MATH/STAT.
T.
SUNDARA ROW S
Geometric Exercises
in
Paper Folding
Edited and Revised by
WOOSTER WOODRUFF BEMAN
PROFESSOR OF MATHEMATICS
IN
THE UNIVERSITY OF MICHIGAK
and
DAVID EUGENE SMITH
PROFESSOR OF MATHEMATICS IN TEACHERS COLLEGE
OF COLUMBIA UNIVERSITY
1
WITH
87
ILLUSTRATIONS
THIRD EDITION
CHICAGO
LONDON
THE OPEN COURT PUBLISHING COMPANY
:::
1917
&
XC?
4255
COPYRIGHT BY
THE OPEN COURT PUBLISHING Co,
1901
PRINTED
IN
THE UNITED STATES OF AMERICA
hn
HATH
EDITORS PREFACE.
OUR
was
attention
rical Exercises in
lesungen
An
first
Paper Folding by
iiber ausgezucihlte
Sundara
attracted to
a reference in Klein
its
Vor-
s
Fragen der Elementargeometrie.
examination of the book, obtained after
convinced us of
Row s Geomet
undoubted merits and
American teachers and students
many
vexatious delays,
of its probable value to
of geometry.
Accordingly we
sought permission of the author to bring out an edition in this
country, wnich permission was most generously granted.
The purpose
of the
book
we need
is
so fully set forth in the author s
it is sure to
prove of
wide-awake teacher of geometry from the graded
the college.
The methods are so novel and the results
introduction that
only to say that
interest to every
school to
so easily reached that they cannot
Our work
fail to
slight modifications of the proofs,
references,
awaken enthusiasm.
as editors in this revision has been confined to
some additions
in the
some
way
of
and the insertion of a considerable number of half-tone
reproductions of actual photographs instead of the line-drawings
of the original.
W. W. BEMAN.
D. E. SMITH.
OK02
CONTENTS.
PAGE
vn
Introduction
I.
II.
III.
m
The Square
The Equilateral Triangle
The Pentagon
V.
The Hexagon
The Octagon
VII.
VIII.
IX.
3
45
the
Dodecagon
The Pentedecagon
5
67
XII. General Principles
82
Sections.
i.
Section n.
Section in.
Section
47
52
XI. Polygons
Section
35
39
The Nonagon
The Decagon and
The Conic
14
,
X. Series
XIII.
9
Squares and Rectangles
IV.
VI.
i
iv.
The Circle
The Parabola
The Ellipse
The Hyperbola
XIV. Miscellaneous Curves
102
115
121
126
131
INTRODUCTION.
book was suggested to me by
The
Kindergarten Gift No. VIII.
Paper-folding.
hundred
colored
consists
of
two
gift
variously
squares
THE
idea of this
of paper, a folder, and diagrams
folding.
The paper
is
The paper may, however, be
both
will
sides.
and instructions
for
colored and glazed on one side.
of self-color, alike
on
In fact, any paper of moderate thickness
answer the purpose, but colored paper shows the
creases better, and
garten
gift is sold
is
more
attractive.
by any dealers
The kinder
in school supplies
;
but colored paper of both sorts can be had from sta
tionery dealers.
Any
sheet of paper can be cut into
a square as explained in the opening articles of this
book, but
ready
2.
it is
neat and convenient to have the squares
cut.
These txercises do not require mathematical
instruments, the only things necessary being a pen
knife and scraps of paper, the latter being used for
setting off equal lengths.
The squares
are themselves
simple substitutes for a straight edge and a
3.
T
square.
In paper-folding several important geometric
processes can be effected
much more
easily than with
INTR OD UC TION.
viii
a pair of compasses and ruler, the only instruments
the use of which
etry
;
for
is
sanctioned in Euclidean
two or more equal
into
and parallels
parts, to
to straight lines.
draw perpendiculars
It
is,
however, not
possible in paper-folding to describe a circle,
number
may
geom
example, to divide straight lines and angles
of points
on a
circle, as well
as other curves,
These exercises
be obtained by other methods.
do not consist merely
of
but a
drawing geometric figures
and fold
involving straight lines in the ordinary way,
ing
upon them, but they require an
intelligent appli
cation of the simple processes peculiarly adapted to
This will be apparent
paper-folding.
mencement
4.
at the
very
com
of this book.
The use of
the kindergarten gifts not only affords
interesting occupations to boys and girls, but also
prepares their minds for the appreciation of science
and art. Conversely the teaching of science and art
on can be made interesting and based upon
proper foundations by reference to kindergarten occu
later
pations.
This
is
particularly the case with geometry,
which forms the basis
of every science
teaching of plane geometry in schools
The
can be made
and
art.
very interesting by the free use of the kindergarten
gifts.
It
would be perfectly legitimate
pils to fold the
give
diagrams with paper.
them neat and accurate
figures,
to require
and impress the
truth of the propositions forcibly on their minds.
would not be necessary
to take
pu
This would
any statement on
It
trust.
INTR OD UC TION.
But what
is
the imagination and ideal
clumsy figures can be seen in the concrete.
would be impossible.
isation of
A
now realised by
ix
fallacy like the following
To prove that every
5.
ABC,
Fig.
Z draw ZO
through
angle ACB by
CO.
is
triangle
be any triangle.
1,
Bisect
isosceles.
Let
AB
and
AB.
perpendicular to
in Z,
Bisect the
*
2
A
Fig.
B
i.
CO and ZO do not meet, they are parallel.
Therefore CO is at right angles to AB. Therefore
(1) If
AC^BC.
If
(2)
Draw
AC.
to
CO
ZO
and
do meet,
OX perpendicular
Join
OA
y
OB.
to
By Euclid
88, cor. 7)* the triangles
*
etry,
let
them meet
in O.
BC and OY perpendicular
YOC
These references are to Beman and Smith
Boston, Ginn & Co., 1899.
I,
and
s
26 (B. and S.,
XOC
are con-
New Plane and Solid Geom
INTRODUCTION.
x
gruent; also by Euclid
156 and
I,
79) the triangles
47 and
I,
8 (B.
AOY and BOX
and
S.,
are con
Therefore
gruent.
AY+ YC=BX+XC,
i.e.,
AC^BC.
shows by paper- folding that, whatever tri
angle be taken, CO and ZO cannot meet within the
Fig. 2
triangle.
Fig.
O
is
2.
the mid-point of the arc
A OB
of the circle
which circumscribes the triangle ABC.
6.
Paper-folding
is
not quite foreign to us.
ing paper squares into natural objects
Fold
a boat, double
IN TR OD UC T1ON.
boat, ink bottle, cup-plate, etc.,
xi
is
well known, as
symmetric forms for pur
In writing Sanskrit and Mah-
also the cutting of paper in
poses of decoration.
to
rati, the paper is folded vertically or horizontally
keep the
lines
and columns
ters in public offices
straight.
an even margin
ing the paper vertically.
is
In copying let
secured by fold
Rectangular pieces of paper
folded double have generally been used for writing,
of machine-cut letter pa
of
various
and
sizes, sheets of convenient
envelopes
per
size were cut by folding and tearing larger sheets, and
and before the introduction
the second half of the paper was folded into an envel
ope inclosing the first half. This latter process saved
paper and had the obvious advantage of securing the
post marks on the paper written upon. Paper-folding
has been resorted to in teaching the Xlth Book of
Euclid, which deals with figures of three dimensions.*
But
has seldom been used in respect of plane
it
fig
ures.
have attempted not to write a complete trea
tise or text-book on geometry, but to show how reg
ular polygons, circles and other curves can be folded
7.
I
or pricked on paper.
I
have taken the opportunity to
known problems of
ancient and modern geometry, and to show how alge
bra and trigonometry may be advantageously applied
introduce to the reader
to
some
well
geometry, so as to elucidate each of the subjects
which are usually kept
* See
especially
in separate pigeon-holes.
Beman and Smith
s
New Plane and Solid Geometry, p. 287.
INTR OD UC TION.
xii
The
8.
first
nine chapters deal with the folding of
the regular polygons treated in the
first
four books of
The paper square of the
has
taken
been
as the foundation, and
kindergarten
Euclid, and of the nonagon.
the other regular polygons have been
thereon. Chapter
is
to
be cut and
I
worked out
shows how the fundamental square
how
it
can be folded into equal right-
angled isosceles triangles
and squares.
Chapter II
deals with the equilateral triangle described on one of
the sides of the square.
Chapter III is devoted to
the Pythagorean theorem (B. and S.,
propositions of the
156) and the
second book of Euclid and certain
puzzles connected therewith.
It is
also
shown how
a
right-angled triangle with a given altitude can be de
This is tantamount to find
scribed on a given base.
ing points on a circle with a given diameter.
9. Chapter X deals with the arithmetic, geometric,
and harmonic progressions and the summation of cer
tain arithmetic series.
lines
In treating of the progressions,
whose lengths form
tained.
A
a progressive series are ob
rectangular piece of paper chequered into
squares exemplifies an arithmetic series.
For the geo
metric the properties of the right-angled triangle, that
the altitude from the right angle
is
a
mean propor
between the segments of the hypotenuse (B.
S.,
270), and that either side is a mean propor
tional between its projection on the hypotenuse and
tional
and
the hypotenuse, are
made
use
of.
In this connexion
the Delian problem of duplicating a cube has been
INTRODUCTION.
xiii
In treating of harmonic progression, the
explained.*
fact that the bisectors of
an interior and correspond
ing exterior angle of a triangle divide the opposite
side in the ratio of the other sides of the triangle (B.
and
esting
This affords an inter
249) has been used.
S.,
method
involution.
of graphically explaining
The sums
of the natural
systems
numbers and
in
of
cubes have been obtained graphically, and the
sums of certain other series have been deduced there
their
from.
Chapter XI deals with the general theory of
regular polygons, and the calculation of the numerical
10.
value of
7t.
The
propositions in this chapter are very
interesting.
Chapter XII explains certain general princi
which have been made use of in the preceding
11.
ples,
congruence, symmetry, and similarity of
chapters,
figures, concurrence of straight
of points are touched upon.
lines,
and collinearity
Chapters XIII and XIV deal with the conic
sections and other interesting curves.
As regards
12.
harmonic properties among others are
The theories of inversion and co-axial circles
the circle,
treated.
its
As regards other curves it is
shown how they can be marked on paper by paperThe history of some of the curves is given,
folding.
are also explained.
and
it is
shown how they were
utilised in the solution
*See Beman and Smith s translation of Klein s Famous Problems of Ele
mentary Geometry, Boston, 1897; also their translation of Fink s History of
Mathematics, Chicago, The Open Court Pub. Co.,
1900.
INTR OD UC TION.
xi v
of the classical problems, to find
between two given
lineal angle.
lines,
and
two geometric means
to trisect a
given recti
Although the investigation of the prop
erties of the curves involves a
knowledge
mathematics, their genesis is easily
of
advanced
understood and
is
interesting.
have sought not only to aid the teaching of
geometry in schools and colleges, but also to afford
13.
I
mathematical recreation
tractive
may
to
and cheap form.
find the
book useful
young and
"Old
old, in
an at
like
mysell
boys"
to revive their old lessons,
have a peep into modern developments which,
although very interesting and instructive, have been
and
to
ignored by university teachers.
T.
MADRAS,
INDIA, 1893.
SUNDARA Row.
THE SQUARE.
I.
The upper
1.
a table
upon
side which
2.
of the
is
side of a piece of paper lying flat
a plane surface, and so
is
is
the lower
with the table.
in contact
The two surfaces are separated by the material
paper. The material being very thin, the other
sides of the paper do not present appreciably broad
surfaces,
and the edges
The two
lines.
of the
paper are practically
surfaces though distinct are insepa
rable from each other.
Look
3.
shown
at the irregularly
in Fig. 3,
and
at this
shaped piece of paper
page which is rectangu
Let us try and shape the former paper like the
lar.
latter.
4
Place the irregularly shaped piece of paper
upon the
table,
and
fold
it flat
be the crease thus formed.
upon
itself.
It is straight.
Let
XX
Now
pass
a knife along the fold und separate the smaller piece.
We
thus obtain one straight edge.
5.
Fold the paper again as before along BY, so
X X is doubled upon itself. Unfolding
that the edge
the paper,
to the
we
edge
see that the crease
X X.
It is
^Kis
at right
angles
evident by superposition that
GEOMETRIC EXERCISES
the angle
YBX
equals the angle
XBY, and
of these angles equals an angle of the page.
Fig.
3-
a knife as before along the second fold
the smaller piece.
that each
Now pass
and remove
IN PAPER FOLDING
Repeat the above process and obtain the edges
CD and DA. It is evident by superposition that the
are right angles, equal to one
angles at A, B, C,
6.
>,
another, and that the sides
BC,
Fig. 4
equal to
DA, AB.
CD
are respectively
.
This piece
of
paper (Fig. 3)
is
similar in shape to the page.
7.
It
can be made equal
in size to the
page by
taking a larger piece of paper and measuring
and
BC equal
to the sides of the latter.
off
AB
GEOMETRIC EXERCISES
4
8.
A
figure like this
superposition
right angles
all
proved that
it is
and
called a rectangle.
is
all
By
(1) the four angles are
equal, (2) the four sides are not
equal. (3) but the two long sides are equal, and so
also are the
9.
and
two short
Now take a
fold
it
sides.
rectangular piece of paper,
A B CD,
obliquely so that one of the short sides,
Fig.
CD,
5-
upon one of the longer sides, DA as in Fig. 4.
Then fold and remove the portion A B BA which
Unfolding the sheet, we find that ABCD
overlaps.
falls
is
,
now
and
square,
all its
10.
i.
e.
,
its
four angles are right angles,
sides are equal.
The
crease which passes through a pair of th&
IN PAPER FOLDING
5
opposite corners B, D, is a diagonal of the square.
One other diagonal is obtained by folding the square
through the other pair of corners as in Fig. 5.
n. We see that the diagonals are at right angles
to each other, and that they bisect each other.
12.
The
point of intersection of the diagonals
is
called the center of the square.
Fig.
13.
Each diagonal
6.
divides the square into two con
gruent right-angled isosceles triangles,
whose
vertices
are at opposite corners.
14.
The two diagonals together
into four congruent right-angled
whose
divide the square
isosceles triangles,
vertices are at the center of the square.
GEOMETRIC EXERCISES
6
Now
15.
one side
fold again, as in Fig. 6, laying
of the square
side.
We
get a crease
upon
opposite
which passes through the center
of the square.
at right angles to the other sides
and
(2)
it
itself
is
its
also parallel to the first
bisected at the center
Fig.
;
(4)
(1) bisects
two sides
it
;
It is
them;
(3)
it is
divides the square
7.
into two congruent rectangles, which are, therefore,
each half of it; (5) each of these rectangles is equal
to
one
of
the
triangles into
which either diagonal
divides the square.
Let us fold the square again, laying the re
maining two sides one upon the other. The crease
16.
IN PAPER FOLDING
now
7
obtained and the one referred to in
15 divide
the square into four congruent squares.
again through the corners of the
smaller squares which are at the centers of the sides
17.
Folding
of the larger square,
we
scribed in the latter.
obtain a square which
(Fig. 7.)
Fig.
18.
the
This square
same
19.
is
8.
half the larger square,
By
joining the mid-points of the sides of the
we
obtain a square which
of the original square (Fig. 8).
to
and has
center.
inner square,
cess,
is in
we can
obtain any
one another as
number
By
is
one-fourth
repeating the pro
of squares
which are
GEOMETRIC EXERCISES
8
1
l
"2*
"4"
1
JL etc
~8~
16
Each square
i.
or
i L
"2"
half
is
2^
:
L
jp
24
*
the next larger square,
of
the four triangles cut from each square are to
e.,
The sums of all these
any number cannot exceed
gether equal to half of
angles increased to
it.
tri
the
original square, and they must eventually absorb the
whole of it.
Therefore
20.
The
-f
^+^+
etc
-
center of the square
to infinity
is
=
1.
the center of
The
its
circumscribed and inscribed
circles.
touches the sides
mid-points, as these are
at
their
latter circle
nearer to the center than any other points on the
sides.
21.
Any
divides
it
crease through the center of the square
into
two trapezoids which are congruent.
A
second crease through the center at right angles to
the
first
divides the square into four congruent quadri
laterals, of
The
each
which two opposite angles are right angles.
quadrilaterals are concyclic,
lie in
a circumference.
i.
e.,
the vertices of
II.
22.
and
THE EQUILATERAL TRIANGLE.
