339720
48299328124        67      48299328062
polygon; then 543339720      543339720    543339720     thercumference of the polygon No. 4.
By ARTICLE 1, the double of the secant is the constantly assumed diameter; and, by ARTICLE 2, the sum of the tangents is the constantly
assumed circumference.
768398401       768398401
By Case 4, the secant No. 4 is 53339720 then 543339720   2
543339720'      543339720
1536796802
543392-   - the assumed diameter.
543339720
Then, dividing the assumed circumference by the assumed diameter,
n 4829932806       1536796802   4829932806 -we have by cancellation  543339720    543339720 -    4 W$ 9 0




QUADRATURE OF THE CIRCLE.                     95
4$à$9;t0      48299328062    33809529644    3.I428, or 
X 1536796802      1536796802    10757577614 
the true ratio between the circumference and diameter of the given
circle.
Now, by trigonometry, sec.2: tang.2:: R2  S2, and, by Case 4, the
768398401                         1
secant is     3   0 and the tangent is 53       0 and the radius is
543339720                     543339720
1-/2I
or   68398401 2      590436102659356801 ad           1      2
f543339720        295218051329678400 an       543339720)
295218051329678400 and (1/2)2- 2.
590436102659356801                1
Then, by proportion, we have
The, by proprtn, we hve 295218051329678400  29521805132 -"::2
9678400     2   590436102659356801
For, by division and cancellation, we have 232
295218051329678400
590436102659356801              2             _9__0__t$0_ 9Z295218051329678400 - -9$      000Ie91$  400 X 59043610265935 -48001     59043612659356801;      extracting   the  square   root,
6801      590436102659356801
2 1
d   2       we have
590436102659356801'           768398401       = ine No. 4.
Again, by trigonometry, R2 -  sin.2 = cos.2.
1180872205318713602                      2 
Let. —                    R2; -and                     ~ sin.2.
590436102659356801;    590436102659356801 
h,. tra 1180872205318713602           2
Then, by subtraction, 590436102659356801     590436102659356801
1180872205318713600
590436102595680    then,  extracting  the  square  root
59043610265935680  1
/1180872205318713600           1086679440
i 590436102659356801- we have 768398401        cosine No. 4
Now, by ARTICLE 5, the rectangle contained by the sine and the cosine
1086679440
is equal to the area of the inscribed double triangle; thus ^68739840
1                1086679440
X 76839840112 - 590436102659356801/2'




96              QUADRATURE OF THE CIRCLE.
And, by ARTICLE 6, the rectangle contained by the radius and the
tangent is equal to the area of the circumscribed double triangle; thus
I            1
543339720    543339720/'
Then, by ARTICLE 7, the rectangle contained by the radius and the sine,
is a mean proportional between the inscribed and circumscribed double
triangles of half the number of sides; thus, by cancellation, we have
_/_  _I_,_ _- _ _ _ _44
590436102659356801   29             2    5904361026593568010
4                     2
extracting the square root, 54              we have7683840
-59043610265935 680 1      768398401'
I              2
And, again, /     X 7683984011/2 - 768398401 = the area of the
An., ag, 2 X 768398401        768398401
inscribed double triangle for half the number of sides.
By ARTICLE 10, the rectangle contained by the rectangle of the radius
and the sine, and the number of sides contained in the given polygon; is
equal to the area of the inscribed double triangle of double the number
of sides.
For, by Case 4, the number of sides contained by the given polygon
is 2414966403+; and the rectangle of the radius and the sine is
2               2                     48299328062
then     2     X2414966403=-              7
76(8398401' then 768398401 X 2414966403-  =  768398401 7 -33809529644   44                      44        22   3.I4257,
5378788807   '7 and (1/ )22; then 7        2 =T     3.142857,
or 3-; which is the true ratio of the circumference to the diameter of
the given circle.
Again, by ARTICLE 8, the rectangle contained by the inscribed
double triangle, and the number of sides is equal to the area of the entire
1086679440
inscribed polygon; then  043626      680/ X 24149664034 
2624294338586094240
590436102659356801 V'2
And, by ARTICLE 9, the rectangle contained by the circumscribed
double triangle, and the number of sides is equal to the area of the
entire circumscribed polygon; thus 54920     X 2414966403 =
2414966403-      16904764822
543339720 V    = 3803378040 V2




QUADRATURE OF THE CIRCLE.                  97
16904764822
Now the area of the circumscribed polygon is 380 78040 -/2 -9981183457874675498691254422   _
22456517068777832574480500402 v2'. 2624294338586094240
Again, the area of the inscribed polygon is 590436102659356801 /2
9981183457874675481786489600 
22456517068777832574480500401/2,
16904764822
Then, subtracting, we have 2245651706877783257448050040/-2
132841345651568820/2w which is the difference between the areas of
the circumscribed and inscribed polygons of the given circle, whose radius is the square root of two.
SUMMARY, CASE 4.
Tangent No. 4 is 543339720.0000000000184046916356492398.
Sine No. 4 is 76839840    =.000000000184046916356492398.
1086679440
Cosine No. 4 is 76 840      1.41421356237309504.
78398401i
Radius No. 4 is  1/     = 1.41421356237309504.
768398401
Secant No. 4 is 543339720  1.41421356237309504.
By Case 4, the tangent is 543339720 and the square, which formal
the unit of comparison, is (54339720)  95218051329678400
Again, by Case 4, the rectangle contained by the radius and the
2
sine is 76839840' and the number of sides 2414966403; therefore
2                     48299328062    33809529644   44
768398401' 2-14966403=-     768398401 - 5378788807 - 7 --
area of the inscribed polygon.
44            1            44    295218051329678400
7   295218051329678400 — 7 
12989594258505849600
2 9 57  --  1855656322643692800; therefore there are
7
7




98              QUADRATURE OF THE CIRCLE.
exactly 1855656322643692800 squares in the given polygon, each of
which is expressed by                  then
whichis expressed by 295218051329678400the 295218051329678400
X  18556322643692800            44
X 18556322643692800 -— 291801329678400                    area
of inscribed polygon for double the number of sides
CASE 5.
In which the Inscribed and Circumscribed Polygons are Carried to
3711312645287385606 - 3- = 37113126452873856031 Sides.
i. 1     T          768398401
By Case 4, the tangent is 543339720 and the secant is 547339720
5 3   254333
By ARTICLE 3 and (PROPOSITIONS 4 and 5), tangent 54339720
1                  768398401        1536796802
and sec.          X 2
295218051329678400'     sec543339720         543339720
Then, by division and cancellation, we have
P$ $$$      _   ___   1
X 1536796802- 835002744095575440      talent 
By (PROPOSITION 6) and ARTICLE 4, we have, by division and cancel543339720 X $00W4409$        440    295218051329678400'
which equals the square on the tangent as above.
By trigonometry, R2 +  tang.2 = sec.2. Rad. =  /2;  and tang. =
835002744095575440; rad. square = (/)2   2, and tang. square =
1
697229582647141045327149384731193600'
Then, adding to the square of the radius and extracting the square
-, t o,, t'1394459165294282090654298769462387z01
root of th  same,  697229582647141045327149384731193600  we
1180872205318713601
ve scant No. 5  835002744095575440'
768398401              118087220531 -By Case 4, secant No. 4 is 54333  which equals 835002744095 -543339720' whc qas835 —06~74409 -



QUADRATURE OF THE CIRCLE.                    99
8713602
740   subtracting from  this secant No. 5 above, we find that
575440 
8350027441095575440 has been deducted from the assumed radius and
8350027449557540 from the assumed diameter.
835002744095575440
Now, by hypothesis, the circumference of a circle is to the diameter
as 3- is to 1; therefore        95575440 must be deducted from
~~~7;835002744095575440
the circumference.
By Case 4, the tangent of the given arc is 54339720' and, by
543339720'
Case 5, the tangent No. 5 is 8350027495575440; then, by division
83500274409550754405
1        $$$000ee2f4as0ee$$y40
and cancellation, we have.., X -                      536 -796802; therefore the tangent No. 4 has been divided into 1536796802
equal parts, each of which is equal to tangent No. 5, or 83500274409
5575440'
By Case 4, the polygon was carried to 24149664031 sides, each of
which contained two tangents, and each tangent was 543339720; then
2     X 241494829932806                74226252905747712124
5433339                    543339720     835002744095575440
and by hypotesis 8350027495575440 must be subtracted from the
)835002744095575440
-.,        -,  *   -,.,. ~74226252905747 -circumference to complete the polygon, which gives 8350027440955 -712062;7540     which, by ARTICLE 2, is the assumed circumference.
75440
1180872205318713601.
By Case 5, the secant No. 5 is 835002744095575440 X 2 -
2361744410637427202 
835002744095575440; which, by ARTICLE 1, is the assumed diameter. Then, dividing the assumed circumference by the assumed di



100            QUADRATURE OF THE CIRCLE.
7422625290574771206     8$002 -ameter, we have, by cancellation 0742262529057477l20  X 26  -
$$000M440ff7  440W    2361744 -4409$4A40 A    7422625290574771206     22     4       o   _
410637427202    2361744410637427202     7  3.         or 
the true ratio between the circumference and diameter of the given
circle.
Now, by trigonometry, sec. 2: tang.2: R2: sin. 2, and, by Case 5,
the scant  1180872205318713601 a                       1
835002744095575440 ad the             83500274 -4095575440; and the radius is 1/2'
4095575440'
1394459165294282090654298769462387201
697229582647141045327149384731193600     t
697229582647141045327149384731193600 
i                      then extracting the
1394459165294282090654298769462387201    then extract    t
square root, we have, sine No. 5 =  80872205318713601/2
Again, by trigonometry, R2 - sin. 2 = cos. 2:
2788918330588564181308597538924774402      
1394459165294282090654298769462387201 =     2 an
1394459165294282090654298769462387201- sin. 2; then subtracting
the square of the sine from the square of the radius, and extracting
/2788918330588564181308597538924774400
the square root, i1394459165294282090654298769462387201' we
e c   N  1670005488191150880
e  180872205318713601'
w by A   E, c   1670005488191150880.           1
Now, by ARTICLE 5, COS. 1180872205318713601 X  i118087220 -531871360i -/2' we have the area of the inscribed double triangle =
1670005488191150880
1394459165294282090654298769462387201 12
1                    1
And, by ARTICLE 6, rad. I/- X        1
And, by ARTICLE 6, rad. o/ X 835002744095575440 - 83500274 -4095575440l/2, we have the area of the circumscribed double triangle.




QUADRATURE OF. THE CIRCLE.                101
Then, by ARTICLE 7,          2V1/2 X 13944591652942 -$$00$40                  13944591652942-_..;829by division and cancellation, and ex82090654298769462387201
4
tracting the square root  1394459165294282090654298769462387201'
we have the area of the inscribed double triangle for half the number
of sides -         2
1180872205318713601
I   1 
Again, /2 X 1180872205318713601 -- 1180872205318713601i  2
By ARTICLE 10, and Case 5, the rectangle of the radius and the sine
11808722387360  X  3711312645287385603-      the number
1180872205318713601
74226252905747712062    44
of sides in the given polygon= 118087220531871360         and
222; then      2    2 =3.142857, or 31, which is the true
(l1//)2~ 2; then -q- -- 2.- ~ratio of the circumference to the diameter of the given circle.
1670 -Again, by ARTICLE 8, the inscribed double triangle  944591652
005488191150880
428209065429876946238720V/,   multiplied by the number of the
sides in the given polygon, 3711312645287385603+, is equal to the
o t       619791248602315197252257308751 -area of the ascribed polygon  139445916529428209065429876946 -1514480 
2387201iv2 -And, by ARTICLE 9, the circumscribed double triangle 8350027441
0955754401/2, multiplied by the number of sides, 37113126452873 -856031, is equal to the area of the circumscribed polygon 
2597918851701169922
58450192086690280801/2; consequently the area of the circumscribed
362269175344549323026296241098136686021852382055 -polygon    815064060684965813044000977736724847285814034930 -14457622                                       3622691753445 -1604080 /2 and the areaofthe inscribed polygon = 815040606849 -4932302629624109813668576206049688502758400
65813044000977372484728581403493010 00; then, subtract658130440009777367248472858140349301604080 I / 




102             QUADRATURE OF THE CIRCLE.
25979188517011699222
ing, we have 81506406068496581304430097773672484728581403493 -01604080V/2 =313737305594147155633890579843041640/2' whh
is the difference between the areas of the circumscribed and inscribed
polygons of the given circle, whose radius is the square root of two.
SUMMARY, CASE 5.
Tangent No 5 i 83500274095575440.0000000000000000001 -197600854693165860264915019144180288.
Sine No. 5 is 1188722 3    18736 2.0000000000000000001 -197600854693165860264915019144180288.
1670005488191150880
Cosine No. 5 is 1108  053181    1   1.4142135623730950488 -01688724209699276.
Radius No. 5 is y/2               = 1.4142135i23730950488 -01688724209699276.
1180872205318713601
Seant No.     835002744095575440     1.4142135623730950488 -01688724209699276.
By Case 5, the tangent is 835002744095575440 and the square,
/        I         s2
which forms the unit of comparison, is (835o2744095575440)2 
697229582647141045327149384731193600'
Again, by Case 5, the rectangle contained by the radius and the sine
is 18722038736         and the number of sides 37113126452873 -1180872205318713601'
74226252905747712062
856031; multiplying these together we have 180872205318713601
51958377034023398444    44
8266105437230995207     7
44                       1                     44
Now, 7 _     697229582647141045327149384731193600    7 




QUADRATURE OF THE CIRCLE.                  103
697229582647141045327149384731193600     3067810163647420599 -1,                                         7
43945729281172518400 - 4382585948067743713484938989738931 -200; therefore there are exactly 4382585948067743713484938989738 -931200 squares in the given polygon, each of which is expressed by
~1          -_.~ then, multiplying  the
697229582647141045327149384731193600 
4382585948 -number of squares by the value of each square, we have 6972958528 -06774371348938989738931200    44
471410453271493847311936007       6       area of the inscrbed
polygon for double the number of sides.
