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Kiselev's
GEOMETRY
Book
II. STEREOMETRY
by
A. P. Kisele. v
Russian
from
Adapted
by Alexander
Givental
Sumizdat ELIHU CENTRAL
BURRITT
CONNECTICUT
LIBRARY STATE
NEW BRITAIN, CONNECTICUT
UNIVERSITY
06050
by Sumizdat
Published
5426 Hillside Avenue, El http://www.sumizdat.org
University of California\177 Berkeley Kiselev, A. (Andrei Petrovich)
Geometriia.Chast
2,
USA
Data
Cataloging-in-Publication
English
Stereometriia.
Geometry. Book II, Stereomerry
Kiselev's
94530,
California
Cerrito,
/
Kiselev;
by A.P.
adapted from Russian by Alexander Givental. E1 Cerrito, Calif.: Sumizdat, 2008. iv, 176 p.: ill.; 23 cm.
Includesbibliographical
and
references
index.
ISBN 978-0-9779852-1-0
1. Geometry. 2.
Geometry,
Alexander.
I. Givental,
Solid.
QA453.K57313 2008
Library of Congress ControlNumber: Alexander
(\177)2008 by All
rights
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part
2008931141
l\177ivental
reserved. Copies or derivative of it may not be produced without
products of
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written
the
Alexander Givental (
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except
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Thomas
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978-0-9779852-1-0
Contents
1
PLANES a plane .....................
1 LINES AND
1 2 3 4 5
2
3
4
Drawing
and
slants
some
other
angles
Perpendiculars
and
Dihedral
3 9
.............
29
POLYHEDRA
1
Parallelepipeds
2
Volumes
3
Similarity
and pyramids
of prisms
4
Symmetries
5
Regular
of polyhedra of space
............
.................. figures ...............
....................
polyhedra
i
Cylinders and cones The
VECTORS
ball
...................
AND
59
75 88
107
FOUNDATIONS
Algebraic operationswith
vectors
2
Applicationsof vectorsto
geometry
3
Foundations o\177 geometry
4
Introductionto
............
107
...........
117
128
................. geometry
non-Euclidean
Isometries ......................... .........................
BIBLIOGRAPHY ....................... INDEX
51 66
..........................
1
AFTERWORD
29 39
75
2
5
..............
and pyramids
SOLIDS
ROUND
16
23
.....................
angles
Polyhedral
1
................. ................
planes
and
lines
Parallel
........
141 154
163 171
172
.............................
iii
the English adaptation of Kiselev's Planimetry, was publishedas [4] (see section Bibliography) and will be referred to as Book I. The reader is directed to Translator's Foreword in Book I for background information on the original work [5, 6]. Preparing Book II, I added about 250 exercises, expanded the sections on Similarity of polyhcdra, Symmetries of space figurcs and Regular polyhcdra, and wrote a new, last chapter.It contains a geometric approach to vectors, followedby a vector approach to logical foundations of geometry, and concludeswith a constructive introduction into non-Euclidean plane geometry. While some accounts of such topics are certainly expected of every modern course in elementary geometry, any The
volume
present
specific choices
have
may
plications. Someof
them
non-obvious mathematical and are explained in Translator's
mathematics,
about
controversies
completes
first volume,
The
Geometry.
pedagogical im-
geometry, and education. Civental
Alexander
of Mathematics
Department
University of California July,
Authors
Plato Euclidof
cited in
Syracuse
of
Archimedes
of Alexandria
Menelaus
Pappus of Alexandria
JohannesKepler
about
about
Cavalieri
Bernhard
287- 212B.C. 70
290
-
1\1770
- 350
-
1826
-
Cantor
18\1775
-
Klein
18\1779
David Hilbert
Hermann Minkowski
Max
1921 1918 - 1925
1862
-
186\177
- 1909
1878
Dehn
Hermann
1895
1835 - 1899 18\1773
Schwarz
Albert
16\1777
- 1866
Hermann Felix
A.D.
1571 - 1630
1821-
Riemann
Eugeneo Beltrami Georg
A.D.
1802- 1860
Cayley
Arthur
B.C.
1792 - 1856
Lobachevsky
Bolyai
Jdnos
3\1777
16\1777- 173\177 1789 - 1857
Augustin Louis Cauchy
Nikolai
-
- 265 B.C.
1598
Ceva
Giovanni
325
1591- 1661
Desargues
G\177rard
Bonaventura
2008
this book: \17727
about
Alexandria
Three
Afterword
19\1773
- 1952
1879- 1955
Einstein
1885-
Weyl iv
1955
Berkeley
Chapter
I
LINES
AND
I
plane
a
Drawing
PLANES
1. Preliminary remarks. In stereometry (called one studies geometric figures not all of whose plane.
also
geometry) fit the same
Geometric figures in space are shown following
certain
conventions,
diagrams
appear
alike.
real objects planes and are
Many metric a
such
distance, It
lelogram.
plane of a diagram the figures and their
around us have surfaceswhich shaped like rectangles:the cover
window pane, the surfaceof a desk, from
the
on
to make
intended
etc.
When
solid
elements
seen
resemble
of a
at an
geo-
book,
a
angle and
appear to have the shape of a paraltherefore, to show a planein a diagram The plane is usually denoted by one letter, e.g. surfaces
is customary,
as a
parallelogram.
\"the
plane
M\"
(Figure
1).
Figure 1
properties of the lowing properties of planes, which considered axioms. 2. Basic
plane. are
Let
accepted
us
point
without
the folproof, i.e.
out
Chapter 1.
2 points 1 of
(1) If two
of the line lies in
given
a
have a commonpoint, then
(3) Through every three points not lying can draw a plane, and such a plane is unique.
3. Coro:]_:]_ar\361es. (1)
can
any two
with
together
on the
such a
same line, through is
plane
Through
such
and
plane,
a
draw
every point
intersect
they
in a
point.
this
through
then
plane,
PLANES
plane.
this
If two planes
(2)
line passing
a line lie in
AND
LINES
a
line
a plane
and
a plane
same
the
line,
points
three
one
outside it, one
a point
is unique. Indeed,the
points on the line form which
on
can therefore
not
point
lying
be drawn, and
unique.
two intersecting lines, one can draw a plane, and is unique. Indeed,taking the intersection point and one more point on eachof the lines, we obtain three points through which a planecan be drawn, and such a plane is unique. (3) Through two parallel lines, one can draw only one plane. Indeed,parallel lines, by definition, lie in the same plane. Sucha plane is unique, since through \177 one of the lines and any point of the other Through
(2)
a plane
such
line,
most
at
one
can be
plane
drawn.
4. Rotating a planeabouta line.Through each line in space, planes can be drawn. Indeed, let a line a be given (Figure2). Take any point A outside it. Through the line a and the point A, a unique plane is passing (\3653). Let us call this plane M. Take a new point B outside the plane M. Through the line a and the point B, too, a unique plane is passing. Let us call this plane N. It cannot coincide with M, since it contains the point B which does not lie in the plane M. Furthermore, we can take in spaceyet another point C outside the planes M and N. Through the line a and the: point C, yet a new plane is passing. Denote it P. It coincides neither with M nor with N, since it contains the point C which lies neither in the plane M nor in the planeN. Proceeding by taking more and more points in space, we will thus obtain more and more planes passing through the given line a. There will be infinitely many such planes. All these planes can be considered as various positionsofthe sameplane which infinitely many
rotates We plane
about may
can
the line
therefore be rotated
a.
one more property of the every line lying in this plane.
formulate
about
\177As in Book I, we will always assume that expressions like planes,\" etc. refer to distinct points, planes, etc.
\"three
plane: a
points,\"
\"two
3
and planes
parallel lines
2.
EXERCISES
1.
many
while
standing
stools
three-legged
why
Explain
stable,
2. Usingthe axiomsfrom \3652, show proof of Corollary1 of \3653 contains,
2
that through be drawn.
$. Prove
4. Prove that Prove
/5.*
same
the
2
planes
many
three
planes
that
lines.
plane
no
Two
can be
can be positioned in space in such a drawn through them. For example, take
lines
lines AB and DE (Figure 3), of which certain plane P, and the secondone intersects
two
C, which doesnot lieonthe
these
two
DE
proofs of Corollaries
Parallel lines and planes Skew
through
given
the
given points two planes can be drawn, through these points can be drawn. that several lines in space, intersecting pairwise,eitherlie plane or pass through the same point. many
5. way
always
described in the line and the
plane
the
indeed,
any two points in spaceinfinitely
through
if
infinitely
then
in
are
3.
and
can
that
complete the
Similarly,
as required.
point
given
on flat floor
ones totter.
four-legged
since
lines, the
and the
first
line.
there
otherwise
the
first
this
one
plane
lies in a at a point
can be drawn through would exist two p!anes passing
No plane
AB and the point C: one (P) intersecting other one containingit, which is impossible
line
the
line
(\3653).
D
N
Of course, two
each other
Figure
2
Figure
no
matter
lines not how
lying
far
in
they
the
same
3
plane
are extended.
do not intersect Howeverthey are
not called being in they
same
are
extended.
Two lines
not
6.
and
line
A
line not
being reserved for those do not intersect each other
plane
same
the
in
lying
are called if they
parallel
4)
Figure
(CD)
AB
pointof
line
the
the same
plane P. Then the lies in the plane R
the AB,
A
and
plane
do not
a
intersect
does that
not lies
lie in
a
in it,
R and assumethat the intersection point, beinga
the plane
draw
CD,
and
intersects
AB
how
extended.
given plane (P) but is parallelto a line then the given line is parallelto the plane. Through
lines which,
skew lines.
to each other.
parallel
a plane
lying in this plane arecalled
each other no matterhow far they are 7. Theorem. If a given line (AB,
line
PLANES
no matter
plane,
the
LINES AND
term
the
parallel,
far
1.
Chapter
4
that containsAB, and at
plane P, of course. Thenthe intersection point,being in both planes R and P, must lie on the line CD of intersection of these two planes. But this is impossible,sinceAB CD by the hypothesis. Thus the assumption that the line AB intersects the plane P is false, and h,ence ABIP. time
it
lies
in the
4
Figure
8.
given (CD)
If a given line (AB, Figure4) is parallel to a plane (P), then it is parallelto the intersection line of the given plane with every plane (R) containing the Theorem.
line.
given
Indeed, firstly the lines AB
secondly AB
would
intersect
9. Corollary. If a plane
plane
must
intersect
a
line
and
through each
(AB, Q),
then
other,
Figure it is
AB and any of the planes
the same plane, since otherwise the
lie in
CD
P, which
plane
the
intersectingplanes(P line (CD). Draw
and
each
intersect
cannot
they
and
line
is impossible. 5) is parallel to each of two' parallel to their intersection
point C ofthe line CD. This P and Q alonga line parallel
2.
to AB
(Book
postulate
one
line
must
be
parallel to a same line.
the
coincides
C. But accordingto the a given point, there is only
the point
through
passing
and
(\3658)
parallel
planes
lines and
Parallel
I, \36575),through
given line. Thereforethe linesof intersection It lies in each of the planesP and Q and thus
line CD of their intersection. Hence
with the
CDIIAB.
C
A A
D
Figure 5
Figure 6
If two
10. Corollary.
to a one (EF), If the three
they are
then
third
sion immediately therefore, the three
from
follows
lines
that
and EF of
these
line
AB.
the line CD
N through are
(\3658).
point
is parallel
11. Parallel planes.
matter
how
lines parallel
respectively
A\177C \177)of
plane
another
The lines AB
and
AC
(P\177,
are
then
to the
parallel
Suppose that the planes P and DE (Figure 7). Then ABllDEand
these
P\177
intersect
ACIIDE
and
N
if they
parallel,
far they
intersecting
two
the
and
lines CD
Since the
to the intersection line A, there is only one
called
are
planes
Two
no
7) of one plane(P) are and
point
the
A.
intersection line of the planesM
to EF, the Thus CD[[AB.
parallel
12. Theorem.If
EF and the point A,
the
through
Since
not intersecteachother
and
of them
each
parallel,
planes
parallel to each other.
the line
M through
plane
the
Draw
plane
are parallel
same plane, then the requiredconcluthe parallel postulate. Let us assume, do not tie in the same plane.
in the
lie
lines
Figure 6)
CD,
and
(AB
lines
is
do
are extended. (AB and to
two
planes
plane P\177 along (\3658).
AC, Figure
lines (A'B' are parallel. (\3657).
a certain
line
Thus,the planeP to the
parallel
and
A
P and
planes
intersected CD)
and
each other.
Figure
Firstly, the lines AB and CDliein they cannot intersect, sinceotherwise
plane
same
the
8
(P and Q, Figure8) are the intersection lines
a
by
(AB
intersect
If two \177arallel planes third plane (R), then are parallel.
Theorem.
\337 13.
lines AB and AC passing through DE, which is impossible. Hence the
7
Figure
PLANES
two
contains
same line not
do
P\177
LINES AND
1.
Chapter
6
(R).
P and hypothesis.
planes
the
Secondly, Q would
intersect each other,thereby contradicting the 14. Theorem. The segments (AC and BD, Figure 9), cut off by parallel planes (P and Q) on parallel lines, are congruent. Draw the plane containing the parallel lines AC and BD. It intersectsthe planesP and Q along the parallel lines AB and CD
respectively.
and
congruent
plane.
same in
Book
I,
Then
they are
AC the
parallelogram,
directed established
with respectively parallel and similarly plane, their congruence has been us assume that the planes P P\177 do not coincide.
in
and
Let
parallel, as has beenshown
in
\36512.
the angles in question are congruent,markon their but respectively congruent segments AB - A\177B \177and = ArC \177,and draw the lines AA t, BB \177,CC \177,BC and B\177C( Since segments AB and A\177B \177are congruent and parallel (and have the
To
sides
is a
Two angles (BACand B\177A\177C\177,Figure 10) whose are parallel and have the same direction, are lie either in parallel planes (P and P\177) or in
same
the
\36579.
ABDC
angles
two
When
sideslie
quadrilateral
BD.
=
15. Theorem. respective sides the
the
Therefore
and henceAC
prove
arbitrary
that
2. Parallellines AA
\177and
and
CC
BB
the figure
direction),
same
-
\177
\177are
CC
hence/XABC
congruent
are
BB'
and
congruent BCC\177B
\177,i.e.
?
planes
and
\177is
ABB'A' is a parallelogram.Therefore and parallel. For the samereason, parallel.
a parallelogram.
=/XA\177B\177C \177(by
the
Therefore
SSS-test).
16. Problem. either
of two
each of the given
\177
\177A
-
Figure 10
Figure 9
on
AA
BB/[[CC \177(\36510) and Thus BC -- B/C \177,and
Therefore
Through
given skew
a
given
point
(M,
lines (a and b),
Figure 11), J\177nd a
line
not
lying
intersecting
ones.
b
Figure
11
Figure
12
Solution. The line in questionmust pass through the point M and intersect the line a, and therefore it has to lie in the plane P passingthrough M and a (since two of its points lie in this plane: M, and the intersection point with a). Similarly, the line in question has to lie in the plane(\177 passing through M and b. Thus this linehas to coincide with the intersection line c of the planesP and (\177. If this intersection line is parallel to neither a nor b, then it will intersect each of them (sinceit liesin the same plane with either of them: a
Chapter1.
8 c lie in
and
line c
the
provide
12) or
the plane P,
b and
c lie in the plane Q). Then the pass through M, and will therefore
solution of the problem. If however the problemhas no solution.
unique
c][b, then
17.
PLANES
AND
b and
and
a and
intersect
will
LINES
clla
(Figure
resemblance between the above problem,and the problems of constructing plane figures using drafting devicessuch as straightedge and compass. For purposes of describing figures in space,the drafting devices become useless. Consequently, in solid geometry we will refrain from formulating construction problems. Howeverthe questionof finding a geometric figure satisfying a certain set of requirementscanbe understood the same way as it has been done in the above solution. Namely, one can seek to find out if the required figure exists, and if it is unique (or, more generally, how many solutions there are, and how the number of solutions depends There
Remark.
on the given 18.
a given
is some
data).
Through
Problem.
plane (P), find
a given
point
parallel
a plane
Figure
(A,
Figure
to the
13)
not
lying in
given one.
13
plane P, draw an arbitrary pairofintersecting Draw two auxiliary planes: Ad through the point A and the line BC', and N through the point A and the line BD. Every plane parallel to P and passing through A must intersect the plane A4\177along the line AC '\177parallel to BC', and the plane N alongthe line AD \177parallel to BD (\36513).There is a uniqueplaneP\177 containing the lines AC '\177and AD \177,and this plane is parallel to P (\36512). Thus the solution exists and is unique. On the
Solution.
lines
BC'
and
BD.
Corollary.
exists
a
unique
Through
plane
each
parallel
point
to the
not
lying on
given one.
a given plane, there
3. Perpendiculars
9
slants
and
EXERCISES
that through every
6. Prove
thereexistsa 7. Derivefrom
allelto a
not lying on a given line, given one. parallel postulate that, in a plane, two lines parare parallel to each other. planes parallel to a third one are parallel to each
the one
third
8. Provethat
two
in space,
point
to the
parallel
line
unique
other.
9.
all lines
that
Prove
point lie in
the same
10. Provethat ing planes,are
11. Can
two parallel
planes
two
parallel
respectively
parallel to a given
plane
and
passing
through
to the given one. parallel lines, lying respectively in two intersectto the intersection line of these planes. intersect, if the first plane contains two lines to two lines contained in the secondplane? the
same
plane
parallel
if a line a is parallelto a plane M, then every line a and passingthrough a point of M lies in M. 13. Provethat for every pair of skew lines a and b, there is a unique pair of planes: one passingthrough a and parallel to b, the other passing throughb and parallel to a, and that these planesareparallel. lJ. Given a pair of skew lines a and b, find the geometric locus of points M for which there is no line passing through M and intersect12.
that
Prove
parallel to
ing
15.
and
a
b.
given lines. 18.
passing through a given
a plane
Find
and
parallel
to two
\177/
a line
Find
point
intersecting two given lines and parallelto a third
one.
locus of midpoints of segmentsconnectinga lying on a given plane. 1$. Let AB and CD be skew lines. Prove that the midpointsof the segments AC, AD, BC, and BD are verticesof a parallelogram, and that its plane is parallel to the lines AB and CD. 19.* Compute the ratios in which the planepassing through barycenters of the triangles ABC, ACD, and ADB,not lying in the same plane, divides the sides AB, AC, and AD. 17.
the geometric
Find
point
given
points
and
Perpendiculars
3
19.
A
line
perpendicular
which
understand
plane,
with
let
us prove
lines
should
slants
to
to lines in a plane. In order be considered perpendicular to a given
the following
proposition.
Chapter1.
10
line (AO, Figure 14) intersecting a lines (OB and OC) its point (0) with the
a given
If
Theorem.
givenplane perpendicular to two drawn in the plane intersection given line, then the givenlineis is
(M)
through
to
perpendicular
in the plane through the
(OD) drawn
line
point.
lines, drawn from
A
and
A
several triangles,which
in the
examine
we
B, D, and
Connect
C.
We thus
obtain
following order.
14
Figure
\177con-
the three
intersecting
line
segments.
straight
\177by
other
any
same intersection
mark the segmentsOA any
some points
O, at
point
the
with
points
these
plane M, draw
On the
to AO.
line AO,
of the
extension
the
On
gruent
PLANES
AND
LINES
15
Figure
consider the triangles ABC and A\177BC. They are congrusince BC is their commonside,BA = BA \177as two slants to the line AA \177whose feet are the same distance away from the foot O of the perpendicular BO (BookI, \36552), and CA = CA \177-- for the same reason. It follows from the congruence of these triangles, that First,
ent,
XABC:
consider
Next,
since BD
ent,
DA
=
Finally,
triangles
the
from
20.
consider
the
in the
A
De\234\361n\361t\361on.
intersects
the
BA
\177,
and
congru-
ZABD
-
of these triangles, that
congruence
the
plane
ADA
triangle
also
say
that
the
to
line
is called
and forms
plane and passingthrough
one would
=
are
They
A\177DB.