Now
fold
it
take this square piece of paper (Fig. 9),
double, laying two opposite edges one upon
the other.
We
obtain a crease which passes through
Fig. 9.
the mid-points of the remaining sides and
angles to those sides.
fold through
it
Take any point on
and the two corners
of the
is at
right
this line,
square which
GEOMETRIC EXERCISES
io
are on each side of
it.
We
thus get isosceles triangles
standing on a side of the square.
23. The middle line divides the isosceles triangle
into two congruent right-angled triangles.
24.
The vertical angle is bisected.
we so take the point on the middle
25. If
Fig.
its
line, that
io.
distances from two corners of the square are equal
to a side of
it,
we
shall obtain
This point
an equilateral triangle
easily determined by turning
the base AB through one end of it, over A A until the
other end, B, rests upon the middle line, as at C.
(Fig. 10).
is
,
26.
Fold the equilateral triangle by laying each
n
IN PAPER FOLDING
of
the sides upon the base.
We
thus obtain the
A A BB CC,
three altitudes of the triangle, viz.:
,
,
(Fig. 11).
27.
Each
of the altitudes divides the triangle into
two congruent right-angled
28.
They
triangles.
bisect the sides at right angles.
Fig. ix.
29.
They pass through
a
common
30. Let the altitudes AA and
Draw BO and produce it to meet
will now be proved to be the third
point.
CC
AC
meet
in
B
altitude.
in
.
O.
BB
From
From
C OA and CO A OC =OA
and A OB, ^OBC ^^A BO. Again
triangles
from triangles ABB and CB B, /_AB B = /_BB C,
the triangles
OC B
,
.
GEOMETRIC EXERCISES
12
i.
e.,
each of them
an altitude
31.
OC
of the equilateral triangle
AC in B
bisects
It
That
a right angle.
is
BOB
is,
ABC.
It
is
also
.
can be proved as above that OA, OB, and
and that OA OB and OC are also
are equal,
,
,
equal.
32. Circles can therefore be described with
O
as a
center and passing respectively through A, B, and
and through
A B
,
,
and C
.
The
latter circle
C
touches
the sides of the triangle.
The
33.
six
set
equilateral triangle
ABC
divided into
is
congruent right-angled triangles which have one
of their equal angles at O, and into three congru
ent,
symmetric, concyclic quadrilaterals.
The triangle A OC is double the triangle
34.
therefore,
AO = 2OA
CO = 2OC
Hence
.
circle of triangle
Similarly,
.
A OC;
BO=r-2OL and
the radius of the circumscribed
ABC
is
twice the radius of the in
scribed circle.
The
35.
right angle A, of the square,
trisected
is
lines AO, AC.
Angle BAC=^\ of
The angles C A O and OAB are each
by the straight
a
right angle.
i
36.
The
37.
Fold through
Then A
six angles at
BC
is
of the triangle
38.
are each
AB BC
ABC.
CA
CA, and halves
O
,
,
of a right angle.
and
an equilateral triangle.
AB BC
,
B and C.
Similarly with the angles at
of a right angle.
of
,
them.
CA
(Fig. 12).
It is
are each parallel to
a fourth
AB, BC,
JN PAPER FOLDING
39.
ACA B
is
a
rhombus.
CB C A
CA
40. A B B C
So are C
13
BA B
and
.
,
,
bisect the corresponding alti
tudes.
CC
41.
= 0.866....
Fig.
42.
The
ratio of
:
1/3
:
rectangle of AC and CC
=
AB IVZ-AB? = 0.433 .... X^
The A ABC=
X \V*
43.
12.
angles of the triangle
1:2:3, and
1/4
.
its
AC C
i.
,
e.
2
-
are in the
sides are in the ratio of 1/1
III.
SQUARES AND RECTANGLES.
Fold the given square as in Fig. 13. This
affords the well-known proof of the Pythagorean the44.
Fig. 13.
FGH being a right-angled triangle, the square
FH equals the sum of the squares on FG and GH.
orem.
on
It is easily
uFA + u DB = n FC.
proved that FC is a square,
and that
PAPER FOLDING
the triangles
15
FEK
FGH, HBC, KDC, and
are con
gruent.
If
the triangles
the squares
two
FA
FGH
and
HBC
are cut off from
and DB, and placed upon the other
triangles, the square
IiAB = a, GA=b,
FHCK is made
FH=c,
and
up.
then a 2 -f
6*
=
c*.
Fig. 14.
Fold the given square as in Fig. 14. Here the
rectangles AF, BG, Cff, and DE are congruent, as
also the triangles of which they are composed.
45.
EFGH
is
a square as also
Let
KLMN.
AK=a, KB = b,
then a*
and
+
NK =
# =^2,
j.
c,
e
.
uKLMN.
GEOMETRIC EXERCISES
i6
Now
square
ABCD
by the four triangles
overlaps the square
AKN, BLK, CML,
and
But these four triangles are together equal
of the rectangles,
i.
Therefore (a -f
46.
e.,
2
)
KLMN
DNM.
to
two
to lab.
= a + P -f 2ab.
2
EF=a
b, auidnFGJ?=(a
EFGHis less than the square KLMN
triangles FNK GKL, HLM, and EMN.
fi)*.
The square
by the four
t
But these four
angles,
i.
e.,
triangles
of the rect
Zab.
Fig.
47.
make up two
The square
15.
ABCD overlaps
the square
by the four rectangles AF, BG, Clf, and
48. In Fig. 15, the square
EFGH
DE.
ABCD = (a-\-
2
)
,
and
IN PAPER FOLDING
the square
= square
EFGH(a
ELCM = a
2
.
<)
2
.
17
Also square
AKGN
KBLF =
Square
square
T ft
Squares
ABCD
EFGH are
and
together equal to
the latter four squares put together, or to twice the
AKGN
square
(a
+
2
)
-f (a
D
and twice the square
2
)
= 2a
2
2
4- 2^
KBLF,
that
is,
.
O
C
N
Fig. 16.
49. In Fig. 16 the rectangle
(
-*).
Because the rectangle
angle
PL =
square
PK
PL
is
equal to (a-\-ft
EK = FM,
square
AE,
therefore rect
i.
e.,
(a-\-
50. If squares be described about the diagonal of
a given square, the right angle at one corner being
common to them, the lines which join this corner with
the mid-points of
the opposite sides of the given
GEOMETRIC EXERCISES
i8
square bisect the corresponding sides of
squares.
(Fig. 17.)
make with
tude
the inner
the diagonal are equal, and their
constant for
is
all
For the angles which these
all
squares, as
may
lines
magni
be seen by
Fig. 17.
Therefore the mid-points of the sides
superposition.
of the inner squares
51.
lie
on these
lines.
ABCD
being the given square piece of paper
it is
required to obtain by folding, the point
(Fig. 18),
^Tin
must
AB, such that the rectangle
the square on
AB-XB
Lay
EB
G such
that
Take
equal to
AX,
BC
upon itself and take
Fold through E and A.
Double
is
upon
EA
and
EG = EB.
AX=AG.
its
mid-point E.
fold so as to get
EF, and
IN PAPER FOLDING
Then
rectangle
19
AB-XB^AX*.
Complete the rectangle
BCHX
and the square
AXKL.
Let Xffcut
EA
in
M.
FY=FB.
Take
Then F = FG = FY=XM m& XM=\AX.
Fig. 18.
Now, because
to
BY
is
bisected in
F
and produced
A,
Y+FY = AF
*
i,
.-.
AB-AY=AG
1
But
^^ =
2
by
+^C
^(9 2
J
49>
2
,
by
44.
GEOMETRIC EXERCISES
20
AY=XB.
AB is
said to be divided in
X
in
median section.*
Also
i.
e.,
AB is
52.
its
will
A
also divided in
Y in
median
can be described with
circle
section.
F
as a center,
It
circumference passing through B, G, and K
touch .ZL4 at G, because FG is the shortest dis
F to
tance from
EGA.
the line
53. Since
subtracting
BK we have
XKNY= square
rectangle
i.
i.
e.,
AX is
e.,
^.AT-KA^^F2
divided in
CHKP,
,
Kin median
section.
X in
median
Similarly j#Fis divided
in
section.
54.
CD-CP
55. Rectangles
rectangle
^F+ square
56.
Hence
57.
Hence
*
rectangle
"
YD
being each
JfY=
rectangle
The term golden section"
and Solid Geometry, p. 196.
Plane
= AB-XB
=
CK=AX^ AB- XB.
Bff and
rectangle
j^A",
i.
y
e.,
^AT= AX-XBBX*.
is
also used.
See Beman and Smith
s
New
IN PAPER FOLDING
Let
58.
AB = a, XB = x.
Then
x^ = ax, by
(a
+
0a
and #
.-.
...
*
a
(^_jc)
The
21
rect.
= .(3
by
54;
1/5).
= 4-a (1/5
= ^(3
z
2
**=:3<i#,
51.
1)
1/5)
= * X 0.6180.
=^
2
X
.
0.3819.
BPKX
=a
2
(1/5
2)
=a
2
X 0.2360
~
4
59. In the
language
of
proportion
AB ^^T=^^: Xff..
straight line AB is said to be divided
:
The
treme and mean
60.
Let
AB
be divided
Complete the rectangle
XA
through
over
in
XN, NB, and
X
in
CBXH (Fig.
MNO.
X so that A
rectangle by the line
laying
"in
ex
ratio."
7V^.
median
19).
section.
Bisect the
N
Find the point
by
falls on MO, and fold
Then
BAN
is
an
isos-
22
GEOMETRIC EXERCISES
celes triangle having
the angle
its
angles
ABN and BNA double
NAB.
AX=XN=NB
Fig. 19.
IN PAPER FOLD IN C
.-.
and
61.
The
AN=AB
= \ of a right angle.
right angle at
A
can be divided into
60.
Then fold^^
<2;
N
Here
equal parts as in Fig. 20.
is
found as
QAB
bisect
Q
p
five
by
in
folding,
<~
Fig. 20.
fold over the diagonal
Q
,
AC
and thus get the point
/"
To
62.
describe a right-angled triangle, given the
hypotenuse AB, and the altitude.
Fold
(Fig. 21) parallel to
EF
of the
Take
ing
AB
at the distance
given altitude.
GB
G
the middle point of
through
G
so that
AB.
B may
fall
Find
Hby fold
on EF.
GEOMETRIC EXERCISES
24
Fold through
AHB is
H and A,
G, and B.
the triangle required.
Fig. 21.
63.
ABCD
(Fig. 22)
to find a square equal to
a rectangle.
is
it
Q
R
It is
required
in area.
P
O
B
M
Fig. 22.
Mark
Find
Q
<9,
the middle point of
AM,
by folding.
IN PAPER FOLDING
Fold OM, keeping
line
O
fixed
and
BC, thus finding P, the vertex
triangle
The square
BMP
.
.-.
M
fall
on
of the right-angled
PB
is
the square
BPQR.
equal to the given rectangle.
BP = QP,
and the angles are equal, triangle
evidently congruent to triangle
QSP.
QS=BM=AD.
.-.
.
.
is
letting
AMP.
Describe on
For
25
triangles
DA T and QSP are
PC=SR and triangles RSA
congruent.
and
CPT are
con
gruent.
rnABCD
can be
fitted
can be cut into three parts which
together to form the square RBPQ.
64.
Take
four equal squares and cut each of
.-.
Fig. 23.
into
them
two pieces through the middle point of one of the
and an opposite corner.
Take also another
sides
26
GEOMETRIC EXERCISES
equal square.
The
eight pieces can be arranged round
the square so as to form a^complete square, as in Fig.
23, the
arrangement being a very interesting puzzle.
The
fifth
may
square
evidently be cut like the
others, thus complicating the puzzle.
made by cutting the
one
corner
and
the trisection points
squares through
65. Similar puzzles can be
of the opposite side, as in Fig. 24.
Fig. 24.
66. If the nearer point is taken 10 squares are re
quired, as in Fig. 24;
if
the remoter point
is
taken 13
squares are required, as in Fig. 25.
67.
upon
The
puzzles mentioned in
the formulas
12
-I-
32
2 2 -f 32
= 10
= 13.
65, 66, are
based
IN PAPER FOLDING
27
The process may be continued, but the number
squares will become inconveniently large.
68. Consider again Fig. 13 in
44.
If
of
the four
triangles at the corners of the given square are re
moved, one square is left. If the two rectangles FK
and KG are removed, two squares in juxtaposition
are
left.
Pi*. 25.
69. The given square may be cut into pieces which
can be arranged into two squares. There are various
ways
of
doing
this.
Fig. 23, in
following elegant method
(1) the
:
The
65,
suggests the
required pieces are
square in the center, and (2) the four con
gruent symmetric quadrilaterals at the corners, to
gether with the four triangles. In this figure the lines
from the mid-points
of the sides pass
through the cor-
GEOMETRIC EXERCISES
28
ners of the given square, and the central square
is
The magnitude of the inner square
fifth of it.
can be varied by taking other points on the sides in
one
stead of the corners.
70.
The given square can be divided
(Fig. 26) into three equal squares
Take
G = hali
as follows
:
the diagonal of the square.
Fig. 26.
Fold through C and G.
Fold
perpendicular
BM
Take MP, CN, and
Fold
Fig. 26.
PH, NK, LF
NL
to
CG.
each
= BM.
at right angles to
CG, as
in
IN PAPER FOLDING
Take
NK=BM,
and
fold
KE
29
at right angles to
NK.
Then
the pieces
1, 4,
and
6,
3 and
form three equal squares.
Now CG =
i
i
?>BG ,
and from the triangles
GBC and CMB
BM _BG
Letting
~BC~~~CG
BC=a, we have
1/3
}
5,
and 2 and 7
IV.
71.
To
THE PENTAGON.
cat off a regular pentagon from the square
ABCD.
Divide
BA
in
the mid-point of
X
median section and take
in
M
AX.
c
D
M
A
X
S
N
Fig. 27.
Then
Take
Lay
AB-AX=XB
1
,
and
AM=MX.
BN=AM or MX.
off
R may lie
NP and J/^ equal
on
BC and
^4Z>
to
MN,
respectively.
so that /* and
PAPER FOLDING
Lay
off
RQ
MNPQR
is
and
PQ = MR and
AN, which
over the distance
Therefore, in Fig. 27,
RP is
J^f
also equal to
AB and
of a
rt.
If
is
it
A
evident that
to Jf.
NR = AB.
/ RMA = |
Similarly
parallel to
/
MP =
it.
.
NMR = f of a /
/ PNM= f of a
/
Similarly
From triangles MNR and QRP, / NMR= RQP
.
=
,
MO.
then
be moved on to BC, and
.A7 will
Ab
MB
AB, has
be moved
equal to
is
the point JV on the perpendicular
^4.Z?
NP.
the pentagon required.
In Fig. 19, p. 22,
on
3:
of art.
The
.
/
rt
.
rt.
.
/.
three angles at J/,
being each equal to
TV,
and
Q
of the
pentagon
remain
of a right angle, the
2
ing two angles are together equal to -^ - of a right
Therefore each of them
angle, and they are equal.
is
| of a right angle.
Therefore all the angles of the pentagon are equal.
The pentagon
72.
\
e.,
The base
to^?-
The
73.
is
also equilateral by construction.
MN of the pentagon
(l/5
is
1)=^X0.6180....
greatest breadth of the pentagon
If/ be the
equal to XB,
altitude,
is
58.
AB.
GEOMETRIC EXERCISES
8
215
0.9510.
.
.
.
=AB cos
Fig. 28.
74. if J? be the radius of the circumscribed circ)e,
x=
AB =
2cosl8
ZAS
T/io
+ 21/5
o
IN PAPER FOLDING
75.
33
r be the radius of the inscribed circle, then
If
from Fig. 28
it is
evident that
20
0.4253....
76.
The
area of the pentagon
the pentagon,
i.
is
5r
Xi
the base of
e.,
10
77. In Fig. 27 let
in
.
Then
PR
be divided by
MQ and
and F.
-.-
MN=
and cos
36=
a
-
l.(i/5
1)
...
72
--^-l)
cos 36
5
_|_
GEOMETRIC EXERCISES
34
RF=MN.
RF: RE = RE
:
i
V
=
snce
.
1/25
+ 101/5,
5
2
the area of the inner pentagon
= EF*
= ^^
The
2) ........... (4)
1/5 2 (1/5
76 the area of the pentagon
By
.