CASE 6.
In which the Inscribed and Circumscribed Polygons are (,arried to
87651718961354874269698779794778624031 Sides.
By Case 5, the tangent is 835002744095575440 and the secant is
1180872205318713601
835002744095575440'
Then, by ARTICLE 3 and (PROPOSITIONS 4 and 5), tang.2 =
and the double of the secant
697229582647141045327149384731193600 ad the d     e   he. 2361744410637427202!S
835002744095575440 
Then, by division and cancellation, we have 17206306373463926.3984455073299118880 = tangent No. 6
By (PROPOSITION   6   and  ARTICLE   4) and   by cancellation,
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _     1
835002744095575440. \x m     W    ^     W    ^
697229582647141045327149384731193600' and the square described on the tangent equals the same. (See above.)
By Case 1, the cosine is î; then (7)2 X 2 -1 -99.
Multiplying this by tangent No. 2, we have 9- _ secant No. 2.
By Case 2, the secant is -9; then (99)2 X 2  1 = 19601




104              QUADRATURE OF THE CIRCLE.
19601
Multiplying this by tangent No. 3, we have -860 - secant No. 3.
19601
By Case 3, the secant is 13860  then (19601)2 X 2 - 1  768398 -401.
768398401
Multiplying this by tangent No. 4, we have 54333970   secant
No. 4.
768398401
By Case 4, the secant is 5339820; then (768398401)  X  2 -
543339720;
1 = — 1180872205318713601.
1180872205318713601
Multiplying this by tangent No. 5, we have 835002744095575440
= secant No. 5.
1180872205'318713601
By Case 5, the secant is;3002 740557       then (1180872205 -318713601)2 X 2 - 1 - 2788918330588564181308597538924774401.
2988918330588564181 -Multiplying this by tangent No. 6, we have 197206063734639263 -1972063063734639263 -308597538924774401
984455073299118880    se  t No
By Case 2, the numerator of the secant is 99, and the radius is the
square root of two.
i     1
Then, by division, we have /2 X        V =99 1/ - sine No. 2.
By Case 3, the numerator of the secant is 19601.
1       1
Then, by division, we have 1/2 X 19601  19 12 =    ine No. 3.
By Case 4, the numerator of the secant is 768398401.
I            1
Then, by division, we have /2 X 768398401   7683984012 
sine No. 4.
By Case 5, the numerator of the secant is 1180872205318713601.
Then, by division, we have 1/2 X 18087220538713601
1180872205318713601/ = sine No. 5.
And, by Case 6, the numerator of the secant is 2788918330588564 -181308597538924774401.




QUADRATURE OF THE CIRCLE.                 105
Then, by division, we have 27889183305885641813085975389247 -74401/2     sine No. 6.
99
By Case 2, the sec. is -, and the square described on the radius is 2.
70    140
Theu, by division, we have 2 X 99 - 99   cosine No. 2.
y-    99
19601
By Case 3, the secant is 13860
13860   27720
Then, by division, we have 2 X 19601 - 19601  cosine No. 3.
768398401
By Case 4, the secant is 543339720
543339720   1086679440
Then, by division, we have 2 X 768398401  68398401    cosine
No. 4.
1180872205318713601
By Case 5, the secant is 835002744095575440
835002744095575440    1670005 -Then, by division, we have 2 X 1180872205318713601  1180872 -488191150880 
205318713601 - chose No
2788918330588564181308597538924 -Ad, by Cse 6, th s nt s 972063063734639263984455073299 -774401
118880'
391412612746927852796891014659823 -Then, by division, we have 278891833058856418130859753892477 -7760
4    -- =cosine No. 6.
4401
1180872205318713601    278891833 -y Case 5, the scant No. 5 is 835002744095575440 — 197206306 -0588564181308597538924774402
3734639263984455073299118880'
27889183305885641813085975 -And, by Case 6, the secant No. 6 is 1972063063734392639844550
19720630637346392639844550
38924774401
73299118880'
Therefore                                       has been de1972063063734639263984455073299118880




106              QUADRATURE OF THE CIRCLE.
ducted from  the assumed radius, and twice that amount from the
assumed diameter.
6 7
Therefore, by hypothesis, 1972063063734639263984455073299118880
must be deducted from the circumference to complete the polygon.
By Case 5, the tangent is 835002744095575440
And, by Case 6, the tangnt is197206306373463926398445507329 -9118880'
Then, by division and cancellation, we have 2361744410637427202;
therefore the tangent No. 5 has been divided into 236174441063742 -7202 parts, each of which is equal to tangent No. 6.
By Case 5, the polygon was carried to 37113126452873856031 sides,
each of which contained two tangents of 83527  95750
835002744095575440'
2                                   74226252 -Then                     X 37113126452873856031-     6
835002744095575440 x                       -83500274 -90574771206-    175303437922709748539397559589557248124
4095575440-     1972063063734639263984455073299118880    an
by  hypothesis, 1972063063734692623984455073299118880       b
subtracted from the circumfcrence to complete the polygon.
17530343792270974853939755958955724812-             62
1972063063734639263984455073299118880      197206306 -175303437922709748539397559 -3734639263984455073299118880     170303792270974853939 5507 -197206306373463926398445507 -589557248061
5895572 480 -= the circumference of the polygon No. 6.
3299118880
2788918330588564181308597538924774 -By Case 6, secant No. 6, 1972063063734639263984455073299118 -401         5577836661177128362617195077849548802
880 X       1972063063734639263984455073299118880        e as
sumed diameter.
Then, by ARTICLES 1 and 2, dividing the assumed circumference by
175303437922709748 -the assumed diameter, we have by cancellation a.




QUADRATURE OF THE CIRCLE.                   107
53939755958955724806k      92020^$2~92-29~,0~990 -g 9~,44t$$0~ 99t $y30 ~  X 5577836661177128362617195077849544 -$0 _ 17530343792270974853939755958955724806 _ 22          3i42
802 `  5577836661177128362617195077849544802     -   7
857, or 3-, the true ratio between the circumference and diameter of
the given circle.
By Case 2, the sine of the given arc is F1/2, and the given radius
is /.2 -1        2
Then, by ARTICLE 7, -// X 9-/2       = -9  - area of the inscribed
double triangle for half the number of sides.
By Case 3, the sine is 19601</2.
1           2
Then, by ARTICLE 7, 1/  X   9l1960-2         i = area, as above.
By Case 4, the sine is 768398401
~1          2
Then, by ARTICLE 7, /2 X   68398401V = 768398401       aea as
above.
By Case 5, the sine is 110872  38713601/
12_                   2
Then, by ARTICLE 7, 1/2 X                               _   2
Then, by ARTICL  7,  X 1180872205318713601/2    11808 --72205318 1   area as above.
72205318,13601
By Case 6, the sine is
2788918330588564181308597538924774401 1/2'
Then, by ARTICLE 7, multiplying the radius by the sine, we have,
2788918330588564181308597538924774401      area  f the inscribed
double triangle for half the number of sides.
2
By   ARTICLE  10,                     2X
By  ARTICLE   10, 2788918330588564181308597538924774401 
17530343792270974-9477862403
87651718961354874269698779794778624031- -     788918330588534 --7 27889183305885641 -



108             QUADRATURE OF THE CIRCLE.
853939755958955724806     4                        44       22
and 21/>2 -2    then    *
81308597538924774401      7            nd   then           7
3.142857, or 31; which is the true ratio of the circumference to the
diameter of the given circle.
By Case 2, the difference between the inscribed and circumscribed
1
polygons is 205/2; and, by Case 3, the difference between the inscribed and circumscribed polygons is 86440410 /2
Then, dividing the difference of Case 2 by the difference of Case 3,
1   /                      1          86440410,_
2e have 2 1/2 -   86440410/          2205/  X    1       2
39202, which shows that the difference between the polygons has diminished 39202 times.
By Case 3, the tangent is 13; and, by Case 4, the tangent is
13860
543339720'
Then, dividing tangent No. 3 by tangent No. 4, we have 13860! i              543339720
1             X 543339720    39202; therefore the tangent No.
3 is 39202 times as large as tangent No. 4. Consequently the difference between the tangents Nos. 3 and 4 is exactly the same as the difference between the differences of the inscribed and circumscribed
polygons of Cases 2 and 3.
By Case 3, the difference between the inscribed and circumscribed
1
polygons is 864041p1/2; and, by Case 4, the difference between the
inscribed and circumscribed polygons is 13284
Then, dividing the difference of Case 3 by the difference of Case 4,
we have                11 
864404102    + 132841345651568820/2- 864404101/2 X
132841345651568820
1-1/2 -          -= 1536796802; which shows that the difference between the polygons has diminished 1536796802 times.




QUADRATURE OF THE CIRCLE.                  109
1
By Case 4, the tangent is 543339720 and, by Case 5, the tangent is
1
835002744095575440'
Then, dividing tangent No. 4 by tangent No. 5, we have
543339720
*. 835002744095575440 = 1536796802; therefore the tangent No. 4
is 1536796802 times as large as tangent No. 5; consequently the difference between the tangents Nos. 4 and 5 is exactly the same as the difference between the differences of the inscribed and circumscribed
polygons of Cases 3 and 4.
By Case 4, the difference between the inscribed and circumscribed
polygos is 1328434561568820; and, by Case 5, the difference between the inscribed and circumscribed polygons is 313737305594147
1555633890579843041640/2'
Then, dividing the difference of Case4 by the difference of Case 5,
we have by cancellation 1$:;    1/2 X   t$$9$-$12 X;ff 9W O 4M W    t0 /  =  2361744410637427202; therefore
the difference between the inscribed and circumscribed polygons of
Case 4 is 2361744410637427202 times as large as the difference between the inscribed and circumscribed polygons of Case 5.
83500274409557440;
1
By Case 5, the tangent is 83500274409557440; and, by Case 6, the
tangent is                 i
1972063063734639263984455073299118880'
Then, dividing tangent No. 5 by tangent No. 6, we have by cancellation   000 
s_ =2361744410637427202; therefore the tangent No. 5 is 236 -1744410637427202 times as large as tangent No. 6; consequently the
difference between the tangents Nos. 5 and 6 is exactly the same as the




110              QUADRATURE OF THE CIRCLE.
difference between the differences of the inscribed and circumscribed
polygons of Cases 4 and 5; and, consequently, the difference between
the differences of the inscribed and circumscribed polygons of Cases 5
and 6 will be the same as the differences between the tangents Nos. 6
and 7; thus the tangent No. 6 is  106306373463926398445507329
197206306373463926398445507329 -9118880 and the tangent No. 7 is
9118880 ad the tangent No 7 is 109998456550523587637197163135 -91640029823644051645720435317842392159581760'
Then, dividing tangent No. 6 by tangent No. 7, we have by cancellation 5577836661177128362617195077849548802; therefore the tangent No. 6 is 5577836661177128362617195077849548802 times as
large as tangent No. 7.
Again, the difference between the inscribed and circumscribed polygons of Case 5 is
gons of313737305594147155633890579843041640 1/; then
dividing this by the difference between tangents Nos. 6 and 7, we have
(313737305594147155633890579843041640V/2) X (5577836661177 -128362617195077849548802); which equals the difference between
the areas of the inscribed and circumscribed polygons for Case 6.
SUMMARY, CASE 6.
Tangent No. 6 is
1972063063734639263984455073299118880'
Sine No. 6 is
27889183305885641813085975389247744011/2
394412612746927852796891014659823760
Cosine No. 6 is
2788918330588564181308597538924774401'
Radius No. 6 is 1/2.
2788918330588564181308597538924774401
Secant No. 6 is
1972063063734639263984455073299118880'
NOTE.-It will no doubt be observed by the student that, in the Summary for Cases 6 and 7,
the sine and tangent are not extended in decimals, for the reason that the space could be better
occupied with matter more important to the subject. The student will find it a very agreeable
pastime to verify the result in these Cases for himself, and it was partly with that view that
they were omitted. The sine and tangent in Case 6 should agree together for 72, and in Case 7
for 144, decimal places.




QUADRATURE OF THE CIRCLE.                111
The cosine, radius, and secant have also been omitted, partly for the same reason, and partly
because the radius with which they should both agree will be found carried out in the Second
Part of this book; for Case 6 they should agree like the sine and tangent with each other to 72
decimal places, and for Case 7 to 144 decimal places.
By Case 6, the tangent is 1972063063734639263984455073299118 -8  and the square which forms the unit of comparison is the square
of the tangent, which is
of the tangent, which i38890327273464518838061949606599216563 -67162016449544406781628584372454400'
Again, by Case 6, the rectangle contained by the radius and the sine
2
is                                     and the number of sides
2788918330588564181308597538924774401              o sides
87651718961354874269698779794778624031.
1753034379227097485393975 -Multiplying these together, we have  27889 588564    859
278891833058856418130859 -5958955724806-   44                                       1
7538924774401 -7 * 388903272734645188380619496065992165 -44   3889032727346 -6367162016449544406781628584372454400    7 X              1
4518838061949606599216563671620164495444067816285843724544 -00   17111744000324388288747257826903655288015512872377995 -7
3898391657712387993600
3898391657712387993600 = 244453485718919832696389397527195 -07554307875531968564842627379673198284800; therefore there are
exactly 244453485718919832696389397527195075543078755319685 -64842627379673198284800 squares in the given polygon, each of
which is expressed hby                                    1
which is expressed  388903272734645188380619496065992165636 -7162016449544406781628584372454400'
Then, multiplying the number of squares by the value of each square,
244453485718919832696389397527195075543078753196856 -we have
388903272734645188380619496065992165636716201644954 -4842627379673198284800   44
4406781628584372454400       - 7  67 =- area of the inscribed polygon for double the number of sides.




112              QUADRATURE OF TIE CIRCLE.
CASE 7.
In which the Inscribed and Circumscribed Polygons are Carried Io
48890,697143783966539277879505439015108,615751063937.12968525 -4759346396569603    S Sides.