\177.
its median DO is perpendicular it
ADB and
is their commonside,BA
follows
It
XA\177BD. DA
XA\177BC.
the plane
\177. It
AA
\177.
to a plane if
perpendicular
a right angle with the
and therefore
is isosceles,
base
intersection
every
point.
line
In this
is perpendicular to the line.
lying
case,
3.
11
slants
and
Perpendiculars
theorem shows that a given line is perpendicular to whenever it is perpendicularto two lines lying in the plane and passing throughits intersection pointwith the given line. A line intersecting a plane, but not perpendicular to it, is called oblique to this plane,or a slant. The intersection point of a line with a planeis calledthe footof The previous
a plane
the
of the
or
perpendicular
slant.
every point 15) lying plane perpendicular to the line
21. Theorem. Through
line (AB), a
a given
Figure
(A,
can
on be
is unique. Draw any two planes M and N through the line AB, and at the point A, erect perpendiculars to AB inside these planes: AC in the plane M, and AD in the plane N. The plane P, passingthrough the lines AC and AD, is perpendicular to AB. Conversely, every plane perpendicular to AB at the point A must intersect M and N along lines perpendicularto AB,i.e. along AC and AD respectively. Therefore every plane perpendicularto AB at A coincides with P. 22. Corollary. All lines perpendicular to a given line (AB, Figure 15) at a given point (A) lie in the same plane, namely the plane (P) perpendicular to the given line at the given point.
Indeed,
plane
the
to AB at
A
is
any
two lines
at the
point
through
passing
to
perpendicular
with the
coincides
a plane
such
and
drawn,
AB
perpendicular and
A
therefore
plane P.
23. Corollary.
point
every
Through
line (AB), one can draw line, and sucha planeis unique. a given
a
plane
(C,
Figure
perpendicular
15) not lying on to the given
plane M through C and the lineAB, and drop from C the perpendicular CA to AB inside the plane Every plane perpendicular to AB and passing through C must intersect the plane M along a line perpendicularto AB, along the line CA. Therefore such a plane must coincidewith the plane P passing through the point A and perpendicular to the line AB. an
Draw
auxiliary
M.
i.e.
24. Corollary. (P), a perpendicular At
line
every (AB)
point can
(A,
Figure
be erected,
15) of a given plane and such a line is
unique.
plane P, draw two arbitrary linesa and b passing through line perpendicular to P at the point A will be perpendicular to each a and b, and therefore will lie in the plane M perpendicular to a at A, and in the plane N perpendicularto b at A. Thus it will coincide with the intersection line AB of the planes M and N. In
the
A. Every
Chapter 1.
12 same
point
and slants. 2
the perpendicular
Comparing
25. the
AB and
a perpendicular
16),
(Figure
A
PLANES
AND
LINES
from
When
a slant AC
the same plane P not passing through A are drawn, the segment BC, connecting the feet of the perpendicular and the slant is called the projection of the slant to the plane P. Thus the segment BC is the projectionof the slant AC, the segment BD is the projection to
of the slant 26.
lying slants (1)
etc.
AD,
Theorem. If from in the given plane (AC, AD, AE, ...)
(2)the slant leg AB, we
gle ABE. into
the
can
superimpose all
Then
same plane,
the
right
and
not any
then:
drawn,
are
projections are congruent;
greater
the
with
Indeed, rotating the
16)
Figure
(A,
(AB),
to this plane
congruent
with
slants
the same point
(P), a perpendicular
triangles
is greater.
projection
ABD around the the plane of the trian-
and
ABE'
their planes with perpendicular AB and
and all their projections will
Then the conclusionsof the theoremfollow in plane geometry ,(Book I, \36552).
all the slants will lie
the
from
in the
fall
same line.
corresponding
results
A
'
P
B
E
c
Figure
16
Figure
17
Remark. Each of the slants AE', AD, AE is the hypotenuse of a right triangle,of which AB is a leg, and hence the slants are greater than the perpendicular AB. We conclude that the perpendicular dropped from a point to a plane (see \36535) is the shortest of all segments connectingthis point with any point of the plane. Consequently, the length of the perpendicular is taken for the measure
of the distance
from
the
point
to the
plane.
2For the sakeof brevity, the terms \"perpendicular\" and \"slant\" are often used instead of \"the segment of the perpendicular between its foot and the given point,\" and \"the segment of the slant between its foot and the given point.\"
13
and slants
Perpendiculars
3.
27. Converse theorems. If from a pointnot lying given plane, a perpendicular and any slants are drawn, (1) congruent slants have congruent projections; the proof (by reductio ad absurdurn) 28. The theorem of the threeperpendiculars. theorem will prove useful later on.
We leave ing
Theorem. The line
then:
reader.
to the
The
follow-
a plane
in
drawn
17)
Figure
(DE,
a
projections.
greater
have
slants
greater
(2)
in
through the foot of a slant (AC) and perpendicular to its projection (BC), is perpendicular to the slant itself. On the line DE, mark arbitrary but congruent segmentsCD and CE, and connect each of the points A and B with D and E by straight segments.Then we have: BD - BE (since B lieson the perpendicular bisector BC of the segment DE in the plane P), and consequently AD - AE (as slants to the plane P with congruent projections BD and BE). The triangle DAE is therefore isosceles, and hence its medianAC is perpendicular to the base DE. This propositionis often called the theorem of the three perpen-
dicularsbecauseit
AB
P,
BC
the
to
Repeat theorem,i.e. the line DE,
perpendicular
pairs:
to
a given
DE.
a slant in
through
passing
(DE)
then
plane,
the
(BC) of
projection
17)
Figure
(AC,
to a line
plane (P) is perpendicular the foot of the slant and lying perpendicular
three
following
the
relates
2_ DE, and AC _h 29. Converse theorem. If 2_
the
line..\177is
the slant.
proof of the direct CD and DE on and connect each of the points A and B with D and E. Then we have: AD - AE (since A lies in the plane of AADE on the perpendicular bisectorofthe segment DE), and consequently BD BE (as projectionsof congruent slants AD and AE). The ADBE is thereforeisosceles, and hence its median BC is perpendicular to segments
congruent
but
arbitrary
mark
in the
performed
constructions
the
the baseDE.
30.Relations
There is a dependence be parallel, the property
and planes. or planes to
be
the
between
and
perpendicular.
parallel,one
be
may
perpendicular,
perpendicular
and
parallel
between
Namely,
if certain
able
to conclude
conversely,
and
if
then certain elements turn betweenparallel perpendicular the theoremsdescribed other
and
below.
of a
elements
properties
lines of lines
line and a plane to of a given figure are
that certain other elementsare certain elements are perpendicular, out to be parallel. This relation lines and planes is expressed by
31. Theorem. If a plane (P, Figure two given parallel lines(AB), to the other (CD).
In
and through the
to BE \177ABF
=
XABE
and
BE
lines
two
any
parallel
and DH
DG
lines
XABE = XCDGand as angles with respectively parallel sides. But are right angles (since AB _L P), and hence
XCDH XABF
ZCDH are alsoright
XCDG and
two
perpendicular
we have:
Then
respectively.
BF
and
D
point
B
the point
through
draw
P,
plane
the
BF,
and
it is
then
to
perpendicular
is
18)
one of
AND PLANES
LINES
1.
Chapter
14
32. Converse theorem. 19) are perpendicular to
(\36520).
19
Figure
18
Figure
_k P
CD
Therefore
angles.
If two lines and CD, the same plane (P), then
Figure
(AB
are
they
parallel.
Let us assume the opposite, i.e. that the lines parallel. Through the point D, draw the line
our assumption,it will
be
line DC
a certain
and
AB
parallel
\177 different
are not
CD
to AB;
under We
DC.
from
have: DC \177:_ P by the direct theorem, and and DC _\177 P by the hypotheses. Therefore two perpendiculars to the planeP are erected at the same point D, which is impossibleby \36524. Therefore our assumption was false, and hence the lines AB and CD are parallel. 33. Theorem. If a line (AA \177, Figure 20) is perpendicular to one of two given parallel and P\177), then it is perpendicular to the other.
planes(P
A\177B \177,the
line
the
Through
tersectsthe
P
planes
other
along
AA
\177, draw
and
P\177
AC and
any
along
two
planes.
A\177C \177. Since
Each of
them in-
one along AB and line AA \177is perpendicand AC, and therefore
lines:
parallel
the
is perpendicular to the linesAB
ular
to P, it
it is
also perpendicular to the lines A\177B \177and A\177C \177respectively to them. Thus AA \177is perpendicular to the plane P\177
allel through
A\177B \177and
A\177C \177.
parpassing
15
slants
and
Perpendiculars
3.
If
34. Converse theorem.
same line
to the
perpendicular
are
20)
(P and
planes
given
two
Figure
are
they
then
(AB),
(\177,
parallel.
Otherwise, i.e.
a point
through
to the sameline,
impossible
is
which
Q intersected, we would have of intersection and perpendicular
P and
planes
the
if
passing
planes
two
(\36523).
From a given point (A,
35. Corollary.
a given plane (P), a perpendicular (AB) given
the
Through
A
from
point
A, draw
AB to
Q
lying on
not
can be
plane
this
the plane (2
perpendicular
the
21),
Figure
to
and such a perpendicular is unique. and erect
21
Figure
20
Figure
to
parallel
dropped, P (\36518), to
According
(\36524).
is perpendicular to P. The uniqueness of such a obvious, since otherwise,i.e. if another perpendic21) were dropped from A to P, we would have a triangle ABB' with two right angles B and B ', which is impossible. AB
line
the
\36533,
perpendicular is ular AB' (Figure
Remark. 24, we
is
a
unique
or not
line
to this
perpendicular
on a given line), there is a
results
or not
(lying
a point
through
with
Corollary
this
Combining
see that:
plane, and
\365\36521,
through
a point
23,
and
(lying
to this
perpendicular
plane
unique
of
on a given plane), there
line.
EXERCISES
20. Prove that thereis a 21. Prove to a line
and
22.
all points
that
given
Prove
tant
that
from the
are
there
passing
plane.
line
unique
given line and passingthrough
a
given
infinitely
in
space,
point
perpendicular
not lying
many lines in
to a
on it.
space perpendicular
a given point on it. line parallel to a \177iven plane
through of a
are
equidis-
Chapter1.
16 20\370. Prove
from the
that
other.
24{.Provethat
line,
25. Provethat Find
27.
Prove
if
a
28.
point the
to
Dihedral
perpendicular to the same
to a plane P intersectsa is perpendicularto b. to two given skew lines.
a parallel
A is plane
b
line
equidistant from S, C, and D, then SCD is the circumcenter of ASCD.
from: (a)
locus of points equidistant given non-collinear points.
(b) three
points,
a line are to each other.
are equidistant
planes
parallel
and
then a
geometric
the
Find
4
line
a
if a
that
of one of two
parallel
perpendicular
line
the projectionof A given
plane
a
to P,
perpendicular
26.
if
are
they
then
all points
PLANES
AND
LINES
and some other
two
angles
36. Dihedral angles. A plane is divided by a line lying in it into two parts, called half-planes, and the lineis calledthe edge of each of these half-planes. A figure in space formed by two half-planes (P and Q, Figure 22) which have the same edge (AS) is called a dihedral angle. The line AS is calledthe and the half-planes P and Q faces of the dihedralangle.Space is divided by the faces into two parts, calledthe interior, and exterior of the dihedral
edge
angle (and defined similarly in plane
angles
angle
dihedral
A
the
edge (e.g.
to
is usually
denoted
a dihedral angleAS).
have the same edge,then
diagram
it was
how
we
If
by the
two
several
dihedral
letters,
of
will
which
the middle
them by
each of
denote
two mark the
and
the
CDE formed by
angle of the dihedral
these
two
is called
rays
outer
23). angle
dihedral
drawn
the angle
in a
angles
Figure
then
for
\36513
marking
letters
edge, two mark the faces (e.g. a dihedralangle SCDR, If from an arbitrary point D on the edgeAS ofa (Figure 24), a ray perpendicular to the edgeis in four
I,
done in Book
geometry).
face,
each
a linear
angle.
of the linear angle of a dihedralangledoesnot dethe position of its vertexon the edge. Namely, the linear angles CDE and C'D'E' are congruent,becausetheir sides are respectively parallel and have the same direction. measure
The
pend
on
The plane containing a linearangleof a dihedral dicular to the edge sinceit containstwo lines
angle
perpendicular
Thus, linear angles of a dihedralangle are obtained both faces by planes perpendicular to the edge.
by
is perpen-
to intersecting
it.
4.
37. Congruence and
oneof themis
comparison of dihedralangles.
Two
become superimposed when the interior of the other, so that their one of them that will lie inside the if they
congruent
are
angles
dihedral
into
embedded
17
angles
other
\177nd some
Dihedr\177l
edgescoincide; the other is consideredsmaller. otherwise,
c
D
angles in plane geometry,dihedralangles
Similarly to
then
If two supplementary dihedralanglesarecongruent to each of them is calleda right dihedralangle.
each
angles
dihedral
angles
other,
congruent
have
dihedral
Congruent
(1)
Theorems.
be
can
etc.
supplementary,
vertical,
Figure 24
23
Figure
22
Figure
linear angles. of
greater
(2)
The
Let
PABQ
linear angle.
two
(Figure 25)
ptA'B'Q'
and
be
two
has the greater angles.
dihedral
the angle A'B' into the angle AB so that the edge coincides with the edgeAB, and the face P' with the face P. Then, if these dihedral angles are congruent,the face Q' will coincide with the face Q. However, if the angle A'B' is smaller, the face Q' will occupy a certain position Q\" in the interior of the dihedralangle AB.
Embed
noticed
Having
this point
through
this, pick on the commonedgeany draw the planeR perpendicular to
tersecting this plane with
their linear
angles.
linear
angles
dral
angles
Q\,")
then
will
the
Clearly,
turn
out
faces if the
to be
of the
dihedral
B,
point
the
dihedral angles, angles coincide,
edge.
and
In-
we obtain then their
the same angle GBD. If the
dihe-
do not coincide (i.e. the face Q' occupies the position the greater dihedral angle will turn out to have the greater
linear angle,namely/CBD >/C\"BD.
18
1.
Chapter
38. Converse
respond
PLANES
angles.
dihedral
angles corresponds to
greater of two linear
(2) The
the
angle.
dihedral
greater
AND
theorems. (1) Congruent linear anglescor-
congruent
to
LINES
These theorems
are easy to prove
by
reductio
ad absurdurn.
Figure 25
39. Corollaries. (1) Right linear angles, and vice versa.
Let a
angle
dihedral
PABQ
dihedral
(Figure
that it is to its supplementary the linear CDE and CDE are also are supplementary,eachof is right. plementary linear angles CDE CDE' corresponding dihedralanglesare congruent angles
\177
them
and
congruent,
them is
(2)
All
right
Similarly, (3)
correspond
to right
26) be right. This means angle P'ABQ. But then congruent, and since they Conversely, if the supare congruent, then the and therefore each of
right.
linear
gruent
angles
Vertical
dihedral
angles
are congruent
because they have con-
angles.
one can dihedral
easily prove that: angles
are congruent.
(4) Dihedral angles with respectively parallel and similarly (or oppositely) directedfaces are congruent. (5) For the measure of a dihedral angle, one takes the degree measure of its linear angle. Thus, a dihedral angle containing n degrees is congruent to the sum of n unit dihedralangles. Two intersecting planes form two pairs of vertical dihedralangles, supplementary to each other. The measure of the smallerof these dihedral angles is used to measure the angle between two planes.
4.
Dihedr\177l
ular
40. Perpendicular planes. Two planesare called perpendicif they intersect and form right dihedral angles. Theorem(test for perpendicular planes). If a plane (Q, Figure
other
\177nd some
angles
26) contains a line(CD)perpendicular to
plane
another
are perpendicular.
planes
two
these
then
P
A
D
Figure
be the intersection line
Let AS plane
P, draw
DE
angle
PASQ.
dihedral the
it is
hypothesis,
therefore
and
right,
Figure
26
to the
perpendicular
/CDE
Then
AS.
2_
Since
the line
planes
P and
(\177.
On
the
is the linear angle of the CD is perpendicularto P by
to DE. Thus the angle CDE is
perpendicular the
the
of
27
angle is
dihedral
right, i.e. the plane(\177 is
plane P.
41. Theorem. The perpendiculardropped from a point Figure 27) of one of two given perpendicularplanes(P and Q), to the other plane (Q) lies entirely in the first plane. Let AS be the perpendicular in question, and suppose that it doesnot lie in the plane P (as shown in Figure 27). Let DE be the intersection line of the planes P and Q. Draw on the plane P the line AC 2_ DE, and on the plane Q the line CF 2_ DE. Then the angle ACF will be right as the linear angle of a right dihedralangle. Therefore the line AC, being perpendicular to DE and CF, will be perpendicular to the plane Q. Thus we will have two perpendiculars dropped from the same point A to the plane Q, namely AS and AC. Sincethis is impossible(\36535), our assumption had to be false, and
thereforethe perpendicular AS lies
in
the
plane
P.
The intersection line (AS, Figure28) of two Q) perpendicular to a third plane (R) is perpendicular
42. Corollary. to
and plane.
(P
planes
this
Indeed,
P and Q,
if from we
drop
any point the
A
of
perpendicular
the
line of the planes plane R, then this per-
intersection
to the
20
pendicular will lie, according to the previous theorem, planes P and Q, and therefore coincide with AB.
Figure 28
in
each
of the
and
direc-
Figure 29
skew lines.
43. Angles between of two skew lines (AB
tions
PLANES
AND
LINES
1.
Chapter
and
CD,
positions
Given
Figure
29), the
angle between
them is as any angle (MON) obtained by picking an arbitrary point O space and drawing from it rays (OM and ON) respectively parallel to skew lines (AB and CD) and similarly directedto defined
in
tWO
th\177
them.
if
for
ZM'O'N'
since
another
directedsides. we
on,
Now
lines
dicular
angle does not depend on the choice of the angle M'O'N' is drawn, then ZMON= angles have respectively parallel and similarly
of this
measure
The
point O,
these
such
use the terms angle between even if the lines do not meet.
44. Orthogonalprojections. As
from
when
a given
point, the
we
have
lines
perpen-
and
in
discussed
perpendicular and a
slant
to
\36525,
a given
then the segmentconnectingthe feet of the perslant is called the projection of the slant to the plane. We now give a more general definition of projection. (1) The orthogonal(or Cartesian) projection of a point to a given plane (e.g. of the point M to the plane P in Figure 30) is defined as the foot (M') of the perpendicular dropped from this point to the plane. plane
are drawn, and
pendicular
For
the
thogonal\"
sake of brevity and say simply
we will usually \"projection.\"
omit the adjective\"or-
(2) The projection of a given figure (e.g.a curve) is defined as the geometric locusof projections ofall
plane this
the
figure.
to
a given points
of
4. Dihedral
and someotherankles if the curve being projected is a straight perpendicular to the plane(P), then
particular,
In
30) not
Figure to this
21 line its
(AB,
projection
plane is also a line. Indeed, if through the line AB and the perpendicular MM' droppedto the plane from a point M of the line\177 the plane Q is drawn, then this plane will be perpendicular to the plane P. Therefore the perpendicular droppedto the plane P from any point of the line AB (e.g. from the point N) will lie in the planeQ (\36541). Thus projections of all points of the line AB will lie in the intersection line (A'B \177) of the planes P and Q. Conversely, every point of the line A'B \177is the projection of some point of the
line AB, because point of the line
A\177B
the
of projections the
of
' will
plane
P erected at any
Q and thereforeintersect the line A\177B \177,is the geometric locus line AB, i.e. it is the projectionof
of the
points
all
the
the plane
lie in
point. Thus
at some
AB
line
to
perpendicular
the
line.