(2)
51) .................... (3)
F(by
:
1/5 = 3
1:3
1/5
-2)...
l
-
1/25+
101/5
1
2
2
-(1/5
~.-
2)
larger pentagon divided
=2
=
1
:
:
I/ 25
5
.
by the smaller
3l/5)
(7
0.145898..
78. If in Fig. 27, angles
+ 101
..
(^^ and LFQ are made
L being points on the sides
then
EFL
will be a reg
QR
QP respectively,
ular pentagon congruent to the inner pentagon. Pen
equal to
ERQ or FQP,
K,
QK
and
tagons can be similarly described on the remaining
The resulting figure
sides of the inner pentagon.
consisting of six pentagons
is
very interesting.
V.
79.
To
THE HEXAGON.
cut off a regular
hexagon from a given
square.
Fig. 29.
Fold through the mid-points of the opposite
and obtain the lines A OB and COD.
On
both sides of
triangles
(
25),
AO
and
OB
sides,
describe equilateral
AOE, AHO; BFO
and
BOG.
GE OME TRIG EXER CISES
36
Draw EF and HG.
AHGBFE is a regular hexagon.
It is
The
80.
unnecessary to give the proof.
greatest breadth of the hexagon
The
altitude of the
hexagon
is
AB.
is
= 0.866....
Fig. 30.
81.
If
R
be the radius of the circumscribed
R=\AB.
82. If r be the radius of the inscribed circle,
r
=
-
4
AB =
433
X A B.
circle,
IN PAPER FOLDING
83.
The area
the triangle
of the
hexagon
is 6
37
times the area of
HGO,
= 6.^-1^.
,
=
3T/ 3
-.4^2^0.6495
8
X^^
2
-
=
Also the hexagon
J -42? CD.
14 times the equilateral triangle on
=
AB.
Fig. 31-
an example of ornamental folding
into equilateral triangles and hexagons.
84. Fig. -30
85.
A
is
hexagon
is
formed from an equilateral
tri
angle by folding the three corners to the center.
The side of the hexagon is i of the side of the
equilateral triangle.
GEOMETRIC EXERCISES
38
The area
of the
hexagon
=f
of
the equilateral
triangle.
86.
The hexagon can be divided
into equal regular
hexagons and equilateral triangles as
in Fig. 31
by
the sides.
folding through the points of trisection of
VI.
87.
To
THE OCTAGON.
cut off a regular octagon from a given square.
Obtain the inscribed square by joining the mid
points A, B, C,
D of the sides of the given square.
Fig. 32.
Bisect the angles which the sides of the inscribed
square make with the sides of the other.
secting lines meet in E, F, G, and
H.
Let the
bi
GEOMETRIC EXERCISES
40
AEBFCGDH is
The
triangles
a regular octagon.
DHA
AEB, BFC, CGD, and
The octagon
congruent isosceles triangles.
is
are
there
fore equilateral.
The
angles at the vertices, E, F, G,
H
of the
same
four triangles are each one right angle and a half,
since the angles at the base are each one-fourth of a
right angle.
Therefore the angles of the octagon at A, B,
are each one right angle and a half.
and
C,
D
Thus
The
the octagon
is
equiangular.
greatest breadth of the octagon
the given square,
R
88. If
is
the side of
a.
be the radius of the circumscribed
and a be the side
circle,
of the original square,
*-f
89.
The
the sides
90.
is
angle subtended at the center by each of
half a right angle.
Draw
the radius
OE
and
let it
(Fig. 33).
Then
AK= OK=
V2
= -=r.
2
KE=OA-OK=
V2
a
2] 2
Now
from triangle
AEK,
cut
AB
in
K
IN PAPER FOLDING
= 4--
41
(4-21/2)
= --.(2-1/2).
=- 1/2-1/2.
91.
The
But C
2
altitude of the octagon
= AC
= *-
is
CE
(Fig. 33).
2
(2
_ 1/2 =
(2
)
+ 1/2)
A
Fig. 33-
a
/
7^"
2
92.
The
angle AOE
area of the octagon
is
eight times the
and
CL
d
d
o
tri
_
GEOMETRIC EXERCISES
42
93.
A
regular octagon
_
may
also be obtained
by
dividing the angles of the given square into four equal
parts.
2
Y
Fig. 34-
It is easily
seen that
EZ= WZ = a,
square.
= WK;
XE
.(21/1).
Now
.-.
Also
a
XZ
a. tf (2
21/2.
1XE
(
1/2).
the side of the
IN PAPER FOLDING
OZ=
Again
= ~(6-4v/2 + 2)
= (2
2
<z
1/2").
HK=KZHZ
=a
-
1/2
(1/2
)
-(1/2
1)
7i/2
and
94.
HA = ~A 1/20
The
141/2.
area of the octagon
is
eight times the
HOA,
area of the triangle
1/2
(6
2
=
2
-
4i/2)
4)
1/2 -(1/2
2
-I)
.
GEOMETRIC EXERCISES
44
95. This octagon
:
= (2
and
the octagon in
1/2 )2
their bases are to
:
1 or 2
:
(j/2
one another as
1/2":
i/2"+
1.
92
+ I)
2
;
VII.
96.
Any
THE NONAGON.
angle can be trisected fairly accurately by
paper folding, and
in this
way we may
construct ap
proximately the regular nonagon.
Fig. 35-
Obtain the three equal angles
equilateral triangle.
(
25.)
at the center of
an
GEOMETRIC EXERCISES
46
For convenience
angles,
A OF, FOC,
of
and
cut out the
folding,
COA.
Trisect each of the angles as in Fig. 35, and
each of the arms =
97.
Each
three
make
OA.
of the angles of a
is -^4-
nonagon
of a
right angle = 140.
The angle subtended by each
J of a right angle or
Half
98.
is
side at the center is
40.
this angle is \ of the angle of the
OA
= %a,
where a
is
nonagon.
the side of the square
;
it
also the radius of the circumscribed circle, R.
The
radius of the inscribed circle
=Jt
.
cos 20
= %a cos 20
= * X 0. 9396926
= aX 0.4698463.
The area
the triangle
of the
nonagon
is
9
times the area of
AOL
= f sin 40
= --X
0.6427876
o
= a* X 0.723136.
7?2
,
-
VIII.
THE DECAGON AND THE DODECAGON
99. Figs. 36, 37
show how a regular decagon, and
a regular dodecagon,
may be
obtained from a penta
gon and hexagon respectively.
Fig. 36.
The main
part of
the process
is
to obtain the
angles at the center.
In Fig. 36, the radius of the inscribed circle of
the pentagon
is
taken for the radius of the circum
scribed circle of the decagon, in order to keep
within the square.
it
GEOMETRIC EXERCISES
48
A
ioo.
follows
also be obtained as
may
regular decagon
:
51, dividing
AB in
Fold XC, MO, YD at right angles to AB.
Take O in
such that YO
AY, or YO
= XB.
Obtain X,
median
Y, (Fig. 38), as in
section.
Take
M
the mid-point of
AB.
=
MO
Fig. 37-
Let YO, and
and
XO
produced meet XC, and
YD
in
C
D respectively.
Divide the angles
parts by
Take
<9For
HOE, KOF,
Off",
DOY into
XOC
and
and
LOG.
OK, OL, OE, OF, and
four equal
OG
equal to
OK
Join X, H, K, L, C, D, E, F, G, and Y, in order.
IN PAPER FOLDING
As
in
49
60,
/
YOX=l
of a
rt.
/ =36.
Fig. 38.
By
bisecting the sides
and joining the points thus
determined with the center, the perigon
into sixteen equal parts.
A
16-gon
is
is
divided
therefore easily
constructed, and so for a 32-gon, and in general
regular 2*-gon.
a.
IX.
101. Fig.
THE PENTEDECAGON.
39 shows
how
the pentedecagon
OD-
is
tained from the pentagon.
Let
to
ABODE
be the pentagon and
O
its
cente r
Draw OA, OB, OC, OD, and OE.
Produce
AB in K.
OF=$ of OD.
Fold GFH
right angles
Make
DO
meet
Take
2,\.
OH= OD.
to
OF.
OG =
PAPER FOLDING
Then GDH.
is
51
an equilateral triangle, and the
angles DOG and HOD are each 120.
But angle DOA
144; therefore
is
angle
GOA
is
24.
That
EOA, which
the angle
is,
is
72,
is
trisected
by OG.
Bisect the angle
and
let
OG
cut
EA
EOG
in
M;
by OL, meeting
EA
in Z,
then
OL = OM.
In
OA
and
<9^
take
Then PM, ML, and
OP and
Z<2
<9<2
equal to
OL
or
are three sides of the
pentedecagon.
Treating similarly the angles
and
DOE, we
decagon.
A OB, BOC, COD,
obtain the remaining sides of the pente
SERIES.
X.
ARITHMETIC SERIES.
IO2. Fig. 40 illustrates
horizontal lines to the
an arithmetic
left of
series.
The
the diagonal, including
the upper and lower edges, form an arithmetic
series.
Fig. 40.
The
initial line
the series
being
is a, a-\- d,
a,
and d the
a-\-2d, a
-\-
common
difference,
3//, etc.
portions of the horizontal lines to the
right of the diagonal also form an arithmetic series,
103.
The
PAPER FOLDING
53
but they are in reverse order and decrease with a
common
In general,
104.
sum
difference.
the
of
be the last term, and
/
if
the
above diagram graphically
the
series,
s
proves the formula
105. If a
term
and
c
are two alternate terms, the middle
is
106.
To
insert n
means between a and
tical line
has to be folded into
common
difference will be
n-\- 1
/,
the ver
equal parts.
The
I- a
Considering the reverse series and interchan
ging a and /, the series becomes
107.
a,
The terms
will
a
d,
"be
and thereafter they
a
2d
.
.
.
.
/.
positive so long as a
will
>(
!)</,
be zero or negative.
GEOMETRIC SERIES.
108.
In a right-angled triangle, the perpendicular
from the vertex on the hypotenuse is a geometric mean
between the segments of the hypotenuse. Hence, if
two alternate or consecutive terms
series are given in length,
of
a
the series can
geometric
be deter-
GEOMETRIC EXERCISES
54
mined as
in
Fig. 41.
Here OPi,
and OP$ form a geometric
being OPi
:
OP
2,
series, the
OP
3,
common
rate
OP
2.
P3
Fig. 41.
If
OP\ be
the unit of length, the series consists of
the natural powers of the
109.
rate.
Representing the series by
These
common
common
lines also
rate
#, ar,
ar 2 ....
,
form a geometric series with the
r.
no. The terms can also be reversed, in which case
the
common
be the unit,
rate will be a proper fraction.
OP
is
the series to infinity
the
is
common
rate.
If OP&
The sum of
IN PAPER FOLDING
55
manner described in
108, one geo
mean can be found between two given lines,
and by continuing the process, 3, 7, 15, etc., means
1 means can be found,
can be found. In general,
in. In the
metric
2"
n being any positive integer.
112.
not possible to find two geometric
It is
between two given
known
the following
given,
It
points.
it is
lines,
means
merely by folding through
can, however, be accomplished in
manner
OP\ and OP being
P% and Ps Take two rect
In Fig. 41,
:
required to find
.
paper and so arrange them, that their
outer edges pass through P\ and P^ and two corners
lie on the straight lines OP* and OP* in such a way
angular pieces of
that the other edges ending in those corners coincide.
The
positions of the corners determine
and OP*.
This process gives the cube root of a given
113.
number,
114.
OP<i
for
if
There
OP\
is
is
the unit, the series
is 1, r,
a very interesting legend in
tion with this problem.*
"The
r2 r 3
,
.
connec
Athenians when suf
fering from the great plague of eruptive typhoid fever
430 B. C., consulted the oracle at Delos as to how
they could stop it.
Apollo replied that they must
double the size of his altar which was in the form of a
in
Nothing seemed more easy, and a new altar
was constructed having each of its edges double that
cube.
of the old one.
The
god, not unnaturally indignant,
*But see Beman and Smith
matics, p.
82, 207.
s
translation of Fink
s
History of Mathe
GEOMETRIC EXERCISES
56
made
the pestilence worse than before.
A fresh dep
whom he in
was accordingly sent to Delos,
it was useless to trifle with him,
utation
formed that
must have
his altar exactly doubled.
as
he
Suspecting a
mystery, they applied to the geometricians.
Plato,
the most illustrious of them, declined the task, but
them
referred
study of the
to
Euclid,
problem."
polation for that of
who had made a special
(Euclid s name is an inter
Hippocrates re
duced the question to that of finding two geometric
means between two straight lines, one of which is
Hippocrates.)
twice as long as the other.
terms of the
succeed
series,
xs
in rinding the
of PJato,
who
If a, x,
He
means.
Menaechmus,
lived between 375
the following three equations
a
From
equations
:
x
=x
this relation
y and 2a be the
= 2a^.
:
we
y
did not, however,
and 325 B.
a pupil
C.,
gave
*
:
y
:
2a.
obtain the following three
:
= ay
y = 2ax
xy = 2a*
x2
(1)
2
(1)
(2)
(3)
and (2) are equations of parabolas and
the equation of a rectangular hyperbola.
(1)
and
(2) as well as (1)
and
(3) give x*
(3) is
Equations
= 2a*.
The
problem was solved by taking the intersection () of
the two parabolas (1) and (2), and the intersection (/?)
of the parabola (1) with the rectangular
*Ibid., p. 207.
hyperbola
(3).
IN PAPER FOLDING
HARMONIC
57
SERIES.
Fold any lines AR, PB, as in Fig. 42, P be
Fold
ing on AR, and B on the edge of the paper.
with
PB.
coincide
both
and
PR
AP
so
that
may
again
115.
PY
Let PX,
be the creases thus obtained,
X
and
Y
being on AB.
Then
range.
the points A, X, B,
That
externally in
is,
Fso
AB
is
Y
form an harmonic
divided internally in
X
and
that
AX: XB = AY: BY.
It is
evident, that every line cutting
PA, PX, PB,
and PYvti\\ be divided harmonically.
R
A
X
Y
B
Fig. 42.
116.
line
Having given A, B, and X,
XPand mark ^corresponding
to find Y: fold
to B.
Fold
Bisect the angle BPR by PY
P
so that /l# and PR coincide.
through
Because XP bisects the angle APB,
.-.AX: XB = AP: BP,
and BP.
= AY:
BY.
any
AKPR,
by folding
GEOMETRIC EXERCISES
58
AX: XB = AY: BY
XYBY=AY:
BY.
orAYXY:
117.
Thus, AY, XY, and BY, are an harmonic
series,
and XY is the harmonic mean between AY and BY.
Similarly AB is the harmonic mean between AX
and
A Y.
118.
If
BY and XYbe given, to find
AY, we have only
XYas
angle on
to describe
the third term
any right-angled
the hypotenuse and
make angle
tri
APX
= angle XPB.
119.
Let
AX
a,
AB = b,
a
or,
ab
-\-
-j-
c
bc=.%ac
ab
or, ^
AY=c.
and
7
~o
b
=
T-
a
When
When
Therefore when
= =
=
b
2a, c=&.
X is
a
fr,
c
b.
the middle point of
Y
is
B. Y approaches
and ultimately the three points
at
an
infinite distance to the right of
B
as
X approaches
it,
AB,
coincide.
As
X
moves from the middle
moves from an
infinite distance
and ultimately X, A, and
120. If
E be
of
on the
Jfand
Y
towards A
to the left,
left
y
Y coincide.
the middle point of
for all positions of
AB
Y with
AB,
reference to
A
or B.
IN PAPER FOLDING
Each
Y is
of the
two systems
59
A
and
called the center
or
X
of pairs of points
called a system in involution, the point
B the
E
and
being
focus of the system.
The two systems together may be regarded
as one
system.
AX
121.
follows
and
AY
XA
and take
being given,
B can
be found as
:
Produce
Take
D the middle
Take
CE = DA
or
AC=XA.
point of
A Y.
AE = DC.
F
A
C
X
B
D
Y
Fig. 43-
A
Fold through
to
so that
AF may
be at right angles
CA Y.
Find
F such
DF=DC.
that
Fold through
at right angles to
EF
and obtain FB, such that
is
the arithmetic
is
the geometric
also the
is
mean between AX and A Y.
mean between AX and A Y.
geometric mean between CD or AE
CD
AF
AFis
FB
EF.
and AB.
Therefore
and
AB is
the harmonic
mean between
AX
A Y.
122.
finding
The following is a very simple method of
the harmonic mean between two given lines.