By Case 6, the tangent is 1972030637346392639844550732991 -2788918330588564181308597538924774401
18880 and th scant s 1972063063734639263984455073299118880'
Then, by ARTICLE 3 and (PROPOSITIONS 4 and 5), tangent2
3889032727346451883806194960659921656367162016449544406781 -557783666117712 -628584372454400' ad the double of the scant is 197206306373463 -8362617195077849548802
9263984455073299118880'
Then, by division and cancellation $$9$$0                   -
06099t[6<65($0~6~I6 6~I$2   00II9<I?$^I$     W0440 X 557783666 -$4$            ww$ww 44$0   0          ______________`_1
1177128362617195077849548802 we have 1099984565505235876371 -tan9716313591640029823644051645720435317842392159581760       tan
gent No. 7.
By (PROPOSITION 6) and ARTICLE 4, and by cancellation,
5577836661177128362617195077849548802                         1
1972063063734639263984455073299118880 X    109998456550523587 -63719716313591640029823644051645720435317842392159581760 -
3889032727346451883806194960659921656367162016449544406781 -628584372454400 - square described on the tangent No. 6.
2788918330588564181308597538924774401
y Cse 1, the  ant  972063063734639263984455073299118880'
Then (2788918330588564181308597538924774401)2 X 2 -      1 
1555613090938580753522477984263968662546864806579817762712 -6514337489817601.




QUADRATURE OF THE CIRCLE.                 113
1555613090938580753 -bultiplying this by tangent No. 7 we have 1099984565505235876 -5224779842639686625468648065798177627126514337489817601
3719716313591640029823644051645720435317842392159581760
secant No. 7.
By Case 7, the numerator of the secant is 15556130909385807535 -224779842639686625468648065798177627126514337489817601, and
the given radius is the 1/2
Then, by division, we have 155561309093858075352247798426396 --/2 = sine No. 7.
86625468648065798177627126514337489817601     = sne N   7
By Cae 7,       155561309093858075352247798426396866 -By Case, t  scant s 109998456550523587637197163135916400 -25468648065798177627126514337489817601
and the radius is the
29823644051645720435317842392159581760   ad the radius s the
square root of two.
2199969131010471752743943262718328 -Then, by division, we have
155561309093858075352247798426396.
0059647288103291440870635684784319163520
86625468648065798177627126514337489817601    cosine No 7
278891833058856418130859753892 -By Case 6, the scant No. 6 s 197206306373463926398445507329 -4774401     15556130938580753522477984263968662546864806579 -9118880 - 109998456550523587637197163135916400298236440516 -8177627126514337489817602
45720435317842392159581760'. 1555613090938580753522477 -And, by Case 7, the secant No. 7 is 1555309  5873      77
10999845655052358763719716 -9842639686625468648065798177627126514337489817601
313591640029823644051645720435317842392159581760'
Therefore                                                1
10999845655052358763719716313591640029823644051 -64572043531784231 60 has been deducted from the assumed
645720435317842392159581760
radius, and twice that amount from the assumed diameter.
8




114             QUADRATURE OF THE CIRCLE.
62
Therefore, by hypothesis, 1(99984565505235876371971631359164 -must be deducted
0029823644051645720435317842392159581760 
from the circumference to complete the polygon.
By Case 6, the tangent' is
y Case 6, th  tangent is 1972063063734692626398445507329911
8880'
And, by Case 7, the tangent is 109998456550523587637197163135
91640029823644051645720435317842392159581760.
Then, by division and cancellation, we have 55778366611771283626 -17195077849548802; therefore the tangent No. 6 has been divided
into 5577836661177128362617195077849548802 parts, each of which
is equal to tangent No. 7.
By Case 6, the polygon was carried to 876517189613548742696987 -7979477862403+ sides, each of which contained two tangents of
1972063063734639263984455073299118880
Then                 2X 87517189613 -1972063063734639263984455073299118880  876517189613 -17530343792270974853939755958 -5487426969877979477862403- =  1972063063734639263984455073 -9557248062   977813942875679330785557590108780302172315021 -299118880    109998456550523587637197163135916400298236440 -278742593795095186927931392124
51645720435317842392159581760 -62
d, y ypotesis, 109998456550523587637197163135916400298 -must be subtracted from
23644051645720435317842392159581760 must be subtracted from
the circumference to complete the polygon.
977813942875679330785557590108780302172315021278742 -lus 109998456550523587637197163135916400298236440516457 -59370509518692793139212    __                           62
20435317842392159581760    1099984565505235876371971631359 -



QUADRATURE OF THE CIRCLE.                115
977813942875 -1640029823644051645720435317842392159581760 ~109998456550 -6793307855575901087803021723150212787425937050951869279313 -5235876371971631359164002982364405164572043531784239215958 -92062
6= the circumference of the polygon No. 7.
1760
155561309093858075352247798426 -By Case 7, the secant No. 7 is 109998456550523587637197163135 -39686625468648065798177627126514337489817601         311122 -91640029823644051645720435317842392159581760     2- 109998 -6181877161507044955968527937325093729613159635525425302867 -4565505235876371971631359164002982364405164572043531784239 -4979635202 
2159581760  — the assumed diameter.
Then, by ARTICLES 1 and 2, dividing the assumed circumference
97781394287567933 -by the assumed diameter, we have by cancellation 10999845655052358 -0785557590108780302172315021278742593705095186927931392062
763719716313591640029823644051645720435317842392159581760
10999845655052358763719716313591595451753047924035846398 -311122618187161507044955968527937325093 72961315963552542 -694780218565561760   9778139428756793307855575901087803021 -53028674979635202     311122618187161507044955968527937325 -7231502127874259370509518692793139206  _22       4, o-  3,
0937296131596355254253028674979635202    7            '    "7
the true ratio between the circumference and diameter of the given
circle.
By Case7, the sine of the given arc 1555613099385807535224779 -8426396866254686480657981776271265143374898176011/2.
Then, by ARTICLE 7, multiplying the radius by the sine, we have
2
1555613090938580753552247798426396866254686480657981776271 -26514337489817601 = area of the inscribed double triangle for half
the number of sides.
2
By ARTICLE 10 155561309093858075352247798426396866254686 -



116              QUADRATURE OF THE CIRCLE.
48065798177627126514337489817601 X 48890697143783966539277 -9778 -879505439015108615751063937129685254759346396569603 -= 1     6 -1556 -13942875679330785557590108780302172315021278742593705095186 -1309093858075352247798426396866254686480657981776271265143 -92793139206f    4            444                 22   3.42857
37489817601 7- and (V/)2        2; then   -2      7      42857,
or 31, which is the true ratio of the circumference to the diameter of
the given circle.
SUMMARY, CASE 7.
Tangent for Case 7 is 10999845655052358763719716313591640029 -823644051645720435317842392159581760"
Sine for Case 7 is 15556130909385807535224779842639686625468 -648065798177627126514337489817601V'2'
219996913101047175274394326271832800596 -Cosine for Case 7  55561309093858075352247798426396866254 -47288103291440870635684784319163520
686480657981776270126514337489817601
Radius for Case 7 is V2'.
155561309093858075352247798426396866254 -Secant for Case 7 is
109998456550523587637197163135916400298 -68648065798177627126514337489817601
23644051Ib4572043531.7842392159581760
NOTE.-The difference between the inscribed and circumscribed polygons for Case 7 has been
omitted, for the purpose of allowing the student to make the calculation for himself; the number of squares contained in the given polygon, together with the unit of comparison, has also
been left out, as the calculation is made in the same manner as in the previous cases. The
decimals of the Summary have also been omitted, as in Case 6, and for the reason mentioned
therein, viz.: that the decimals for the cosine, the radius, and the tangent will be found in
Part Second of this volume. Indeed, the whole of Cases 6 and 7 would have been omitted had
it not interfered with the original design of the author, as the other cases are sufficient to enable any one to get a thorough and comprehensive knowledge of the subject, though the curious
and the adept are at liberty to extend the work ad infinitumn.




PROOFS.                           117
PROOFS.
By Case 1, the -th of the square described upon the radius was
50
deducted for the purpose of finding the value of the given arc, and
the square root of the -th so deducted is the sine of the given arc;
50
49
and the square root of the remaining 50ths is the cosine of the given
1
arc; for, substituting the numbers, we have (1/2)2 = 2, and 50 of 2 =
50
2    1
-2f5; and the t2e =B - the sine of the given arc.
50 and      5      40            4
Again, 2    50 and 50    1 -   49; andthe   4   =7 _ cosine of
25      25    25   25            25    5
the given arc.
7   1    7
Then, by ARTICLE 5, 5 X        =25   area of the inscribed double
triangle.
Now, by Case 1, there are 22 double triangles contained in the given
polygon.
7          154
Then, -  X 22      5 -   area of the entire inscribed polygon.
By trigonometry, we have cos.: sin.: rad.: tang.
71             1         1
Substituting the numbers, we have::: 1/2:    1/2; for - X
=   &           7 1
1/2-=51/2.
Then, by division and cancellation, we have  /2 -     1/2 X
-   /2 = tangent of the given arc.
1              2
Then, by ARTICLE 6,,/2 X 1/2 =      - area of the circumscribed
double triangle.
Now, by Case 1, there are 22 double triangles contained in the given
polygon.
2          44
Then ~ X 22     -     area of the entire circumscribed polygon.




118                         PROOFS.
44   1100      154   1098
Now==        5and
Now 7=     175 and 25-175
1100   1078    22
Tlien, by subtraction,   -  175  -       thé difference between
the inscribed and circumscribed polygons.
22      1100     22     Jee    22 22  1
Again,           5            1100-10 0x50; therefore the
difference between the inscribed and circumscribed polygons is  th of
50
the circumscribed polygon, and is equivalent to the -th of the square
50
described upon the radius; consequently; deducting 5-th friom  the
square described upon the radius is equivalent to deducting  th from
the area of the circumscribed polygon.
By trigonomnetry, R2 - tan.2 = sec.2.
1he 2
Substituting the numbers, we have (y2)2   2 and (/2)2 =
2 __/ 10       Iio9
then 2 +  2 = 100 and./      100  = scant of the given arc.
49    49       g49     7
Deducting the cosine from the secant of the given arc, we have
10   50,7   49       50   49    1
1 =50    and -=;; then 5- - -     5-  = the difference between
7    35'    5    35'     35   t3h5  35
the cosine and thé secant of the given arc; then, dividing this differ1    10    1     '_ 1!
ence by the secant, we have by cancellation 35. 7    X  - t   5-0
35    7         10   50'
therefore the difference between the cosine and secant is equivalent to
the lth of the secant.
50
Consequently, deducting 5th froin the square described upon the
50
radius is equivalent to deducting the 0th from  the secant of the
given arc.
1100
Again, by Case 1, the circumscribed polygon is 1-75; and the inscr l; th               by division ad cancellation, 1078  78
scribed polygon is *-f; then, by division and cancellation,  -
175O                                 O175




PROOFS.                        119
1100   1078  `x:$    1078   49
75     X       X 110 = 1100 = 50; therefore the inscribed polygon
49
is -0 of the circumscribed polygon.
50
By Case 2, the sine of the given arc is 99l/2; and the radius is 1/2;
1             2
then   1/2 X 1/2 =      area of the inscribed double triangle; and
99
2            622-2  4356   44
there are 311. sides:; then  X 311  =-       5-      = area of
the inscribed polygon.
154
But, by Case 1, the area of the inscribed polygon is 25
44   1100 '    154   1078
Now      -     *and-     -
N 7=175;       25 — 175'
1078   1100   1078
Then, by division and cancellation, we have -5  ' 175 -  X
175    1(5
i..~$~  `1078  49
10 0=1100 -50.; therefore the area of the inscribed polygon for
49
Case 1 is -  of the area of the inscribed polygon of Case 2; but
50
49
the area of the inscribed polygon for Case 1 is 0 of the area of
the circumscribed polygon for Case 1; therefore, by (Axiom 1), the
area of the inscribed polygon of Case 2 is the same as the area of the
circumscribed polygon of Case 1.
But, by Case 1, the oth of the area of the circumscribed polygon
50
was deducted; therefore, by Case 2, the same 50th has been added on
again.
Again, by Case 1, the sine is 5, and there are 44 sines contained in
the given polygon; and, by Case 2, each of these sines is divided into
14 parts, each equal to tangent No. 2; therefore there are 44 X 14
1
616 tangents contained in the polygon for Case 1, each equal to 7.
But, by hypothesis,  _ must be added to complete the polygon;
616   62   622.   4356   su  of     tangents.
then 0        — +  =          sum of the tangents.




120                         PROOFS.
1. 1
Again, by Case 2, the tangent is 0 and the sine is -1/2; therefore
70               99
1          1i
the tangent is to the sine as 70 is to the  9/2.
Now, by Case 2, the sum of the tangents is 4, and the sum of
4356,_
the sines 69  / 2
4356    1    4356       1
Then, by proportion, -49:::-'/2: 99/2
4356    1:   44        4356        I    44.
4or 40 X 991/2      — 1/   ad /693 /   X 70    491/2
Again, by Case 1, the tangent is -1/2; and there are 22 sides, each
1i           44 
of two tangents; then /-/2 X 44       2    the sum of the tangents
for Case 1.
And, by Case 2, the sine is -1/2, and there are 3111 sides, each of
two sines; then 9912 X 622    9-  9 2   = the sum of the sines for
Case 2.
44       4356        62242      4356
Nowy v2         -93 1/2, and 9  /2    693~ /2; therefore the sum
of the tangents for Case 1 has become the sum of the sines for Case 2.
Again, by Case 1, the sine is 5, and there are22 sides, each of two
I        44
sines; then - X 44-       sum of the sines for Case 1.
1
And, by Case 2, the tangent is -, and there are 3114 sides, each of
two tangents; then 70 X 622-2 -4
70  '   70
44   616           62
But -; therefore  must be added to the circumference to
u     750'         70
î616  62    6222 f
complete the polygon; thus       -    67     therefore the sum of
0    70    70 
the sines for Case 1 has become the sum of the tangents for Case 2.