A
B
N
M
Figure 30
The
45. (AB,
Figure
angle a slant 31), oblique to
them is defined as the
its projection to
the leastof all
acute
plane.
the
which
angles
Figure
\177'
31
makes with a plane. a plane (P), the angle
Given
given
angle
This the
(ABC)
angle given
formed
by
a
line
between this line with
has the property of being slant AB forms with lines
the foot B of the slant. Let us prove, is smaller than the angle ABD shown in Figure 31. For this, we mark the segment BD = BC and connect D with A. The triangles ABC and ABD have two pairs of respectively congruent sides, but their third sidesare not congruent, namely: AC < AD, since the perpendicular to a plane is shorter than any slant dropped from the same point (\36526).Therefore ZABC < ZABD (BookI, \36550).
drawn
for
in
example\177
the
plane that
P through
the
angle
ABC
AND PLANES
LINES
1.
Chapter
22
EXERCISES
29. Provideproofsof the theoremsstated
a
$1. Prove that
$2. Provethat a
to
\36539:
Vertical
and
a line
to each
to
a right angle
angle, the sides of any faces of the angle.
linear
the
perpendicular to a given
line and
a given
plane
plane makes
a given
other
each
angles with
making right
both
dihedral
in
are congruent.
angles
plane.
to the
$$. Prove that an acute angleare projections of 34. Find a containing
\36538.
dihedral
other.
to
parallel
line
perpendicular
line
any
plane
parallel
are
line
another
3 in
Corollary
Prove
$0.
in
plane.
35. a
and
36. it,
plane P and a line a]]P,find a plane containing the lin\177 a given angle with the plane P. a plane P and two points A and B on the same side of point C on the plane P such that the sum AC + BC is
Given a making Given
a
find
,
minimal.
$7. Find the greatestamong plane and all planescontaining $$. Prove that if the intersection
dihedral
are
angle
same
the
a
line with faces of a awaY from the edge, then the of a
points
distance
and the facesare congruent. \1779. Can four lines in space (not necessarilypassingthrough point) be pairwise perpendicular? between
angles
the line
4 0. Provethat the degreemeasures with a lineand a plane perpendicular 41. Prove parallelogram.
of
the projection of
that
a parallelogramto a
four
=
AC
and
DB
same
plane
is
a
barycenter of a projection of the
of the
plane
of the
triangle.
45.* In space,
other
each
the
by a given line add up to 90\370.
formed
angles to
42. Prove that the projectionto a given given triangle coincides with the barycenter
AB
a given
to it.
slant
given
between
angles
dihedral
the
points A, B, C, and D are given such that - DC. Provethat the lines AD and BC are
perpendicular.
44.* Prove that a line making congruent intersecting lines of a plane,is perpendicular 45.* Compute the angle betweena line
the anglesof 45\370and plane.
60
\370with
two
angles to and
with the
a plane
perpendicular
three
pairwise
plane. if the line forms lines lying in the
23
5. Polyhedralangles
5
Polyhedral angles
32): ASB, BSC, CSD,..., ES/w, which being consecutively adjacent to each other, liein the same plane around a common vertex S. Rotatethe plane of the angle ASB about the side SB (common with the angle BSC) so that it forms some dihedra]anglewith the plane BSC. Then, keeping the dihedral angleunchanged, rotate it about the line SO so that the planeBSC forms some dihedral angle with the plane CSD. Proceed with such consecutive rotations about each common side. If in the endthe first side SA turns out to coincide with the lastsideSF, then the geometric figure thus formed (Figure 33) is calleda polyhedral angle.The angles ASB, BSC, ..., ESA are calledplaneangles,or 46.
their
faces;
Take
Definitions.
sides
SA, SB,
several
angles (Figure
..., SE edges;
and
their
common
vertex
vertex of the polyhedralangie. Every edge of a polyhedral angie is at the same time the edge of some dihedral angie formed by two adjacent faces. Thus a polyhedral angie has as many edges, or dihedral angles, as it has plane angles. The smallest number of faces a polyhedral angie can have is equal to three, and such angles are called trihedral. There exist tetrahedral, pentahedral,etc.
S the
angles.
D
Figure
32
Figure 33
We will denote a polyhedralangie eitherby a single letter (S) marking its vertex, or by a string of letters (SABCDE) of which the first one denotes the vertex and the others label the edgesin their consecutive
order.
A polyhedral of
the
planes
angle is called'convex,if it lies entirely on one side of its faces. Thus, the polyhedralangle shown
of each
1.
Chapter
24
in Figure 33 is convex.On the 34 is not convex,sinceit is situated
the
contrary, on
angle sides
both
or of the planeBSG.
AND PLANES
LINES
shown in Figure of the plane ASB
If we intersect all faces of a polyhedral angle by a plane, then in this plane a polygon is formed.In a convex polyhedral angle, such a polygon is also convex.In the sequel, unless the opposite is specified, all polyhedral angles we consider will be assumed convex. s
s
A D
A
C
Figure 34
In a
be
angles
35), let the greatest of the interior of this angle,markthe XASB, and draw any line AC intersecting
to
ASD
triangles
+ DC
ASB
and
sides, and henceAD AD and AB on each
DC < BC.
We
note
B
with
A and
C.
=
side now
< AB
+ BC.
are congruent since their anglesat the between respectively congruent Discarding the congruent summands
enclosed
and
congruent
are
connect
have:
AD
The vertexS
each plane angleis plane angles.
In the
ASC.
some point D. Mark SB -- SDand
In AABC, we
35
(Figure
SABC
angle
the
congruent
ASD
a trihedral angle, of the other two
angle
trihedral
plane SD at
sum
the
than
angle
In
Theorem.
47. smaller
Figure
\177
AB.
of the above inequality we concludethat in the triangles BSC and DSC, two
that
of them are respectively congruentto two sides of the the third respectivesidesare not congruent. In this case, the angle oppositeto the smaller of these sides is smaller, i.e. sides
of
one
Other, while
ZCSD < ZCSB. the
Adding
to
right
side the
left side of this inequality the angleASD, and to the angle ASB congruentto XASD, we obtain the required
25
angles
5. Polyhedral inequality:
< ZASB
ZA$C
Subtracting
Corollary.
the angleASB or the
we obtain:
CSB,
angle
XASB < XCSB
XASC-
XASC- XCSB< from
inequalities
these
Reading
accountthat the angle ASC, angles, is also greater the
right
that
difference
in
to left,
and also taking into
greatest
of the three plane other two, we con-
of the
difference
each plane
angle,
a trihedral
other two plane
of the
ZASB.
the
being
than
clude
angle is greaterthan
D
Figure
48. Theorem. plane
angles
36
In a convex polyhedralangle,the sumof all is smaller than 3 4d.
Intersect all the facesof a convex polyhedral 36) by any plane. Then a convex n-gon formed in the plane of the cross-section. Applying orem to each of the trihedral angleswhose vertices and E, we find: < XABS
XABC
all these inequalities term-wise. the sum of all angles of the polygon Add
3Recall
from
and
Book
on
the
right
I, that the
will
ABCDE are
A, B,
be the-
previous
the
C, D,
+ XCBS;
ZBCD < XBCS+ XDCSand
- 4d,
SABCDE
angle
(Figure
to 2dn
the
angles. s
A
the
latter inequality
of the
sides
both
from
the
Onthe
so
left
measure of a right
hand
which
ABCDE,
sum of
on.
we obtain is congruent
side,
angles of the triangles angle
is denoted
by d.
Chapter 1. LINESAND
26 Denoting
S.
on, except those angles,which the sum of the latter angles by the
and so
BCS,
ABS,
after the summation:
2dn -
<
4d
2dn
order
in
it is
one,
for
4d and 2dn - x, first difference to
the
necessary that the first
second subtrahendx, i.e. 4d
or
by
planes.
Let us
find
be smallerthan 4d
second
the
be greater
than the
x.
angles.
As we already know, ver-
as they are angles formed by lines if this statement still holds true for
long
out
angles.
polyhedral
Extend all the vertex
past
as
congruent,
vertex
the minuends are the same.
subtrahend
polyhedral
Symmetric
49.
tical anglesare
>
the
x, we find
- x.
In the differences2dn -
Thus,
have
letter
PLANES
formed, which
edges of a S, so that
polyhedral angle SABCDE (Figure
a
new
polyhedral
can be calledverticalto the
hard to seethat
angle original
SA'B'C'D'E' one.
It is
37)
is
not
these polyhedral angles, all plane angles of as well are respectively congruent, but the anglesof either kind are positioned in the opposite order. Indeed, if we imagine an observer who is looking from the exterior of the first polyhedral angle at its vertex, we find that the edges SA, SB, SC, SD, SE of the first angle will appear to him ordered counter-clockwise. However if he is looking at the angle SA\177B\177C\177D\177E\177 from its exterior, then the edges SA', SB\177,SC \177,SD \177,SE \177will appear ordered clockwise. as
dihedral
in
ones
Figure 37
angles whose plane and dihedralangles are respecbut are positioned in the oppositeorder cannot, generally speaking, be superimposed onto each other, and are therefore not congruent. Such polyhedral anglesareoften called symmetric. Symmetry of figures in space will be discussed in more detail in Section 4 of Chapter 2. Polyhedral
tively
congruent
5.
Polyhedr\177l
27
\177n\177le\177
50. Theorem (tests for congruence of trihedral angles). Two trihedral anglesare congruent if have: (1) a pair of congruent dihedral angles enclosed they
between
positioned
similarly
and
congruent
respectively
two
plane
angles, or a
(2)
angles enclosed between two
plane
of congruent
pair
respectivelycongruentand similarly
an-
dihedral
positioned
gles.
S'
S
C'
A
B'
B
Figure
S and S' be two
(1) Let
/ASB -
',/ASC
/A'S\177B
38
trihedral
--
/A\177S\177C
38) such
(Figure
angles \177(and
these
respectively
that conangle
gruent angles are also positioned similarly), and the dihedral AS is congruent to the dihedralangle A'S'. Insert the angle S' into the angle S so that the vertices S' and S, edges S'A' and SA, and planes A'S'B' and ASB coincide. Then the edgeS'B' will merge with SB (since ZA'S'B' = ZASB), the plane A'S'C' will merge with ASC (since the dihedral angles are congruent), and the edge S'C' will merge with the edge SC (since ZA'S'C' = ZASC). Thus
all respective edgesofthe trihedralangles become and therefore the anglesare congruent. (2) The second test is proved similarly by
superimposed
onto
each other,
superimposing.
EXERCISES
\1776.
Show
\1777.
Can
every
that
a non-convex
smaller than 2d? \1778.
\1779.
an
Give
of plane How
many
be obtuse?
sum
of plane
angles
of a non-convex tetrahedral angle whose sum is: (a) greater than 4d; (b) smaller than 4d; (c) 4d. of the planes anglesof a convex tetrahedral angle can
example
angles
angle is convex. polyhedral angle have the
trihedral
Chapter 1. LINESAND
28 50.
that:
Prove
two of
if a
trihedral angle has two
its dihedralanglesare right.
51. Prove that
plane
every
of the
sum
the
angle
of a
of
its
angles
then
trihedral angles are
right.
plane if a
Conversely,
has two right dihedral anglesthen two than
right
PLANES
plane
angle
angle is smaller
tetrahedral
other three.
52. Can symmetricpolyhedral angles be congruent? Prove that two trihedral angles are congruent if: (a) all their plane angles are right, or (b) all their dihedral angles are right. 5\177.* Prove that a tetrahedral angle can be intersectedby a plane in such a way that the crosssectionisa parallelogram. 55. In the interior of a trihedral angle, find the geometric locus of 53.
points equidistant from the faces. 56. Suppose that two dihedral angleswith parallel edges have respectively perpendicular faces. Provethat eitherthesedihedral angles are congruent, or their sum is congruentto 2d (i.e. the sum of two right dihedral angles). 57. Supposethat from a point in the interior of a dihedralangle, perpendiculars are dropl\177ed to its faces. Prove that the anglebetween the perpendiculars is congruent to the angle supplementaryto the
to
angle.
linear
55. the angle
Suppose that edges of one trihedral angle are perpendicular faces of another trihedral angle. Prove that facesof the are perpendicular to the edges of the second.
59.*
to first
sum of all dihedralangles pair of dihedralangles is smaller than the sum of the third dihedral angleand 2d. 60.*Prove that in a (convex) polyhedral angle with \177 faces the sum of all dihedral angles is greater than 2d\177 - 4d (i.e. than the angle that
Prove
is greater
than
sum of a 61.*
the
of two
other. than
greater
true
convex
One
if
one
Which one?
in
2d,
a trihedral and
the
angle, the of each
sum
\177-gon).
polyhedral angles with the
same
vertex
lies
inside
the sum of plane anglesof the latter oneis the sum of plane angles of the former. Doesthis remain of the polyhedral angles is not required to be convex?
Prove that
2
Chapter
POLYHEDRA
I
Parallelepipeds and
pyramids
geometric solid bounded a polyhedron are calledits faces.A common faces is calledan edge of the polyhedron. When several faces meet at their commonvertex, they form a polyhedral angle, and the vertex of the angleis calleda vertex of the polyhedron. A straight segment connecting any two vertices, which do not lie in the same face, is called a diagonal of Polyhedra.
51.
by
\177
A
is a
polyhedron
boundary polygons of side of two adjacent
The
polygons.
the polyhedron.
The
of faces
number
smallest
a polyhedron
can have is four. Such
can be cut out of a trihedralangleby a plane. consider only those polyhedra which are convex,i.e.lie on one side of the plane of each of its faces. 52. Prisms. Take any polygon ABE'DE (Figure 39), and through its vertices,draw parallel lines not lying in its plane. Then on one of the lines,take any point (A f) and draw through it the planeparallelto the plane ABE'DE, and also draw a plane through eachpair of adjacent parallel lines. All these planes will cut out a a polyhedron
We will
ABCDEAfB\177C\177D\177E
polyhedron
\177called
a prism.
planes ABE'DE and AfB\177C\177D\177E \177are the lateral planes along parallellines (\36513), and therefore The parallel
laterals
hand,
AA\177B\177B, in
the
are congruent \177 Polyhedra
polygons
etc.
BB\177C\177C,
ABE'DE
(as opposite (or polyhedrons)
are
and
parallelograms.
by
intersected
the
quadri-
On the
A\177B\177C\177D\177E \177, corresponding
other sides
sides of parallelograms),and correspondis the
plural
29
of polyhedron.
Chapter
30
ing angles are Thus, a
are
faces
are
parallelograms
are
and
congruent.
two of whose faces parallel sides, and all other parallel sides.
a polyhedron
as
respectively connecting the
with
polygons
parallel
respectively polygons
these
prism can be defined
congruent
with
angles
(as
congruent
similarly directedsides).Therefore
2. POLYHEDRA
B
D
A
E
Figure
39
Figure 40
The faces (ABCDE and A'B'C'D'E') lying in parallel planes are called bases of the prism. The perpendicular OO'droppedfrom any point of one base to the planeof the other is called an altitude of the prism. The parallelograms AA'B'C, BB'C'C, etc. are called lateral faces, and their sides AA', BB', etc., connecting corresponding vertices of the bases, are called lateral edges of the prism. The segment A'C shown in Figure 39 is one of the diagonalsofthe prism. A
is
prism
(and
the
bases
are
rectangles,
if its if they are
right
called oblique
and a
lateral edges are perpendicular to not). Lateral facesof a right prism
lateral edge can be considered
A right prism is called regular Lateral faces of a regularprismare Prisms what
the
can be bases
are:
triangular, triangles,
if
its
as
are
bases
regular
altitude.
polygons.
rectangles.
congruent
quadrangular, quadrilaterals,
the
etc.
etc. dependingon
1. parallelepipedsand pyramids 53.
whose bases are parallelograms Just like general prisms, parallelepipeds can be right or right parallelepiped is called rectangular 2 if its base is
40).
(Figure
are prisms
Parallelepipeds A
oblique.
31
(Figure 41). It follows from the definitions that: (1) All six faces of a parallelepiped are parallelograms; (2) The four lateral sides of a right parallelepiped are rectangles,
a rectangle
are parallelograms;
its bases
while
(3) All
The tex are
six
three
of a
rectangular
parallelepiped are rectangles.
edges
of a
rectangular
its
dimensions;
one
parallelepiped meeting at a verof them can be consideredits
and
the third
faces
called
another
length, A
width,
gruentto
each
a cube.
is called
other
height. faces
All
are all concube are squares.
dimensions
three
whose
parallelepiped
rectangular
of a
!
D
c
A
Figure
41
Figure 42
54. Pyramids.
A
.(calledits base) lateral) triangles In orderto angle S 42), edges, take the
can
are
(Figure
is a
pyramid be
any
meeting
cut
finite
of which one face all other faces (called
polyhedron,
polygon, and at a common
a pyramid,
construct
and
B
it suffices
it by a plane part SABCD.
vertex. to take any polyhedral
ABCD intersectingall the
The common vertexof the lateralfaces is called the vertex, the edges adjacent to it lateral edgesofthe pyramid, and the perpendicular SO, dropped from the vertex to the base, its altitude. A plane drawn through the vertex and any diagonal of the base (e.g.the diagonal BD, Figure 44) is called a diagonal plane of the 2Recte[ngule[r
pe[re[llelepipeds
are also
known
as
cuboids
or boxes.
Chapter 2.
32
by names of its vertices,one usually e.g. SABCD (Figure 42).
vertex,
the
with
begins
a pyramid
Identifying
pyramid.
POLYHEDRA
Pyramids can be quadrangular, base is: a quadrilateral, mid is also calleda tetrahedron tetrahedron are triangles.
on
depending A triangular
etc.,
triangular,
what the
etc.
triangle,
(Figure 44)
four
All
43).
(Figure
pyrafaces of a
its base is a through the center pyramid, all lateral edges are congruent to each other (as slants with congruent projections). Therefore all lateral facesof a regular pyramid are congruent isosceles triangles. The altitude SM (Figure44) of each of these triangles is called an apothem. All apothems of a regular pyramid are congruent. is
pyramid
A
regular
called
regular polygon, and secondly, of this polygon.In a regular
s
if,
firstly,
passes
altitude
the
s
c B
Figure
Figure 44
43
Figure 45
of a pyramid. The of a pyramid (Figbetweenits base (ABCDE) a section plane parallel to the base is called a pyramidal frustum. faces are called bases, and the segment OO of the dropped from any point of one baseto other, an
(A\177B'C'D\177E
part
and
\177)
parallel
\177
perpendicular
the
of the pyramidal frustum. A frustum is called a regular pyramidal frustum.
altitude
56.
(1) opposite all
four
prisms. The two
intersecting
regular
pyramid
faces are congruent and parallel; intersect at their midpoints.
diagonals
(1) The basesABCD
and congruentby
of a
a parallelepiped:
In
Theorem.
(2)
C
B
55. Frustum ure 45) contained
The
D
A
D
A
the
very
faces
lateral
lines
BB'
and
BB'C'C
(Figure 46) are parallel of parallelepipeds as a kind of and AA'D'D are parallel because
A'B'C'D'
definition
and B'C'
in one of
them are respectively
1.
33
pyramids
and
Parallelepipeds
to two lines AA' and A\177D being parallelograms,are congruent
parallel
in
which
to
bisect
parallelogram
ZAA\177D' sides).
AA
\177
(as
Figure 48
Figure
BD',
and
the
edges
AB
47), and and D'C'
DC, they are parallel and configure ABC'D' is a parallelogram
are diagonals,and diagonalsin any Thus the diagonals_BD'and AC' But the same applies to any pair of
each other.
midpoints.
their
at
faces,
\177=
edge
the
each other. Therefore the the lines AC' and BD'
intersect
\177=
directed
Figure 47
are parallel and congruent to
A\177D \177,BB
ZBB\177C
similarly
(2) Pick any two diagonals (e.g. AC' draw the auxiliary linesAD' and BC'. Since gruent
(\36512).These
other B\177C\177=
and
angles with respectivelyparalleland
Figure 46
the
since
of parallelograms)
sides
opposite
(as
\177in
and any of the remaining diagonals B'D or A'C. Thereforeeachof thesediagonals also intersects AC' at the midpoint and is itselfbisectedby the point of intersection. Thus all the diagonalspassthrough the same point -- the midpoint of and are bisectedby it. 57. Theorem. In a rectangular parallelepiped, the square of any diagonal (C'A, Figure 48) is equal to the sum of the e.g.
diagonals,
squares of
to AC'
dimensions.
the
AC of the base we obtain the triangles They are both right triangles, the first one because the parallelepiped is right, and thereforethe edge C'C is perpendicular to the base, and the second one because the parallelepiped is Drawing
ACC'
diagonal
the
and ABC.
rectangular,and henceits baseis a
From
rectangle.
these
we find:
C'A 2
=
AC
2+C'C
2 and
AC2
=
AB 2
+ BC 2.
triangles,
Chapter 2.
