MATH/STAT.
T.
SUNDARA ROW S
Geometric Exercises
in
Paper Folding Edited and Revised by
WOOSTER WOODRUFF BEMAN PR...
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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 ,
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
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,
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
,
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
B\,
D\ and B^ t
2,
D
= 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
,
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,