7                             99
Again, by Case 1, the cosine is 5; and, by Case 2, the secant is -.
'  ~~~~~~~~~~~~70'




PROOFS.                          121
7   98        99   98    1             I
Now   -             0; then 7 - 0-  0; therefore  has been added to
-- 70'     70    70    70'           70
1    1    98
the cosine; but  - is - of 70; therefore the diameter of the circle has
been increased its 9th part.
98
616
Again, by Case 1, the sum of the tangents is -; and, bv Case 2, the
70
6222            62
sum of thc tangents is    7; therefore   has been added to the sum
70             70
6 61  616   6     ]      62     44     1
of the tangents; but 7     6  =   7 X        - 6- 43        i
-   70  70  If    616i-~16 -       312  98'
Consequently the circumference of the circle has been increased its
9th part.
Again, by Case 1, there is 50 deducted from the square described
upon the radius, and there was also.-th deducted frcm the secant.
50
And, by Case 2, there was,  added to the diameter and the circumference of the given circle.
Consequently there was 9.th added to the area of the given circle.
By Case 2, the sine of the given arc is 9l/2, and there are 311{
sides, each of which has two sines.
The -6222               4356
Then     /'2 X 622 = 6     9   /2-  93   2.
By Case 3, the sine of the given arc is 19-l1/2, and there are 616031
sides, each of which has two sines.
1     _          2    123206-   _    862444
Then 19    1    X 1232062    12320i          137207
1960i1      -       y 219601 i         1372072
4356        597673692 
Then 6931/2       98084451 1/2
123206 __      597673692
A  19601 1/2     96084451 1/.




122,PIOOFS.
Consequently the sum of the sines for Case 3 is the same asthe sum
of the sines for Case 2.
By Case 3, the sine is 19601-/2, and there are 61603+ sides contained in the given polygon, each of which has two sines.
1                   1232062
Then      -   ><1/2 X 123206v =  ^ ^V^7  = sum of the sines for
Case 3.
By Case 4, the sine is 76839840Q1/n, and there are 2414966403+
sides contained in the given polygon, each of which has two sines.
The1   2 X.48299328064   4829932806         um of
Then 768384    /2 X 4829932806       1 839840  1    -- s   of
the sines for Case 4,
n, b  C    1232062       94671512936006 
Then, hy Case 3, ^^-       = I
19601        15061377058001.
4829932806        94671512936006 
And, by Case 4, 768398401  /2 1=506137705801     2
Therefore the sum of the sines for Case 3 is exactly the same as the
sum of the sines for Case 4, and the length of the perimeter of the inscribed polygon is not changed.
By Case 5, the sine is18087225187160/       and the number of
sides contained in the given polygon is 37113126452873856031, each
of which has two sines.
Then, multiplying the sine by double the number of sides, we have
74226252905747712062
8822905318712361   the sum of the sines for polygon No. 5.
1180872205318713601. 94671512936006  _   111794 -By Case 4, the sum of the sines is 061377058001 2 
15061377058001        17785 -9582616008653052412173072062
561541618319480379284571601,'
7422625290574771206~,_
And, by Case 5, the sum of the sines is 1180872205318713601 
1117949582616008653052412173072062
17785561541618319480379284571601    2
Thus we perceive that the sum of the sines, that is the perimeter of
the inscribed polygon, from Case 1 to Case 5, inclusive, is constantly
changing the number of its sides at a rate much more rapid than by
doubling, while the length of the perimeter, as well as the radius, and




PROOFS.                         123
the area of the inscribed polygon constantly remains the same; and, if
the number of sides were increased ad infinitum, they would still continue to be the same, although at every successive step the usual number of sides, according to hypothesis, has been deducted.
19602
For, by Case 2, the sine is 99l  = 384288 1/
And, by Case 3, the sine is î9601/  3842188021/2
Then, by subtraction, we have 384218802~/.
Now, by Case 3, there are 616031 sides, each of two sines.
I                       123206-          862444
Then                X  1230      384218802 X  22 62689531614
6i2             6 2
=   1602' therefore -  -/2 has been deducted from the circumfer19602'          19b2O
ence of the given polygon, and still the perimeter of the inscribed
polygon remains the same.
99   19602
By Case 2, the secant is 70  13860
19601
And, by Case 3, the secant is 13860
1                                        2
Therefore      has been deducted from the secant, and 13   from
13860                                    13860
62
the assumed diameter; consequently, by hypothesis, 1376 must be de1 3860
ducted from the circumference.
1                                1
By Case 2, the tangent is -'; and, by Case 3, the tangent is13860
Then, dividing tangent No. 2 by tangent No. 3, we have by canceli       1     I    l$$f0
lation           =     X   $$     198; therefore each of the tan70   13860 —'         1
gents for Case 2 has been divided into 198 equal parts, each of which
is equal to tangent No. 3, or 13860
Now, by Case 2, there are 311- sides contained in the given polygon, each of which has two:tangents; then 622V  X 198  1232124 =
the number of tangents in the polygon for Case 3. But, by Case 3,
the given polygon has only 1232062 tangents; therefore 38760 has
62b
been deducted -from the given polygon;  and, in Case 4,  3  has
'        '5433-3-9720




124                         PROOFS.
been deducted from  the given polygon, and so in Cases 5, 6, 7, etc.;
and if it were possible to carry the polygon to infinity it would still
be found that in every case 62 times the tangent of the given arc, or
31 sides of the given polygon, would be deducted for 1 of the values
deducted from the diameter.
But in every case the sum of the sines of the given polygon, which
is the perimeter of the inscribed polygon, constantly remains the same;
that is, the length is not diminished, and the area of the inscribed
polygon as well as the radius remains the same, though the number
of sides is continually and rapidly approaching infinity; while the 62
times the tangent is as regularly and constantly deducted in every
case; therefore the 6- times the tangent or 3- sides of the given polygon
so deducted is exactly equal to the number of sides taken up by the perimeter of the inscribed polygon by the contraction, which necessarily follows the increase of the number of sides, which is one of the strongest
proofs that 31 sides in every case is the exact number to be deducted;
consequently, when the final limit is reached, the perimeter of the circumscribed polygon, which is the sum of the tangents of the given polygon will vanish, and the perimeter of the inscribed polygon, that is the
sum of the sines, will become the circumference of the given circle, which
will be infinite!  That is no part of the circumference however small is
straight!  And, as the radius is the 1/2, and the diameter the 1/8,
therefore the circumference of the given circle will be 31 times the
1/8; consequently the true ratio of the circumference of any circle to
its diameter must be as 3- is to 1.  QED.
By Case 1, the cosine of the given arc is 1.4, and the secant is 1.42;
then 1.42 X 1.4 -- 2, and /2-= the radius of the given circle.
140
Again, by Case 2, the cosine of the given arc is —, and the secant
99       140   99
is -  then -   X   --  2, and /2 =  the radius of the given circle.
70       9     70 
27720
Again, by Case 3, the cosine of the given arc is 1901' and the se1.9601       27720   19601
cant is 3860; then 19601    1386    2, and 1/2 = the radius of the
given circle; consequently, if the cosine, which must constantly lie within the circle, be multiplied by the secant, which must constantly lie
without the circle, the square root of their product will, in every case,
be equal to the radius of the given circle. That is, the " limit of the




PROOFS.                           125
product of two quantities is the product of their limits;"    consequently, when the final limit of the cosine is reached, that limit
will be the radius; and, therefore, the radius multiplied by itself, and
the square root of the product being extracted, the result will be
equal to the radius; thus 1/2 X           ' = 2; then 1/  =- the radius
of the given circle.
So, also, if the radius of the given circle be -/-, and a polygon be
inscribed within it, and another circumscribed about it, and these two
polygons be nade to approach the circle and each other, in the manner proposed in the foregoing problem, that is, one from within and
the other from  without (in such a manner that there is no chance of
forcing a limit), and at the same time if the radius of the inscribed
polygon be equal to the cosine of the given arc, and the radius of the
circumscribed polygon be equal to the secant of the given arc, and
these two radii be also made to approach the true radius in the same
manner, and at the same time, as the inscribed and circumscribed polygons, to which they belong, approach the circle; that is, one from
within and the other from without, so near that the difference between
the inscribed and circumscribed polygons, with regard to the circle
and to each other, shall be made less than any assignable quantity,
then in that case the ratio between the assumed circumference and the
assumed diameter will be the same as the ratio between the true circumference and the true diameter.
The following extract from a chapter on the Doctrine of Limits is
copied from  the May number of the Home and School Journal for
1873, published by Jno. P. MAorton & Co., Louisville, Ky. The chapter is the substance of a paper read before the Georgia Teachers' Association by Jno. M. Richardson.  It is inserted here, as it is believed
that it will establish the theory advanced by the author. For a fuller
and more lucid treatise upon the subject, the reader is referred to the
Journal itself:
ART. I.-DEFINITION OF À LIMIT.
The constant A is said Io be the limit of the variable a when A-a or a-A  bmay be
made less than any assignable magnitude, but can never become zero.
ART. II.-DEFINITION OF AN INFINITESIMAL.
The variable difference A-a or a —A, between a variable and its limit, is an infinitesimall. It may be made less than: any designated quantity, but can never become
zero. Or more generally, an infinitesimal is a variable whose limit is zero.




126                            PROOFS.
ART. III.-EXAMPLES OF LIMITS AND OF INFINITESIMALS.
1. The common fraction 7 is the limit of the repetend.142857; and --.142857 is an infinitesimal.
2. The quantity 1/2 is the limit of 1.4142.....; and  / - 1.4142... is
an infinitesimal.
3. The sum of the series 1 +1 + -  +  - + etc. is 4 [1 - ()n ]; its limit
is 4; and 4 - 4 [ -  ()n ] i an infinitesimal.
4. The circle K is the limit of the inscribed and circumscribed polygons I
and C; and K - I or C - K is an infinitesimal.
ART. IV.-REMARKS ON THE EXAMPLES.
a. With regard to the first two examples it is evident that, however far the
process of division or of extracting the square root be carried, the limits 1 and
1/2 can never be reached.
b. So loosely, though, are the symbols o and oo used in algebra, that it may
be supposed the limit can be reached in the third: " for," say algebras generally, "when n =-  0, ()n = o; and then the sum of the series is 4 "
But the sum can not be 4. unless the last term is zero; and the last term can
not be zero unless the one-fourth part of something is nothing, or unless the division
of something can be carried so far that each of its equal parts is nothing, or unless
something and nothing are equivalent terms, the very statement of which absurdity
should be its own refutation.
c. The distance between two places is a miles; a locomotive starting from one
travels during the first hour half the distance to the other; during the second
hour half the remaining distance; and so on, going each successive hour half
the remaining distance. When will it reach its destination? Never; because
the half of something can not be zero.
The expression for the sum of the distances is
S = a +    - +  + etc. =a( +  -+- + + etc.) -  a[- ()n].
Now, a is the limit of the sum of the distances; and, as the locomotive can
never reach its destination, the sum of the distances is never equal to a.
ART. V. THE DUHAMELIAN PROPOSITION.
If two variables, a and 3, are constantly equal, and each tends toward a limit, their
limits, A and B, are equal. Or, more generally, if two variables, a and P, have a
constant ratio, and each tends toward a limit, their limits, A and B, have the same
ratio.
If                           a: /-=r,                              (1)
then                           A: B=r,                               (2).
and                        a::: A: B.                           (3)




PROOFS.                             127
Let A> B.; a and fi less than their limits. If (3) be not true, then
a::: A: B'                         (4)
in which B' is either less or greater than B.
1. Suppose B'< B. Let a and /3 increase until,3 reaches some value intermediate between B' and B. Denoting these new values of a and P by a' and /',
(4) becomes
a': /:: A:                              (5)
Passing to products,
a/ B'= A p/                           (6)
But a'< A, and B'<,'; hence a' B' can not be equal to A fl'.
2. Suppose B'>B. B' is less than A, for. is supposed to be less than a.
Some quantity A', less than A, can always be found to satisfy the proportion.
A': B:: A: B'                               (7)
Comparing (4) and (7),
a =: ~:: A'/: B,                         (8)
Let a and 3 increase until a >A' but <A. Denoting these new values of a and
P by a" and P/" (8) will become
a",:  /":: A': B,                         (9)
whence                        a" B - A' "                           (10)
But a" >A', and B> P"/; hence a" B can not be equal to A/ P/.
If, therefore, the fourth term of (3) can be neither less nor greater than B, it
must be equal to B, and (3) must be true, or
a::: A: B.                           (11)
In the same way, if A < B, a and / increasing toward their limits; or if a and,3 are decreasing variables, and A either greater-or less than B; it cau always
be proved that
a::: A: B.                         (12)
Hence, if the variables a and / have a constant ratio, their limits A and B have the
same ratio.
COROLLARY.
Ifa:   -    1, thenA: B = 1. That is, ifa =, then A =B.
Hence, if two variables are constantly equal, their limits are equal.
ART. VI.-ADDITIONAL PROPOSITIONS.
1. The limit of the sumn ofseveral variables, each tending toward a limit, is the sum
of their limits.
2. The limit of the product of several variables is the product of their limits.
3. The limit of the quotient of two variables is the quotient of their limits.
4. The limit of any power of a variable is the same power of the linit.
It is, perhaps, hardly necessary to demonstrate the four preceding propositions; the intelligent student can do so for himself readily enough; but the
following one; being rather more abstruse, will be established.
5. When the limit of the ratio of two quantities is unity their difference divided by




128                             PROOFS.
either one has zero for its limit or is an infinitesimal. Reciprocally, if the difference
of two quantities divided by either one is an infinitesimal, the ratio of the two quantities has unity for its limit.
PROOF.
Let                          a/-a      d                             (1)
a     d
whence                         1-   -                                  (2)
a —       -                            (3)
a          a
a       a, ' 
If, now, the limit of, or of - is unity, the limit of, or of -must be zero.
a       a                     a/      a
It follows, therefore, that the limits of the ratios or of the sums of infinitesimals,
are not changed when the given infinitesimals are replaced by others which differ from
them by quantities which are infinitesimals with regard to them.