34
POLYHEDRA
Therefore C\177A
+ BC 2
+
2.
U\177C
all diagonalsare con-
rectangular parallelepiped,
In a
Corollary.
AB 2
2 =
gruent.
of pyramids.
sections
cross
Parallel
58.
Theorem. If a (Figure parallel to the base, then:
(1)lateral
ilar
(2) the cross to the base;
divided
(SM) are
altitude
the
and
edges
by
this
parts;
proportional
into
plane
plane
by a
is intersected
49)
\177pyramid
section itself is a
(A\177B\177C\177D\177E \177) sim-
polygon
(3) the areas of the crosssectionand the base are proto the squares of the distancesfromthemto the
portional vertex.
Figure
Figure
49
lines
(1) The
lines of
plane
third
\177BC,
Thales' theorem
(2)
planes
Therefore
ASB.
we have: B\177C
BSC
A\177B \177and
parallel
two
that SA\177
SB
A\177A
B\177B
From similarity of and B\177SC \177, we derive: AB
....
A\177B \177
BS
B\177S '
BS
can
(the
base etc.,
SC
t
and
and the A\177M
It
follows
A'SB
\177,and
\177AM.
SM
the trianglesASB
and
hence
from
'
AB
BC
reason,
\177
M\177M
C\177C
B\177C\177 '
the same
(\36513). For
\177
and B\177S
as intersection cross section) by a
be considered
AB
A\177B \177AB
C\177D \177CD,
50
-
A\177B \177
BC B\177C\177\337
then
35
pyramids
and
Parallelepipeds
1.
Similarly, we obtain: BC \337B\177C'\177= CD \337C\177D \177,as well as the proportionality of all other sidesof the polygons ABCDE and A\177B\177C\177D\177E \177. These polygons also have congruent corresponding angles (as angles with parallel and similarly directed sides), and are thereforesimilar. (3) Areas of similar polygons are proportional to the squaresof
corresponding sides
I,
Book
(she
AB A\177B \177
\365251).
Since
AS
MS
A\177S
M\177S
of AASM and AA\177SM\177), Area of ABCDE AB2
to similarity
(due
of
Area
-
A\177B\177C'\177D\177E \177
1. If
(A\177B\177)
that
conclude
we
MS
2
M\177S2'
2
pyramids with congruent altitudes are insame distance from the vertices, the areas of the cross sections are proportional to the areas of the bases. Let a\177 and a2 (Figure 50) be the areas of the basesof two pyramids, h the altitude of each of them, and a\177 and a[ the areas of the cross sectionsparallel the bases and drawn at the same distance h\177from the vertices. According to the theorem, we have:
Corollary
tersected by
two
the
at
planes
to
al _
(h')\177
with
the
and
are
altitudes
are also
vertices
a-!\177
a2
a\177
a2 then
=
=
al
hence
a2
a\177
congruent
from
equidistant
If
2.
Corollary
pyramids
a\177 ,
h2
al
a\177
=
i.e.
a\177,
equivalent
of two
bases
if
cross sections
3, then
equivalent.
59. Lateral surface area of prisms. Theorem.
The
area
surface
lateral
the productof a lateral dicular cross section.
edge
and
the
of a prism perimeter
is equal to
of a
perpen-
cross section (Figure 51) of a prism, we abcde obtained by intersecting all lateral facesof the prism by a plane perpendicular to the lateral edges.Sidesofthis polygon are perpendicular to the lateral edges (\365\36531, 20). The lateral surface area of the prismis equalto the sum of areas of parallelograms. In each of them, a lateral edge can be considered as the base, and oneofthe sides of the perpendicular cross section as the altitude.Therefore the lateral surface area is equal to AA \177.ab 4B B \177 . bc + C C \177 . cd + D D \177 . de + E E \177 . ea = AA \177 . ( ab + bc + cd + de ) . a
By
mean
the
perpendicular
polygon
+ea
aRecall
that plane
figures
are
called
equivalent
when
they
have
equal
areas.
Corollary. to
2. POLYHEDRA
Chapter
36
the
lateral
The
of the
product
surface
prism is
of a right
area
perimeter of the base and the
edges of such a prism are congruent base can be considered as the perpendicular lateral
and its
altitude,
the
to
equal
because
altitude,
section.
cross
S
E'
D'
surface
Lateral
60.
lateral
The
Theorem.
equalto the
Figure
51
Figure
product
52
area of regular pyramids. surface area of a regular pyramid is of an apethem and the semiperimeter
of the base.
52) be a regular surface areaofthe pyramid
(Figure
SABCDE
Let
The lateral
apethem.
is
the
its
and
SM
sum
of areas
pyramid,
of congruent isoscelestriangles.The area of one of them, e.g. ASB, is equalto \253AB. SM. If n is the number of triangles, then the lateral \177 AB. n is the semiperimeter surface is equalto \253AB. n. SM, where \177 of the base, and SM is the apethem.
Theorem.The lateralsurface of a regular pyramidal is equal to the productof an apethem half the sum of the perimeters of the bases. The surface area (Figure 52) of a regular frusturn abcdeABCDE is the sum of areas of congruentisosceles trapezoids. area of one of the trapezoids, e.g. AabB, is equal to area
frustum
and
lateral
pyramidal
The
+ ab). Min. If n is the number surface area is equal to
2
\177.
(AB
+ ab
AB
2
where
AB
\337 n
and
\337 ]Vim.
ab. n
n:
of
Mm.
trapezoids,
AB
\337 n
then
+ ab.
2
are perimeters of the bases.
n
'
the lateral
1.
37
\177nd pyramids
p\177r\177llelepipeds
EXERCISES
62. Show
that in a tetrahedron,or a parallelepiped,
be chosen
for its base.
63. the angle cube.(Consider the Compute
first
6J. Provethat
face of a polyhedron has an odd number of the facesis even. that in every polyhedron all of whose facesare triangles edge such that all planeanglesadjacentto it are acute. that in every tetrahedron, there is a vertexall of whose
65.
\177Prove
is an
there
66.
\177Prove
equidistant
a point
Find
69.\177Prove
70. Prove that hedra are
is convex
faces,crosssections, and
if
only
given tetrahedron. only if every segment
of a and
lies
of convex
projections
in
entirely
poly-
of
that
perpendicular,
two
diag-
and an edge? if two diagonals of a rectangular parallelepipedare then its dimensions are congruent .to the sidesof a versa.
vice
and
triangle,
the
cube with the edge 1 cm. angles is greater: between two
a diagonal
or between Prove
of the
diagonal
the which
right
faces
if and
congruent
in the interiorofthe polyhedron
72.Ina cube, 73.
all
are
convex polygons.
Compute
onals,
from
a polyhedron
that
with the endpoints the interior.
71.
of
acute.
that in a tetrahedron,all faces of oppositeedgesare congruent.
67. Prove if all pairs 68.
every
if
are
angles
plane
meet,
that
the number
then
sides
diagonals
can
of two adjacent facesof a then skew ones.)
diagonals
between
face
each
7\177. Compute the length of a segment if its orthogonal three pairwise perpendicular planeshave lengths a, 75. \177Is a polyhedron necessarily a prism, if two of
gruent polygons with
respectively
are parallelograms?
(First
allow
parallel non-convex
projections b, and
to
c.
are conall other faces
its faces
sides, and polyhedra.)
diagonals in a prism are concurrent (i.e.pass then this prism is a parallelepiped. 77. Prove that in a pyramidal frustum with quadrilateral bases, all diagonals are concurrent, and vice versa, if in a pyramidal frustum, all diagonals are concurrent, then its bases are quadrilateral. 78. \177 Find a parallelepiped three of whose edges lie on three given lines, of which no two lie in the same plane. 79. Compute the total surface area of a right prism whose altitude equals 1 cra, and the base is a right triangle with legs 3 cra and 4 cm. 76.
through
that
\177Prove
the
same
if all
point),
38
POLYHEDRA
2.
Chapter
area of a rectangular is equal dimensionsof the base 25 m and 14 m. Compute the lateral surfacearea the lateral edge. 81. In a rectangular parallelepiped a square base and the altitude h, a section through two opposite lateral edges is drawn. 80.
to
The total surface 1714 m 2, and the
parallelepiped are
and
with
Cross
Compute 82.
the and
total
parallelepiped,
of the
if
of
area
the
section equals S. hexagonal
pyramid
a. Compute surface area.
the lateral
regular
A
base
area
surface
total
the
the cross
has the altitude h and the sideof edge, apothem, lateralsurfacearea,
Compute the total surfaceareaof the tetrahedron edges have the same length a. 84{. Compute the angle between lateral facesand the 85.
all
base
of whose
of a
regular
pyramid whose lateral surfaceareais twice the area of the base. 85. Prove that if all lateral edges of a pyramid form congruent angles with the base, then the base canbe inscribed into a circle. 86. Prove that if all lateral faces of a pyramid form congruent angles with the base, then the base canbe circumscribed about a circle. 87. A regular hexagonal pyramid, which has the altitude15cm and the side of the base 5 cm, is intersectedby a plane parallel to the base. Compute the distancefrom this plane to the vertex, if the area
of the crosssectionis 88.
The altitude
is h, and the areas area
of the
to
equal
of a regularpyramidal of
bases
the
2.
\177-v/\177 cm
are a
with
frustum
a square
base
and b. Find the total surface
frustum.
89. The bases of a pyramidal frustum is intersectedby
36 and 16. The bases and bisecting the altitude.Compute the area of the cross section. 90.* Througheachedge of a cube, draw outside the cube the plane making 45 \370angles with the adjacent faces. Compute the surfacearea ofthe polyhedron bounded by these planes, assuming that the edges of the cube have length a. Is this polyhedron a prism? 91. Prove that if all altitudes of a tetrahedron are concurrent,then each pair of opposite edges are perpendicular, and vice versa. 92.* Prove that if one of the altitudesof a tetrahedron passes through the orthocenter of the opposite face,then the same property holds true for the other three altitudes.
93.*
From
the
planes
oneof
the
frustum
a
plane
parallel
have
areas
to the
to
in the interior of a polyhedron,perpendiculars faces are dropped. Prove that the foot of at least perpendiculars lies in the interior of the face. a
point
of the
2. 2
prisms
of
Volumes
of
Volumes
39
pyramids
\177nd
and
prisms
pyramids
61.Main assumptionsabout volumes. vessel of a cercan hold a amount of water. The amount can be measuredand expressed in \"cubic\" units, such as A
tain shape
certain
quantitatively
cm3, m3, ft \177, etc.
measuring
Such
geometric of presstherefore, 'in theory
an
polyhedra into several
and, more polyhedra.
properties
hold
we
by
expressed
are
form,
idealized
capacity of vessels.Namely
ric solids
Our
volumes.
will
positive
generally, for Furthermore,
of solid shapes leads us to the assumptions about volumes exproperties of the water-holding assume that volumes of geometnumbers, and are defined for all
all solidsthat canbe partitioned we assume that the following
true.
solids have equal The volume of a solid subdivided
(1) Congruent
volumes.
(2) into several parts is equal to the sum of the volumes of these parts. (3) The volume of the unit cube (i.e. the cubewhose edge is a unit of length) is equal to i (in the corresponding cubic units). Two solids that have equal volumes are called equivalent. 62. Theorem. The volume of a rectangularparallelepiped is equal to the product of its dimensions. Let a, b, c be three numbers expressing the three dimensionsof a rectangular parallelepiped in a certain unit of length. Let V be the number expressing the volume of the parallelepipedin the corresponding cubic unit. The theorem says that V - abc. In the proof, we
consider
the
following
three
cases.
(i) The dimensionsare expressed by whole numbers. Let for example the dimensionsbe (Figure 53) AB = a, BC = b, and BD = c, where a, b, c are whole numbers (in Figure 53,a = 4, = 2, and c unit squares. b
- 5). Then the baseof the parallelepiped On each of them, a unit cube can be
contains
ab
placed. Thus a Figure 53) containing ab congruent to one unit of contains c such units, c such layers. Therefore
layer of such cubesis obtained(as shown in unit cubes. Sincethe altitudeofthislayer is length, and the altitudeof the parallelepiped the whole parallelepipedcanbe filled with the volume of the parallelepipedis equalto abc (ii) The dimensions are expressed by fractions. Let the dimensions of the parallelepiped be
m a = --, n
b
=
p -, q
c=
r
-. s
cubic
units.
Bringing the fractionsto a common
a=
Picka (auxiliary) of original unit. pressed means of
b
nqs
nqs,
denominator,
=
we have:
nqr nqs
nps
c = \177
--,
nqs
of length congruent to the 1/nqs-th part the dimensions of the parallelepiped,exthis new unit, are given by whole numbers,and the result of case (i) the volume is equal to their product
new
the
by
by
therefore
mqs --,
POLYHEDRA
2.
Chapter
40
unit
Then
(qr) if
new cubic unit. in the original unit
the
by
measured
units
contained
The numberof such cube is equal to (nqs)
cubic
new 3,
i.e.
the
is equal to the 1/(nqs)3-thpart of the original one. Therefore the volume of the parallelepipedexpressed in the original cubic units is equal to new cubic unit
1 .(mns)(r\177ps)(nqr)=raqs
nps
nqs
nqs
3
(nqs)
,4',
Q,',,;
c
nqr
nqs
D '\177
D
'\177.LL ....... \177_-_ _-_ _-_-_.
b
a
A
B
53
Figure
(iii) The
B
A'
A\"A
Figure 54
dimensions are expressed by
arbitrary
real
numbers.
dimensions of the rectangularparallelepiped Q (Figure 54) be AB = a, BC = b and BD - c, where a, b and c are positive real numbers, possibly irrational. Eachof these numbers can be represented by an infinite decimal fraction. Take the finite decimal ! ! ! fractions: firsfly C\177n,/\177,\177and \"/,h approximating a, b and c from below Let the
with
the
a, b and
precision
c
from
of 1/10 above
n \177 and
with
the
then
same
II C\177n,/3
nII and
precision.
II \"/n, approximating
On the lines AB,
=
?,\177,
segments: firstly
mark the
BD,
and
BG
BD'
=
BA\"
then
and
41
and pyramids
of prisms
Volumes
2.
C\177n, \"
' B(7' =
ozn,
=
BA' =/\177 \" and
BC\"
BD\"
=
with the
homothety
the
by
We
respectively.
will
the negative
I (including
<
k
S'
same coe\177cient
k
assume
with
that
values) is very
will be left to the reader as an exercise. In the triangles BAB', which lie in the same plane, the angles at the vertex congruent (as vertical), and the sides adjacent to the these are proportional. Indeed, since BS \337AS - k - B'S \177\337AS', and
similar
SAS' and are
A
angles
have:
we
BA :
similar, and in the
Thus
particular
BB
segment
\177is
the length of
1 times
-
k
I-
SA - k -
the directionof SS'. (In
= ZBSS
parallel
to SS', the length of it the direction of it is
SS \177,and case
\177,and
k <
where
are
\177 : k - 1. is equal to opposite to 1, the direction will be
ZB\177BS
the
the
the triangles
Therefore
: S'A.
B'A
BB'
\337 SS
same.) -B
\177
$
Figure
69
We conclude that by taking the point B homothetic to A with to the center S, then moving it in the direction parallel and opposite to the direction of the segmentconnectingS with S \177,and placing it at the distance (k - 1)\337SS \177from B, we obtain the point B' homotheticto A with respect to the center S(
respect
In this argument, if (I)
and
coe\177cient
(I)\177are k
>
be any point of the given figure. Thus, homothetic to the given one with the same 1 but with respect to two different centers S and S \177, A
the
figures
could
2.
Chapter
54
then the figure (k can (k as a whole in the
be
onto the
superimposed
POLYHEDRA
figure
\177'
by
moving
direction paralleland oppositeto SS' through
- 1) \337SS'. Thus \177 and \177' are congruent. 72. gemark on trans:iat\361ons. The operation of moving all points of a geometric figure by a given distancein the direction parallel to a given segment is a geometric transformation called translation. It generalizes to the caseof spacegeometry the concept of translation on the plane describedin Book I, \177101. For example, translating a given polygon ABCDE (Figure 70) in a direction not parallel to its plane, we obtain another polygon A'B'C'D'E', congruent and parallel to the given one. The segments AA', BB', etc. are parallel,congruent and similarly directed. Therefore the quadrilaterals AA'B'B, BB'C'C, etc. are parallelograms, and thus the given and translated polygons are two bases of the same prism AE'. Using the conceptof translation, our proof of the lemma can the
(k
distance
be summarizedas the following statement mations of figures: two \177homotheties with centers
different
differ
by an
appropriate
73. Corollaries. (1) A one with a positive homothety
coefficient
same
the
translation. angle
polyhedral
transforcoefficient but
geometric
about
given
to a
homothetic
is congruent to
it. Indeed,
is the vertex, the homotheticangle however, according to the lemma,the choice of another rise to a congruent polyhedral angle. (2) A polygon homothetic to a given one is similarto it. Indeed, this is true in plane geometry, i.e. when the center of homothety lies in the plane of the polygon. Therefore this remainstrue for any center due to the lemma (since polygons congruentto similar ones with
coincides
are
of homothety
center
the
when
the
given one; center gives
similar).
(3) A polyhedron
obtained from a given one
by
It is
a
with
homothety
a positive coefficient is similarto it. obvious that corresponding elements of homothetic polyhedra are positioned similarly with respect to each other, and it follows from the previous two corol-
laries that congruent
the polyhedralanglesof such polyhedra faces similar.
are
respectively
and corresponding
74. If two polyhedra them is congruentto a Theorem.
polyhedron
are similar, homothetic
then each of to
the
Since polyhedral anglesat homologousverticesof similar are congruent, we can superimpose one of the polyhedral of the first polyhedron onto the homologouspolyhedral angle second. Let SABCDE (Figure 71) be the given polyhedron dra
other. polyheangles
of the
P,
and
3.
of
$imilari\177y
55
polyhedra
to it
let SA'B'C'D'E' be the polyhedron P' similar and positioned in such a way that one pair of homologous polyhedralanglesofboth edges
homologous
SA and
Q homotheticto P with
SA\177. We to
respect
with
k coincides
coefficient
the polyhedraQ and P'
P\177.
faces
Q
of
angles
SAaB
polygons
the
of both
A\177B \177and
to the
Similarly Q
and
P\177
the
center
this,
note since
edges
homologous
\177,SCUD
of
S with the homothety respective edges of they are proportional Therefore
of P.
edges SA', SB \177,SC the polyhedron Q as
\177,SBiC
SA
that
of the of
\177:
the polyhedron
SD'
\177, and
of
well. Conse-
\177,and
SD\177A \177are
with
A\177B\177E \177 and
common
adjacent
B\177C\177coincide
B\177C\177E \177.
and P' with the vertex B \177coincide. way we compared edgesand facesofthe polyhedra to their common polyhedral angle S, we can now
polyhedral
the
the
that
polyhedra. Furthermore, sinceall homologousdihedml and P\177 are congruent, the planes of the faces of Q adja-
cent to the edges Thus
For
k be the ratio SA prove
will
congruent,
are
with the sanhe coefficient k to the endpoints A \177,B \177,C \177,and D' the polyhedron P\177 are vertices quently
S. Let
vertex
the
at
coincide
polyhedra
angles
of Q
their polyhedral angle B '. For example, the ray B\177E \177 of this polyhedral angle in both polyhedra,and since their respective edges are congruent, the endpointE\177of the edge B\177E \177is their common vertex. So, we have found that the polyhedraQ and P' coincide since they have the sanhe vertices. (Should Q and P\177 have more elements than those shown in Figure 71, we could consecutively compare their respective faces, edges, and vertices, and conclude that they all coincide.) with
proceed is an
edge
A'\177
D'
S
Figure 70
Thus the polyhedronsimilarto the given gruent to the polyhedron Q homothetic to P.