GEOMETRIC EXERCISES
6o
Take AB,
CD
the given lines.
sides
AC,
BD of
on the edges of the square equal to
Fold the diagonals AD, ^Cand the
the trapezoid ACDB. Fold through
E, the point of intersection of the diagonals, so that
PEG may
be
at right angles to the other sides of the
square or parallel to
AB
and CD.
A
Let
EEG
cut
AC
B
Fig. 44.
and
BD in F and
between
AB
Then
G.
EG is
the harmonic
and CD.
For
EG
CD
EE
AB
1
~AB
__
~~
_
EE
AB
FE
CD
__
~~
__ EB_
~~
~C~B
EF
CE
CB
CD
1
+ CD
CE
CB
~=
H
"
1
~~~
~FE
EB_
CB
_2_
~EG
~
mean
IN PAPER FOLDING
123.
and
The
BD is
124.
= FG.
To
line
HK connecting the mid-points of AC
the arithmetic
Fold ZJ/at
AB and CD.
take HL in HK
mean between
find the geometric
mid-point of
6t
mean,
right angles to
HK. Take O
the
HK and find M in LM so that OM^OH.
HM\s the geometric mean between ^^ and CD as well
as
between
The geometric mean between
is thus seen to be the geometric mean
arithmetic mean and harmonic mean.
FG and
two quantities
between their
ZfA".
Fig. 45-
SUMMATION OF CERTAIN
125.
To sum
1
SERIES.
the series
+ 3 + 5.... + (2
1).
Divide the given square into a number of equal
Here we have 49 squares, but
squares as in Fig. 45.
the number may be increased as we please.
GEOMETRIC EXERCISES
62
The number
of squares will evidently
number
number, the square of the
be a square
of divisions of the
sides of the given square.
Let each
of the small squares
unit; the figure formed
gnomon.
The numbers
A
\yy
-\-
be considered as the
O -\- a
being called a
of unit squares in each of the
mons AOa, BOb,
are respectively
etc.,
3,
5,
gno
7, 9,
11, 13.
Therefore the
13
is
7
sum
of the series
1, 3,
5, 7, 9, 11,
2
.
Generally,
1
+ 3 + 5 -f
.
.
.
.
-f (2
1)
=
;/
2
.
Fig. 46.
126.
To
find the
sum
of the
cubes of the
first
n
natural numbers.
Fold the square
into 49 equal squares as in the
IN PAPER FOLDING
63
preceding article, and letter the gnomons. Fill up the
squares with numbers as in the multiplication table.
The number
The sums of
in the initial
2
The sum of
row is the sum
Let us
call
Then
=2
3
3
,
3
4
,
in the
gnomons Aa, Bb,
3
5
,
numbers
the
I3.
is 1
numbers
the
+ 4-f
etc., are 2
=
square
3
6
,
3
,
and
73
.
in the first horizontal
of the first seven natural
numbers.
it s.
the
sums
of the
numbers
in
and
Is.
rows
a, b, c, d,
etc., are
2s, 3s, 4s, 5j, 6s,
sum
numbers
is
s(l + 2 + 3 + 4 + 5 + 6 + 7) =
s*.
Therefore the
Therefore, the
natural numbers
of all the
sum
is
of the
cubes
of the first
equal to the square of the
those numbers.
Generally,
For
I3
+2
[( + I)]
3
3
-f 3 ____ -f
2
2
(>
Putting
=
1, 2,
4-l 3
3
!)]*
[(
=--
;z
-f
2
)
(>
3 .... in order,
)2
=
we have
= (l-2)2_ (0-1)2
seven
sum
of
GE OME TRIG EXER CISES
64
Adding we have
1
4.5 /* 3
= [( + I)]
2
2
127- If
be the sum of the
*
To sum
128.
1-2
first
n natural numbers,
the series
+ 2-3 + 3-4. ... +
!)-.
(
In Fig. 46, the numbers in the diagonal
cing from
1,
are the squares of the natural
commen
numbers
in order.
The numbers
in
one gnomon can be subtracted
from the corresponding numbers
By
gnomon.
this process
+ 2[(
=* +
1
Now
(
!)
1)2
4. 3(n
the succeeding
we obtain
+
(
2)
+
(
l)n.
3
(;?
1)
93
Hence, by addition,
3)....
+ .2[; + 2.. ..+(*!)]
3
.
in
=
1 -f-
p^!
3(
3 9-1
.
1),
+1]
IN PAPER FOLDING
65
Therefore
To
129.
find the
sum
of the squares of the first n
natural numbers.
+ 2-3.. + (_!)
= 2 2 + 3 3.. + ^ a
= 12 22 + 3 ____ +
(1 4- 2 -f 3 ____ -f n)
1-2
..
2
2
2
..
2
rc
_|_
2
Therefore
To sum
130.
the series
c
_i
s
)
==:W
.
2
= (2
.
.
^ (2
I)
+ (_!) + (.!),
2
12_|_32_|_ 52
8_
2
1)
by putting n
=
-(^
1, 2, 3,
!)-,
....
13_o3= i2__o.l
2
Pr=3 2 1-2
z
3
33
23
=5
2
2-3
by
128,
GEOMETRIC EXERCISES
66
Adding, we have
n*
=
I 2 4.
32
+5
2
2
.
.
.
.
-{-
(2
[1-2 -f 2-34-3-4..
I)
..
+ (_ !)],
XI.
O the
Find
I3lt
Bisect
diameters.
POLYGONS.
center of a square by folding
the
its
angles at the center,
right
then the half right angles, and so on. Then we
obtain
equal angles around the center, and the
2"
each of the angles
of
magnitude
is
^ of
a right angle,
being a positive integer. Mark off equal lengths on
each of the lines which radiate from the center. If
the extremities of the radii are joined successively,
we
get regular polygons of
2"
sides.
Let us find the perimeters and areas of these
In Fig. 47 let OA and OA\ be two radii
polygons.
132.
each other.
at right angles to
OA
OA
S,
4
Draw AA\, AA-
4, 8 .... parts.
OA-2
radii
at
,
B
AA^,
,
of 2 2
OA^ OA
,
23
,
Then
at B\, B-2
B\,
,
B^
the respective chords.
points of
AA
2,
angles.
right
Let the radii
etc., divide the right angle
,
24
.
.
AA
B
3
A\OA
.... cutting the
respectively,
3
3
.
OA^
into 2,
.
.
.
mid
are the
Then AA\,
AA<t,
.are the sides of the inscribed polygons
sides respectively,
and
OB^ OB*
are the respective apothems.
Let
a
OA = R,
(2")
represent the side of the inscribed polygon
GEOMETRIC EXERCISES
68
of
2"
sides,
perimeter,
(2")
and
For the
the corresponding apothem,
.4(2")
/(2")
its
its area.
square,
/ (22)= ^-22- 1/2;
A,
Fig. 47-
For the
in the
octagon,
two triangles
and
OA
or
AAv =
(1)
PAPER FOLDING
2
1/2
(2)
= 1*1/2+1/2
23 )
6g
...(3)
= J perimeter X apothem
1/2
-
Similarly for the polygon of 16 sides,
2*)
^(2
and
tor the
)
= ^-2
the
-1/2
1/2
+ 1/2;
1/2;
of 32 sides,
5
general law
-
1/2
is
thus clear.
^2==-
Also
As
4
= ^2- 22-1/2
polygon
25)
The
4
= *-2
number
of sides is increased indefinitely
GEOMETRIC EXERCISES
7o
the
apothem evidently approaches
Thus the limit of
its
limit, the ra
dius.
2
for
if jc
+ I/ 2 + 1/2..
represent the limit, #
which gives x
=
2,
or
1;
..is 2;
= 1/2
-J-
#,
a quadratic
the latter value
is,
of
course, inadmissible.
133.
If
perpendiculars are drawn to the radii at
we
their extremities,
get regular
polygons circum
scribing the circle and also the polygons described as
in the
preceding
article,
and
of the
same number
of
sides.
F
C_
In Fig. 48, let
gon and
FG a
Then from
AE
E__
_G
D
be a side of the inscribed poly
side of the circumscribed polygon.
the triangles
OE
.-.
FIE
FE
FG = R AE
and EIO,
IN PAPER FOLDING
The values
AE
FG is
of
previous article,
The
as
areas of
FG AE
1
1
:
i.
,
Of
and
being
71
known by
the
found by substitution.
the two polygons are to one another
as
e.,
&
:
Of*.
preceding articles it has been shown
.2*
how regular polygons can be obtained of 2 2 2 3
it
is
sides. And if a polygon of m sides be given,
easy
134. In the
to obtain
polygons of
135. In
Fig. 48,
2"-m
.
.
,
.
sides.
AB and CD
are respectively the
sides of the inscribed and circumscribed polygons of
Take
n sides.
AE and
BE.
E the mid point of CD and draw AE,
BE are the sides of the inscribed poly
of 2n sides.
gon
Fold AF,
ing
CD
in
Then
BG at
Fand
FG
is
right angles to
A C and BD,
meet
G.
a side of the circumscribed polygon
2n sides.
of
OG and OE.
P be the perimeters
Draw OF,
Let
/>,
of the inscribed
and
circumscribed polygons respectively of n sides, and
B
A,
their areas,
and /,
P
the perimeters of the in
scribed and circumscribed polygons respectively of 2n
sides,
and
A B
,
their areas.
Then
p
= n-AB,
Because
to
P=n-CD, p
OF
bisects
= 2n-AE, P = 2n-FG.
/ COE, and
CD,
CO
CJ^__CO__
~~
~~
FE
~OE
~AO
_
"
CD
~
AB
is
parallel
GEOMETRIC EXERCISES
72
CE
_CD+AB
~
~EE
or
AB
n-CD+n-AB
4n-C
EE
n-AJ3
2P _P+p
.
.
=
P
Again, from the similar triangles
ET^
_
EIE and
AffE,
EE
~Aff~ AJE
or
A
or
2
p
=.2.
= V P p.
Now,
The
tude,
triangles
A Off and AOE
are of the
same
alti
COE
and
AH,
&AOE
__
=
~
OH
~OE
Similarly,
_OA
~
~OC
AB
Again because
A
Now
to find
B
.
||
CD,
A Off
A
AOE
Because the triangles
IN PAPER FOLDING
FOE
angle
have the same
altitude,
and
73
OF
bisects the
EOC,
&COE CE
&FOE~ FE
and
OE =
OC+OE
OE
OA,
,__ ~ &AOH
OA
~~
_
Off
&AOE +
&COE
&AOH
From
136.
this
equation
we
Given the radius
ular polygon, to find
easily
y obtain
.
IT
B
.
A
R
and apothem r of a reg
and apothem r of
the radius
R
a regular polygon of the same perimeter but of double
the
number
Let
OA
of sides.
AB be
O
its
center,
the radius of the circumscribed circle, and
the apothem.
OB.
a side of the first polygon,
On OD produced
take
OA
and
Draw AC, BC.
Fold
OC=OA
OB
OD
or
perpen-
GEOMETRIC EXERCISES
74
AC
dicular to
BC
and
respectively, thus fixing the
Draw A B cutting OC in D
Then
points A B
the chord A B is h^f of AB, and the angle B OA is
half of BOA.
OA and OH are respectively the ra
.
,
dius
R
.
and apothem
OH
Now
OD, and OA
and
OD
137.
Then
r of the
the arithmetic
is
is
second polygon.
mean between OC and
mean proportional between
the
OC
.
Now, take on
H being
A
bisected by
ED
A
OE=OA
<9C,
less
than
^
C,
and draw
and /
DA
^ ^.
C being
E,
is less
than
.. R\
\CH
,
r\ is less
As the number
i.
e., less
than
than
\{R
\CD
*")
of sides is increased, the
polygon
approaches the circle of the same perimeter, and R
and r approach the radius of the circle.
That
is,
= the diameter of the circle =
.
7t
Also,
_/?]== Rr\ or
and
~ = -^,
R
_-
=-f?\
and so on.
Multiplying both sides,
*
-
-
-
-
tne radius of the circle
=
-.
IN PAPER FOLDING
The
138.
radius of the circle lies between
r nJ the sides of the polygon being
and
TT
between
lies
2
2
and
-
n can
in
4-2"
Rn and
number;
The numerical value
-.
Rn
rn
of
75
therefore be calculated to any required de
gree of accuracy by taking a sufficiently large
number
of sides.
The
othems
following are the values of the radii and apof the regular
polygons of
4,
16.... 2048
8,
sides.
= 0-500000
=
-603553
n
R = r\/^=
4-gon, r
8-gon,
2048-gon,
139. If
polygon
of
r<>
R"
-707107
RI = 0-653281
= 0-636620
^
9
= 0-636620.
be the radius of a regular isoperimetric
4^ sides
_ _
2
or in general
T~\ ^^
140.
and the
The
radii R\,
R^. .... successively diminish,
ratio -77-12 less
than unity and equal to the
cosine of a certain angle a.
^=\
RZ
\l
+ cos a
a
2--= cos 2-
^
GEOMETRIC EXERCISES
76
R
i,\
of.
,
we
multiplying together the different ratios,
J^
jt+l
The
-
=J?i Cosa- cos
limit of cos or-cos
O
-
is
141.
,
a result
It
known
get
cos
^?
.
.
.
.
cos
as Euler
s
when
,
=
Formula.
was demonstrated by Karl Friedrich Gauss*
(1777-1855) that besides the regular polygons of
3-2",
oo,
5-2",
15
-2"
sides,
the
2",
only regular polygons
which can be constructed by elementary geometry
are those the number of whose sides is represented
by the product of
bers of the form
of 5
polygons
The
and one or more
2"*-f-l.
We
shall
different
num
show here how
and 17 sides can be described.
following theorems are required
(1) If (7
ference
2"
and
ACDB,
Z>
are
and
if
spect to the diameter
:f
two points on a semi-circum
C
C with
be symmetric to
R
AB, and
re
the radius of the
circle,
.
AC-BC=R-CC
.............
Let the circumference
(2)
into an
odd number
iii.
of a circle
of equal parts,
and
let
be divided
AO be
the
*Beman and Smith
s translation of Fink s History of Mathematics, p.
see also their translation of Klein s Famous Problems of Elementary
245;
Geometry pp.
,
t
16, 24,
and their
New Plane and Solid
These theorems may be found demonstrated
Problemes de Geomttrie Elementaire.
Geometry,
p.,
212.
in Catalan s Theortmes et
IN PAPER FOLDING
77
diameter through one of the points of section A and
the mid-point O of the opposite arc.
Let the points
of section
At,
A
s
.
.
.
on each side of the diameter be named A\,
.A n beginning next
.A n and A i, A 2 A 3
.
,
,
.
.
to A.
Then OAi OA* OA 3 ____ OA n
= R ......
= R*.
n
iv.
n
and OAi OA^
-
OA
....
142. It is evident that
mined, the angle
divided into
i<<.
2"
Let us
A n OA
if
OA
the chord
found and
is
it
OA n
is
deter
has only to be
equal parts, to obtain the other chords.
take the pentagon.
first
By theorem
iv,
By theorem
i,
R(OA
l
OA )=
2
GEOMETRIC EXERCISES
1),
and t?^ 2
=
(i/5
1).
Hence the following construction.
and draw the tangent
Take the diameter
AF. Take D the mid-point of the radius OC and
^C6>,
On OC as
diameter describe the circle
FD cutting the inner circle
Then FE =OA, and FE= OA
in
Join
144.
2
E
AE
and
CE.
.
.
Let us now consider the polygon
of
seven
teen sides.
Here*
OA OA OA -OA r OA
Az
-
6
and
By
theorems
i.
and
OA OA
-
l
%
= R*.
O
ii.
=
Suppose
principal steps are given. For a full exposition see Catalan s ThtoProbltmes de Gtomttrie Elententaire. The treatment is given in full
in Beman and Smith s translation of Klein s Famous Problems of Elementary
*The
rtmes
et
Geometry, chap.
iv.
IN PAPER FOLDING
MN=R*
Then
and
PQ = R*.
Again by substituting the values
Q
in the
79
of
M,
IV,
P
and
formulas
MN=R\ PQ=R^
and applying theorems
i.
and
ii.
N)~ (P
(M
we
Q^)
get
= R.
P
and Q in
ALSO by substituting the values of M, N,
the above formula and applying theorems i. and ii.
we
get
M
MN,
(
Hence
N} (P Q} = 4^ 2
P Q, J/, N, P and Q
.
are deter
mined.
Again
Hence
145.