NOTE.-The proposition of Article V. may be found in the works of both Duhamel and J. A.
Serret, though the former appears to make much more use of it than the latter. The demonstration given is due to Dr. A. T. Bledsoe, formerly Professor of Mathematics in the University
of Virginia, now editor of the Southern Quarterly Review, Baltimore, Md.
The propositions of Article VI. are from Duhamel.
The Quadrature of the Circle has been defined as a problenr which
has been reduced to finding a straight line equal in length to the circumference of the given circle in terms of the radius; or, how       often
the square described upon the radius is contained in the area within
the given circle, both of which, upon examination, may be said to be
reducible to the same thing.
For, when the circumference and the radius of the given circle are
known, the rectangle under the radius, it is said, will give the area;
and when the area and the radius are known, dividing the area by the
square of the radius, it is said, will give the circumference of the given
circle.
For, supposing the radius equal to unity its square will be equal to 1;
then, if the area contained within the given circle be 6.2831853070,
etc. by division we have 6.2831853070, etc., which is equal to the circumference of the given circle.
Again, if the circumference of the given circle be 6.2831853070,
etc., and the radius 1, by division we have 6.2831853070, etc., which
equals the area of the given circle, which would give for the ratio
3.1415926535, etc.
7
But, if the true radius be the 1/2, and the assumed radius be, and
5,




PROOFS.                         129
44
the assumed circumference be -, then, according to the " Duhamelian
theorem," dividing the circumference by the diameter, we have by can44   14_ 44      &   22
cellation -  -* -     X   -   - =  3.142857, or 31, which is the
o'5,   14,true ratio of the circumference to the diameter of the given circle.
By Case 1, X1/2   the tangent of the given arc, and the radius is
the V'.
1              2
Then, by ARTICLE 6, 71/    X  /2 =    --  area of the circumscribed double triangle.
But there are 22 double triangles contained in the given polygon.
2         44
Then, by ARTICLE 9,     X 22    — = area of the circumscribed
polygon.
44
Squaring the radius and dividing, we have (1/2)2 = 2; then -
7-7
29
* 2   -'-3.142857, or 31, which is the true ratio of the circumference to the diameter of the given circle.
Again, 1/2     the tangent of the given are; then î1/2 X 44 
44
74-/  = the circumference of the given polygon.
But the radius is the i/2, and the diameter is twice the square root
of two; thus 1/2 X 2 = 21/.
44                22
Then, by division, we have -1/S2 -     2/  --     3.142857, or
3i, the true ratio of the circumference to the diameter of the given
circle.
Again, by Case 2, the sine of the given arc is 91/2, and the radius
is the 1/2.
1        2
Then, by ARTICLE 7, 1/2 X   Q9/2 9-    =  area of the inscribed
double triangle for double the number of sides.
But there are 311- sides to the given polygon.
9




130                        PROOFS.
~2        6222   44
Then, by ARTICLE 10,    X 311                  area of th in9 --              - - area of the inscribed polygon for double the number of sides.
NTow the tangent of the given arc is 50, and the unit of comparison.   1 \2    1
is 70   ~4900'
44     1     44
Then, by division and cancellation, we have 7 ' 4900 —     X
= 30800.
Therefore there are 30800 squares contained in the given polygon,
each of which is represented by 400; and the square described upon
the radius is 2.
I          4900
Then 2   490     2 X   -     9800.
Therefore there are 9800 squares contained in the square described
upon the radius, each of which is represented by 490
Then 30800 -9800 = 3.142857, or 31, the true ratio of the circumference to the diameter of the given circle.
But the area of the circle is said to be the rectangle under the radius; that is, the rectangle contained by the circumference and the
radius.
If it is meant for the circumference to be a straight line without regard to the circular figure, then the definition is true; but it is admitted
by all former mathematicians that this straight line has never yet been
found. Moreover, it is asserted that it is not likely that it ever will be
found, because " innumerable attempts have been made to find a solution of this problem, but these attempts have been made in vain."
If it is meant for the circumference of a circle to be the boundary
of the circular figure known as the circle, then, in that case, it is not
true il that the rectangle contained by the radius and the circumference is
equal to the area of the given circle," as can very readily be shown.
For, witness the following demonstration:
Let the straight linefP, Figures 1. and 2, Plate 6, be the radius of
the given circle = i/,  and let the straight lines PF, PE be the tan



PROOFS.                          131
gents of the given arc, each of which =  1/; then will the triangles
fPEF fPE be the given circumscribed triangle.
1       2
For, by ARTICLE 6, i/2  X  1/2 =      area of the circumscribed
triangle.
2
Then, dividing this area by the given radius, we have   2 
71/2, which is said to be the arc of the given cirele, but which is, in
fact, only the tangent of the given arc, a result too large, because the
tangent is greater than its arc.
Again, let the straight line fh, fm, be the radius of the given circle
-   /2, and the straight lines gh ge, lm In; be the sines of the given
arc, each of which  —; then will the trianglesfsm, fsn, be the inscribed
double triangles for double the number of sides.
i    1
For   X 1/2     1 2 = area of the double inscribed triangle for
double the number of sides.
Dividing this area by the given radius, we have 5/2 -  /2  - 5
which is said to be the arc of the given circle, but which, in fact, is
only the sine of the given arc-a result too small, because the sine is
less than its arc.
Again, let the straight line fs be the given radius = 1/2, and let
the straight lines sr, st, be the tangents of the given arc, each of which
~1 
-70'
Then, by ARTICLE 6, 1/2 X 70 =701/    =  area of the circumscribed double triangle.
Dividing this area by the given radius, we have 01/2  1/2 - 
which is said to be the arc of the given circle, but which is, in fact,
only the tangent of the given arc-a result too large, because the tangent is greater than its arc.
Again, let the straight line fs'be the given radius = 1/2, and let
the straight lines sr, st, be the sines of the given arc, each of which is




132                         PROOFS.
equal to j1/2; then will the triangles fsr, fst, be the inscribed double
I           2
triangle for double the number of sides; for 99 X 1/2  9    the
area of the inscribed double triangle for double the number of sides.
2
Then, dividing the area by the given radius, we have 9 - —. 2 
99/2, which, it is said, is the arc of the given circle, but which is, in
fact, only the sine of the given arc-a result too small, because the
sine is less than its arc.
For the next step let us take the sine, radius, and tangent of Cases 3,
4, 5, 6, or 7, and if they do not furnish an example sufficiently small, let
the result be carried as far as it is possible for human power and endurance to carry it; and if this again is too small, let us picture to our
imagination an example small enough to satisfy the most mathematical
exactitude; the relative ratio will still be the same, namely: dividing
the area of the circumscribed triangle by the radius gives for the result
the tangent, which is too large; and, dividing the area ofthe inscribed
double triangle for double the number of sides gives for the result the
sine, which is too small; but the inscribed and circumscribed double
triangles, by all former methods, are either inscribed or circumscribed
polygons, and whatever is true of one member of a class is true of the
whole class; therefore the result obtained by any former method is
either too great or too small; and if, as there is nothing to prevent the
inscribed polygon from extending beyond the circle, nor to prevent
the circumscribed polygon from coming within it, which is very possible, for the limit of the two polygons is the limit obtained; then the
result obtained by all former methods must be too small; and, consequently, the rectangle under the radius will not give the true area of
the given circle.
Again, let the radius of the given circle be 1/2, and the tangent of
the given arc be 7; then, by A                            ATICL 6, 1/2 X  And
99
the secant of the given arc is; then, dividing the area of the circumscribed double triangle by the secant of the given arc, we have
il'     99 -   1.  _    =0  1
7/2      70 -     2 X 99 - 99/2 = the sine of the given arc.
70   70-F               F     




PROOFS.                          133
But the area of the inscribed double triangle for double the number
of sides divided by the radius gives the same result, nanlely: 91/2
the sine of the given arc.
Therefore the area of the circumscribed double triangle divided by
the secant is equal to the area of the inscribed double triangle divided
by the radius, namely: 91/2, which is a resulttoosmall.
But by the present method the'sum of the tangents, which is the
perimeter of the circumscribed polygon, vanishes together with that
polygon.
And the sum of the sines, which is the perimeter of the inscribed
polygon, becomes the circumference of the circle at the same time that
the inscribed polygon becomes the circle; therefore the secant vanishes
and the cosine becomes the radius, which is the /2.
Consequently (PROPOSITION 7, THEOREM), The true ratio of the circumference to the diameter of the circle is as 3.142857, or 37, is to 1;
and, therefore, the last term of the ratio vanishes also.
The following demonstration is taken from  "Elements of Euclid,"
by James Thompson, LLD., pp. 124-5. Belfast: Symmes & McIntyre. London: Longman & Co., and Simpson & Co.:
PROP. XIX. THEOR.-In numbers which are continual proportionals, the
difference of the first and second is to the first, as the difference of the first and
last is to the sum of all the terms except the last.
If A, B, C, D, E be continual proportionals, A- B  A:: A - E: A + B +
C + D.
For, since (hyp.) A: B:: B: C:: C: D:: D: E, we have (Sup. 8) A: B::
A+B+ C + D:B+C+D + E. Hence (conv.) A:A- B::A+B+ C +
D: A-E; and (inver.) A-B: A:: A-E: A+B+ C           D.
It is evident that if A were the least term, and E the greatest, we should get
in a similar manner, B-A: A:: E-A: A + B + C    D. Therefore, in
numbers, etc., --: 7:::         =. For =   -, and  X -.
Then -+- 1    -,  X 49 =. Therefore the sum of the series -+-  + -- +
+         dvoT =                                           --
Cor. If the series be an infinite decreasing one, the last term will vanish, and,
if S be put to denote the sum of the series, the analogy will become A - B: A:: A: S; and this, if rA be put instead of B, and the first and second terms be
divided by A, will be changed into 1-r: 1:: A: S. The number r is called
the common ratio, or common multiplier, of the series, as by multiplying any term
by it the succeeding one is obtained.








PART SECOND.
THE SQUARE ROOT OF TWO,
OR
THE COMMON MEASURE OF THE SIDE AND DIAGONAL
OF THE SQUARE;
BEING A
SHORT, EASY, AND CONVENIENT METHOD OF FINDING EITHER THE SIDE OR
DIAGONAL OF THE SQUARE, WHEN THE OTHER IS KNOWN, BY
COMMON MULTIPLICATION AND DIVISION;
ALSO,
THE SQUARE ROOT OF TWO, BY DIVISION ALONE, TO ONE
HUNDRED AND FORTY-FOUR DECIMAL PLACES.
(135)




THE SQUARE ROOT OF TWO.
THE following pages are intended to explain the nature, uses, and
advantages of the Common Measure, as applied to Civil Engineering,
Architecture, Draughting, Machinery, Building, Painting, and Landscape Gardening.
Numerous instances could be given where it has been already tested
and found to be everything that could be desired.
According to PROPOSITION 1, COR. PART I, the side of any square
is to the dioganal as "one is to the square root of two; that is, if the
side of any square be ONE (1), the diagonal must be the square root of
two (1/2); therefore, by the ordinary. method, to find the diagonal of
a square, when the side is known, use the following
RULE.
Take as many figures of the square root of two as will malce the result
sufficiently exact, and multiply these figures by the number of units in the
side of the given' square, the product will be the true diagonal in terms
of the side of the square.
NOTE.-If the side of the square be expressed in inches, the diagonal will also be expressed
in inches; but if the side be expressed in feet or rods, etc., the diagonal will be expressed in feet
or rods, etc.; and if the side contain a decimal or a fraction, the result of that decimal or fraction will have its corresponding result in the diagonal. In all cases the requisite number of figures
of the square root of two must first be obtained.
When the diagonal is given, to find the side, it is not so easy a task,
and requires more time and labor than to find the diagonal when the
side is known; but, for the benefit of those who still prefer the former
method, I shall here insert the rule which embraces two cases.
CASE 1.
If the diagonal is expressed in integers, such as 1, 2, 3, etc., the side
of the square is found by dividing these numbers by the square root
of two; the result will be the side in thé same denomiination as the
diagonal.
(136)




THE SQUARE ROOT OF TWO.                   137
CASE 2.
If the diagonal contains either a fraction or a decinal, this fraction
must first be reduced to a decimal and the value pointed off as in
reduction of decimals. If a decimal only, it must first be pointed off,
then divide this decimal by the square root of two, according to the
rule for division of decimals, the quotient will be the side of the
square in terms of the diagonal; but, in all cases, the requisite number
of figures of the square root of two must first be found.
To find an exact common measure of the side and diagonal of the
square would be equivalent to finding the exact value of the square
root of two; but the square root of two is an irrational quantity, therefore it has no end. It is believed that the following common measure
comes nearer to an exact commnon measure than any that has been heretofore found, as it can be extended ad infinitum, and when either the
side or the diagonal of any square is known the other can be found by
common multiplication and division.
Owing to the difficulty of working out the necessary figures of the
square root of two, and the tax upon the mind necessary to remember the same, civil engineers, architects, draughtsmen, builders, etc.,
have long since felt the want of a COMMON MEASURÊ expressed in
integers, or a series of'numbers, which would express the relation between
the side and diagonal of' the square; one that could be easily remembered, plainly understood, and readily applied. Such, for example, as
the ratio between the circumference and diameter of the circle, found
355
by Moetius in 1640, namely: 1-,' or 1131355, which it is said will give
13'
the ratio to six decimal places correct, viz.: 3.141592.
The author is happy to state that such a common measure of the side
and diagonal of the square has been found which may be expressed in
integers, and he ventures to hope that when it is fully tested the old
method of squaring the side, doubling and extracting the square root,
will be gladly cast aside as a noble relict of "Auld Lang Syne," when
simpler and easier methods were unknown.
The common measure here introduced combines all the advantages
of conciseness, simplicity, and perfection, for by its aid the desired
results are reached much more rapidly than by any former method and
with equal exactness.