B'
B
Figure 71
polyhedron
P
is con-
Chapter 2.
56
figures.
geometric
of arbitrary
Similarity
75.
POLYHEDRA
One can give
definition of similarity: two geometric figures are similar if one of them is congruent to a figure homotheticto the other. The lemma of \36572 shows that the position of the center of homothety in this definition is irrelevant: when a given figure becomes congruentto another after application of homothety with one center, it becomes congruentto it after homothety with any other center, provided that the coefficient of homothety remains the same. Thus, this definition fully expresses the idea of similarity as \"being of the sameshape but possibly different scale.\" general
following
the
consistencywith the definition of similar polyhedra to assume that the homothety coefficientsin the arbitrary similar figures are positive. Then the previous
To maintain given in
\36569,
of
definition
need
we
are similar in
3 of
Corollary
with
together
theorem
sense, and vice
are
they
two similar
that
show
\36573
this new sense whenever
polyhedra in the old
versa.
In a pyramid (SABCDE, Figure72), if a cross to the base is drawn, then it cuts off another pyramid (SA\177B\177C\177D\177E\177f similar to the given one. Theorem.
76.
parallel
section
According to part (1)ofthe k. Then the resultingfigure A \177, B \177, C \177, D \177, E \177,and
Set
a polyhedron
sides
lateral
the
\177\337$A,
center
the
with
be
will
\36558,
$A
-
k
and
apply
of to
S and coefficient with the vertices
it will be the pyramid SA\177B\177C\177D\177E( are similar, the theorem follows.
i.e.
S,
polyhedra
homothetic
in
theorem
the pyramids are proportional. the given pyramid the homothety
Since
77. Theorem.Surfaceareasof similar have the ratio as the squares of homologousedges. A2,..., An denote the areas of faces of oneof the similar and a2,..., an the areas of the homologousfaces the other. Let L and I be the lengthsof two homologous edges. Then, due to similarity of homologousfaces proportionality of polyhedra
same
Let
A\177,
polyhedra,
a\177,
of
any
and
edges_,___.we have:
homologous
a\177 A\177
properties
From
l \177
L 2'
A2
L 2'
'\"'
al +a2+...+an 78.
the
+
Volumes
Theorem.
ratio as
cubes
of
L 2'
An
ratios, it follows:
of equal
A\177
l \177
as
12
a\177
A2 +
.\" +
12 An
of similar
homologous
polyhedra have the same edges.
of polyhedra
57 Consider the case of pyramids. Let SABCDE (Figure 72) be of the given pyramids, L be the lengthof oneofits edges,e.g. and ! < L be the length of the homologous edgeofthe similar to it. On the altitude SO of the pyramid, take the point O' such that SO': SO= 1: and draw through O' the cross section parallel to the base. the pyramid SA'B'C\"D'E', cut off by this plane, is homotheticto the given pyramid with the homothety coefficient equal to 1: and is therefore congruent to the secondgiven pyramid. Let denote by V and v the volumes of the SABCDE and SA'B'C\"D'E' respectively, and prove that v: V: : Z 3. For this we note that, according to theorem of the altitudes of these pyramids are proportional to their edges, and the bases are similar polygons. Therefore, a and a' denote the areas of the bases ABCDE A'B'C\"D'E', then 3. Similarity
first
one
SA,
pyramid
given
L,
Then
first
L
us
pyramids
\1773
the
\36558,
lateral
if
and
Since V =
\253a.
12
1
SO
a L' and -r
v =
and
SO
a'
SO'
we find:
\177 ' SO', \177a \337
12
SO'
a'
v
= L2.
Figure 72
Suppose now
umes lengths
V
L
and
S (Figure planes
SAB,
polyhedron.
Figure 73
that we are given
1. In
the interior of
73), connectit SBC,
etc.
with
through
all
the
first
A, B, and
Then the planespartitionthe polyhedron
a point
pick
polyhedron,
the point S
vol-
homologous edgesof
a pair of
vertices
with
polyhedra
similar
two
and with
v respectively,
and
D
E
C
B
13
1
C, etc.,
and
edge
each into
draw
of the
pyramids
2. POLYHEDRA
Chapter
58
SABCF, SBCD, etc.,which
bases are
of the
volumes
vertex
common
the
have
4 Let
polyhedron.
given
the
of
faces
V\177,
pyramids. If one appliesthe homothety
the
V\177,. be
with
!:
whose
S, and
V2,...,
the
cem
and the coefficientequalto L, a polyhedron congruent to the second given one is obtained,pm%itioned into pyramids homothetic to SABCF, SBCD, etc. previously constructed. Let vl,v2,...,v\177 be the respective volumes of these pyramids. Then ter S
V1
\2343'
V2
\2343'
Vn-
''\"
\2343'
therefore,
and
vl q- v2
v
--'
q-
-4-
13
vn
EXERCISES
111. Prove if
that
n-gonal
regu\177lar
two
at
angles
plane
their
tl\177e
are
vertex
pyramids are congruent.
similar if and only
which regularprismsare similar. Prove that two pyramids are similarif'the baseand a lateral of one of them and the baseand a lateral face of the other are
112. Find out 113.
face
respectivelysimilar, respectively are similarly positioned.
114. The
---
same
115. Show that
to a
to
parallel
a
figure
11
not lying in
a
the
if one of two figures is other, then vice versa, the homothetic m the first one. that
118. Prove that similarpolyhedra area) are congruent.
119.
Given
a
cube
with
the edge
whose volume is twicethe volume 4Recall our begin
of
lemma
to the
figure
convex
line (or a plane) with respect it, is a line(respectively a plane)
it.
Prove
7.
and
to a
homothetic
116. Provide the proofof thety coefficientk < 1. thetic
angles,
two prisms.
about
homothety
of
center
dihedral
congruent
form
of
a, of
the
case
the
congruentto a other
the
given
edge
of the
homo-
figure
homo-
is congruent
figure
(or equal
volume
equal
find
in
\17771
x of
to
surface
another cube
one.
convention to consider only convex polyhedra. polyhedra, the theorem holds true, but the partitioning with dividing into convex polyhedra.
In the case of nonprocedure should
4. Symmetdesofspacefigures is easily
uity, a x
problem of doubling the cube, known solved by computation(namely, x = ), but it cannot be solved by a straight.
This
Remark.
59
1.259921...
antiq-
since
=
32v/\177
\177/\177a:
and com-
edge
pass construction. parallel to the baseof a pyramid the volumes of the parts into which the plane divides the pyramid have the ratio m: n? 121. A pyramid with the altitude h is divided by two planes parallel to the base into three parts whose volumes have the ratio 1: m: n.
120. In
its
divide
a plane
should
ratio
what
so that
altitude
Find the distancesof
the
from
planes
these
vertex.
122. Compute the volumes of two similar polyhedra, volume is V, and their homologousedgeshave the
4
of space
Symmetries
79.Central
to
corresponds a point
the
the segment
of
of
symmetry
Thus,
are called symone of the figures there other figure such that the midpointof O. The point O is calledthe center
point
figures
A of
point
every
figures.
the
in order
of
A'
the
\177is
AA
total
n.
m:
figures geometric
Two
symmetry.
metric about a pointO if
if their ratio
to
the
find
figure
\177
symmetric
about
the
center
a given figure (I), one needs for every point A of the figure (I), to extend the line AO past the center O and mark on the extension the
O to
OA
segment
locus of
We have when
\36549,
one
to
\177congruent
all points
about
homothety
the figure
(I)\177 is
the
geometric
examples of centrally symmetricfigures in angle symmetric to a given the vertex. Also, central symmetry is a specialcase of with negative coefficients: the figure homothetic with the coefficient ]c = -1 to a givenfigure is centrally symmetric the polyhedral
described
to it about the centerof In
Then
encountered
we
homothety
AO. obtained.
A \177thus
symmetric
centrally
as segments, the as wholes speaking cannot plane
figures
they
angles,
are be
homothety.
certain homologous elements, such angles, are congruent. However not necessarily congruent, because generally superimposed onto each other. We have seen figures,
or dihedral
example of symmetric polyhedralangles. Yet sometimes centrally symmetric figures are congruent, but the elementsbeing superimposed are non-homologous. For instance, consider a trihedralangle with the vertex O and edges OX, OY and OZ (Figure 74) all of whose plane anglesare right. Consider the angle
this
phenomenon
in the
2. POLYHEDRA
Chapter
60
OX'Y'Z' symmetricto it.
Rotating
line XX
\177coincides
the
\177until
ray
OZ
angle
the
' about
OX\177Y\177Z
rotating the resultingangleabout the line OZ,
OZ, and
ray
the
with
can
we
the
then
superimpose
the ray OX \177then merges and OY \177with OX. However, if we rotate the angle OX\177Y\177Z \177about the line ZZ \177until the rays OX \177and OY \177coincide with OX and OY respectively, then the rays OZ and OZ \177turn out to have opposite directions. angle OX\177Y\177Z with the ray 0\276,
the
\177 onto
However
OXYZ.
z
z z'
z
Y
X
X
X'
,
Y'
Y
Z'
Z'
Figure 74
If the geometric figure symmetric to a given one center coincides with the given figure, one says that
has a centerof symmetry. center of symmetry, namely
For
example,
the
intersection
about
a certain
the given figure any parallelepiped has a point of the diagonals
(\36556).
Figure
75
Two geometric figures are called P, if to every point A of one of the figures there corresponds a point A \177of the other, such that the segment AA \177is perpendicular to the plane P and is bisectedby the intersection point with it. The plane P is calledthe plane of symmetry of the figures. 80.
Bilateral
symmetric
about
symmetry. a
given
plane
4. Symmetriesofspacefigures In
angles
dihedral
of a given figure, and A about a planeP (Figure
any two points symmetric to them
are
AB and
A\177B \177 are
sectionof
points
AA
(because by
\177 and
lines
BB
symmetric, about Q, to the points A \177and
and
P
=
to this
perpendicular
\177 are
points
the
\177are
segments
the
BB
\177 and
AA
in
this
Inside
P.
\177 are
and
other,
to
are
B
and
A
AB
Therefore
it).
then
same plane, Q, perpendicular
planes
the
B
75),
they are parallelto each
plane P,
lie in the
plane, the
\177and
since the
Indeed,
congruent.
to the
perpendicular
particular
about a plane, correspondingsegments, and are congruent. For example,if A and B
symmetric
figures or
plane
61
the line of interB
\177respectively
line and
are bisected
A\177B \177.
central symmetry, figures symmetricabout a congruent. Examples of symmetric figures are obtainedby reflecting any object in a mirror: every figure is symmetric to its mirror reflection with respect to the plane of the in
As
are
plane
the
case of
not
necessarily
mirror.
figure coincides with the certain plane (or, in other words,
geometric
If a
about a
parts symmetricabout this
planeof
or
symmetry,
body has a plane sides.
to have a
this provides a
the
in
found
it into
the
figures which are not congruent. Namely, hands are symmetric, but cannot be superimposed,
metric
the as
left it is
left
and
of sym-
example
convincing
house-
the human
For instance,
dividing
symmetry
bilateral
of
By the way,
into two
bilaterally.
is symmetric
Bilaterally symmetric objects are frequently hold (e.g. chairs, beds, etc.) and in nature. right
be divided figure is said
the
then
plane),
to it
symmetric
figure
can
and right clear from
the fact that the sameglove does not fit both hands. Symmetry about a line. Two figures are called symmetric about a line ! if to every point A of one of the figures there correspondsa point A \177of the other such that the segment AA \177is perpendicular to the line I, intersectsit, and is bisected by the point of intersection. The line I is calledthe axis of symmetry of the 2nd
order.
It
follows
metric
about
this line
plane
from
the
a line
(at some point O)
figures
symmetric
then
about
if two
that
definition,
are intersected the
the
by
cross
point
geometric figures symplane
any
sections
perpendicular
of the
to
figures are
O.
Furthermore, it follows easily that two solids symmetric about a line can be superimposedby rotating one of them 180 \370 about the line. Indeed, imagine all possible planes perpendicular to the axis of symmetry. Each of these planes contains two cross sections sym-
2. POLYHEDRA
Chapter
62
metricabout the point of intersection of plane with the axis. If one moves the plane itself by rotating it in space 180 about the axis, then the first figure becomes superimposedonto the second. holds true for every cross section perpendicular to the axis. simultaneous rotation of all cross sections through the angleof 180 is equivalent to rotating the whole figure 180 about the
along
\370
This
Since
\370
the
\370
follows.
our statement
axis,
If a figure, obtained from a given one by rotating 180\370about a line, coincides with the given figure, one says that it has an axis of symmetry of the 2nd order. The name reflects the fact that in the processof rotationby 360 \370about an \"axis of symmetry of the 2nd order\" the rotatedfigure will occupy its original position twice. Here are someexamples of solids possessing axes of symmetry of
certain
the 2nd order:
(1)
A
axisof
the
is
an even
with altitude.
pyramid
regular
symmetry
number of lateral faces.The
parallelepiped. It has three axes of symmetry namely,the linesconnectingthe centersofopposite
(2) A rectangular
2nd order,
of the faces.
the prism
has n + i
axesof
number n of lateralfaces the
of
symmetry
of opposite
midpoints
the
connecting
lines
If the
prism.
A regular
(3)
is
namely
order,
2nd
lateral edges,and has
symmetry of the 2ndorder,namely of a lateral edgeand the center
face.
of the
line
the
opposite lateral and
bases.
1 If
u axes of the midpoint
connecting
81. Relations betweencentral,bilateral
\253n
\253\177 +
lines connecting the centers of oppositefaces,including the number u of lateral faces is odd, then the prism each
then
even,
sym-
axial
metries.
through
passing
(O, Figure 76), the
a point
about
figures are symmetric to
If two
Theorem. one
then
it\177
they
perpendicular to the planeat Let
A about
A be
a point of
A \177 the
symmetric
this
point. \177the
about the line
point
symmetric
point
J\177gure\177 (P)
are
the given figure, A
the center O, and
plane P. Denoteby
a given other abouta plane symmetric
to
to A about
the
point of the segmentAA \177 plane through the points A, A \177and A \177. This plane is perpendicular to the plane P sinceit contains the line AA \177perpendicular to this plane. 'In the plane AA\177A \177,draw the line 1 passing through the point O and parallelto AA\177( This line is perpendicular to the plane P and to the line BO. Let C be the
with
the
plane
P.
B
the
intersectlob
Draw the
intersectionpointof
the
lines
A\177A \177and
1.
of
Symmetries In the A'A\".
triangle BO
But
midpoint
of
2_
the
sp\177ce
AA'A\",
l,
and
segment
63
figures
segment BO is the midline parallel to A'A\" 2_ 1. Furthermore, since O is the AA', and the line 1 is parallel to AA\", we the
hence
= CA\". We conclude that the points A' and A\" are symmetric about the line 1. Sincethe same holds true for any point A of the given figure, we conclude that the geometric locusofpoints A' symmetric to points of the given figure about the centerO, and the geometric locus of points symmetric to pointsofthe given figure about the plane P, are symmetric to eachother about the line 1. find
that
A'C
A'
C
A
Figure
Corollaries.
figure about follows
from
two
the
(1) different
lemma
centers
of
\36571,
are since
congruent figures
to a
symmetric
centrally
figures
Two
76
to each
centrally
given
other. This
symmetric
to
the given one are homothetic to it with respect to two different centers and the samecoefficient k = -1. (2) Two figures symmetric to a given figure about two different planes are congruent to each other. Indeed,replace each figure, symmetric to the given one about a plane, by a congruent figure, namely by the figure symmetric to the given one about a center lying on the planeof symmetry. Then the problem reduces to the previous one about figures symmetric to the given one about different centers. 82. Axes of symmetry of higher orders. If a figure possesses an axis of symmetry of the 2nd order,it is superimposed onto itself by the rotation about this axis through the angle of 180 \370 . It is possible however that a figure is superimposedonto itselfafter the rotation about a line through a certain anglesmallerthan 180 \370. Thus in the process of rotating the figure about this line,it occupies its original position several times. The number of times this happens(including the original position) is called the order of symmetry, and the line is called an axis of symmetry of higher order. For example, while
2. POLYHEDRA
Chapter
64
a regulartriangular pyramid does not have axes of symmetry of the 2nd order,its altitude serves as an axis of symmetry of the 3rd order. Indeed, after rotation through the angleof 120 \370about the altitude (Figure 77), the pyramid occupiesits originalposition.In the process of rotation about this axis, the pyramid becomes superimposed onto itself three times (after the rotationthrough the angles of 0 \370,120 \370, 240\370).
and
easy to
It is
an axisof
prisms,
nth
centersof
axis
the
2nd
of
faces are examples of solids with symmetries The altitude (respectivelythe line connectingthe
order.
is the
bases)
the
of an even order is also Regular pyramids, or regular
of symmetry
order.
lateral
n
with
of the
see that any
symmetry
axis.
Figure 77
Figure 78
83. Symmetries of the diagonals is the center of symmetry
cube. The of
There are nine planesof symmetry: DBFH) and three planespassing quadruple of parallel edges.
The
the
and
faces.
axes
nine
has
cube
connecting
midpoints
In addition,
diagonal the
the
78).
planes midpoints
(such as of each
of symmetry of the 2nd order: six lines of opposite edges (e.g. of AD and FG),
lines connecting the The latter lines are in fact
three
6 through
(Figure
cube
the
of
point
intersection
centersof each
pair
axes
of symmetry
of
the
of the
opposite
4th order.
four axes of symmetry of the 3rd order, AG). Indeed, the diagonal AG, obviously, makes congruent angles with the edges AB, AD, and AE, and these angles make the same (right) angles with each other. If we connect the points B, D, and E, then we obtain a regular triangular pyramid, ABDE, for which the diagonal AG is the altitude. When the cube is rotated about this diagonal throughthe angleof 120 \370 , the pyramid returns to its original position. The same is true about the pyramid GHFC, centrally symmetric to the pyramid ABDE. Thus, as the the
namely
its
diagonals
cube
has
(e.g.
4. Symmetries of spacefigures is not
It
whole cubereturns to its original
rotation, the hard to see that
of this
result
65
the cubedoes
position.
axes of
other
any
have
not
symmetry.
Let us find order gives only
has
cube
the
rotation through
the trivial
3rd order gives
of the
2nd
of the
of symmetry
axis
(excluding
4th order three. Since
of the
the
a whole. An
way
such
one
of symmetry
axis
An
0\370).
it as
preserve
that
there
are
the cube
different ways of rotating
how many
now,
out
and
ways,
such
two
of symmetry of
six axes
order, four of the 3rd order, and three of the 4th order, that there are 6 x 1 d- 4 x 2 d- 3 x 3 = 23 ways, excluding trivial one, to superimpose the cubeonto itselfby rotation.
we
2nd
find
It
is not
other (e.g. by
from each
etc.
see directly that all the
hard to
some
that
noting
differently).
positions
their
change
23
are different
rotations
of the Together
the
vertices A, B, C, with the trivial
rotation (leavingthe positionofeach vertex unchanged) ways of superimposing the cubeontoitself.
give 24
they
EXERCISES
123.
12,{. The same-- for symmetry 125. Prove that the figure symmetric plane is congruentto it.
126.
ular
a given
to a line(or a plane),
line
a plane.
about to
and
axes,
centers,
Determine
formed by
symmetric
a plane).
(respectively
a line
centrally
figure
the
that
Prove
is
a dihedral
planes
a given
intersecting
angle about any
the figure plane, but not perpendic-
of symmetry of
it.
to
127. Determine centers, axes, and planesof symmetry consisting of two intersecting lines.
128.*
has a
a prism
that
Prove
for
center of symmetry if
the
figure if its
only
and
base does.