OA
By
S
is
solving the equations
M N= ^R
P
Q = R(
OA 8 = IR\_
2
determined.
1
get
(1 -f 1/17).
1
+ 1/17).
1+1/17 + 1/34
iXl7+3i/17+ 1/170
= J7?[
we
26
V 11
2i/17
4
|/34+2v
/
17 ]
+ /I7 + 1/34 21/17
+ 31/17 1/170 + 38J/17],
GEOMETRIC EXERCISES
8o
146.
Let
center.
OA
in
The geometric
BA
construction
be the diameter
OA
Bisect
in C.
and take AD = AB.
CD
is
as follows
of the given circle
;
:
O
its
Draw AD at right angles to
Draw CD. Take E and E
CE = CA.
of C so that
C=
and on each side
Fig. 51.
Bisect
^Z>
pendicular to
Draw
FG
Take
H
CD
in
it is
G
and
>
.
and take
and /r
GH=EG and
Then
in
in
G
Draw
.
Z>^
per
DF= OA.
.
7^7 and
.# ==
H
G D.
evident that
in
^
produced so that
IN PAPER FOLDING.
81
also
FH = P,
(FH DE FH= DF* = &.
Again in DF take A such that FK=FH.
Draw KL perpendicular to DF and take L in KL
.-
such that
FL
)
DL.
perpendicular to
is
Then FL* =DF- FK=RN.
Again draw
J7W perpendicular
to
FH
and take
H N= FL. Draw NM perpendicular to NH
M in NM such that ZT J/ perpendicular to
.
is
Draw MF
Then
perpendicular to
FH
F H FF = ^ J/
But
FF
2
.
Find
FM.
GENERAL PRINCIPLES.
XII.
147. In the
preceding pages we have adopted sev
eral processes,
e.
bisecting and trisecting finite
g.,
angles and dividing them
lines, bisecting rectilineal
into other equal parts, drawing perpendiculars to a
given
Let us now examine the theory
line, etc.
of
these processes.
148.
The general
Figures and straight
that of congruence.
is
principle
lines are said to be congruent,
they are identically equal, or equal in
all
In doubling a piece of paper upon
tain the straight edges of
each other.
This line
if
respects.
itself,
we ob
two planes coinciding with
also be regarded as the
may
intersection of two planes
if
we consider
their posi
tion during the process of folding.
In dividing a finite straight
number
of
Equal
line, or
an angle into a
number
of
con
lines or equal angles are
con
equal parts,
gruent parts.
we
obtain a
gruent.
149.
Let
XX
be a given
A
any two parts by
ling the line on itself.
.
Take O
finite line,
divided into
the mid-point by
Then OA
is
doub
half the difference
PAPER FOLDING
A
X
A
between
and
XA
OX corresponding
in
ence between
A
X
and
I
to
Jf
1
X
XX
Fold
.
over O, and take
A Then A A is the differ
^ and it is bisected in O.
.
1
1
A
83
1
X
A
O
Fig. 52-
As
-4
taken nearer O,
is
same time
property
line
is
AA
made use
by means
150.
angle.
A O
diminishes, and at the
This
diminishes at twice the rate.
of the
of in finding the
mid-point of a
compasses.
The above observations apply also to an
The line of bisection is found easily by the
compasses by taking the point
of intersection cf
two
circles.
X X,
In the line
segments to the right of
may be considered positive and segments, to the
eft of O may be considered negative.
That is, a
151.
point
point
moving from O
moving
negatively.
both members
in
to
A moves
positively,
the opposite direction
AX=OXOA.
AX
OA = OX
and a
OA moves
,
of the equation being negative.*
OA, one arm of an angle A OP, be fixed and
be considered to revolve round O, the angles
152. If
OP
which
it
makes with
*See Beman and Smith
s
OA
are of different magnitudes.
New Plane and Solid
Geometry,
p. 56.
GEOMETRIC EXERCISES
84
All such angles
formed by
OP revolving
hands
tion opposite to that of the
The
of a
in the direc
watch are
re
OP revolving
angles formed by
garded positive.
an opposite direction are regarded negative.*
in
one revolution,
153. After
Then
the angle described
is
OP coincides
half the revolution,
OAB.
Then
angle,
which
When OP
it
the angle described
evidently equals
which
When OP
evidently equals four right angles.
completed
with OA.
called a perigon,
in
is
called a straight
is
two
right
angles. f
has completed quarter of a revolution,
perpendicular
magnitude.
to
OA.
So are
has
with
a line
it is
All right angles are equal in
all
straight angles
and
all
peri-
gons.
154.
Two
lines at right angles to
four congruent quadrants.
Two
each other form
lines otherwise in
clined form four angles, of which those vertically op
posite are congruent.
155.
The
mined by
its
The
above.
position of a point in a plane
distance from one line
allel to the other.
ties of
is
deter
distance from each of two lines taken as
is
measured par
In analytic geometry the proper
plane figures are investigated by this method.
The two
lines are called axes
;
the distances of the
point from the axes are called co-ordinates, and the
intersection of the axes
* See
Beman and Smith
t/*.,p.
5-
s
New
is
Plane
called the origin.
and
Solid Geometry, p. 56.
This
IN PAPER FOLDING.
85
method was invented by Descartes in 1637 A. D.*
has greatly helped modern research.
X X, YY
156. If
be two axes intersecting at O,
distances measured in the direction of
the right of
to the left of
to
YY
,
O are positive,
O are negative.
OX,
i.
e., to
while distances measured
Similarly with reference
distances measured in the direction of
positive, while distances
OY
It
measured
6>Fare
in the direction of
are negative.
symmetry is defined thus If two fig
same plane can be made to coincide by
157. Axial
ures in the
:
turning the one about a fixed line in the plane through
a straight angle, the
two figures are said
to
be sym
metric with regard to that line as axis of symmetry. f
symmetry is thus defined If two fig
the same plane can be made to coincide by
158. Central
ures in
:
turning the one about a fixed point in that plane
through a straight angle, the two figures are said to
be symmetric with regard to that point as center of
symmetry. J
In the
first
case the revolution
plane, while in the second
If in
of
one
outside the given
in the
it is
same plane.
the above two cases, the two figures are halves
figure, the
whole figure
is
with regard to the axis or center
or center of
symmetry or simply
*Beman and Smith
t Beman and Smith
t/.,p.
is
183.
s
s
said to be symmetric
these are called axis
axis or center.
translation of Fink s History of Mathematics, p. 230.
New Plane and Solid Geometry, p. 26.
GEOMETRIC EXERCISES
86
159.
Now,
in the
quadrant
XOVmake
a triangle
its image in the quadrant VOX
by
on
the
YY
axis
and
the
folding
pricking through
paper at the vertices. Again obtain images of the two
PQR.
Obtain
triangles in the fourth
and third quadrants.
It is
seen
that the triangles in adjacent quadrants posses axial
Fig- 53-
symmetry, while the triangles
in alternate
quadrants
possess central symmetry.
Regular polygons of an odd number of sides
possess axial symmetry, and regular polygons of an
even number of sides possess central symmetry as
160.
well.
AV PAPER FOLDING.
161.
If
a figure has two axes of
87
symmetry
at right
angles to each other, the point of intersection of the
axes
is
symmetry. This obtains in reg
an even number of sides and certain
a center of
ular polygons of
curves, such as the circle, ellipse, hyperbola, and the
lemniscate
;
regular polygons of an
odd number
of
Fig. 54-
sides
may have more
axes than one, but no two of
If a sheet
be at right angles to each other.
of paper is folded double and cut, we obtain a piece
them
will
symmetry, and if it is cut fourfold, we
obtain a piece which has central symmetry as well, as
which has
in Fig. 54.
axial
GEOMETRIC EXERCISES
88
162.
Parallelograms have a center of symmetry.
A
quadrilateral of the form of a kite, or a trapezium
with two opposite sides equal and equally inclined to
has an axis of
either of the remaining sides,
sym
metry.
163.
The
termined by
position of a point in a plane
its
is
also de
distance from a fixed point and the
inclination of the line joining the two points to a fixed
drawn through the
line
OA
If
length
be the fixed
OP and
fixed point.
line
and
P the
/_AOP, determine
given point, the
the position of P.
FiR. 55-
O
is
called the pole,
radius vector, and
and
^_AOP
164.
The
the prime-vector,
OP
the vectorial angle.
the
OP
are called polar co-ordinates of P.
The image
OA may
OA
/_AOP
of a figure
symmetric
to the axis
be obtained by folding through the axis OA.
radii vectores of
corresponding points are equally
inclined to the axis.
165.
ABC
BC to
Let
CA, AB,
person to stand
at
be a triangle.
Z>,
A
E,
F
Produce the sides
respectively.
with face towards
D
Suppose a
and then to
IN PAPER FOLDING.
A
proceed from
B to
to B,
C,
and
successively describes the angles
Having come
8g
C to A. Then he
DAB, EBC, FCD.
to his original position A,
he has corn-
Fig. 56.
pleted a perigon,
e.
i.
,
four right angles.
We
may
therefore infer that the three exterior angles are to
gether equal to four right angles.
The same
inference applies to any convex polygon.
Suppose the man
161.
towards
C,
to stand at
A
with his face
AB
then to turn in the direction of
and
proceed along AB, BC, and CA.
In this case, the
i.
e.,
man completes
two right angles.
the angles
He
successively turns through
CAB, EBC, and FCA.
+ Z FCA -f / CAB (neg. angle) =
This property
on the railway.
its head towards
towards F.
backwards
on
to
AD.
is
made use
An engine
A
is
a straight angle,
Therefore
/_EBF
a straight angle.
of in turning engines
standing upon
driven on to CF, with
DA
with
its
head
The motion is then reversed and it goes
EB. Then it moves forward along BA
The engine has successively described
to
GEOMETRIC EXERCISES
90
the
ACB, CBA, and BAG.
angles
Therefore the
three interior angles of a triangle are together equal
to
two right angles.
The property
167.
that the three interior angles of
two right angles
a triangle are together equal to
illustrated as follows
Fold
CC
AC
in
and
meeting
BC
by paper folding.
perpendicular to
M. Fold
and
,
AC m A
and
B C A,
,
Bisect
C B in N
t
perpendicular to
B
.
Draw A
NA MB
t
C of the
BC A and A
find that the angles A, B,
to the angles
AB.
NA MB
folding the corners on
By
is
and
AB,
C BC
,
.
A B we
,
triangle are equal
CB
respectively,
which together make up two right angles.
168.
to
Z>,
Take any
line
ABC,
Draw
perpendiculars
ABC at the points A, B, and C. Take points
E, F m the respective perpendiculars equidistant
IN PAPER FOLDING.
from their
Then
feet.
91
easily seen
by superposi
and proved by equal triangles that DE is equal
AB and perpendicular to AD and BE, and that
it is
tion
to
EF is equal to BC and
Now AB (=DE} is the
lines AD and BE, and
perpendicular to
BE and
CF.
shortest distance between the
it is
Therefore
constant.
AD
Fig. 58.
and
BE can never meet,
lines
i.
e.,
they are parallel.
which are perpendicular
to the
same
Hence
line are
parallel.
The two
to
angles
BAD and EBA
are together equal
two right angles. If we suppose the lines AD and
to move inwards about A and B, they will meet
BE
and the
angles.
This
is
interior angles will be less than
They
meet
will not
embodied
much
in the
two right
produced backwards.
abused twelfth postulate
if
of Euclid s Elements.*
169.
in
If
AGffbe
any
line cutting
BE in G and CF
H, then
*For
historical
sketch see
History of Mathematics,
p. 270.
Beman and Smith
s
translation of Fink s
GEOMETRIC EXERCISES
92
/
GAD= the alternate
each
.
is
/_HGE
/_AGB,
complement of /_BAG; and
the interior and opposite / GAD.
the
they are each
Also the two angles
.
.
= / A GB.
GAD
EGA
and
are together
equal to two right angles.
170.
Take
a line
AX
and mark
on
off
it,
from A,
AB, BC, CD, DE.. ..Erect perpen
Let a line AF cut
B, C, D, E.
Then AB
C D E
perpendiculars in B
equal segments
diculars to
the
AE
at
.
,
B C CD D E
,
,
..
A
.
.are
C
B
.
,
all
.
.
,
.
.
.
,
equal.
D
F
E
Fig. 59-
If
DE be unequal,
AB:BC=AB :B C
BC: CD = B C\
171.
may
then
AB, BC, CD,
If
ABCDE.
.
.
.
CD
,
and so on.
be a polygon, similar polygons
be obtained as follows.
Take any point O within the polygon, and draw
OA, OB, OC,....
Take any point A in OA and draw A B B C
,
C D,
parallel
to
AB, BC,
CD
,
respectively.
IN PAPER FOLDING.
Then
the
ABCD
mon
lie
.
.
.
polygon
.
AB
CD
The polygons
will
be similar to
so described around a
point are in perspective.
outside the polygon.
93
The
point
It is called
com
O may
also
the center of per
spective.
172.
To
Let
parts.
divide a given line into
AB
be the given
at right angles to
AC=BD.
Now
^AC
Draw CD
produce
or
in PS, -A,
AB
BD.
A,
Then from
AC
2, 3, 4, 5.
line.
.
.
.equal
BD
Draw AC,
on opposite sides and make
cutting
AB in
and take
Draw DE, DF,
....
similar triangles,
P*.
Then
CE = EF= FG
.
.
.
.
DG ____ cutting AB
GEOMETRIC EXERCISES
94
.-.
P .B: AB = BD: AF
9
=
1 :3.
Similarly
and so
on.
If
AB = \,
P =
A
-
3-.
4,
*(
But
A Pi
-\-
P P + ^ A -f
2
is
3
+ *)
ultimately == AB.
Or
1
1
""
2"
"3
=
_
1
2~-
3
1
1
1
n
n-\-\
n(n-\- 1)
Adding
F2 + ^3 +
"-
+
IN PAPER FOLDING.
95
J_
""
-"1
The
limit of
*"
1
-- when
n
is
co is
1.
173. The following simple contrivance may be
used for dividing a line into a number of equal parts.
Take
and mark
a rectangular piece of paper,
off
n
equal segments on each or one of two adjacent sides.
Fold through the points of section so as to obtain
Mark
perpendiculars to the sides.
and the corners
tion
0, 1, 2,
quired to divide the
the points of sec
......
Suppose
it is
of another piece of
edge
re
paper
AB into n equal parts. Now place AB so that A or
B may lie on 0, and B or A on the perpendicular
n.
through
In this case
AB
must be greater than ON.
smaller side of the rectangle
the
But
may be used
for
smaller lines.
The
points where
AB
crosses the perpendiculars
are the required points of section.
174.
tains
that
of
Center of mean position.
(m
4-
)
equal parts,
AC contains m
and
it
a line
If
is
of these parts
AB
divided at
CB
and
con
C
so
contains n
from the points A, C, B perpendicu
AD, CF, BE be let fall on any line,
them
lars
;
then
if
m-BE -f
Now, draw
and
AD in H.
sion
AB lines
n-AZ>
= (m -f
BGH parallel
to
)
ED
CF.
cutting
CFin G
Suppose through the points of divi
are
drawn
parallel to
BH.
These
lines
GEOMETRIC EXERCISES
g5
will divide
AH into (m-{-ri)
equal parts and
CG
into
n equal parts.
and since
DH and BE
are each
= GF,
Hence, by addition
n-l -f
^4Z>
C
mean
m
is
(/
called the
center of
and
+ m-BE = (m +
+ 0* .#
-f
Z>
A
B
)
mean
center of
and
)
for t.he
CF.
position, or the
of multiples
system
n.
The
principle can be extended to any
Then
points, not in a line.
if
number
P represent
the feet of
the perpendiculars on any line from A, B, C, etc.,
a, b, c
be the
of
...be the corresponding multiples, and
mean center
if
if
M
c-CP....
If
the multiples are
equal to
all
we
a,
get
a(AP+BP+CP+.. ..}=na-MP
n being the
175.
number
The
of points.
center of
mean
points with equal multiples
position of a
is
the line joining any two points A,
third point
C
and divide
GC
D
in
number
obtained thus.
B in
G, join
71 so that
HD
of
Bisect
G
to a
GH=\GC\
K
in
so that
and divide
join If to a fourth point
will be
found
last
and so on: the
point
HK=\HD
the center of
mean
position of the system of points.
IN PAPER FOLDING.
The notion
176.
is
position
of
mean
97
center or center of
mean
derived from Statics, because a system of
material points having their weights denoted by
c
.
.
.
