Geometric Exercises in Paper Folding

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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
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

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
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
.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
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
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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