129. Determinethe number with
of symmetry
planes
of
130. Determine the numberof planesofsymmetry mid with n lateral faces.
131.
figures
three
Let
plane
P, and
that
\1772and
132.
What
problem
of a
regular prism
faces.
n lateral
(I)\177 and
\1772 \177are
about
symmetric
(I)\" be a plane
about
symmetric: (\177 perpendicular
the intersection
a regular
(I) and to
pyraa
(I)\177about
P.
Prove
line of P and
(\177.
said about the figures (I) and (I)\177 of the previous planes P and Q makethe angle:(a) 60\3707 (b) 45\3707
can be
if the
(I)\177,and
(I), (I)\177
of
133.* Prove that
angle the
2. POLYHEDRA
Chapter
66 has
figure
a
if
intersection
their
then
180\370/n,
two symmetry planes making line is an axis of symmetry
an of
order.
nth
154. Describe the crosssectionofa cube by to one of the diagonalsat its midpoint.
135.*
correspond to24 ing its diagonals.
the
(including
ways
different
the cube onto itself trivial one) of permut-
of superimposing
24 ways
the
that
Show
perpendicular
plane
the
four
5
polyhedra
Regular
84. Definition. Let us of its plane angles are congruent
regular if all dihedral angles are
angle
a polyhedral
call
all
and
of its
regular if all of its facesare conof its polyhedralanglesare regular and congruent. Thus a cube is a regular polyhedron. It follows from the definition, that \177in a regular polyhedron: all plane angles are congruent,all dihedral angles are congruent, and all edges are
congruent.
A
regular
gruent
is called
polyhedron
and all
polygons,
congruent.
85. Classificationofregularpolyhedra. Let
count that
a convexpolyhedral
and that their sum
has
to
angle
be smaller
has
than 4d
us
into
take
three plane
at least
ac-
angles,
(\36548).
Since in a regular triangle, every angle is repeating it 3, 4, or 5 times, we obtain the anglesum smallerthan 4d, but repeating it 6 or more times, we get the angle sum equal to or greater than 4d. Therefore convex polyhedral angles whose faces are anglesofregular trianglescan be of only three types: trihedral, tetrahedral, or pentahedral.Angles of squares and regular pentagons are respectivelyd and \177d. Repeating these angles three times, we get the sums smaller than 4d, but repeating them four or more times, we get the sums equal to or greater than 4d. Thereforefrom angles of squares or regular pentagons, only trihedral convexanglescanbe formed. The with angles of regular hexagons are 4d 3 , and of reg\177ular polygons more than 6 sides even greater. The sum of thre\177 or more of such angles will be equal to or greater than 4d. Therefore no convex polyhedral \177-d 3 ,
angles can be formed
from
It follows that only the can occur: those whose faces four
or
squares,
five
triangles
or regular
such following
are
angles. five
regular
types
triangles,
polyhedra meeting by three,
of regular
at each vertex, or those whose pentagons, meetingby three faces
faces
are
at each
either
vertex.
5. Regular polyhedr\177
67
Regular polyhedra of each of the five types do exist and are often called Platonic solids after the Greek philosopher Plato. They are: (i) regular tetrahedron whose surface is formed by 4 regular triangles (Figure79); (ii) octahedron whose surface is formed by 8 regular triangles 80);
(Figure
(iii)
(Figure
icosahedron
whose surface
is formed by
20
surface
consists
triangles
regular
81);
(iv) cube (or hexahedron) whose (Figure 82);
(v) dodecahedron tagons(Figure83).
Figure
Figure
79
Figure
80
Figure 82
86. Constructions of Platonic that regular polyhedra, if they it does not prove that regularpolyhedra
exist. In orderto establish which
81
Figure 83
shows
constructionof eachof cube, was defined
squares
regular pen-
by 12
is formed
surface
whose
of 6
their the
five
as a
solids. The exist,
existence Platonic
rectangular
of
fall
each
above
five
into
of the
it suffices solids. In
argument
types,
but
five types
to point out a the case of the
parallelepiped all of whose
68
2.
Chapter
POLYHEDRA
a construction is
three dimensions are congruent, such We will show here how each of the remaining
be constructedfrom A regular
of
in
Figure
84.
a
pick any
square faces adjacentto this
oppositeto are
wise
A.
diagonals
us.
solids can
vertices of
vertex A
vertex,
of
take
four of the
the tetrahedronas shown the
cube,
the
vertices
and in the
B, C,
three
and D
connecting the vertices A, B, C, D paircube's faces (onediagonalin each face), and
edges
six
The
by taking
constructed
be
can
for the
cube
Namely,
to
familiar
Platonic
cube.
a
ietmheclror\177
eight vertices
four
of the
congruent. This shows that all facesof the tetrahedron regular triangles. Rotating the cube 120 \370about any of its diagonals (e.g the diagonalpassingthrough the vertex A) will keep one of the vertices(A) of the tetrahedron in its place, but cyclically permute the edges, adjacent to it, and the otherthree vertices (e.g. move B into C, C into D, and D into B). The corresponding polyhedral angles becomesuperimposed ontoeach other (the angle B onto the angle C, etc.) This shows that all polyhedral angles of the tetrahedron are congruent, and all dihedral angles (in each of them) are congruent. Thlm the tetrahedron is a regularpolyhedron. are therefore congruent
are
D
A
Figure
84
Figure 85
A regular octahedron can be constructed by of cube's faces (Figure85) for the vertices. Each
polyhedronconnects
of
centers
as
easily
computed,
has
the
two
length
adjacent \1772
a
where
the six' centers edge of the resulting faces of the cube and, taking
a denotes
the cube's
dimension. In particular,all edges have the same length, and hence all facesof the octahedronare congruent regular triangles. To prove that all dihedral and all polyhedral angles of the octahedron are congruent, we notethat by rotating the cube (say, about axes passing through centersof opposite faces) one can move any face (e.g. P) of
5.
69
polyhedra
Regular
onto
the cube
(e.g.
other
any
(\177).
Since
rotation
the
preserves
the
cube, it also preservesthe setof eight centers of the faces. Therefore the rotation preservesthe octahedron asa whole, but moves the edges (CA to C*/\177) and vertices (A to B). Thus the corresponding dihedral and polyhedral angles of the octahedron become superimposed. A dod\275cahcdro\177 can be constructed by drawing 12 planes, one through each of the 12 edges of a cube (Figure86),and choosing the slopes of these planes in such a way that the resulting polyhedron is regular. The fact that it is possibleto achieve this is not obvious at all, and proving it will require some preparation. Let us begin with examininga regular pentagon. All of its diagonals are congruent,and can be assumed to have the same length as the dimension of the cube. The angles of the pentagoncontain 108\370each. If we place two copies of the pentagonin one plane so that they have an edge in common,the anglesat the common vertices will add
216 \370which
up to
is smaller
fore we can rotate the
edge
87),
Figure
(AB,
by 144 \370than in space
the
pentagons
two
so that
their planes
form
full
angle.
about their a
dihedral
There-
common angle,
OAF decreasesfrom 144 \370to 108 \370. Since the figure two regular pentagonsin space is symmetric about the plane perpendicular to the edgeAB at the midpoint, the angle DBE symmetric to the angle CAF will also contain 108 \370. This means that two more regular pentagons congruentto the original ones can be attached, one to the edgesFA and AC (shown in Figure 87 as GCAFH), the other to the edges EB and BD. If we now draw the diagonalsC'D,DE, EF, and FC' in these pentagons, they will form a square. Indeed,C'D[[EF (since these diagonals are parallel to the commonsideAB of the pentagons) and thus C'DEF is a rhombus,and XFC'D - XEDC' (as angles symmetricabout the until the angle formed by the
plane perpendicular to Let us
in Figure 87. From baseC'DEF, draw the
then
and diculars
ON
to
and
CD
to
spectively
the
(\36528),
base. OM
line
the
the
examine
now
_1_
vertex
two
AB
at the
midpoint).
polyhedron ABC'DEF shaded A, drop the perpendicular AO to its slams: AM and AN perpendicular re-
tent-like
and finally draw their projectionsOM Then, by the theoremof the threeperpenC'D, ON 2_ FC', and therefore XAMO and
FC,
XANO are linear anglesof the dihedralangles formed by the base C'DEF with the lateral facesC'ABDand AFC' respectively. Since the base of the tent-like polyhedron is a square, we can attach polyhedra congruentto it, to each of the faces of the cube as shown in Figure 86. We claim that in the resultingpolyhedron, the faces (triangles and quadrilaterals) of the attached tent-linepolyhe-
dra will agree in their slopesand thus form In order to show that the slopesagree,
the dihedral angleformedby
CDEF is base
it
suiTices \370,
and FAC
that
OM : of a
legs
are
they
that
and the
base
the extension
the cube adjacentto
P of
face
= 90\370.
For
with the hypotenuse
right triangle
the
by
\177DEF, add up to this, we will compute = ON: AN, and show
cos(ZANO)
and
AM
base,
the
plane
same
the
with
+ ZANO
ZAMO
that
i.e.
cos(ZAMO) -
to check
ABCD
the plane P is perpendicularto instead that the dihedralangles, formed
to check
ABeD
faces
suiTices
formed by
angle
the
and
FA\177
it
Since
\177DEB.
the
90
triangle
the
dihedral
the
to
congruent
of
_FHG\177
pentagons.
regular
face
lateral
the
POLYHEDRA
2.
Chapter
70
equal
to
1:
d-
XACM = 72\370, and
Note that
=
ON As
we
= AC. cos36\370, = A,C . cos 72\370,
= NC
OM
MC
in Book
found
at the vertex36\370(and the base to the lateral From
the
cos 72
\370
I,
\365223,
side
-
\177
- AC
AN
=
AC
have:
we
. sin 72\370, . sin 36\370.
triangle
72 \370at
angles
with the angle the ratio of
base),
the
is equal to the golden mean (v/\177triangle, we find:
cos 36\370=
1
AM
isosceles
an
in
the
hence
of this
geometry
= 36 \370, so
XACN
- 2 cos 2
1
72 \370
v/\177
+
1)/2.
1
4
4
/
Using
the
cos 2 c\177+ sin 2 c\177= 1, we compute:
identity
cos 2
72 \370=
cos 2 36 \370=
3-V\177
sin272
3 + v/\177 8 '
sin2
5+v\177
o_
36 o =
5
cos
2 72
\370
sin
2 36
\370
-
8
Therefore
OM
cos 2 36 \370
2
2
AN)
=
sin 2 72 \370 (3
d-
+
-
v/\177)(5
v/\177)
+
(3
-
x/\177)(5
d-
+
+
+
20
-
20
--1.
5.
71
polyhedra
Regular
by-product, we have computed the ratioOR//: AA//, i.e. the the angle OAM, which is a half of the linear angle of the dihedralangleAB. Notethat we used only the fact that plane angles of the trihedral angle A contain 108\370each. This shows that all trihedral angleswith this property have congruent dihedral angles and are thereforecongruent to each other (\36550). Thus the constructed As a
sine of
polyhedron with 12 regularpentagonal
faces
is regular.
D
\234
Figure
Figure
86
87
existence of the dodecahedronis established, a regular be constructed by taking the centersof 12 faces of dodecahedron for the vertices. $?. Theorem. Any regular polyhedron is similar to one of
Once the
can
icosahedron
the
the five Platonic solids.
In
same
\36585,
type
proved
we as
one
of the
any regular polyhedron R falls into the five Platonic solids P. Replace now the poly-
that
a polyhedron Q, homothetic to it and such that the edges of Q have the same length as the edges of P, and prove that Q is congruent to P. For this we need to establishfirst that Q and P have congruent polyhedral angles.We know that these polyhedral angles are regular and have the samenumber of congruent plane angles. Let the regularpolyhedralangle S be onethem, with n plane angles c\177 (see Figure 88, where n = 5). It is easy to see that S has an axis of symmetry of the nth order. \177It can be located as the intersection line SOof two planes of symmetry, e.g. of the bisector plane ASO of the dihedral angle, and of the plane HSO passing through the bisector SH of the plane angleASE and perpendicular to its plane. Draw the plane perpendicularto the axis of symmetry and passing through any point O on it in the interior of the polyhedral angle S. Thenthe crosssectionofthe polyhedral angle by this plane will be a regular n-gon ABCDE.In the faces of the dihedral angle
hedron
R with
Chapter 2.
72 AG and
perpendiculars
SB,
drop
tices
A and
C of the n-gon,and
n-gonand the regular
measure
c\177.
by the the
But
-
\370
the
is
isosceles by the
determined The
c\177)/2.
is determined
and
n-gon
determined
is
AGC
= (180
ZSBA
by
CG to the edgeSB from
consider
The length of its lateral sideAG
by its
number angles,
the
ver-
AGC.
triangle
side AB of the
diagonal of side AB. Thus the angie base
AC
is a
number n of the plane anglesand their is the linear angleof the dihedral
AGC
angle
This proves that regular polyhedral angleswith and measure of their plane angleshave congruent and therefore are congruent to each other.
SB.
angle
POLYHEDRA
the
same
dihedral
this, we can pick one vertex in each of the polyhedraQ and P superimpose their polyhedral angles at this vertex. Since the edgesof polyhedra are congruent, the adjacent vertices will also coincide. Sinceall angles of both polyhedra are congruent to eachother, polyhedral angles at these adjacent vertices also becomesuperimposed. the edges adjacent to these vertices, and proceedingthis to other vertices, we conclude that the polyhedraQ and P superimposed. Using
and
these
dihedral
the
Examining way
\177e
a very demanding definition of reguup to scale, there are only five such polyhedra. One may ask if the same conclusion can be derived from milder requirements of regularity. It turns out that the answer is \"yes\": in order to conclude that a polyhedron is regular, it suffices to require that all of the faces are congruent regular polygons and polyhedral anglesare congruent (but assume nothing abo\177t dihedral angles). In fact, many attempts to relaxthe definition even further lead to mistakes. First, merely assumingthat all faces are congruent regular polygons is not enough (to constructa counterexample, attach two congruent regular tetrahedra to each other by their bases). Next, the class of polyhedra, all of whose polyhedral angles are congruent and faces regular, includesregularprisms with square lateral sides. This class was first systematically explored by a German mathematician and astronomerJohannes Kepler. In 1619,he found that in addition to the prisms, it also includesantiprisms(Figure 89), and 15 Archimedean solids (13, if symmet88.
lar
polyhedra
We accepted
Remark.
that,
found
and
ric polyhedra are not which were described in the 4th century A.D. a Greek mathematician Pappus and attributed by him to Archimedes. regularly shaped solids of various kinds were thoroughly studiedand classified the 20th century, the ancient symmetry patterns of the Platonic solids still play the most fundamental role modern mathematics and physics. distinguished),
by
Although
in
five
in
5. Regular
73
polyhedr\177
Figure
Figure
88
89
EXERCISES
those pyramids 137. Checkthat the numbers 136. Describe
are
to
respectively
equal
the
numbers
edges edges,
vertices,
of faces,
are congruent.
of a cube edges and verticesof and faces
octahedron.
an
--
same
The
138.
for
that
Prove
139.
tetrahedron
of a
that
Prove
1\1770.*
octahedron
an
and
icosahedron
to establishthis resultwithout
of
of whose
all of
Is there
dodecahedron.
a way
counting?
the polyhedron whose verticesare centers of faces is a tetrahedron again. the polyhedron whose vertices are centersof faces (or icosahedron) is a cube (respectively a dodeca-
hedron).
1J1. 1\1772.
of the
Describe
all
rotations,
self
by
the
trivial
1\1773.
one). the
four
many
How
Realize
planes
a regular tetrahedron
symmetry
rotations
there
cor-
the set of vertices. does a regular tetrahedronhave?
of the
four
planes
that an octahedron has as many symmetry of each orderas a cube
Prove
axes of
of symmetry
all permutations
hedron by reflectionsin 1\177 6.
of the 12 rotations of and that to different
vertices,
different permutations of
respond
1\1775.*
each
that
Show
permutes
1JJ.
five Platonic solidshave a center of symmetry? ways to superimpose a regular tetrahedrononto itand show that there are 12such rotations (including
Which
tetra-
of symmetry
and
rotations.
planes does.
regular
of a
vertices
and
POLYHEDRA
2.
Chapter
74
147. The same-- about icosahedron and dodecahedron. 148. Show that a cubehas nine planes of symmetry. 149. Locate all 15 planesof symmetry of an icosahedron. 150.* Find all axes of symmetry (of any order) of an icosahedron, and show that there are in total 60 ways (includingthe trivial one) to superimpose the icosahedron onto itself by rotation. 151. Describe cross sections of the icosahedronor dodecahedron by the plane passing through the center and perpendicular to one of the of symmetry.
axes
152. Do there exist six
making
15o\370. Show
diagonals into the
that inscribed
cubes
of a dodecahedron's dodecahedron.
154.* Showthat eachofthe
the
rotations
of a
faces are edgesof five permutes
dodecahedron
that to different of the set of five
it, and
permutations
different
correspond
60
into
inscribed
cubes
five
point and
other?
each
to
the same
through
passing
lines
angles
congruent
there
rotations
cubes.
155. Give an accurate construction of an n-gonal antiprism (referring to Figure 89 as an e\177ample with n -- 5)\177 a polyhedron which has two parallel regular \177z-gons as bases, 2\177z regular triangles as lateral faces, and all of whose polyhedral angles are congruent. Prove that when \177z= 3, the antiprism is an octahedron.
Hint: Seethe
cover.
front
an icosahedron
18\177. Cut
into two regular pyramids
187.* Find and compare the numbers' of planes (of each order) of: (a) a regular \177z-gon in space, obtained from two copies of a regular \177z-gonal each other by their bases, (c) a regular r\177-gonal
and
and
antiprism.
an
axes of symmetry the polyt\177edron
(b)
to
attached
pyramid
an
(d)
prism,
r\177-gonal
antiprism.
volumes
1/\1778. Compute the
1/\1779.* equal
160.*
of regular
tetrahedron and octahedronwith
a.
edge
Prove to
Prove
to
volume
the
that
\1772(3 4-
x/\177)a
that
of an
icosahedron with the edgea is
3.
the volume of
a dodecahedronwith
edge
the
a is
+
Hint: Representthe dodecahedron as a cube like\" solids attached to each of its sixfaces, and of the solid first.
with
congruent
compute
the
\"tent-
volume
3
Chapter
SOLIDS
ROUND
I Cylindersand cones 89.
by
generatrix
of revolution is the Figure 90), called a
line (AB), calledthe axis
a fixed
about
of revolution.
(MN,
curve
any
rotating
generator)
(or
A surface
revolution.
of
Surfaces
surfaceobtained
M
M
A
B
On the
generatrix, take any
dicularPO to
the
about
each
point
axis,
position
the
and
axis.
the
generatrix dicular,
Figure
90
Figure
P and in the
point
Obviously,
the
of its
drop from it the perpenprocess of rotation of the
the length of the perpenfoot O remainunchanged. Therefore
angle APO,
generatrix describes a to the axis of revolution,
of the
perpendicular
91
the intersectionof this planewith
the
75
circle,
plane
the
center
the
and
axis.
of which
is
of which
is
Thus, pendicular Any the
and
meridians
because
of a
section
cross
the
surface of revolution
90. Cylindrical surfaces. A
paralleltoa AB
a plane
per-
other meridian.
swept by a line (AB,
line
by
axis consistsof one or several circles. plane containing the axis of revolution is calledmeridional, cross section of the surface by this planea meridian. All of a given surface of revolution are congruent to each other, in the process of revolution eachof them assumes the posito the
of every
tions
the
3. ROUND SOLIDS
Chapter
76
is a
cylinder
A
planes
two
surface
so that it remains curve (MN). The
in space a given
(or generator), cylindrical surface.