.
,
and placed
mean
the
center
at
M,
A, B,
if
C
.
.
.
.
a, b,
would balance about
free to rotate
about
M under
the action of gravity.
The mean
center has therefore a close relation to
the center of gravity of Statics.
The mean
177.
is
center of three points not in a line,
the point of intersection of the medians of the
angle formed by joining the three points. This
is
tri
also
the center of gravity or mass center of a thin
tri
angular plate of uniform density.
178. If
M
is
the
mean
center of the points A, B,
C, etc., for the corresponding multiples a,
and
if
P is
=a
Hence
b, c, etc.,
any other point, then
in
A M* + b BM* + c- CM
any regular polygon,
or circum-center and
P is
AB^
Similarly
O
is
+
.
.
.
.
the in-center
any point
4- BP* + ....= OA*
Now
if
2
+
OB^ + ....+
OP
2
GE OME TRIG EXER CISES
98
Adding
The sum of the squares of the lines joining
the mean center with the points of the system is a
minimum.
If J/be the mean center and P any other point
179.
not belonging to the system,
2P^ = 2MA +2PM
2
sum
2PA
P is the
..
when
2
is
the
mean
(where
2 stands for
"the
type").
minimum when PAf=Q,
i.
e.,
center.
Properties relating to concurrence of lines
180.
and
*,
of all expressions of the
collinearity of points can be tested
ing.*
(1)
by paper fold
Some instances are given below:
The medians of a triangle are concurrent. The
common
(2)
point
The
is
called the centroid.
altitudes of a triangle
are
concurrent
The common point is called the orthocenter.
(3) The perpendicular bisectors of the sides of a
The common point is called
triangle are concurrent.
the circum-center.
(4)
The
concurrent.
(5)
point.
bisectors of the angles of a triangle are
The common
point
is
called the in-center.
ABCD be a parallelogram and P any
Through P draw GT and EF parallel to BC
Let
*For treatment of certain of these properties see
Neiu Plane and Solid Geometry, pp. 84, 182.
Beman and Smith
s
IN PAPER FOLDING.
and
AB respectively.
and the
(6)
line
Then
the diagonals
99
EG, HF,
DB are concurrent.
two similar unequal rectineal figures are so
If
placed that their corresponding sides are parallel, then
the joins of corresponding corners are concurrent.
The common
point
is
called the center of similarity.
two triangles are so placed that their corners
(7)
are two and two on concurrent lines, then their corre
If
sponding sides intersect collinearly. This is known
The two triangles are said
as Desargues s theorem.
to be in perspective.
The
point of concurrence and
line of collinearity are respectively called the center
and axis
(8)
of perspective.
The middle
points of the diagonals of a
com
plete quadrilateral are collinear.
(9) If
from any point on the circumference
of the
circum-circle of a triangle, perpendiculars are dropped
on
its
produced when necessary, the feet of
This line is called
sides,
these perpendiculars are collinear.
Simson
s line.
Simson
s
line bisects the join of the orthocenter
and the point from which the perpendiculars are
drawn.
(10) In any triangle the orthocenter, circum-center,
and centroid are collinear.
The mid-point
circum-center
is
of the join of the orthocenter
and
the center of the nine-points circle, so
called because it passes through the feet of the alti
tudes and medians of the triangle and the mid-point
GEOMETRIC EXERCISES
ioo
of that part of
each altitude which
orthocenter and
vertex.
The
lies
center of the nine-points circle
is
between the
twice as far
from the orthocenter as from the centroid.
known
This
is
as Poncelet s theorem.
If
(11)
A, B,
D, E, F, are any
C,
six points
on a
which are joined successively in any order, then
the intersections of the first and fourth, of the second
circle
and
fifth,
and
duced when
as Pascal
s
and sixth
of these joins
This
necessary) are collinear.
is
joins of the vertices of a triangle with the
points of contact of the in-circle are concurrent.
same
pro
known
theorem.
The
(12)
of the third
The
property holds for the ex- circles.
The
(13)
internal bisectors of two angles of a
tri
angle, and the external bisector of the third angle in
tersect the opposite sides collinearly.
The
(14)
external bisectors of the angles of a
tri
angle intersect the opposite sides collinearly.
(15) If
any point be joined to the vertices
triangle, the lines
of a
drawn through the point perpen
dicular to those joins intersect the opposite sides of
the triangle collinearly.
(16) If
triangles
CO
and
on an axis
ABC, A
fi
C
of
symmetry
a point
intersect the sides
O
of the
be taken
BC, CA and
congruent
A
O,
AB
B
O,
collin
early.
(17)
The
points of intersection of pairs of tangents
to a circle at the extremities of
chords which pass
IN PAPER FOLDING
101
through a given point are collinear. This line is called
the polar of the given point with respect to the circle.
(18)
lines
The
isogonal conjugates of three concurrent
CX with
AX, BX,
a triangle
ABC
respect to the three angles of
are concurrent.
lines
(Two
AY
AX,
are said to be isogonal conjugates with respect to an
angle
BAC, when
they
make equal
angles with
its
bisector.)
(19)
a triangle
If in
drawn from each
ABC,
AA\ BB CC
the lines
,
of the angles to the opposite sides
are concurrent, their isotomic conjugates with respect
to the corresponding sides are also concurrent.
lines
AA A
,
A"
with respect to the side
BA
the intercepts
(20)
The
current.
a triangle
three
(The
is
(The
are said to be isotomic conjugates,
and
BC of the
CA"
triangle
are equal.)
symmedians
of a triangle are
isogonal conjugate of a
called a
ABC, when
symmedian.)
median
con
AM oi
XIII.
THE CONIC SECTIONS.
SECTION
181.
A
I.
THE
CIRCLE.
piece of paper can be folded in numerous
ways through a common
point.
Points on each of the
be equidistant from the
on the circumference of a circle,
lines so taken as to
point will
lie
common
the
point
is
The
the center.
common
of
which
circle is the
locus of points equidistant from a fixed point, the
centre.
182.
drawn.
183.
Any number of concentric circles can be
They cannot meet each other.
The center may be considered
concentric circles described round
it
as the limit of
as center, the
radius being indefinitely diminished.
184. Circles
with equal radii are congruent and
equal.
185.
The
curvature of a circle
out the circumference.
to slide along itself
Any
A circle
is
uniform through
can therefore be made
by being turned about
figure connected with the circle
may
its
center.
be turned
about the center of the circle without changing
lation to the circle.
its re
PAPER FOLDING
186.
A
103
straight line can cross a circle in only
two
points.
187.
Every diameter
the circle.
It
is
is
bisected at the center of
equal in length to two
All
radii.
rjiameters, like the radii, are equal.
188.
The
center of a circle
is
its
center of
sym
metry, the extremities of any diameter being corre
sponding points.
Every diameter
and conversely.
189.
circle,
190.
systems
191.
The
is
an axis of symmetry of the
propositions of
188, 189 are true for
of concentric circles.
Every diameter divides the
circle
into
two
equal halves called semicircles.
192.
Two
diameters at right angles to each other
divide the circle into four equal parts called quadrants.
193.
bisecting the right angles contained by
By
the diameters, then the half right angles, and so on,
obtain 2 n equal sectors of the
we
between the
Of
27T
circle.
The
angle
4.
radii of
each sector
is
-
of a right angle
7t
.
2*
194.
2-i-
As shown
in the
preceding chapters, the right
3, 5, 9, 10, 12, 15 and
angle can be divided also into
17 equal parts.
And each
can be subdivided into
2"
of the parts thus obtained
equal parts.
GEOMETRIC EXERCISES
io4
A
195.
and a
circle
circle
can be inscribed
in a regular
can also be circumscribed round
former circle
will
touch the sides
polygon,
it.
The
at their mid-points.
Equal arcs subtend equal angles at the cen
and conversely. This can be proved by super
196.
ter;
If
position.
a circle be folded
two semicircles coincide.
a diameter, the
upon
in
Every point
one semi-
circumference has a corresponding point in the other,
below
it.
Any two
197.
radii are the sides of
angle, and the chord which joins
an isosceles
tri
their extremities
is
the base of the triangle.
A
198.
radius which bisects the angle between two
radii is perpendicular to the base
sects
chord and also
Given one fixed diameter, any number
may be drawn, the two radii of each
199.
pairs of radii
being equally inclined
it.
The
to the
all
of
set
diameter on each side of
chords joining the extremities of each pair of
The chords
radii are at right angles to the diameter.
are
bi
it.
parallel to one another.
200.
The same diameter
well as arcs standing
bisects
all
upon the chords,
the chords as
i.
e.,
the locus
of the mid-points of a system of parallel chords
is
a
diameter.
201.
The perpendicular
circle pass
bisectors of
through the center.
all
chords of a
IN PAPER FOLDING
105
202.
Equal chords are equidistant from the
203.
The
center.
extremities of two radii which are equally
inclined to a diameter on each side of
are equi
it,
distant from every point in the diameter.
number
Hence, any
can be described passing through
In other words, the locus of the cen
of circles
the two points.
through two given points
ters of circles passing
straight line
which bisects
is
the
at right angles the join of
the points.
204. Let
dius OA.
CC
Then
be a chord perpendicular to the ra
the angles
AOCand AOC
are equal.
Suppose both move on the circumference towards A
with the same velocity, then the chord CC is always
parallel to itself
the points C,
A
and perpendicular to OA. Ultimately
and C coincide at A, and CA C is
A
perpendicular to OA.
is
the last point
the chord and the circumference.
becomes ultimately
205.
common
CAC
to
produced
a tangent to the circle.
The tangent
is
perpendicular to the diameter
through the point of contact; and conversely.
206.
If
two chords
of a circle are parallel, the arcs
joining their extremities
equal.
towards the same parts are
So are the arcs joining the extremities
of either
chord with the diagonally opposite extremities of the
other and passing through the remaining extremities.
This
is
easily seen
by folding on the diameter perpen
dicular to the parallel chords.
GEOMETRIC EXERCISES
io6
The two chords and
207.
ities
the joins of their extrem
towards the same parts form a trapezoid which
viz., the diameter perpen
has an axis of symmetry,
The diagonals
dicular to the parallel chords.
trapezoid intersect on the diameter.
It is
of the
evident by
folding that the angles between each of the parallel
chords and each diagonal of the trapezoid are equal.
Also the angles upon the other equal arcs are equal.
The angle subtended
208.
by any
arc
is
at the center of a circle
double the angle subtended by
it
at the
circumference.
Fig. 61.
An
Fig. 62.
Fig- 63.
inscribed angle equals half the central angle
standing on the same arc.
Given
A VB
an inscribed angle, and
on the same
To prove
AOB
the central angle
arc. AB.
that /
A VB =
*-
/
A OB.
Proof.
1.
Suppose VO drawn through center O, and pro
duced to meet the circumference at X.
IN PAPER FOLDING
2.
Then
And
Z VBO,
/_XVB = \_
.-.
3.
4.
XOB= /.XVB +
107
A VX= \i_AOX (each=zero in Fig. 62),
and .-. LAVB = tAOB.
The proof holds for all three figures, point A hav
/
Similarly
moved
ing
to
X (Fig.
62),
and then through
X (Fig.
63).*
209.
The angle
at the center
angles subtended by an arc at
being constant, the
all
points of the cir
cumference are equal.
210.
The angle
211.
If
chord
be a diameter of a
at right angles to
lateral of
angles
AB
in a semicircle is a right angle.
which
BCA
AB
is
it,
circle,
\\\tn.ACBD
an axis of
ADB being
angles DBC
and
DC
and
is
a
a quadri
The
symmetry.
each a right angle, the
CAD are together
If A and B be any other
equal to a straight angle.
points on the arcs DAC and CBD respectively, the
/_CAD=-/_ CA Dand /_DBC=Z.DB C, and ^CA D
remaining two
-\-DB
C=
+ / A DB
a straight angle.
Therefore, also, /_B
CA
== a straight angle.
Conversely,
site
and
if
a quadrilateral has
angles together equal to
two
of its
oppo
two right angles,
it is
inscriptible in a circle.
*The above figures and proof
and Solid Geometry, p. 129.
are from
Beman and Smith
s
New
Plane
GEOMETRIC EXERCISES
io8
212.
The angle between
the tangent to a circle
and
a chord which passes through the point of contact
is
equal to the angle at the circumference standing upon
that chord and having
its
vertex on the side of
posite to that on which the
Let
chord.
OB.
AC
first
angle
op
lies.
be a tangent to the circle at
Take O the center
it
of the circle
A
and
AB
a
and draw OA,
Draw OD perpendicular to AB.
Then ^_BAC^=L AOD = % / BOA.
213. Perpendiculars to diameters at their extremi
ties
The
touch the circle
at these extremities.
(See Fig. 64).
line joining the center and the point of intersection
IN PAPER FOLDING
109
two tangents bisects the angles between the two
It also bisects
tangents and between the two radii.
of
the join of the points of contact.
The tangents
are
equal.
This
is
seen by folding through the center and
the point of intersection of the tangents.
AB
Let AC,
line
be two tangents and
the center O, cutting the circle in
in
ADEOF
the
A
and
BC
through the intersection of the tangents
D
and
F
and
mean
of
AD and
E.
Then
AF\ AE
AC or AB is
is
the geometric
the harmonic
mean;
andAO
the arith
metic mean.
AD-AF_~ 2AD- AF
~OA~
AD-^AF
any other chord through A be ob
tained cutting the circle in P and R and BC in Q,
then AQ is the harmonic mean and AC the geometric
mean between AP and AR.
Similarly,
214.
if
Fold a right-angled triangle
the perpendicular on the hypotenuse.
such that
OD= OC (Fig.
65).
Then O
and
OA OC=OC: OB,
OA OD=OD\ OB.
:
:
OCB
and
Take
D in AB
CA
GEOMETRIC EXERCISES
no
A
can be described with
circle
OC or OD
as radius.
The
A
points
and
B
O
as center and
are inverses of each other
O and
with reference to the center of inversion
the
CDE.
circle of inversion
Fig. 65.
Hence when
the center
is
taken as the origin, the
foot of the ordinate of a point on a circle has for its
mverse the point
of intersection of the tangent
and
the axis taken.
215.
line
to
Fold
FBG is
FBG
Then the
perpendicular to OB.
A with reference
called the polar of point
the polar circle
called the pole of
CDE
FBG.
and polar center
Conversely B
O
is
;
and
A
is
the pole of
2N PAPER FOLDING
CA
CA
and
same
is
the polar of
Produce
perpendicular to
The
points A, B, F,
is,
two points and
That
cyclic
FBG in
F, and fold
AH
points.
F
the polar of H.
is
217.
meet
the polar of 7% and the perpendicular at
AJfis
<9^
OC to
OC.
F and Zf are inverse
Then
H,
are concyclic.
their inverses are con-
and conversely.
;
Now
and
reference to the
circle.
216.
to
B with
Mi
take another point
AK
fold
G
on
FBG.
perpendicular to OG.
Draw OG,
Then
K
are inverse points with reference to the circle
The
218.
points F,
J3,
G
and
G
CDE.
are collinear, while their
polars pass through A.
Hence, the polars of collinear points are concur
rent.
on the polar
of the other are called conjugate points, and lines so
related that each passes through the pole of the other
219. Points so situated that each lies
are called conjugate lines.
A
and
F are
conjugate points, so are
A
and B,
A
and G.
The
points
point
is
220.
of
intersection of
the polars of two
the pole of the join of the points.
As A moves towards D, B also moves up to it.
and B coincide and FBG is the tangent at B.
Finally A
GEOMETRIC EXERCISES
ii2
Hence
the polar of any point on the circle
the
is
tangent at that point.
A moves
As
221.
The
infinity.
polar center
to O,
B moves
forward to
the line at infinity.
is
The angle between
222.
is
back
polar of the center of inversion or the
the polars of two points
equal to the angle subtended by these points at the
polar center.
The
223.
BC as
a radius cuts the circle
AB
224. Bisect
to
AB.
will
Then
CDE
circles.
circles passing
AF
and
AG
Hence
ally the extremities of
if
a center
and
orthogonally.
LN perpendicular
through
These
this line.
orthogonally.
circles.
spective
CDE
and fold
the quadrilaterals
scribing
such
all
L
in
have their centers on
the circle
B as
circle described with
The
ABFH
A
circles
and
and
B
circles cut
circum
ABGK
are
are diameters of the re
two
circles cut
any diameter
orthogon
con
of either are
jugate points with respect to the other.
225.