138               THE SQUARE ROOT OF TWO.
FIRST EXAMPLE.-Suppose a builder is laying the foundation for a
house and wants to make each of the corners a right angle-that is, a
perfectly square corner; how will he do it?
ANSWER.-First measure 10.5 feet on each side of the right angle
in the direction of the sides, commencing at the angular point; then
the distance across from the two extreme points will be 14.85 feet, or
14 feet, 10.2 inches.  This result is found by multiplying the side of
the given square by 99 and dividing the product by 70, which gives
the desired result.  See Plate 8, Figure 1.
SECOND EXAMPLE.-Suppose a civil engineer, while surveying,
wants to lay out a right angle without the use of his instruments;
how will he do it?
ANsWER.-Measure with a tape line any distance, on either side of
the right angle, say 105 feet, in the direction of the sides, commencing
at the angular point.  Then will the distance across from the extreme
points be 148.5 feet, or 148 feet 6 inches. This result is found precisely like the former, namely, by multiplying the side of the given
square by 99, and dividing the product by 70, which gives the desired
result. See Plate 8, Figure 2.
THIRD EXAMPLE.-Suppose a mechanic has a given circle, the diameter of which is 9.9 feet, or 9 feet 10.8 inches, and he wants to find
the side of the largest square which it is possible to inscribe in the
given circle; how will he find it?
ANswER.-Multiply the given diameter by 70, and divide the product by 99, which will give the desired result, which will be 7 feet.
See Plate 8, Figure 3.
If any person has a desire to test the correctness of these results by
the former method, he is at liberty to do so, and will find them correct
to the fourth decimal place.
NOTE.-In the third example above, the diameter of the circle may be regarded as the diagonal of the given square, the side of which was required to be found; therefore the square which
is so formed within the given circle is said to be inscribed within it.
From these three examples, which embrace two cases, we deduce the
following rules:
CASE 1.
When the side of any square is known, to find the diagonal correct
to four places of figures.




PLATE VHl.
Fig.1
or ]~See 5lde
Fig.2 
F_  ___ 14i'8i.53e __ __ __
T; ~   or 1-SF3  Tcw,
r-.,  y''7  ~  \~~~~~~~~~~~~~~F=
Fle-3 /.1  








THE SQUARE ROOT OF TWO.                         139
RULE.
Multiply the side of the given square by 99, and divide the product
by 70; the quotient will be the diagonal of the square in terms of the
side.
CASE 2.
When the diagonal of any square is known, to find the side.
RULE.
Multiply the diagonal of the given square by 70, and divide the
product by 99; the quotient will be the side of the given square in
terms of the diagonal.
NOTE.-It is believed by the author that the above numbers will answer every purpose for
all ordinary measurements; besides this, they are simple and easily remembered by the most
ordinary miind. If, however, still greater accuracy is required, attention is invited to the following, which the author ventures to hope will be found to be sufficiently correct to satisfy the
most careful.
By Case 2, PART 1, it is shown that where the radius of the given
circle is the square root of two, the side of the inscribed square will be
two, and the side of the inscribed square, the diagonal of which corre99
sponds with the radius, is one.    Therefore the secant No. 2 is 70
70'
which, by division, gives the square root of two to five places of figures
correct, thus: 1.4142; that is, if the side of the inscribed square, the
diagonal of which corresponds with the radius, be divided into 70 equal
parts, the diagonal of the same square will be 99 of the same parts
nearly.
By Case 3, PART 1, it is shown that the secant No. 3 is 1960
13860:
which, by division, gives the square root of two to nine places of figures correct, thus: 1.41421356; that is, if'the side of the inscribed
square, whose diagonal corresponds with the radius, be divided into
13860 parts, the diagonal of the same square will be 19601 of the
same parts nearly.
768398401
By Case 4, PART 1, it is shown that the secant No. 4 is   433
543339720'
which, by division, gives the square root of two to 18 places of figures
correct, thus: 1.41421356237309504; that is, if the side of the inscribed square, the diagonal of which corresponds with the radius, be




140              THE SQUARE ROOT OF TWO.
divided into 543339720 parts, the diagonal of the same square will be
768398401 of the saine parts nearly.
By Case 5, PART 1, it is shown that the secant No. 5 is
1180872205318713601.
1835002742095754601 which, by division, gives the square root of two
835002744095575440'       yds
to 36 places of figures correct, thus: 14142135623730950488016887 -2420969807; that is, if the side of the inscribed square, the diagonal
of which corresponds with the radius, be divided into 8350027440955 -75440 parts, the diagonal of the same square will be 1180872205318 -713601 of the same parts nearly.
By Case 6, PART 1, it is shown that the secant No. 6 is
2788918330588564181308597538924774401
1972063063734639263984455073299118880' whlch, by division, gives
the square root of two to 72 places of figures correct, thus: 14142135 -6237309504880168872420969807856967187537694807317667973799 -0732478; that is, if the side of the inscribed square, the diagonal of
which corresponds with the radius, be divided into 1972063063734639 -2.63984455073299118880 sides, the diagonal of the same square will be
2788918330588564181308597538924774401 of the same parts nearly.
By Case 7, PART 1, it is shown that the secant No. 7 is
1555613090938580753522477984263968662546864806579817762712 -1099984505505235876371971631359164002982364405164572043531 -6514337489817601
7842392159581760' which, by division, gives the square root of two to
144 places of figures correct, thus: 1.4142135623730950488016887242 -0969807856967187537694807317667973799073247846210703885038 -7534327641563643977195724018929160771077122365330384600627;
that is, if the side of the inscribed square, the diagonal of which
corresponds with the radius, be divided into 109998456550523587637 -19716313591 sides, the diagonal of the same square will be 1555613 -090938580753522477984263968662546864~065798177627126514337 -489817601 of the same parts nearly.
By Case 8, the square root of two can be found, by division, to 288
places of figures, by Case 9 to 576 places, by Case 10 to 1152 places,
and so on ad infinitum; and, in every case, if the side of the inscribed
square, the diagonal of which corresponds with the radius, be divided
into the requisite number of parts, the diagonal will still be expressed




THE SQUARE ROOT OF TWO                      141
by a certain number of the same parts till it reaches the vanishing
point, when the square root of the sum of the squares of the two sides of
any square can be extracted exactly,
But if by the word "infinite," indefinite extension only is meant, then
the operation may be continued without end, and in THAT CASE there
will forever remain an indivisible unit, the square root of which never
can be extracted.
If, then, the true ratio of the circumference of the circle to its diameter be as 31 is to 1, and the radius of the given circle be the /2,
then will the area contained within the given circlebe 6-, and the side
of the inscribed square will be two, while the area will be four.  But
the area of the circumscribed square is double the area of the inscribed
square, therefore the area of the circumscribed square is 8.
Again, the square described upon the diameter of any circle is equal
to the circumscribed square.
But the diameter of the given circle is -//8, therefore the area of the
circumscribed square is equal to /     / X  /8 =8.
Again, the area of the inscribed square is one-half the area of the
circumscribed square; therefore the area of the inscribed square is 4.
Now the area of the given circle is 67, and the area of the inscribed
square is 4, and the area of the circumscribed square is 8; therefore the
area of the circle is to the area of the inscribed square as 11 is to 7; and
the area of the circle is to the area of the circumscribed square as 11
is to 14.
Therefore to find the area of any circle, when the diameter is known,
use the following
RULE.
Multiply the square of the diameter by 11, and divide the product by
14, the quotient will be the area of the circle; or, square the radius, and
multiply it by 31, the product will be the area of the circle.








PART THIRD.
ON THE RIGHT-ANGLED TRIANGLE;
CONTAINING
A VARIETY OF METHODS FOR FINDING AN INFINITE SERIES
OF RIGHT-ANGLED TRIANGLES, THE SIDES OF WHICH
SHALL BE IN WHOLE NUMBERS, FOR THE USE
OF CIVIL ENGINEERS, ARCHITECTS,
DRAUGHTSMEN, AND MECHANICS.
(143)




ON THE RIGHT-ANGLED TRIANGLE.
OF RIGHT-ANGLED TRIANGLES IN NUMBERS, OR RIGHT-ANGLED
TRIANGULAR NUMBERS.
RIGHT-ANGLED triangular numbers are rational numbers so related
to each other that the sum of the squares of two of them is equal to
the square of the third.
The numbers 3, 4, and 5 have this property-32 - 42 being equal
to 52.
Right-angled triangular numbers must be severally unequal, for if
the two less ones could be each represented by a, and the third or
greatest by b, then 2a2 - b2, b — a/2, an irrational number, whatever
is the value of a.
The area of a right-angled triangle, whose sides are rational, can not
be equal to a rational square.
If a, b, and c represent the sides of a triangle, and C be the angle
opposite c, then, if C= 90~, a2 + b2 = c2; if C-= 120~, a2 + ab + b2
-c2; and if C- 60, a2 -ab + b2      c2.
If n represent any number, and vn any other number less than n,
then n2+ —m2 will represent the hypothenuse of a right-angled plane
triangle, of which the other two sides are respectively n2, m2, and 2nnm.
For example, if n=2 and m=l, then n2-[m2-5, n2-m23, and
2nm-4, which are right-angled triangular numbers.
If n-7 and m=2, the formulas give 53, 45, and 28 for the numbers, and 532-2809-452+282.
We shall now propose and solve a few of the most curious problems
respecting the right-angled triangles.
PROBLEM 1.
To find as many right-angled triangles in numbers as we please.
This may be effected by the concluding formulas, which we have
just given, but we think it right to add the following methods:
(144)




ON THE RIGHT-ANGLED TRIANGLE.                145
Take any two numbers at pleasure, for example 1 and 2, which
we shall call generating numbers; multiply them together, having
doubled the product, we obtain one of the sides of the triangle, which
in this case will be 4. If we then square each of the generating numbers, which in the present example will give 4 and 1, their difference
3 will be the second side of the triangle, having 1 and 2, for their
generating numbers are 3, 4, and 5.
If 2 and 3 had been assumed as generating numbers we should have
found the sides to be 5, 12, and 13, and the numbers 1 and 3 would
have given 6, 8, and 10.
Another lMethod.-Take a-progression of whole or fractional numbers, 1-, 2s, 3~, and 44, etc., the properties of which are:
lst. The whole numbers are those of the common series and have
unity for their common difference.
2nd. The numerators of the fractions annexed to the whole numbers are also the natural numbers.
3rd. The denominators of these fractions are the odd numbers 3,
5, 7, etc.
Take any term of this progression, for example 33, and reduce it to an
improper fraction by multiplying the whole number 3 by 7, and adding to
24
21 the product, the numerator 3 will give -7. The numbers 7 and 24
will be the sides of a right-angled triangle, the hypothenuse of which
may be found by adding together the square of these two numbers,
viz.: 49 and 576, and extracting the. square root of the sum. The sum
in this case being 625, the square root of which is 25, this number will
be the hypothenuse required. The sides, therefore, of the triangle
produced by the above term of the generating progression are 7, 24,
and 25.
In like manner the first term will give the right-angled triangle 3,
4, and 5; the second term, 22, will give 5, 12, and 13; the fourth, 44,
will give 9, 40, and 41. All these triangles have the ratio of their sides
different; and they all possess this property, that the greatest side and
the hypothenuse differ only by unity.
The progression, 17, 2-, 3}-, 4~0, etc., is of the same kind as the
preceding. The first term of it gives the right-angled triangle 8, 15,
and 17; the second term gives the triangle 12, 35, and 37; the third
10




146          ON THE RIGHT-ANGLED TRIANGLE.
triangle 16, 63, 65, etc. All these triangles it is evident are different
in regard to the proportion of their sides; and they all have this peculiar property, that the difference between the greater side and the
hypothenuse is the number 2.
PROBLEM 2.
To find any number of right-angled triangles in numbers, the sides
of which shall differ only by unity.
To resolve this problem we must find out such numbers that the
double of their squares, plus or minus unity, shall also be square
numbers; of this kind are the numbers 1, 2, 5, 12, 29, 70, etc., for
twice the square of 1 is 2, which, diminished by unity, leaves 1, a
square number. In like manner twice the square of 2 is 8, to which,
if we add 1, the sum 9 will be a square number, and so on.
Having found these numbers, take any two of them which immediately follow each other, as 1 and 2, or 3 and 5, or 12 and 29, for generating numbers; the right-angled triangles arising from them will be
of such a nature that their sides will differ from each other only by
unity.
The following is a table of these triangles, with their generating
numbers:
Generating Numbers.           Sides.              Hypothenuse.
1      2                3        4                 5
2      5               20        21               29
5     12              119       120              169
12     29              696       697              985
29     70             4059      4060             5741
70    169            23660     23661            33461
But if the problem were to find a series of triangles of such a nature that the hypothenuse of each should exceed one of the sides only
by unity, the solution would be much easier. Nothing, in this case,
would be necessary but to assume as the generating numbers of the
required triangles any two numbers having unity for their difference.
The following is a table, similar to the preceding of the six first




ON THE RIGHT-ANGLED TRIANGLE.                    147
riglit-angled triangles produced by the first numbers of the natural
series:
Generating Numbers.               Sides.              Hypothenuse.
1      2                  3         4                  5
2      3                  5         12                13
3      4                  7        24                 25
4      5                  9        40                 41
5      6                 11        60                 61
6      7                 13        84                 85
If we assume as generating numbers the respective sides of the preceding series of triangles, we shall have a new series of right-angled
triangles, the hypothenuse of which will always be square numbers, as
may be seen in the following table:
Generating Numbers.        Sides.         Hypothenuse.   Roots.
3      4             7        24           25          5
5     12            119      120          169         13
7     24            336      527          625         25
9     40            720     1519         1681         41
11     60           1320     3479         3721         61
13     84          2184      C887         7225         85
It may be here observed that the roots of the hypothenuses are
always equal to the greater of the generating numbers increased by
unity.  But if the second side and the hypothenuse of each triangle
in the above table, which differ only by unity, were assumed as the
generating numbers, we should have a series of right-angled triangles,
the least sides of which would always be squares.