91. Cylinders. surfaceand parallel atrices. The part of
is the
surface
cylindrical
moving intersects
generatrix
the
of the
directrix
and
direction
given
called
is
91)
Figure
and the curve
solid bounded by 92) intersecting
(Figure
MN
a cylindrical the gener-
surface contained betweenthe parts of the planescut out by this surface bases of the cylinder. A perpendicular dropped from any point of one be\177se to the plane of the other is calledan altitude ofthe cylinder. A cylinder is called right, if its generatrices \177
planesis
called
the
the cylindrical surface,
lateral
are perpendicular to
the
bases,
and
and the
oblique
otherwise. A
C
Figure
Figure
92
\177
93
A right cylinder whose bases are disksis called a right circular cylinder (Figure 93). Such a cylinder can be considered as a solid obtained by rotating a rectangle (OAA'O') about one of its sides (OO'). The (AA \177) describes then the lateral surface, and the other two sides the bases. A segment BC (see Figure 93) parallel to OA also describes a disk whoseplaneis perpendicular to the axis OO \177.
oppositeside
Generatrices
is used
as the
plural
for
generatrix.
i.
and cones
Cylinders
Thus, crosssectionsofa right
cylinder
circular
to the bases are disks.
In
cylinders,
we will
exposition,
elementary
our
by planes
consider only
parallel
right circular
the sake of brevity refer to them simply as cylinwe will deal with prisms whosebasesare polygons inscribed into the bases of a cylinder, or circumscribed about them, and the altitudes are congruent to the altitudeof the cylinder. We will call such prisms inscribed into (respectivelycircumscribed for
and
Sometimes
ders.
the
about)
cylinder.
surfaces.
Conical
92.
surface
conical
A
surface ob-
is the
so that it passesthrough fixed point (S) and intersects a given curve (MN). The line AB called a generatrix, the curve MN the directrix, and the point the vertex of the conicalsurface. tained
a line
moving
by
(AB,
Figure 94)
a
is
S
$
93.
Cones.
Figure
94
Figure
A cone
is a solidenclosed between
95
a conical
surface
plane intersecting all generatriceson onesideof the vertex 95). The part of the conicalsurfaceenclosed betv\177een the vertex and the plane is called the lateral surface,and the part of the plane cut out by the conical surface the base of the cone. The perpendicular dropped from the vertex to the plane of the baseis and a (Figure
the
called A
cone
altitude is called
of the
cone.
right
circular
if the
base is a disk,and
the
al-
passes through its center (Figure96). Such a cone can be obtained by rotating a right triangle (SOA) about one of its legs (SO). Thenthe otherleg(OA) describes the base, and the hypotenuse (SA) the lateralsurface of the cone. A segment BC (Figure 96) parallel to OA also describes a disk perpendicular to the axis SO. titude
Thus, cross sections of a the base are disks.
parallel to
by planes
cone
circular
right
SOLIDS
ROUND
3.
Chapter
78
only right circular cones and referto them simply sake of brevity. Sometimeswe will consider pyramids whose vertices coincide with the vertexof a given cone, and bases are inscribed into, or circumscribed about its base.We will call such pyramids inscribed into the cone,and respectively circumscribed will
We
as
cones
consider
for the
it.
about
',4
94.
Conical
Figure
96
Figure
frusta.
and
a
cross
disks (the base of the cone,and bases of the conicalfrustum.
parallel
A
conical
frustum
axis OO', of a pendicular to the an
(Figure
trapezoid
97) can OAA\177O
section the
part of.a cone to i\177. The section) are called
is the
frustum
conical
A
enclosedbetweenthe base
97
cross
parallel
be obtainedby
\177whose
lateral
about
rotation
side
OO
\177is
per-
bases.
95. Surface area of conesand cylinders. The lateral surface cylinder or cone is curved, i.e. no part of it can be superimposed onto a plane. Therefore we need to define what is meant by area of such surfaces when it is expressed in the units of area of plane figures. We will accept the following definitions. of a
cylinder,
For the lateral surface area of a we to which the lateral surface areaof a regular prism into the cylinder tends as the numberof its lateralfaces indefinitely (and hence the area of each lateral facetends (1)
limit
take
the
inscribed increases
to zero).
l.
(2) For the we take
79
cones
and
Cylinders
lateral surface area
of a cone,
(or
frus-
conical
the limit to which the lateral surface areaof an inscribed regular pyramid (or regular pyramidal frustum) tends as the number of its lateral faces increases indefinitely (and hencethe area of each lateral face tends to zero). 96. Theorem. The lateral surface area of a cylinder is
turn)
equal to the productof the circumference the
base
of
an
and
altitude.
Into a cylinder (Figure98), inscribe and h the perimeterof the baseand respectively. Then the lateralsurface area
by the productp. h. Suppose now of the prism increases indefinitely.
limit,taken to be
the
that Then
c of
circumference
altitude remains unchanged. of the prismtendsto limit taken to be lateral surface
Therefore
h
prism.
regular
any
by p
the
of the
altitude
prism
prism is expressed number of lateral faces perimeter p tends to a
of the the the
the cylinder's base, and the the lateral surface area p-h
c.h. By definition (1), area s of the cylinder.
the
Denote
this limit is
Therefore
the
Figure
98
(1) If r denotes the radiusof the cylinder's base, 27rr, and hence the lateral surfaceareaof the cylinder is by the formula:
Corollaries.
c:
then
expressed
s=
to
if t
surface area of the cylinder,it
(2)
To
obtain
the total
add
the
lateral
surface area
denotes the
2\177rrh.
and the areasof the
total area of the cylinder, t =
2\177rrh
+
\177rr2
+
\177rr\177=
we
2\177rr(h
have: + r).
two
bases.
suffices
Thus,
3. ROUND SOLIDS
Chapter
80
97.
surface
lateral
The
Theorem.
area
to the productof the circumference of
the
of a cone base
and
is equal a half
of a generatrix.
Into a
99), inscribe any regular pyramid, and denote numbers expressing the perimeterof the base and the length of the apethem of this pyramid. Then (\36560) the lateral surface area of the pyramid is expressedas !p 2 \337 a. Suppose now that the number of the lateral facesof the pyramid increases indefinitely. Then the perimeter p tends to a limit, taken for the circumference c of the cone%base,and the apethem a tends to the generatrix 1of the cone.(Indeed,the generatrix SA is the hypotenuse of the right triangle SAL and is greater than the leg SL which is an apethem of the pyramid. The other leg AL is a half of the side of the regular polygon in the base, and tends to zero asthe number of sides increases indefinitely. Since SA- SL < AL, we conclude that a tends to 1.) Thereforethe lateralsurface area -\177p \177. \337 a of the pyramid tends to the limit \253c. 1. By definition (2) the limit is taken to be the lateral surface area s of the cone. Thus by
and
p
(Figure
cone
a the
s-c.l/2.
II
A
Figure
Corollaries.
base,the lateral
Figure
99
(1) Since c
=
r is the radius of the expressed by the formula:
where.
2\177r,
of the
area
surface
100
cone is
1
-
s =
\337 2\177r.
1 =
\177rl.
2
(2) lateral area,
The total surface one by adding the
area of the cone is obtainedfrom area of the base. If t denotesthe
we have:
t =
\177rrl
q- \177rr2 =
7rr(1 +
r).
the total
81
1. Cylinders and cones
98.
to
equal
the product
of the circumferencesthe Into a conical frustum frustum. Denote by p
bases, and
frusand half the sum
of a conical
of a generatrix bases.
of
dal
area
surface
lateral
The
Theorem.
is
turn
(Figure 100),inscribeany
and p'
of
perimeters
regular
lower
the
pyrami-
and upper
the length of an apothem. Then (\36560) the lateral the pyramidal frustum is equal to \253(p q-p')a. As the number ofjateral faces increasesindefinitely, the perimeters p and p' tend to the limitstaken for the circumferences c and c' of the bases, and the apothem a tends to the generatrix I of the conicalfrustum. Therefore the lateral surface area of the pyramidal frustum tends to a limit equal to \253(c + c')l. By definition (2), this limit is takenfor the lateral surface area s of the conicalfrustum, i.e.
surface
a
by
area of
1 (c + c')l. s=\177
(1)If r
99. Corollaries. and upper bases,then by the
expressed
the
the radii of the area of a conical frustum
denote
r'
and
surface
lateral
1 (2\177rr
+
=
2\177rr')l
\177r(r
+ r')l.
(2) Considering the conicalfrustum (Figure tained by rotation of the trapezoid OAA\177O f about the midline BC of the trapezoid.We have:
Therefore s = is
frustum
of the
+ O'A')
2\177r.
equal
BC.
to the
middle cross
= 1 (r + r'),
hence
the
r +
r' =
2BC.
section. t =
100. Nets of
and
as a solid obaxis OO f, draw
100)
l, i.e. the lateral surface area of a conical product of a generatrixand the circumference
(3) The total surfacearea
101), inscribe
is
formula: s=\177
1 (OA
lower
\177r(r
2 +
of
a
r'2
conical
+ rl
frustum
+
cylindersand cones.
any regular prism, and
is
Into
then
imagine
a
cylinder
(Figure
the lateral lateral faces breaking it or
that
surface of the prismis cut alonga lateral edge. Rotating about the edges one can develop the surface (without distorting the faces)into a plane figure. The resulting figure is called the net (or development) of the lateral surface of the prism. The
82
net is a rectangle tfLMN consistingof as many as there are lateral facesof the prism.Thebase/F/N is congruent to the perimeterof the base,and the altitude of the prism.
SOLIDS
ROUND
3.
Chapter
rectangles
smaller of the
the
rectangle to
altitude/fN
/(
h/
Figure
101
Imagine now that the numberof lateral faces of the prism increases indefinitely. Then the net ofits lateralsurface becomes longer and longer, and in the limit, tends to a certain rectangle/fPQN, such that the altitude aiSd base are congruent respectively to the altitude of the cylinder and the circumference of its This rect-
base.
angle
the
is called cylinder.
the net
(or development) of the lateralsur.,face
of
\177
$
$
Figure
102
Similarly, into a cone (Figure 102),inscribe any regular pyraWe can cut its lateralsurface along an edge and then, rotating lateral faces about the edges,develop it to the plane into a net shapedas a polygonal sector StfL, which consists of as many isosceles trianglesas there are lateral faces in the pyramid. The segments Stf, Sa, are congruent to lateral edges of the pyramid (or to generatrices ofthe cone),and the length of the broken line tfab... L
mid.
Sb,...
1.
83
cones
and
Cylinders
base. to
is equal to the perimeterof the pyramid's lateral faces increases indefinitely, the net tends whose radius SK is congruent to a generatrix of
length of the arc KM is equal
This sectoris
It is easy to see
lateral
the
that
101.Volumesof cylinders (1)
Definitions.
to which
limit
the
a
of the lateral surfaceof
cones.
and
the
of the cone.
area of the correspondingnet.
to the
is equal
frustum
conical
cone's base.
part PIfMQ of an annulus. area of a cylinder, cone, or
as the surface
102)
(Figure
frustum
the net
construct
similarly
can
One
conical
of the
lateral surface
of the
net
the
called
and the
cone,
the
circumference
the
to
SKM
sector
the
of
number
the
As
of a cylinder,
the volume
For
we
take
inscribed into the
prism
a regular
of
volume
cylinder tends as the number of lateralfacesof the
increases
prism
indefinitely.
of a cone
(2) For the volume limit to which the volume
regular
the
we take
frustum),
conical
pyramid (or
regular
of lateral faces increases
the number
as
tends
frustum)
pyramidal
(or
inscribed
an
of
indefinitely.
(1)
Theorems.
The
productof the area
of
of a
volume
and
base
the
cylinder is equal to the
the
altitude.
(2) The volume of a cone is equal to the product of the area of the base and a thirdof the altitude. Into a cylinder, inscribe any regular prism, and into a coneany regular pyramid. Then, if we denote by B the areaof the base of the prism or pyramid, their altitudeby h, and the volume by V, then we find
\36568):
and
(\36565
for prisms
V =
Imagine now that the number indefinitely.
increases
Then
Bh; of
for
lateral
B tends
pyramids
V =
--1Bh.
3
faces of the prism or pyramid to a limit equal to the area
b
cylinder or cone,and the altitude will remain unchanged. Therefore the volume V will tend to the limit bh in the case of prisms, and \253bh in the case of pyramids. Therefore the volume v of the cylinder or cone is given by the formula: of
the
of the
base
the
for
Corollary. cone, then
b
v =
and
\177rr2h,
=
cylinder
v = bh;
for the
cone
v
=
-lbh.
3
If r denotesthe radius of the baseof a cylinder and hence for the volume of the cylinder we for the volume of the cone v = \177rr\177'h/3. \177rr2,
or have
3.
Chapter
84
ROUND
SOLIDS
102. Similar cones and cylinders. According to the general of similar figures (\36575), a solid similar to a cone (or cylinis congruent to the solid homothetic to this cone(respectively
definition der)
cylinder).
the
Consider
cone (Figure
103) obtained
about the axis SO, and
by
a right
rotating
tri-
triangle homothericto/\177SOA with respect to the center S. Rotating/\177SO'A \177 about the axis SO \177,we obtain a cone homothetic to the given one. Sinceany solid similar to the given cone must be congruent to one of these cones (with an appropriate choiceof the homothety coefficient), we conclude that a figure similar to a coneisa cone, and that angle SOA
cones are similar if they are triangles about homologouslegs. two
let
obtained
SO\177A'
be the
by rotating
similar
right
AJ
Figure
103
Figure
104
Likewise, considering the cylinder (Figure104)formedby rotata rectangle SOAB about its side SO and applying a homothety with the center S, we obtainthe cylinder formed by rotating the rectangle SO\177A\177B \177,homothetic to the given one, about the side SO( We conclude from this that a solid similar to a cylinder is a cylinder, and that two cylinders are similarif they are obtained by rotating similar rectangles about homologoussides. ing
Let 103
104)
OA and SA \177of triangles
by h and h' the altitudes SO and SO' (Figures of the similar cones or cylinders,by r and / the radii ' of their bases, and by 1 and 1\177the generatrices SA and similar cones (Figure 103). From similarity of the right
denote
us
and
OA
the
SOA and
SO'A' (or rectanglesSOABand
SO'A'B'),
we
1.
Cylinders and
85
cones
find:
Theorem.
103.
conesor
r+l
r
\177q-h
proportions
these
Using
we derive:
ratios,
r+h
r
\177
and
r\177
we obtain Lateral and
r
r\177q-l\177
r\177.
the following results. total surface areas
radii or altitudes,and their cubes of the radii or altitudes.
of the to
proportional
are
volumes
of similar
squares
the
to
proportional
are
cylinders
-l
and -- =
--
of equal
properties
Applying
r
h
r
-- =
the
be respectively: the lateral surface,totalsurface, cone or cylinder, and st, t \177,and v \177be those of the other solid similar to the first one. For the cylinders, we have: s,
Let
and
and v of one
t,
volume
h\1772'
For
we have:
cones,
the
r
\177rl
7rr(r + 7rr\177(r
\177 +
h2
r2
1
l)
r r +l
l \177) -
r \177r \177+ l \177
r
h 2
2.
r\1772
-
h\1772,
v \253Trr2h
_
( r
)2
h
_
r3
h3
EXERCISES
161. Show that the surfaceobtainedby direction of a given lineis cylindrical.
162.
Find
cular)
all
planes,
axes,
and centers
of symnietry of a
(right
in the cir-
cylinder.
Prove that an oblique cylinder (\36591) is cylinder with the same generatrixand the 163.
a curve
translating
equivalent base
perpendicular cross sectionof the obliquecylinder.
to the
congruent
right to the
SOLIDS
ROUND
3.
Chapter
86
164. Use Cavalieri'sprincipleto prove that an oblique cylinder is equivalent to a right cylinder with the same base and the generatrix congruent to the altitude of the oblique cylinder. 165. A cross section of a (right circular) cylinder, not intersecting the bases,makes the angle c\177with them. Compute the area of the crosssectiongiven the radius r of the base. 1 66. Find a triangular pyramid which admits a square as a net, and compute its volume, assuming that the side of the squareis a. 16'7.On the surface of a regular tetrahedron, find the shortest path between the midpoints of two opposite edges. 168. Can a unit cube be wrapped into a square pieceof paper with 3?
sides
169. The axial cross the axis) of a cone has
section (i.e.
vertex.
the
angle at the vertexof the cone'snet.
1'70.
the plane passing through a generatrix of a cone and perpendicular to the planeof axialcrosssection this genOatrix does not have other commonpoints
that
Prove
cylinder)
(resp.
through
passing
the
with
cone
(resp.
cylinder).
This
plane
is called
l\177emark:
tangent plane? 1'72. Two cones
a
with
that
\177Prove
to the cone (resp.cylinder). vertex have: (a) a common planes?
tangent
common
vertex which have a common genplane passing through it are called tangent cones have a common symmetry
common
tangent
same
the
and
a common
many
infinitely
(b)
eratrix tangent.
tangent
of cones with
a pair
Can
1'71.
through Compute the
passing
section
cross
the
of 60 \370at
angle
the
two
plane.
of
surface area of a cylinder is equal to half of area. Compute the ratio ofthe altitudetothediameter
lateral
The
1'73.
surface
total
the
base.
the
174{. Generatrices Compute the ratio
of a cone makethe angle of 60 \370with of the lateralsurfaceareato the area
1'75. Compute the volume and the frustum that has the generatrix 15cm
18 cm
27
and
axis
passing
by
area
surface
long,
and
the
of
radii of
base.
base.
a conical
the bases
cm.
1 '76. In a coneof volumeV, two cross drawn dividing the altitudeinto three volume of the conicalfrustum enclosed 1'77. Compute the volume and the
obtained
lateral
the of the
rotating through
a regular its
vertex
sections congruent between
parallel parts. these
to the base are Compute the cross sections.
total surface area of the solid triangle with the edge a about the and parallel to the oppositeside.
1.
cones
and
Cylinders
the volume and the total surface area of the solid by rotating a squarewith the side a about the axis passing through oneof the vertices and parallel to one of the diagonals. Compute
178.
obtained
area
of the
180. Compute the
obtained its sides.
181.
of a
area
the area of the cross sectionby A. Compute the volume of the cone.
Compute the volume of
182.
a coneif
184.*
is:
of a coneis equal to
the
from
of a
volume
the
and 60
the
same
the
\370.
the
of
third
center
conical frustum is equal to
cones, all having
of three
volumes
is
the axis
of the base is r (a) 90 \370,(b) 120 \370,(c)
one surface area and the distance
that
Prove
of
base,
radius
the
net
of the lateral base to a generatrix.
product of the
sum
area of the solid side a about oneof
containing
plane
the
cone's
the
the volume
that
Prove
18S.
with the
cone is twicethe areaof the
and
the angleat the vertexof
surface
total
hexagon
rotating
sides.
of its
the
and
surface
a solidobtainedby
area of one
about
a regular
lateral
The
A
volume
rotating
by
surface
total
the
Compute
17\177/.
a thombus
the
as
altitude
frustum, and whose bases are respectively: the lower base of the frustum, the upperbaseof the frustum, and a disk with the area equalto the geometric mean between the areas of the other two the conical
bases.
the ratio
Compute
18/5.*
cone by
ing a
the base
congruent
chord
a
along
of volumesof two passing
plane
the
186.* Two perpendiculargeneratrices divide cone in the ratio 2: 1. Compute the radius of its baseis r.
of a
187.*
cross sectionif of the
altitude
and
Four
18\177/.*
to
gent
to
the
each
of the
surface
area
cone if the
only
cone. and
lateral about
two
surface faces
area of the cylinder of a given octahedron
a.
edge
the
lateral
cross sections, passing throughthe vermaximal area, and prove that it is the axial if the radius of the base doesnot exceed the
188. Compute the volume whose basesare circumscribed with
the
volume
by divid-
and intersecting
of the
has the
a cone,
of
which
out
Find
tex
obtained
solids
through the vertex to the radius.
congruent
cones
other.