A,
The
K being
points O, A,
//and
K are concyclic.
inverses of points on the line
inverse of a line
is
being the extremities of a diameter
226.
If
H,
the
a circle through the center of in
version and the pole of the given
D and D
FBG,
DO produced
line,
;
cuts the circle
are harmonic conjugates of
these points
and conversely.
A
CDE
and B.
in
D
,
Sim-
IN PAPER FOLDING
if
circle
CDE
conjugates
through B cuts AC in A and the
d and d then ^/ and */ are harmonic
line
any
ilarly,
k
in
of
,
A
and B.
227. Fold any line
pendicular to
Then
LM=LB = LA,
and
LM meeting AB produced in
the circle described with center
O Mcuts orthogonally
center L and radius LM.
dius
Now
and
OL
2
Z
2
<9
O (O M).
By taking other
J/6>
O
O and
per
.
ra
the circle described with
= O* + L
= O M* ZJ/
2
,
ZTVis the radical
.-.
11.3
2
.
axis of the circles
O (OC}
and
points in the semicircle
repeating the same construction as above,
infinite
systems of circles co-axial with
O\O M),
we
get two
O(OC} and
one system on each side of the radical
point circle of each system is a point,
or B, which may be regarded as an infinitely small
axis,
A
AMB and
viz.,
LN. The
circle.
The two infinite systems of circles are to be re
garded as one co-axial system, the circles of which
range from infinitely large to infinitely small the
radical .axis being the infinitely large circle,
limiting points the infinitely small.
co-axial circles
If
is
two
is
and the
This system of
called the limiting point species.
circles cut
each other their
their radical axis.
Therefore
all
common chord
circles passing
GEOMETRIC EXERCISES
ii4
through
A
and
B
are co-axial.
axial circles is called the
228.
to
and
B
in
Then
OPQ.
OPQ
point species.
OAB
OAB
draw AP,
AP and BQ
P and Q.
From two
and OPQ.
BQ
perpendicular
A
circles described with
centers and
at
common
lines
Take two
A
points
This system of co
and
B
as
as radii will touch the line
OA-.OB = AP:BQ.
Then
This holds whether the perpendiculars are towards
the same or opposite parts.
The tangent is in one
case direct, and in the other transverse.
In the
ond
it is
first
case,
between
A
O
is
outside
and B.
AB, and
In the former
the external center of similitude and
in
in the sec
called
it is
the latter the
internal centre of similitude of the two circles.
229.
The
allel radii of
line joining the extremities of
two par
the two circles passes through their ex
ternal center of similitude,
the radii are in the
if
direction, and through their internal center,
drawn
are
230.
if
same
they
in opposite directions.
The two
radii of
one
circle
drawn
to its points
any line passing through either
center of similitude, are respectively parallel to the
of intersection with
two
radii of the other circle
with the same
drawn
to its intersections
line.
231. All secants passing through a center of simil
itude of two circles are cut in the
circles.
same
ratio
by the
IN PA PER FOLDING
232..
section,
If
B\>
B\,
D\ and B^
t
2,
D<t
11
5
be the points of inter
and D\, DI being corresponding
points,
= OD
Hence
-
l
OBi
=
the inverse of a circle, not through the cen
ter of inversion is a circle.
Fig. 66.
The
center of inversion
of the original circle
The
and
original circle,
is
its
its
the center of similitude
inverse.
inverse,
and the
circle of
inversion are co-axial.
233.
The method
of inversion is
one of the most
It was discov
important in the range of Geometry.
ered jointly by Doctors Stubbs and Ingram, Fellows
of Trinity College, Dublin, about 1842.
ployed by
proof of
Sir
some
William Thomson
of the
most
mathematical theory of
SECTION
234.
A
parabola
which moves
in a
is
It
in giving
difficult
was em
geometric
propositions in the
electricity.
II.
THE PARABOLA.
the curve traced by a point
plane in such a manner that
its dis-
GEOMETRIC EXERCISES
n6
tance from a given point
always equal to
is
its
dis
tance from a given straight line.
235. Fig. 67
on paper.
rix,
O
shows how a parabola can be marked
The edge
of the square
the vertex, and
and obtain the
axis.
F the focus.
MN
is
the direct
Fold through
OX
Divide the upper half of the
Fig. 67.
square into a number of sections by lines parallel to
the axis.
These
lines
meet the
directrix in a
number
Fold by laying each of these points on the
focus and mark the point where the corresponding
of points.
horizontal line
is
cut.
The
points thus obtained
lie
on a parabola. The folding gives also the tangent to
the curve at the point.
IN PAPER FOLDING
236.
FL
which
117
right angles to
is at
OX
is
called
the semi-latus rectum.
237.
When
points on the upper half of the curve
have been obtained, corresponding points on the lower
half are obtained by doubling the paper on the axis
and pricking through them.
238.
When
the axis and the tangent at the vertex
are taken as the axes of co-ordinates,
and the vertex
as origin, the equation of the parabola
becomes
2
j>
= 4:ax
or
Y
The parabola may be denned as the curve traced
which moves in one plane in such a manner
by
a point
that the square of
its
distance from a given straight
from another straight line
the mean proportional between the
line varies as its distance
or the ordinate
is
;
1 1
GEOME TRIG EXERCISES
8
and the latus rectum which
abscissa,
is
equal to 4- OF.
Hence the following construction.
Take O T in FO produced = 4 OF.
Bisect TN m M.
Take Q in OYsuch that MQ = MN=MT.
Fold through
to
OY.
of
N.
Let
P
so that
QP may
be
the ordinate
a point on the curve.
FPFG = FT
The subnormal^ 2 OF and
239.
at right angles
QP meets
be the point where
P is
Then
Q
.
These properties suggest the following construc
tion.
Take ^Vany point on the
On
the side of
Fold
NP
such that
Then
A
N
perpendicular to
circle
a point
and
find
P in NP
on the curve.
can be described with
The double
Amoves
F as center and FG,
as radii.
ordinate of the circle
ordinate of the parabola,
240.
OG
FP = FG.
P is
FP and FT
as
axis.
remote from the vertex take
i.
e.,
is
also the double
P describes
a parabola
along the axis.
Take any point
N
between
RN P at right angles to
Take R so that OR = OF.
Fold
Fold ^/^perpendicular
to
O and F (Fig.
69).
OF.
OR,
N being on the axis.
IN PAPER FOLDING
119
NP perpendicular to the axis.
Now, in OX take OT=OW.
Take P in RN so that FP = FT.
Fold through P F cutting NP in P.
Fold
P and .P
Then
are points on the curv
N
/F
N
Fig. 69.
241.
A
r
and
N
coincide
when
PFP
the latus
is
rectum.
As
N
recedes from
F to
O, 7^ moves forward
from
infinity.
At the same time,
moves
242.
To
Amoves toward
in the opposite direction
O, and
T (OT =
toward
find the area of a parabola
infinity.
bounded by
the axis and an ordinate.
Complete the rectangle
ONPK.
Let
OK
be
di-
GE OME TRIG EXER CISE S
\20
vided into n equal portions of which suppose Om to
contain r and mn to be the (r -f- 1) *.
Draw mp, nq at
OK meeting
right angles to
The
angles to nq.
at right
the limit of the
structed as
sum
mn on
But mi/
:
I
q,
curvilinear area
and pn
OPK is
of the series of rectangles
con
the portions corresponding to mn.
NKpm
I
the curve in p,
mn PK- OK,
:
and, by the properties of the parabola,
pm\PK=Om*\ OK*
and mn
.
.
OK=
OK= r*
r
ipn = w x en A^1
:
n.
:
pm mn\PK
:
;z
3
.
l
.
Hence
the
p
_|_
.
sum
22
-f-
of the series of rectangles
+
32
1)(2
(
2
(
I)
1)
1-2-3-w 3
=
.
.
The
-J
of
in
curvilinear area
parabolic area
243.
nn^VA
^^TV^^f
The same
line
the limit,
OPK=^
of
of
I
i.
of
e.,
when
is oo.
cZi^VA and the
,
\NK.
proof applies
when any
diameter and an ordinate are taken as the boundaries
of the parabolic area.
IN PAPER FOLDING
SECTION
An
244.
moves
THE ELLIPSE.
III.
ellipse is the curve traced
in a
plane
by a point which
such a manner that
in
from a given point
121
is
its
distance
in a constant ratio of less in
from a given straight line.
Let Fbe the focus, OYthe directrix, and XX the
Let FA \AObe the
perpendicular to O Y through F.
equality to its distance
Fig. 70.
constant
ratio,
FA
being less than
AO.
A
is
a point
on the curve called the vertex.
As
in
116, find
A
in
XX
such that
FA :A O = FA :AO.
Then A
is
another point on the curve, being a
second vertex.
Double the
line
AA
on
itself
point C, called the center, and
sponding
to
F and
O.
and obtain
mark
F
Fold through
O
and
its
O
middl e
corre
so that
OY
1
GEOMETRIC EXERCISES
122
may be
at right angles to
XX
Then F*
.
the sec
is
Y the second directrix.
A
A
obtain the perpendicular through C.
By folding
ond focus and O
,
FA :AO = FA A O
-.
= FA-\-FA
= AA OO
= CA CO.
:AO + A O
:
:
B and B in the perpendicular through
on opposite sides of it, such that FB and FB
are each equal to CA.
Then B and B are points on
Take points
C and
the curve.
AA
is
called the major axis,
and
BB
the minor
axis.
245.
point
To
E in
find other points
on the curve, take any
E and A,
the directrix, and fold through
and through E and A
Fold again through E and F
and mark the point P where FA cuts EA produced.
.
Fold through PF and
points on the curve.
Fold through
P
P on EA
and
P
so that
are perpendicular to the directrix,
the directrix and
FL
Then
.
KPL
L and L on EL.
A FP,
tLFP = LPLF
and
FP\PK=PL\PK
= FA
:
AO.
And
FP -.PK =P L
and
K and K
bisects the angle
.-.
P and P
-.PK
are
KLP
being on
IN PAPER FOLDING
123
= FA :AO.
If
EO = FO, FP is at right angles
PP is the latus rectum.
FP=FP
246.
to
FO, and
.
When
number
a
of points
on the
left half of
the curve are found, corresponding points on the other
marked by doubling the paper on the
and pricking through them.
half can be
minor
axis
247.
If
An
\AN-NA
of
P from
may also be
move in such
ellipse
a point
is
P
which
a
manner
AA
is
an
2
PN being the distance
A and
a constant ratio,
,
:
/W
that
the line joining two fixed points A,
N being between A and A
of
defined as follows
the locus of
,
P is an ellipse
axis.
248. In the circle,
In the ellipse
This ratio
PN AN-NA
2
:
may
the former case
is
a constant ratio.
be less or greater than unity.
APA is obtuse, and the curve
within the auxiliary circle described on
eter.
is
In the latter case,
outside the circle.
/
major, and in the second
249.
The above
APA
In the
it is
the vertex
is
the origin.
lies
diam
as
acute and the curve
first
case
AA
is
the
the minor axis.
definition corresponds to the
tion
when
is
AA
In
equa
GEOMETRIC EXERCISES
124
250.
nate
AN* NA
QN
the
of
is
equal to the square on the ordi-
auxiliary
circle,
and
PN QN =
:
BC-.AC.
251. Fig. 71
shows how the points can be deter
mined when the constant
Thus,
lay off
E any point
CD=-AC,
of
ratio
is
less
than unity.
the semi-major axis.
^Cdraw
DE and
produce
it
Through
to
meet
Q
Draw
the auxiliary circle in Q.
to
meet the ordinate
when
the ratio
E
and produce
it
in P.
Then is PN: QN
The same process is appli
QN
&C\DC=BC\AC.
cable
JB
is
greater than unity.
When
points in one quadrant are found, corresponding points
in other
252.
quadrants can be easily marked.
If
P and P
are the extremities of two conju
gate diameters of an ellipse and the ordinates
MP
IN PAPER FOLDING
and
MP
angle
meet the auxiliary
QCQ
Now
is
circle in
Q
and
Q
,
the
a right angle.
take a rectangular piece of card or paper and
mark on two adjacent edges beginning with the com
mon corner lengths equal to the minor and major
axes.
By
turning the card round
C mark
correspond
ing points on the outer and inner auxiliary circles.
Let Q, R and Q R be the points in one position.
and RP and R P
Fold the ordinates
and Q
r
,
M
QM
perpendiculars to the ordinates.
,
t
Then
P and P
are
points on the curve.
Fig. 72.
253. Points on the curve
mined by the application
may
also be easily deter
of the following property of
the conic sections.
The
focal distance of a point on a conic
is
equal
GEOMETRIC EXERCISES
126
to the length of the ordinate
produced
to
meet the
tangent at the end of the latus rectum.
Draw A A
From any point D
A A produced draw DR perpendicular to AD. Take
254. Let
A
and
and produce the
in
A
be any two points.
both ways.
line
RA and RA Fold AP
perpendicular to AR, meeting RA in P. For different
positions of R in DR, the locus of P is an ellipse, of
which AA is the major axis.
any point
R in DR
and draw
.
Fig. 73-
PN perpendicular to A A
Now, because PN parallel to RD,
PN:A N=RD:A D.
Again, from the triangles, APN and DAR,
PN\AN=AD\ RD.
PN* .AN A N=AD\A D, a constant
Fold
.
is
.-.
less
than unity, and
tion that
jVmust
lie
IV.
An hyperbola
which moves
in
is
ratio,
evident from the construc
between
SECTION
255.
is
it
A
and
A
.
THE HYPERBOLA.
the curve traced by a point
a plane in such a manner that
its
IN PAPER FOLDING.
distance from a given point
is
in a
greater inequality to its distance
127
constant ratio of
from a given straight
line.
256.
The
construction
is
the
same
as for the el
As
the position of the parts is different.
lipse, but
119, X, A lies on the left side of the
explained in
directrix.
the foci
Each
lie
directrix lies
A
between
without these points.
and
A
,
and
The curve con
two branches which are open on one side.
The branches lie entirely within two vertical angles
sists
of
formed by two straight lines passing through the cen
These are tan
ter which are called the asymptotes.
gents to the curve at infinity.
The hyperbola can be defined thus If a point
move in such a manner that PN^ AN NA is a
257.
P
:
-
:
constant ratio,
line joining
PN being
the distance of
P from
the
two fixed points A and A and TV not
A and A the locus of P is an hyper
being between
bola, of which
,
,
AA
is
the transverse axis.
This corresponds to the equation
where the origin
hyperbola.
Fig. 74 shows
is
at the
how
right-hand vertex of the
points on the curve
found by the application of
Let C be the center and
may be
this formula.
A
the vertex of the curve.
GEOMETRIC EXERCISES
12 S
CA
Fold
Fold
CD
any
= CA = CA =
line through
DN perpendicular
to
a.
C and make
CD.
Fold
NQ
= CA.
perpen
CA and make NQ = DN. Fold Q
CA in S. Fold ^ S cutting (Win P.
dicular to
ting
C>
A"
cut
Fig. 74-
Then
/>
a point
is
For, since
diameter
DN
is
on the curve.
tangent to the circle on the
AA
DN* = AN- (2CA + AN),
or since
QN=DN,
IN PAPER FOLDING.
QN
Squaring,
or
y
QN=b
If
AC
and
BC
^
then
the asymptotes.
A"C
_=_
^
=
2
is
,
+x
(2ax
2
2
the focus and
we complete
If
za?
the asymptote
is
).
CD
is
one of
the rectangle on
a diagonal of the rect
angle.
258.
The hyperbola can
259.
An hyperbola
also be described
said to be equilateral
is
the transverse and conjugate axes are equal.
a
=
fr,
when
Here
and the equation becomes
In this case the construction
nate of the hyperbola
tween
by the
253.
property referred to in
AN and A N,
is itself
and
is
is
simpler as the ordi-
the geometric
mean be
therefore equal to the tan
gent from TVto the circle described on A A as diameter.
The polar equation to the rectangular hyper
when the center is the origin and one of the axes
260.
bola,
the initial line,
is
r 2 cos
or r 2
Let OX,
VOX into
a
OYbe
number
26
=a
= cos26
-pj
2
a.
the axes; divide the right angle
of equal parts.
Let
XOA, A OB
GEOMETRIC EXERCISES
130
be two of the equal angles.
to
XB
at right angles
BO and take OF= OX. Fold OG
BF and find G in OG such that FGB
Produce
OX.
perpendicular to
is
Fold
a right angle.
Take
OA =
Then
OG.