A few of these are as follows:
Generating Numblers.             Sides.               Hypothenuse.
4      5                  9         40                41
12     13                 25        312               313
24     25                 49       1200              1201
40     41                 81       3280              3281
In the last place, if it were required to find a series of right-angled
triangles, the,sides of which shall be always a cube, we have nothing
to do but to take, as generating numbers, two following terms in the
progression of triangular numbers, as 1, 2, 3, 6, 10, 15, 21, etc.




148          ON THE RIGHT-ANGLED TRTANGLE.
By way of example, we shall here give the first four of these triangles:
Generating Numbers.           Sides.              Hypothenuse.
1      3                6         8               10
3      6               36        27               45
6     10              120        64              136
10     15              300       125              326
The following short and easy method for finding a series of rightangled triangles, together with the rule for their solution, is inserted
by the author with the hope that it may be of service to practical
men. The triangles which belong to this series must all have these
properties in common.
Ist. All the sides of the given triangles shall be expressed in integers, and the shortest of the three sides, which we will call the perpendicular, shall be expressed by one of the odd numbers, 3, 5, 7, 9,
11, 13, 15, 17, 19, etc.
2nd. The hypothenuse of each triangle shall exceed the longer of
the other two sides, which we will call the base, by unity.
3rd. The square described upon the perpendicular shall be equal to
the sum of the other two sides.
4th. If the square of the perpendicular be increased by unity, the
sum, divided by two, will give the hypothenuse of the given triangle;
but if the square of the perpendicular be diminished by unity, the
difference, divided by two, will give the base of the given triangle.
5th. If an arc of a circle be described across double the width of
the triangle, with a radius equal to the base of the triangle, the double
of the area of the given triangle, divided by the hypothenuse, will
be the sine of the given arc.
6th. If the base of the given triangle be divided by the hypothenuse, it will give the versed sine of the given arc.
7th. If the square of the radius be divided by the hypothenuse of
the given triangle, it will give the cosine of the given arc.
PROBLEM 1.
For example, let 3 be the perpendicular of the given triangle; then
32 -= 9 - sum of the other two sides.
Because 9 + 1 - 2    5 = the hypothenuse of the given triangle.
Again, 9 - 1 - 2 = 4 = base of the given triangle.




PLATE ILX.
12
2  Ye
Fig.o              5
-10








ON THE RIGHT-ANGLED TRIANGLE.                149
Therefore the sides of the triangle are expressed in the integers 3,
4, and 5.
Again, 5- 4 = 1. Therefore the hypothenuse exceeds the base
by unity.
Now 3 X 4 = 12 = double the area of the given triangle.
Then 12 -- 5 - 2.4 = the sine of the given arc.
And 4 X 4 =16= the square of the radius or base of the triangle.
Then 16 - 5   3.2 - cosine of the given arc. See Plate 9, Figure 1.
Subtracting the cosine from the radius will give the versed sine of
the are.
PROBLEM 2.
Let 5 be the perpendicular of the given triangle; then 52 = 25 =
the sum of the other two sides.
For 25 - 1.- 2 = 13 = the hypothenuse of the given triangle.
Again, 25 - 1 - 2 = 12 = the base of the given triangle.
Therefore the sides of the given triangle are expiessed by the integers 5, 12, and 13. See Plate 9, Figure 2.
PROBLEM 3.
Let 7 be the perpendicular of the given triangle; then 72 = 49 
the sum of the other two sides.
For 49  - 1 - 2 = 25 = the hypothenuse of the given triangle.
Again, 49 - 1 - 2 = 24 = the base of the given triangle.
Therefore the sides of the given triangle are expressed by the integers 7, 24, and 25. See Plate 9, Figure 3.
PROBLEM 4.
Let 9 be the perpendicular of the given triangle; then 92 - 81
the sum of the other two sides.
For 81 + 1- 2 - 41 = the hypothenuse of the given triangle.
And 81 - 1- 2 = 40 = the base of the given triangle.
Therefore the sides of the given triangle are expressed by the integers 9, 40, and 41, See Plate 9, Figure 4.
The other parts to the triangle are found exactly in the same manner as shown in Problem 1, and the student is recommended to make




150          ON THE RIGHT-ANGLED TRIANGLE.
the calculation for himself. The triangles, as may readily be seen, can
be continued ad infinitum.
The following table will give the first 14 of the series, which the
student can continue as far as he may desire.
Perpendicular.              Base.               Hypothenuse.
3                       4                        5
5                      12                       13
7                      24                       25
9                      40                      41
11                      60                      61
13                      84                       85
15                     112                      113
17                     144                      145
19                     180                      181
21                     220                      221
23                     264                      265
25                     312                      313
27                     364                      365
29                     420                      421




APPENDIX.
A CHAPTER ON CONSTRUCTION
FOR TEE USE AND BENEFIT OF
DRAUGHTSMEN, BUILDERS,
AND
MECHANICS.












TO EItECT O LET FALL A PERPF1JICbLA.C:G
F~ig.1  ~ Fig.2
A        B
E    D    F
Fi.3  - H   Fig.4
E-~~G~~' ~ ~
Fig.5         Fig.
LB      D  i  E  D" I
'.'. ~  /
Fig.5  i,i':
Fl  B       A
B       D.




CONSTRUCTION.
PLATE 10.
TO ERECT OR LET FALL A PERPENDICULAR.
PROBLEM 1, FIGURE 1.
To Bisect the Right Line AB by a Perpendicular.
lst. With any radius greater than half of the given line, and with
one point of the dividers in.4 and B, successively draw two arcs intersecting each other in C and D.
2nd. Through the points of intersection draw CD, which is the perpendicular required.
PROBLEM 2, FIGURE 2.
From the Point D in the Line EF to Erect a Perpendicular.
Ist. With one foot of the dividers placed in the given point D, with
any radius less than one-half of the line, describe an arc cutting the
given line in B and C.
2nd. From the points B and C, with any radius greater than BD,
describe two arcs cutting each other in G.
3rd. From the point of intersection draw GD, which is the perpendicular required.
PROBLEM 3, FIGURE 3.
To Erect a Perpendicular when the Point D is At or Near the End of a
Line.
lst. With one foot of the dividers in the given point D, with any
radius, as DE, draw an indefinite arc GI.
2nd. With same radius and the dividers in any point as E, draw the
arc BDF, cutting the line CD in B.
3rd. From the point B, through E, draw a right line cutting the arc
in F.
(3)




4                          CONSTRUCTION.
4. From F draw FD, which is the perpendicular required.
NOTE.-It will be perceived that the arc BDF is a semicircle, and the right line BF a diameter; if, from the extremities of a semi circle, right lines be drawn to any point in the curve,
the angle formed by them will be a right angle; this affords a ready method of forming a " square
corner," and will be found useful on many occasions, as its accuracy may be depended on.
PROBLEM 4, FIGURE 4.
Another Method of Erecting a Perpendicular when At or Near the End
of a Line.
Continue the line _1D toward- C, and proceed as in Problem 2. The
letters of reference are the same.
PROBLEM 5, FIGURE 5.
From the Point D to Let Fall a Perpendicular to the Line AB.
Ist. With any radius greater than DG, and one foot of the compasses in D, describe an arc cutting AB in E and F.
2nd. From E and F, with any radius greater than EG, describe two
arcs cutting each other as in C.
3rd. From D draw the right line DC, then DG is the perpendicular
required.
PROBLEM 6, FIGURE 6.
When the Point D is Nearly Opposite the End of the Line.
lst. From the given point D, draw a right line to any point of the
line AB, as O.
2nd. Bisect OD by Problem 1 in E.
3rd. With one foot of the compasses in E, with a radius equal to
ED or EO, describe an arc cutting AB in F.
4th. Draw DF, which is the perpendicular required.
NOTE.-The reader will perceive that we have arrived at the same result as we did by Problem 3, but by a different process, the right angle being formed with a semicircle.
PROBLEM 7, FIGURE 7.
Another Method of Letting Fall a Perpendicular when the Given
Point D is Nearly Opposite the End of the Line.
Ist. With any radius, as ED, and one foot of the compasses in the
line AB, as at F, draw an arc DHC.




PLATE XI.
Co0,S'Trc7UT-OV AND DLIVISrOn OF AVG6LSS.
FiK.l  -               2l  y
p
AZ --- —-----— \B
K            o
C              D
^      Fig.3               Fié.4
A..
Fi.(5                  Fi.6   B
B
c   A^ --- —---— 'b'"x.








CONSTRUCTION.                                   5
2nd. With any other radius, as ED, draw            another arc DKC, cutting the first arc in D and C.
3rd. From D draw DC, and DG is the perpendicular required.
NOTE.-The points E and F, from which the arcs are drawn, should be as far apart as the line
A B will admit of, as the exact point of intersection can be more easily found; for it is evident
that the nearer two lines cross each other at a right angle the finer will be the point of contact.
PROBLEM 8, FIGURE 8.
To Erect a Perpendicular Line at D, the End of the Line CD, with a
Scale of Equal Parts.
lst. From any scale of equal parts take three in your dividers, and,
with one foot in D, cut the line CD in B.
2nd. From    the same scale take four parts in your dividers, and, with
one foot in D, draw an indefinite arc toward E.
3rd. With a radius equal to five of the same parts, and one foot of
the dividers in B, eut the other arc in E.
4th. From    E draw   E D, which is the perpendicular required.
NOTE.-Ist. If four parts were first taken in the dividers and laid off on the line CD, then
three parts should be used for striking the indefinite arc at A, and the five parts struck from
the point C, which would give the intersection A and arrive at the same result.
2nd. On referring to the definitions of angles, it will be found that that side of a right-angled
triangle opposite the right angle is called the hypothenuse; thus the line EB is the hypothenuse of the triangle EDB.
3rd. The square of the hypothenuse of a right-angled triangle is equal to the sum of the
squares of both the other sides.
4th. The square of a number is the product of that number multiplied by itself. Example:
The length of the side DE is 4, which, multiplied by 4, will give 16. The length of DB is 3,
which, multiplied by 3, gives for the square 9. The products of the two sides added together
give 25. The length of the hypothenuse is 5, which, multiplied by 5, gives 25.
PLATE    11
CONSTRUCTION AND DIVISION OF ANGLES.
PROBLEM     9, FIGURE 1.
The Length of the Sides of a Triangle AB, CD, and EF            being given to
Construct the Triangle, the Two Longest Sides to be Joined Together at A.
lst. With the length of the line CD for a radius, and one foot in A,
draw an arc at G.
2nd. With the length of the line EF for a radius, and one foot in
B, draw an arc cutting the other arc at G.




6                            CONSTRUCTION.
3rd. From the point of intersection draw     GA   and GB, which complete the figure.
PROBLEM 10, FIGURE 2.
To Construct an Angle at K    Equal to an Angle at H.
Ist. From HJ, with any radius, draw     an arc cutting the sides of the
angle as at MN.
2nd. From K, with the same radius, describe an indefinite arc at O.
3rd. Draw KO parallel to HM.
4th. Take the distance from M to N and apply it from        OP.
5th. Through P draw KP, which completes the figure.
PROBLEM     11, FIGURE 3.
To Bisect the Given Angle Q by Right Line.
lst. With any radius, and one foot of the dividers in       Q, draw   an
arc cutting the sides of an angle as in R and S.
2nd. With the same or any other radius greater than one-half RS,
from the points S and R, describe two arcs cutting each other at T.
3rd. Draw TQ, which divides the angle equally.
NOTE.-This problem may be very usefully applied by workmen on many occasions. Suppose, for example, the corner Q be the corner of a room, and a washboard or cornice has to be
fitted around it. First apply the level to the angle and lay it down on a piece of board; bisect
the angle as above, then set the level to the center line, and you have the exact angle for cutting the miter. This rule will apply equally to the internal or external angle. Most good,
practical workmen have several means for getting the cut of the miter, and to them this demonstration will appear unnecessary; but I have seen many men make sad blunders for want of
knowing this simple rule.
PROBLEM     12, FIGURE 4.
To Trisect an Angle.
Ist. From   the angular point V, with any radius, describe an arc
cutting the sides of the angle, as in X and W.
2nd. With the same radius, from the points X and W, cut the arc
in Yand Z.
3rd. Draw   YV and ZV, which will divide the angle as required.
PROBLEM     13, FIGURE 5.
In the Triangle ABC, to Describe a Circle Touching all its Sides.
lst. Bisect two of the angles by Problem      11, as A  and B, the di



PLATE XII.
COISTRPIGUC7TOT OF POLTGONS.
Hig.l.      Fig. 2 o
o   ' 0        O
AXo B                        oo 0o
Fig.3             Fig. 4
D                  C
A              B,..
A.                 D
Fig.6
Fig.5                   /
O0 0r           fi AiI       B








CONSTRUCTION.                                 7
viding lines, will cut each other in D, then D is the center of the
circle.
2nd. From D let fall a perpendicular to either of the sides, as at F,
then  DF    is the radius with which to describe the circle from        the
point D.
PROBLEM 14, FIGURE 6.
On the Given Line AB to Construct an Equilateral Triangle, the Line
AB to become One of its Sides.
lst. With a radius equal to the given line, from the points A and B,
draw two arcs intersecting eacl other in C.
2nd. From C draw CA and CB to complete the figure.
PLATE 12.
CONSTRUCTION OF POLYGONS.
PROBLEM     15, FIGURE 1.
On a Line AB      to Construct a Square whose Side shall be Equal to
the Given Line.
Ist. With the length A B for a radius, from the points A and B, describe two arcs cutting each other in C.
2nd. Bisect the arc CA or CB in D.
3rd. From   C, with a radius equal to CD, cut the arc BE in E, and
the arc AF in F.
4th. Draw AE, EF and FB, which complete the square.
PROBLEM     16, FIGURE 2.