A
is
a point
on the curve.
Fig. 75-
Now,
the angles
XOA
OB=
and
AOB being each
0,
COS 26
a
And O4 2 =OG^=
COS29**"
The
261.
minous
of
a
points of trisection of a series of conter
circular arcs
lie
which the eccentricity
means
of trisecting
* See
Taylor
with footnote.
s
Ancient
two hyperbolas
This theorem affords
on branches
is 2.
of
an angle.*
and Modern Geometry of
Conies,
examples
308,
390
MISCELLANEOUS CURVES.
XIV.
262.
I
propose in
this,
the last chapter, to give
hints for tracing certain wall-known curves.
THE
263. This
CISSOID.*
word means ivy-shaped curve.
It is
de
OQA (Fig. 76) be a semicircle
be two
on the fixed diameter OA, and let QM,
nned as follows: Let
RN
ordinates of the semicircle equidistant from the cen
Draw
ter.
of
P
If
is
OR
cutting
QM
in P.
Then
the locus
the cissoid.
OA=2a,
the equation to the curve
/(2a
Now,
let
PR
is
x)=x*.
cut the perpendicular from
C
in
D
AP cutting CD in E.
RN:CD = ON: OC=AM~AC=PM:EC,
and draw
.-.
But
If
RN-,PM=CD:CE.
RN\ PM=ON\ OM=ON: AN^ON* NR*
CF be
:
the geometric
*See Beman and Smith
mentary Geometry, p. 44.
s
mean between
translation of Klein s
CD and
CE,
Famous Problems of Ele
GEOMETRIC EXERCISES
132
CD:CF=OC:CD
.
.
tween
CD and CF
OC and CE.
M
are the two geometric
C
N
means be
A
F
Fig. 76.
264.
The
cissoid
was invented by Diocles (second
century B. C.) to find two geometric means between
OC and
two lines in the manner described above.
P
CE
was determined by the
being given, the point
aid of the curve, and hence the point D.
PD and DR are each equal
AOQ is trisected by OP.
265. If
the angle
to
OQ, then
IN PAPER FOLDING.
Then
Draw QR.
QR
is
parallel to
133
OA, and
THE CONCHOID OR MUSSEL-SHAPED CURVE.*
266. This curve
150 B. C.).
fixed
O
be a
its
dis-
Let
a
point,
tance from a fixed
DM,
was invented by Nicomedes
(c.
line,
and let a pencil of
DM.
rays through
O
On
these rays
each
of
cut
each way from its
lay
intersection with DM, a
off,
segment
b.
The
locus
of the points thus deter
mined
is
the conchoid.
According as
or
<#,
b
>,
the origin
=,
is
a
node, a cusp, or a con
The
jugate point.
fig-
ure| represents the case
when
b
>
a.
267. This curve also
was employed
Fig. 77-
for finding
two geometric means, and
for the trisection of an angle.
*See Beman and Smith s translation of Klein s Famous Problems of Eltmentary Geometry, p. 40.
tFrom Beman and Smith s translation of Klein s Famous Problems of
Elementary Geometry,
p. 46.
GEOMETRIC EXERCISES
134
OA
Let
be the longer of the two lines of which
two geometric means are required.
Bisect
OA
in
with
B\
as a center
and
Place a chord
radius describe a circle.
circle
O
OB
BC
equal to the shorter of the given lines.
AC and
produce
collinear with
AC and BC
O and
to
such that
D and E,
DE
OB,
as a
in
the
Draw
two points
or
BA.
Fig. 78.
Then
ED and CE are the two
mean
proportionals
required.
Let
OE
cut the circles in
By Menelaus
s
E and
G.
Theorem,*
BC-ED OA=CE OD -BA
... BC OA=CE-OD
BC ~~_ OD
-
OA
~CE
BE
CE
See
Beman and Smith
s
~
OD + OA
GE
OA
OA
New Plane and Solid
Geometry,
p. 240.
IN PAPER FOLDING.
.-.
GE EF= BE -EC.
GE -OD =
.-.
OA -OD =
But
The
position of
135
E is
found by the aid of the con
the asymptote, O the focus, and
choid of which AD
DE the constant intercept.
is
The
268.
trisection of the angle is thus effected.
In Fig. 77, let
On
M
OM
lay
= / MOV,
<
OM=b,
off
and
as a center
O draw
any arbitrary length.
With
a radius b describe a circle,
M perpendicular
through
the angle to be trisected.
to the axis of
and
X with origin
a vertical line representing the asymptote of
Construct the con
the conchoid to be constructed.
choid. Connect
O with A,
the intersection of the circle
Then
and the conchoid.
is
/
A OY one
third of
<p.*
THE WITCH.
OQA
269. If
ordinate of
to
it,
(Fig. 79) be a semicircle and
and
NP be taken
ON, OA and (M7 then
,
Fold
the locus of
AM at right angles to
NQ an
a fourth proportional
P is
the witch.
OA.
Fold through O, Q, and M.
Complete the rectangle
PN\
NAMP.
QN=OM: OQ
= OA\ON.
"-Beman
and Smith
tary Geometry,
p. 46.
s
translation of Klein s
Famous Problems of Elemen
GEOMETRIC EXERCISES
136
Therefore
Its
P is
equation
a point on the curve.
is,
Fig- 79-
This curve was proposed by a lady, Maria Gaetana
Agnesi, Professor of Mathematics at Bologna.
THE CUBICAL PARABOLA.
270.
The equation
Let
OX and OYbe
and
to this
curve
is
a*y
=
x*.
the rectangular axes,
OA=a,
OX=x.
OY take OB = x.
Draw BA and draw AC at right
In the axis
ting the axis
O Kin
C.
angles to
AB cut
IN PAPER FOLDING.
Draw CX, and draw XYat right
Complete the rectangle XOY.
P is
a point
137
angles to CX.
on the curve.
Fig. 80.
THE HARMONIC CURVE OR CURVE OF
271. This
vibrates
is
SINES.
the curve in which a musical string
when sounded.
The
tional to the sines of angles
ordinates are propor
which are the same frac
tions of four right angles that the corresponding ab
scissas are of
Let
some given
length.
AB (Fig. 81) be the given length.
Produce
BA
i
GEOMETRIC EXERCISES
38
to
C
and
AD perpendicular to AB.
DAC into a number of equal
fold
right angle
Mark on each
four.
then
1
PP QQ RR
,
,
sines of the angles
Now,
twice the
C ?
angle.
VV,
S,
Q,
,
T,
bisect
Q
V
fold perpendiculars to
and
DA
A C;
are proportional to the
PAC, QAC, RAC, DAC.
divide
T U
AE and EB into
chosen
for the right
V
the successive ordinates
etc., equal to
U,
R
of equal parts
S
am
ACAP=AQ = AR = AD.
AB in E and
number
Draw
,
parts, say,
radius a length equal to the
plitude of the vibration,
From points/
Divide the
SS
,
PP, QQ RR DA,
,
,
TT UU\
,
Then
etc.
are points on the curve, and
V
is
the
VV
and pricking
through S, T, U, V, we get corresponding points on
The portion of the
the portion of the curve VE.
highest point on
it.
By
curve corresponding to
folding on
EB
is
equal to
A VE
but
lies
on the opposite side of AB. The length from A to
is half a wave length, which will be repeated from
E
E
PAPER FOLDING
to
B
on the other side
of
E
AB.
139
is
a point of inflec
tion on the curve, the radius of curvature there be
coming
infinite.
THE OVALS OF
272.
When
product of
is
plane
The
its
a point
moves
CASSINI.
in a
distances from two
constant,
it
traces out one of Cassini
fixed points are called the foci.
X
plane so that the
fixed points in the
MA
s
ovals.
The equation
of
A
B
Fig. 82.
rr =k?, where r and r are the distances
on
the curve from the foci and k is a con
any point
the curve
of
is
stant.
F be the foci. Fold through F and
Bisect FF in C, and fold BCB perpendicular to
FF Find points B and B such that FB and FB
are each =k. Then B and B are evidently points on
Let
F
and
jF".
f
.
the curve.
1
GEOMETRIC EXERCISES
40
FK perpendicular to FF
Fold
and on
FF
A
A
and
FK=k,
each equal to CK. Then
are points on the curve.
For
CA*
Produce
point
CA and CA
take
and make
FA
=CK* = CF* -f
AT=FK.
and take
J/and draw MK.
MK meeting FA
in
J/
Fold
A ^/
In
AT
take a
perpendicular to
.
FM-FM =&.
center F and radius FM,
Then
With
center
F
the
and radius
each other
When
a
FM
,
and with the
describe two arcs cutting
Then P is a point on
number of points between
in P.
the curve.
A
and
points in the other
found, corresponding
B
are
quadrants
can be marked by paper folding.
=V*k
When FF
sumes the form
When FF
of
and rr
of a lemniscate.
is
greater than
= \k*
curve as
279.)
(
V 2k,
the
the curve consists
two distinct ovals, one about each focus.
THE LOGARITHMIC CURVE.
273.
The equation
The
ordinate at the origin
If
to this
curve \sy
is
=
a*.
unity.
the abscissa increases arithmetically, the ordi
nate increases geometrically.
The
values of y for integral values of x can be ob
tained by the process given in
108.
IN PAPER FOLDING.
The curve extends
141
to infinity in the
angular space
XOY.
If
y= a x
x be negative
The negative
increases numerically.
OX is
and approaches zero
as
x
side of the axis
therefore an asymptote to the curve.
THE COMMON CATENARY.
274.
The catenary
is
the form assumed by a heavy
inextensible string freely suspended from two points
and hanging under the action
The equation
of gravity.
of the curve is
the axis of y being a vertical line through the lowest
point of the curve, and the axis of x a horizontal line
in the plane of the string at a distance c
lowest point
;
c is
below the
the parameter of the curve, and e
the base of the natural system of logarithms.
When
when x
x
275.
From
=
= -e^C
c
2c,
y
=
(fi -f-
the equation
,=(<*--,-
e
can be determined graphically.
e~ 2
~)
and so on,
I
GEOMETRIC EXERCISES
42
2
I/}-
f 2 is
tween y
-j-
c
found by taking the geometric mean be
and jv
r.
THE CARDIOID OR HEART-SHAPED CURVE.
276.
draw
From
a fixed point
a pencil of lines
O
on a
and take
off
circle of radius
a
on each ray, meas
ured both ways from the circumference, a segment
equal to 2a. The ends of these lines lie on a cardioid.
Fig. 83.
The equation
The origin is
is
to the curve
is
r
= 0(1 -f cos #)
a cusp on the curve.
The
cardioid
the inverse of the parabola with reference to
its
focus as center of inversion.
THE LIMACON.
277.
From
a fixed point on a circle,
ber of chords, and take
of these lines
off
draw a num
a constant length on each
measured both ways from the circum
ference of the circle.
IN PAPER FOLDING
the constant length
If
the circle, the curve
is
is
143
equal to the diameter of
a cardioid.
be greater than the diameter, the curve
If it
is
altogether outside the circle.
If
it
curve
If
be less than the diameter, a portion of the
lies inside
the circle in the form of a loop.
the constant length
the curve
is
exactly half the diameter,
called the trisectrix, since by
is
its
aid any
angle can be trisected.
The equation
The
first
is
sort of
and the second
r
= acos6-\-
liir^on
sort
is
is
b.
the inverse of an ellipse
;
the inverse of an hyperbola,
with reference to a focus as a center.
The loop
is
the
inverse of the branch about the other focus.
278.
Let
The
trisectrix is applied as follows
AOB be
O
Take OA, OB equal
Describe a circle with the
the given angle.
to the radius of the circle.
center
:
and radius
OA
or
OB.
Produce
AO
in-
GEOMETRIC EXERCISES
i 44
definitely
that
OB
beyond the
O may
correspond
produced
C.
Draw
trisectrix so
to the center of the circle
and
Let the outer curve cut
AO
the axis of the loop.
in
Apply the
circle.
BC
cutting the circle in D,
Draw OD.
Fig. 85.
^ACB
Then
is
For
\ of
CD
/_CJ3O
/_ODB
THE LEMNISCATE OF BERNOULLI.
279.
The
polar equation to the curve
r
Let
O
2
=a
2
Take the
OA=a.
OD
at right angles to
angle A OP =8 and A OB =--26.
AB perpendicular to OB.
AO produced take OC=^OB.
Draw
In
cos26.
be the origin, and
Produce AO, and draw
is
OA
IN PAPER FOLDING
D in OD such
Find
CD A
is
a right angle.
OP = OD.
Take
P is
that
145
a point on the curve.
= OB-OA
=a
As stated above,
2
cos 2 6.
this curve is a particular case of
the ovals of Cassini.
Fie. 86.
It is
the inverse of the rectangular hyperbola, with
reference to
its
its
center as center of inversion, and also
pedal with respect to the center.
The area
of the curve is a 1
.
THE CYCLOID.
280.
The
cycloid
is
the path described by a point
on the circumference of a
roll
upon
Let
point
A
circle
which
is
supposed
to
a fixed straight line.
and
when
in
A
be the positions of the generating
contact with the fixed line after one
GEOMETRIC EXERCISES
1 46
Then
complete revolution of the circle.
to the circumference of the circle.
The circumference
length in this
gift
No.
JI.,
Wrap
way.
circular object,
of a circle
may be
a strip of
AA
is
equal
obtained in
paper round a
the cylinder in Kindergarten
e. g.,
and mark
off
Un
two coincident points.
paper and fold through the points. Then the
straignt line between the two points is equal to the
circumference corresponding to the diameter of the
fold the
cylinder.
By
to
proportion, the circumference corresponding
any diameter can be found and
A
vice versa.
G
D
A
Fig. 87.
Bisect
AA
,
AA
in
and equal
D and
draw
to the
diameter of the generating
DB
at right angles to
circle.
Then A, A and
Find
O
Fold a number
through
O
B
are points on the curve.
the middle point of
BD.
of radii of the generating
circle
dividing the semi-circumference to
the
right into equal arcs, say, four.
Divide
AD into
the
same number
of equal parts.
IN PAPER FOLDING
Through the ends
right angles to
and
a radius,
GA
one of these
let
G
F being
lines,
or to the length of arc
P is
Then
AD may be
The curve
off
FP equal
BF.
on the curve.
a point
Other points corresponding
tion of
the end of
be the corresponding point of sec
AD, commencing from D. Mark
tion of
to
diameters fold lines at
of the
BD.
EFP be
Let
147
marked
to other points of sec
same way.
the axis BD and corre
in the
symmetric to
the other half of the curve can be
on
sponding points
marked by folding on BD.
is
The length
of the
curve
is
4 times
BD and
area
its
3 times the area of the generating circle.
THE TROCHOID.
281. If as in the cycloid,
plane of the circle but
circumference traces out the curve called a
straight line,
not on
its
a circle rolls along a
any
point in the
trochoid.
THE EPICYCLOID.
282.
An
epicycloid
on the circumference
is
the path described by a point
of a circle
which
rolls
on the
circumference of another fixed circle touching
it
on
the outside.
THE HYPOCYCLOID.
283.
If
the rolling circle touches the inside of the
fixed circle, the curve traced
cumference
of the
former
is
by a point on the
a hypocycloid.
cir
GEOMETRIC EXERCISES
148
When
the radius of the rolling circle
is
a sub-
multiple of the fixed circle, the circumference of the
has to be divided
latter
in the
same
These sections being divided
ratio.
into a
number
of
equal parts, the position of the center of the rolling
and
of the generating point
corresponding to
each point of section of the fixed circle can be found
by dividing the circumference of the rolling circle into
circle
the
same number
of equal parts.
THE QUADRATRIX.*
284. Let
OACB
be a square.
If
the radius
a circle rotate uniformly round the center
position
OA
same time
through a right angle to
a straight line
move uniformly
OA
to
BC
OB
O
and
if
drawn perpendicular
parallel to itself
OA
of
from the
in the
to
OB
from the position
the locus of their intersection will be the
;
quadratrix.
This curve was invented by Hippias of Elis (420
B. C.) for the multisection of an angle.
If
P
and
P
A OP and A OP
are points on the curve, the angles
are to one another as the ordinates
of the respective points.
THE SPIRAL OF ARCHIMEDES.
285.
If
OA
the line
revolve uniformly round
O
as
P moves uniformly from O along
P will describe the spiral of Archi
center, while point
OA, then the point
medes.
*
Beman and Smith
tary Geometry,
p. 57.
s
translation of Klein
s
Famous Problems of Elemen-
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