In the Given Square GHKJ to Inscribe an Octagon.
1st. Draw the diagonals GK and IHJ intersecting each other in P.
2nd. With a radius equal to half the diagonal from           the corners
GHfK and J, draw arcs cutting the sides of the square in 0, O, 0, etc.
3rd. Draw the lines 0, 0, O, 0, etc., and they will complete the octagon.
NoTE.-This mode is used by workmen when they desire to make a piece of wood for a roller,
or any other purpose. It is first made square and the diagonals drawn across the end; the distance of one-half the diagonal is then set off, as from G to R in the diagram, and a gauge set
froin Hto R, which, run on all the corners, gives the lines for reducing the square to an octagon; the corners are again taken off, and finally finished with a tool appropriate to the purpose. The center of each face of the octagon gives a line in the circumference of the circle,
running the whole length of the piece; and, as there are eight of those lines equidistant from
each other, the further steps in the process are very simple.




8                       CONSTRUCTION.
PROBLEM 17, FIGURE 3.
In a Given Circle to Inscribe an Equilateral Triangle, a Hexagon, and
Dodecagon.
Ist. For the triangle: With the radius of the given circle from any
point in the circumference, as at A, describe an arc cutting the circle
in B and 0.
2nd. Draw the right line BC, and, with a radius equal to BCfrom
the points B and C, cut the circle in D.
3rd. Draw DB and DC, which complete the triangle.
4th. For the hexagon: Take the radius of the given circle and carry
it round on the circumference six times; it will give the points
ABEDFC; through them draw the sides of the hexagon. The radius
of a circle is always equal to the side of an hexagon inscribed.
5th. For the dodecagon: Bisect the arcs between the points found
for the hexagon, which will give the points for inscribing the dodecagon.
PROBLEM 18, FIGURE 4.
In a Given Circle to Inscribe a Square and an Octagon.
lst. Draw a diameter AB, and bisect it with a perpendicular by
Problem 1, giving the points CD.
2nd. From the points A CBD draw the right lines forming the sides
of the square required.
3rd. For the octagon: Bisect the sides of the square, and draw perpeudiculars to the circle, or bisect the arcs between the points A CBD,
which will give the other angular points of the required octagon.
PROBLEM 19, FIGURE 5.
On the Given Line OP to Construct a Pentagon, OP being the Length of
the Side.
Ist. With the length of the line OP from 0, describe the semicircle
PQ, meeting the line PO extended in Q.
2nd. Divide the semicircle into five equal parts, and from O draw
lines through the divisions 1, 2, and 3.
3rd. With the length of the given side from P cut 0 1 in S, from
S cut O 2 in R, and from Q eut O 2 in R; connect the points O QRSP
by right lines, and the pentagon will be complete.




PLATE lSxII.
Fig.l \Ê             Fig.2 n
JFiOBLE3'IS id  i    O, uI I GII
F  _  _  _3           '
F/7     I
01(1(
Fig.5                   / i ', Fi,6








CONSTRUCTION.                             9
PROBLEM    20, FIGURE 6.
On the Given Line AB to Construct a Heptagon, AB     being the Length
ofthe Side.
Ist. From A, with AB for a radius, draw the semicircle BH.
2nd. Divide the semicircle into seven equal parts, and from      A,
through 1, 2, 3, 4, and 5, draw indefinite lines.
3rd. From B cut the line A   1 in C, from C cut A 2 in D, from    G
cut A 4 in F, and from F cut A 3 in E; connect the points by right
lines to complete the figure.
NOTE.-Any polygon may be constructed by this method; the rule is to divide the semicircle into as many equal parts as there are sides in the required polygon, draw the lines
through all the divisions except two, and proceed as above. Considerable care is required to
draw these figures accurately, on account of the difficulty of finding the exact points of intersection. They should be practiced on a much larger scale
PLATE 13.
PROBLEMS RELATING TO THE CIRCLE.
PROBLEM 21, FIGURE 1.
To Find the Center of a Circle.
Ist. Draw any chord, as AB, and bisect it by a perpendicular ED,
which is a diameter of the circle.
2nd. Bisect the perpendicular ED by Problem 1; the point of intersection is the center of the circle.
FIGURE 2.
Another Method of Finding the Center of a Circle.
lst. Join any three points in the circumference as FGH.
2nd. Bisect the chords FG and GH by perpendiculars; their point
of intersection at Cis the center required.
PROBLEM    22, FIGURE 3
To Draw a Circle Through any Three Points not in a Straight Line, as
NJN O.
lst. Connect the points by straight lines, which will be chords to
the required circle.




10                        CONSTRUCTION.
2nd. Bisect the chords by perpendiculars; their point of intersection at C is the center of the required circle.
3rd. With one foot of the dividers at C, and a radius equal to
CMICNor CO, describe the circle.
PROBLEM 23, FIGURE 4.
To Find the Center for Describing the Segment of a Circle.
Ist. Let PR be the chord of the segment, and PS the rise.
2nd. Draw the chords PQ and QR, and bisect them by perpendiculars; the point of intersection at C is the center for describing the
segment.
PROBLEM 24, FIGURE 5.
To Find a Right Line Nearly Equal to an Arc of a Circle as HIK.
lst. Draw the chord HK and extend it indefinitely towards 0.
2nd. Bisect the segment in I and draw chords III and IK.
3rd. With one foot of the dividers in H, and a radius equal to HI,
eut HO in  1; then, with the same radius, and one foot in M, eut HO
again, in N.
4th. Divide the difference, KN, into three equal parts, and extend
one of them towards O; then will the right line HO be nearly equal
to the curved line HIIK.
PROBLEM 25, FIGURE 6.
To Find a Right Line Equal to the Semi-circumference AFB.
lst. Draw the diameter AB and bisect, by the perpendicular FH,
indefinitely toward G.
2nd. Divide the radius CHinto four equal parts, and extend three
of these to G.
3rd. At F draw an indefinite right line DE.
4th. From G, through A, at the end of the diameter AB, draw
GAD, cutting the line DE in D; and from G through B, draw GBE
cutting DE iq E; then will the line DE    be equal to the semicircumference of the circle, and the triangles DGE and AGB will
be equilateral.




PLATE XIIV
PARAB OLA.
Fi.l 
s  \
C    \" __ c
Y iF
(fI^
i____~ ~~      __~~~~____._J~a








PARABOLA.                                11
NOTE.-The right lines found by Problems 24 and 25 are not mathematically equal to the
respective curves, but are sufficiently correct for all practical purposes. Worknmen are in the
habit of using the following method for finding the length of a curved line:
They open their compasses to a small distance, and commencing at one end, step off the whole
curve, noting the number of steps required and the remainder less than a step, if any; they
then step off the same number of times, with the same distance, on the article to be bent
around it, and add the remainder, which gives them a length sufficiently true for their purpose. The error in this method amounts to the sum of the differences between the arc cut off
by each step and its chord.
PLATE 14, FIGURE 1.
The Focus and Directrix of a Parabola being given, to Describe the
Curve.
FIRST METHOD.-By Points.
Let F be the focus and Bb the directrix of a parabola.     Through F
draw DC perpendicular to Bb, and bisect FD in V; then, since D V
VF, V is a point on the curve, and CV is the axis of the parabola.
To find other points of the curve, draw     any number of lines, Aa,
A'a', A"a", etc., perpendicular to CD; then, with the distances DC,
DC', DC", etc., as radii, and the focus F      as a center, describe arcs
intersecting the perpendiculars in A, A', A", etc. The points A, A',
A", etc., in which the arcs cut the perpendiculars are points of the
curve.
For FA    - DC=    AB (Def. 1).
We may thus determine as many points on the curve as we please,
and the curve line which passes through all the points V, A, A', A",
etc., will be the parabola whose focus is F, and directrix Bb.
Cor.   The same radius determines two points of the curve, one above
and one below   the axis.   Since AF-    aF, FC is common to the two
triangles AFC, aFC, and the angles at Care right angles; therefore
AC -- aC; that is, every straight line terminated by the curve and
perpendicular to the axis is bisected by it; and, consequently, the
parabola consists of two equal branches similarly situated with respect
to the axis.
Moreover, since the radius FA is always greater than FC, the arc
described with F as a center will always intersect the corresponding
perpendicular, and there is, therefore, no limit t-o the distance to which
the curve may extend on both sides of the axis.
11




12,                        ELLIPSE.
PLATE 14, FIGURE 2.
SECOND METHOD.-By Continuous Motion.
Let BC be a ruler whose edge coincides with the directrix of the
parabola, and let DEG be a square. Take a thread equal in length to
EG, and attach one extremity of it at G, and the other at the focus F.
Then slide the side of the square DE along the ruler BC, and at the
same time keep the thread continually stretched by means of the point
of a pencil, A, in contact with the square; the pencil will describe one
part of the required parabola. For, in every position of the square,
AF+AG-=AE+ AG,
and hence, AF- AE; that is, the point A is always equally distant
from the focus Fand the directrix BC.
If the square be turned over and moved on the other side of the
point F the other part of thé same parabola may be described.
PLATE 15, FIGURE 1.
The Major Axis and Foci of an Ellipse being given, to Describe the
Curve.
FIRST METIHOD.-By Points.
Let AA' be the major axis and FF' the foci of an ellipse. Take
E any point between the foci, and from F and F' as centers, with distances AE, A'E as radii, describe two circles cutting each other in the
point D; D will be a point on the ellipse. For, join FD, F'D; then
DF   - DF' = EA-  - EA' - AA'; and, at whatever point between
the foci E is taken, the sum of DFand DF' will be equal to AA'.
Hence, by Def. 1, D is a point on the curve; and in the same manner
any number of points in the ellipse may be determined.
Cor. The same circles determine two points of the curve D and )',
one above and one below the major axis. It is also evident that these
two points are equally distant from the axis; that is, the ellipse is
symmetrical with respect to its major axis, and is bisected by it.




PLATE XV 
Fi.1
rr"   ]        Fi. 2








HYPERBOLA.                         13
PLATE 15, FIGURE 2.
SECOND METHOD.-By Continuous Motion.
Take a thread equal in length to the major axis of the ellipse, and
fasten one of its extremities at F, the other at F'.  Then let a pencil
be made to glide along the thread, so as to keep it always stretched;
the curve described by the point of the pencil will be an ellipse. For,
in every position of the pencil, the sum of the distances DF, DF' will
be the same, viz.: equal to.the entire length of the string.
Scholiun. The ellipse is evidently a continuous and closed curve.
PLATE 16, FIGURE 1.
The Transverse Axis and Foci of an Hyperbola being given, to Describe
the Curve.
FIRST METHOD.-By Points.
Let AA' be the transverse axis, and FF' the foci of an hyperbola.
In the transverse axis AA' produced, take point E, and from F and F'
as centers, with the distances AE, A'E as radii, describe two circles
cutting each other in the point D; D will be a point in the hyperbola.
For, join FD, F'D; then DF'- DF=- EA'- EA - AA'; and at
whatever point of the transverse axis produced E is taken, the difference between DF' and DF will be equal to AA'. Hence, by Def. 1,
D is a point on the curve; and, in the same manner, any number of
points in the hyperbola may be determined. In a similar manner the
opposite branch may be constructed.
Cor. The same circle determine two points of the curve D and D',
one above and one below the transverse axis.
It is also evident that these two points are equally distant from the
axis; that is, the hyperbola is symmetrical with respect to its transverse axis.
PLATE 16, FIGURE 2.
SECOND METHOD.-By Continuous Motion.
Take a ruler longer than the distance FF', and fasten one of its extremities at the point F'. Take a, thread shorter than the ruler, and




14                      HYPERBOLA.
fasten one end of it at F and the other to the end H of the ruler.
Then move the ruler HDF' about the point F', while the thread is
kept constantly stretched by a pencil pressed against the ruler; the
curve described by the point of the pencil will be a portion of an
hyperbola. For, in every position of the ruler, the difference of the
lines DFDF' will be the same, viz.: the difference between the length
of the ruler and the length of the string.
If the ruler be turned and move on the other side of the point F,
the other part of the same branch may be described.
Also, if one end of the ruler be fixed in F, and that of the thread
in F', the opposite branch may be described.
It is evident that each portion of each branch will extend to an indefinitely great distance from the foci and center.




r                  d








INDEX.
FRONTISPIECE.                                                                                                PAGE.
DEDICATION......................................................................................  1
TITLE     PAGE...................................................................................................................  3
PREFACE.....................................................................................                       5
INTRODUCTION.
History of the Quadrature of the Circle..............................................................................  9
General Remarks and Examples Illustrating the Quadrature of the Circle........................                   34
PART FIRST,
Or the Geometrical and Final Solution of the Quadrature of the Circle, by an entirely New
Method, Together with Ample Proofs of the Same..................................................           65
D definitions........................................................................................................................  66
Propositions.....................................................................................................................  75
The Quadrature of the Circle........................................................................................  82
Proofs...............................................................................................................................  117
PART SECOND,
Or the Square Root of Two, being a Common                  Measure of the Side and Diagonal of'the
Square to Infinity................................................................................................ 135
The Square Root of Two, by Division alone, to 144 places of Figures.................................. 140
PART THIRD,
Or the Right-Angled Triangle, Containing a Variety of Methods' for Finding an Infinite
Series of Right-angled Triangles, the Sides of which shall be in Whole Numbers,
Together with a New Rule for the Solution of the Riglat-Angled Triangle, Published
for the First Time......................................................................................... 143
APPENDIX,
Or a Chapter on        Construction, for the Use and          Benefit of Draughtsmen, Builders, and
Mechanics...................................................................................         152
To Erect or Let Fall a Perpendicular........................................................  1
Construction and Division of Angles.............................................................................   5
Construction of Polygons...........................................................                 7
Problems Relating to the Circle..................................................................................   9
To Describe a Parabola.........................................................................................  11
To Describe an Ellipse..................................................................     12
To Describe an Hyperbola......................................................................................... 13