Compute
generatrix.
with a
common vertex are pairwisetan-
the ratio of
the altitude of eachcone
Chapter
88
2
3. ROUND SOLIDS
The ball
104.Balls
spheres.
and
Definitions.
is called
diameter
called a
A solid obtained by rotating a half-disk a ball, and the surfaceswept by the
spherical surface or simply
One
sphere.
about
the
is also say that specified distance semicircle
can
a sphereis the geometric locusofpointslying at a from a fixed point (calledthe center of the sphere and the ball). The segment connecting the center with any point on the sphere is called a radius, and the segment connecting two points on the sphereand passing through the center is called a diameter of the ball. All radii of the same ball are congruent to eachother, and to a half of a diameter. Two ballsofthe sameradius are congruent, because they become superimposed when their centersare placed at the same point. 105. Cross sections of a ball. Theorem. Any cross section of a ball by a plane is a disk. Suppose at first that,the plane of the crosssection(p, Figure 105)
passes
the center
through
tersection curve with
center. Therefore
the
Figure
the cross
spherical section
O
surface is a
All points are equidistant
ball.
the
of
disk centered at
Figure
105
of the in-
from the the point O.
106
now the case when the planeof the cross section (Q, does not pass through the center. From the center O, drop the perpendicular 014to the plane Q, and take any point M on the intersection ofthe plane with the sphere. Connecting M with O and K we obtain a right triangle OKM and find: Consider Figure 105)
MK = x/OM 2
-
OK2.
(,)
2.
89
ball
Tl\177e
of the point M
position
As the
segments
of the
lengths
the
varies,
OM and OK donot change, and therefore the distance MK remains constant. Thus the crosssectionof the sphere lies on a circle, and the point K is its center. It follows from (,) that, conversely,every point M of this circle lies on the sphere. Corollaries. Let _R and r denote the lengths of the radii of a ball and a cross section, and d be the distancefrom the center to the plane of the cross section.Then the equality (,) assumes the form r - v/-R 2 - d 2. It follows that:
(1)
radius of the cross section is obtained if d passes through the center of the ball. Then r obtained as such a cross sectionof the sphereis called greatest
The
i.e.
if the
The
circle
plane
O, R.
a
circle.
great
(2) The R.
=
-
In
this
smallest radius of the crosssectionis obtained -- 0, i.e. the disk turns into a point.
d
when
----
case, r
(3) Cross
sections equidistantfrom the center are congruent.
(4) Out of two cross sections, the one that of the ball has the greaterradius.
to the
closer
is
center
106.Greatcircles. Any
Theorem.
center of a ball
Figure
(P,
plane
106) passing through two
into
surface
its
divides
symmetric
congruent parts (calledhemispheres). On
the
the
spherical
perpendicular
from the
the line AB past the point B until
point C.
any point A,
consider
surface,
dropped
it
point A meets
to
the
the
and let AB
plane
spherical
the and be
Extend surface at a P.
the center O, we obtain two and CBO. (They have a common leg BO and hypotenuses AO and CO congruent as radii of the ball.) Therefore AB - BC, and hence to each point A of the spherical surface,therecorresponds another point C of the same surface symmetric to the point A about the plane P. Thus the two hemispheres into which the plane P divides the sphere are symmetric. congruent
Connecting
right
A,
and
B
C with
ABO
triangles
The hemispheres are not only symmetric but also congruent, since by cutting the ball along the plane P and rotating one of the parts through 180 \370about any diameter lying in the plane P we will superimpose the first hemisphere onto the second.
Theorem.Through
any
two
points
of
a spherical
other than the endpointsof the samediameter, draw a great circle, and such a great circleis unique.
surface, one
can
ROUND
3.
Chapter
90
SOLIDS
On a sphericalsurfacewith the center O (Figure 107), let two and B be given not lying on the sameline with the point O. Then through the points A, B, and O, a unique plane can be drawn. This plane, passing through the center O, intersectsthe spherealong a great circle containing the points A and B.
point A
Figure
great circle l\177assing
Another
deed,
is obtained
circle
great
any
107 A
through
as a
and
B cannot
,...exist.In-
cross sectionof the \177phere
by
through the center O. If another great circle were passing through A and B, we would have two distinct planes passing through the same three points A, B, and 0 not lying on the same passing
a plane
line,
impossible.
is
which
bi-
(Figure 107)liesin each of the planes of the two hence lies in the intersection linc of theseplanes. linc is a diameter of each of the gmat circles,and
and
circles, this
Therefore
hence
same sphere
center O
The gmat
circles of the
other.
each
sect
two great
Every
Theorem.
the
bisects
circles.
tangent to a
107. Planes
Definition.
ball is established called
Theorem.
dius at is tangent (OA)
to
the
the
common point with a Existence of tangent planesis
only one
theorem.
following
The plane
ball. has
the ball.
to
tangent
by
that
plane
A
(P, Figure 108)perpendicularto a ralying on the surface of the ball,
endpoint,
it.
On the plane P, take any point B and draw the line OB. Since OB is a slant,and OA is the perpendicular to P, wc have: OB > OA. Therefore the point B lies outside the sphericalsurface.Thus the plane P has only one common point A with the ball, and hence the
plane P is tangent
to
it.
2.
91
The ball
Converse theorem.
A
tangent
pendicular to the radius
(OA)
plane drawn
(P, Figure 108) is to the point of
pertan-
gency.
Sincethe point .4 of plane P the ball, any
is the only common point of the other point B of the plane P liesoutside the spherical surface, i.e. is farther from the center than the point .4. Therefore O\1774 is the shortest segment from the point O to the plane, O\1774 is perpendicular to P. tangency
and
i.e.
A line
that has only
gent to it.
It
is
easy
one
common
to
see that
point
with a
there are
ball is calledtanmany
infinitely
lines
to the ball at a given pointof the spherical surface. Namely, every line (AC, Figure 108) passingthrough the given point (.4) and lying on the plane,tangent to the ball at this point, is tangent to the ball at this point. tangent
Figure 108 108.
Spherical
109) intersecting
segments and frusta. a ball divides it into two solids
\177igure
A
109
(P,
plane
(U
and
Figure
V), either
of which is calleda domeora spherical segment.The disk found in the cross section is calledthe baseofthe spherical segment. The segment KM of the diameter perpendicular to the base is called the altitude of the sphericalsegmentU. The surface of a spherical segment consists of two parts: the base,and a part of the sphere, which will be calledthe lateral surfaceofthe spherical segment. The part of a ball (W, Figure 109)enclosed between two parallel planes (P and Q) is called a zoneor spherical frustum.The disks found in the cross sections of the ball by these planes are called the bases of the sphericalfrustum. The part of the spherical surface enclosed betweenthe basesis called the lateral surface of the spherical frustum, and the part KL of the diameter perpendicular to them, the altitude.
Chapter 3.
92 as
MN, the spherical segment surfaces, we will first establish diameter
of the
faces of
surfaces the
about
can be considMABN is rotated arcs MA and AB describelateral U and frustum W. To determine
surfaces of spherical segments and frusta of revolution. When the semicircle
Lateral ered
SOLIDS
ROUND
such
109.
surface
lateral
The
Lemma.
sur-
areas
the following lemma. area of each of the three and cylinder, is equal to the
solids:a cone,conical product of the altitudeofthe solid circumference of a circle whose radius is the perpendicularto a generatrix frustum,
the
and
its
from
/Figure110) AB,
then
rotating If D is the
AC.
leg
the
about
surface area s of the
the lateral
the right
by
formed
be
cone
axis.
to the
up
midpoint
/1) Let a
midpoint
of
is (see
cone
triangle ABC the
generatrix
Corollary
i in
\36597):
s =
Drawing DE
we
\177_AB
(they are right and rive: BC : ED = into
Substituting
AC
(,), s =
two similar
obtain
have :
(,)
AD.
BC.
2\177r.
triangles ABC and EAD A), from which we deBC. AD -- ED. AC.
a common angie AD, and therefore
we\177find: 2\177-.
'\"'
A
A
as required.
AC
ED.
B
F
B
C
,0
C
D
D
F
Figure
(2) Let a conicalfrustum ABCD (Figure 111)about we find the lateral surface 2 in
Figure
110
obtained
be the
area s
side
Figure 112
111
AD.
by rotating the trapezoid Drawing the midline EF,
of the conicalfrustum
(see
Corollary
\36599):
s =
2\177r.
EF.
BC.
(**)
EG
Draw
93
ball
2. The
BH
and
BC
2_
and thereforeEF. BC into (**), we conclude:
s=
:
2\177-.
BH
\337 EG
EG.
AD
(3) The lemmaholds true for
in the
lemma
the
of
formulation
= AD
This ical the
and line
broken
diameter
111.
Theorem.
the surface, formed by
(BCDE)
same diameter) a regular brokenline tends as the sides of the brokenline decrease therefore their number increases indefinitely).
is to
to the lateral surface area of a spherentire spherical surface. In the latter case,
applies
to the
the entire semicircle. surface area of a spherical seg-
be inscribed into The
lateral
ment (or sphericalfrustum)is to the product altitude of the segment (respectively,of the equal
fvustum)
the
circumference
Let
the
the
arc AF
of the
any arc (AF), is
arc,
also
definition
segment,
the
about
area of
rotating
the this
(and
indefinitely
The area by
formed
semicircle
into
this
since the circle to the circle of the
parts.
their
limit
(about
BC,
as well,
is congruent
spherical
inscribed
to
= EG:
Substituting
\337 EG.
cylinder
a
lateral surfaceof a frustum, (BE, see Figure112)ofa defined to be the to which the rotating
BH
EF:
right
as required.
and
spheres
of
similar
perpendicular
are
cylinder'sbase.
110.Areas
two
obtain
We
DC.
_1_
triangles EFG and BHC (sidesofoneofthem the sides of the other). Fromthem, we derive:
of the
of the and
great circle.
surface of a spherical segment be formed by rotat(Figure 113). Into this arc,inscribe a regular broken line ACDEF with any number of sides. The surface obtained by rotating this broken line consists of the parts formed by rotating the sides AC, CD, DE, etc. Eachofthe parts is the lateral surface of a cone (when the rotatedsideis AC), or conical frustum (when the rotated sides are CD, EF, etc.),or a cylinder (if the rotated side is DE, and DEIIAB ). Therefore we can apply the previous lemma. For this, noticethat each perpendicular erected at the midpoint of the generatrixup to the axis is an apothem a of the brokenline.Denote by SAC, SCD, SDZ, etc. the lateral surfaceareaobtained respectively by rotating the side AC, CD, DE, etc. Then we have: ing
lateral
s\1774c
=
AC
$CD
:
C\177D \177 '
SDZ =
\177' 2\177ra,
27ra,
D\177E \177' 27ra,
etc.
we find that the area SACDEF the broken line ACDEF is given
by rotating
formed
surface
term-wise,
equalities
these
Adding
the
SOLIDS
3. ROUND
Chapter
94
F --
SACDE
AF
of by:
\177. 2\177-a.
sides of the regular brokenline increases indefia tends to a limit congruent to the radius R of the sphere,and the segment AF', which is the altitude h of the sphericalsegment, remains unchanged. Therefore the limit s, to which the areaof the surface obtained by rotating the broken line tends, and which is taken for the definition of the lateral surface areaof the spherical segment, is given by the formula: As
the
nitely,
number of the apothem
s = h. 27rR
case when the broken line is inscribedinto any than AF), and thus generatesunder rotation the
(rather
CF
2\177rRh.
the
now
Consider arc
=
lateral surface the same and
of a spherical frustum. The above leads to the sameconclusion about
area s \177of
spherical
the
h
the
27rR =
h'.
altitude
the
\177denotes
C\177F \177 of
frustum.
spherical
the
area of a sphereis to of the great circleandthe surface area of a sphere is equal
the
equivalently,
or
to four times
the
of
area
the
formed by previous
the
rotating
27rR. AD
\177 +
27rR. (1)
Corollaries.
the squares of
the areasof two
D\177B
disk. the
sum
DB. Therefore, applying for the area s of the sphere:
=
27rR(AD
The areas
\177 +
of radii
=
D\177B)
of spheres have
their radii, or diameters, spheres
the semicircle ADB of the surface areas
rotating
by as
AD and
arcs
we obtain
theorem,
equal
great
The area of the sphere,obtained (Figure 113), can be considered
s =
27rRh\177,
112. Theorem. The surface product of the circumference
diameter,
surface
lateral
f\177ustum:
s \177=
where
remains
argument
the
r and
since
2\177R. 2R the
the
=
4\177R
2.
ratio as by s and S
same
denoting
R, we find:
s'S=47rr 2'47rR2=r 2'R 2=(2r)2' (2) The area of a sphere is equal to the cylinder circumscribedabout it (Figure of the. cylinder'sbasecoincides with the
the
(2R)
2.
lateral surface 114), because the
radius
R of
area of
radius the sphere, the
2. The
ball
altitude
with 2R
2\177rR.
=
95
the diameter 4\177rR
2R, and hencethe lateralsurface
area
is
2.
Moreover, on the circumscribed cylinder (Figure114),the part lateral surface, enclosed between any two plane sections perpendicular to the axis, has the same area as the part of the sphere enclosedbetween these planes. Indeed, if h denotes the distance between the planes, then the area enclosed by them on the cylinder is equal to 2\177rRh, which coincides (according to \365111) with the area of (3)
of
the
the corresponding
Figure
part on the sphere.
Figure
113
114
Remark. In fact, the coincidence betweencorresponding areas on and on the circumscribed cylinder holds true for spherical regions of arbitrary shape. More precisely, to a point C (Figure ll4) on the lateral surface of the circumscribed cylinder, associate a sphere
the
from
C' on the sphericalsurfaceby C to the axis AB of the cylinder
point
intersection
with
Applying
sphere.
the
dropping
the
perpendicular
and taking the point this construction to
of its each
surface of the cylinder, we obtain and vice versa, any region on corresponds this way to a certain region on the cylindrical surface. Taking into account Corollary 3, as well as the the rotational symmetry of both the cylinder and the sphere about the axis AB, it is not hard to show that corresponding regions are equivalent. This fact is useful in cartography. Namely, one way of exhibiting the surfaceofthe Globeona geographical map -- the so-called Lainbert's equal-areacylindrical projectionconsists in projecting the sphere to the cylindrical surface as explained above (i.e. away from
point of any
region
on
the corresponding the sphere
region
the
on
lateral the
sphere,
3.
Chapter
96
the axis passingthrough net
without
\365100). While
(see
surface
the
this has the advantage the areasofregions. instance, cylinder
but
in
the
map
fact
the
as the
proportions between may appear in this (because the South pole is circle of the bottom base),
size
large by
then unrolling the cylinder it is generally impossible of a sphere on a plane map,
Antarctic
the
For
have
SOLIDS
of preserving
method
map to disproportionately representedon the
and
poles)
the
into the rectangular to exhibit distortion
ROUND
the
whole
of the Antarctic occupiesthe samefraction continent itself of the Earth'ssurface.
image
of
A
C
D
Figure 115
113. Spherical sectors. Thesolidobtained by rotating a sector Figure 115) of a disk about a diameter (AB) not intersecting the arc of the sector is called a spherical sector. This solid is boundedby the lateral surfaces of two cones and the lateralsurface of a spherical frustum. The latter is calledthe base of the spherical sector. When the axis of rotation (BO) coincides with one of the radii of the disk sector (e.g.ofBOE),the resulting spherical sector is bounded by the lateralsurfacesof a cone and of a spherical segment. As an extreme caseof a spherical sector, the entire ball is obtained by rotating a semi-disk about its diameter. To evaluate the volume of a sphericalsector,we need to prove the following lemma. 114. Lemma. If a triangle ABC (Figure 117) is rotated about an axisAM, lying in the plane of the triangle and Passing through the vertex A, but not intersecting the side BC, then the volumev of the solid of revolution thus obtained is equal (COD,
to the productof the surface
area
side BC and
\253
of
the
altitude
s
formed
h dropped
by
rotating
from the vertex
the A.
The
97
b\17711
the
Consider
three cases.
following
(1) The axis coincides
case
v
volume
the
by rotating the right
cones obtained
is
first volume
sum of the
to the
v=
The product CD \337BA
to BC
is equal
the area of AABC. Therefore
\337 h
DA,
each
since
two
The and hence
and
= \177vrCD.CD.
+ DA)
\177vrCD2(DB
2.
\253zrCD
the
of
volumes
BCD
triangles
the second
DB,
2.
\253zrCD
116). In this
(Figure
AB
side
the
with
is equal
DCA.
BA. expresses
twice
1 v
=
But, accordingto Corollary lateral surface v = sh/3.
\3657rCD. 1 in
BC. h.
\36597,
the
vrOD
product
area of the cone obtainedby
rotating
\337 BO
is the
Thus
ABCD.
B
C
D
C
A
Figure 116
(2) BC
(Figure AAMB.
not coincide with AB and is not parallel to this case the volume v is the difference of the VAMB of the solids obtained by rotating AAMC the result of the first case, we find:
and
Using
h \177$MG,
\177AMG--
where sMc
117
axis does 117). In
The
volumes VAMC and
Figure
h \177AMB =
and sMB are surfaceareasobtained
MB. Therefore
h v
=
(suc-
by
h suB)
rotating
MC
and
Chapter 3.
98
axis is parallel to the side BC (Figure 118). Then the equal to the volume vBCDE of the cylinder, obtained by
The
(3)
v is
volume
SOLIDS
ROUND
rotating the rectangleBCDE, minus the sum of the volumes VAEB and VADC of the cones, obtained by rotating /\177AEB and /\177ADC. Since bases of these solids have the same radius equal h, we find:
1 =
v\177CDZ
' ED,
\177rh2
VAZ\177
=
\365\177rh2
VADC = l\177rh2 3
' AE,
' AD,
and therefore
v
=
The by
\177rh2
_ \177AE_
(ED
product
2\177rh
the
rotating
Thus
BC.
\177rh2
v
=
_
(ED
the area
expresses
\337 ED
side
=
\177AD)
\177ED
=
)
\177rh
2
2 .ED.
s of the surfaceobtained
sh/3. '
B
A
o A
c D
C
Figure
F
Figure
118
119
115. Volumes of spherical sectors. Given a spherical sector, obtained by rotating a disk sector(AOD Figure 119) about a diameter (EF), we inscribeinto the arc (AD) of the disk sector a regular brokenline (ABCD), and form a solid of revolution by rotating about the same diameter the polygon (OABCD), bounded by the broken line and by the extreme radii (OA and OD). The volume of the spherical sector is defined to be the limit to which the volume of the solidof revolution tends as the number of sides of the regular broken line increases indefinitely. 116. Theorem. The volume of a spherical sector is equal to the product of the of its base and a third of
surfacearea
the radius.
Let AOD
the
(Figure
spherical
119)
sector
about
be obtained
the
diameter
by rotating
the disk sector
EF, and let
ABCDbe a
99 any number of sides inscribedinto the arc volume of the solidobtainedby rotating the This volume is the Sum of the volumes obtained
V the
by
Denote
AD.
with
line
broken
regular
polygon OABCD. .by rotating the triangles
OAB, OBC, etc.
each of these volumes,we apply the the altitudes of the trianglesare congruent
regular brokenline. Thus
we
where we denote by
obtained ABCD
by
the
a -- S
+'''
SBC\177
3,
S the areas of surfaces BC, etc. and the whole brokenline and by
etc.
8BC,
AB,
sides
rotating
To that
a of
apothem
the
a
8AB\177 q-
8AB,
to
note
and
\365114
have:
a --
V
of
EF.
axis
the
about
lemma
respectively.
of sides of the brokenline increases tends to the radiusR, and the area S tends to a limit s (see \365110), equal to the lateral surface area of the sphericalfrustum (or spherical segment) formed by rotating the are AD. Thelatter surface is the base of the spherical sector. Therefore the volume v of the spherical sector is given by the formula: the number the apothem a
that
now
Imagine
Then
indefinitely.
V
of V
limit
---
R
= s--.
3
Notice that this remains valid of the disk sector with the axis of 117. The volume of the ball. the extreme case of a semi-disk rotated argument
coincide
if one or even revolution.
corollary.
the following
Corollary 1. surfacearea a and
where
frustum.
As
of the
seen
spherical
R denotes the radius Using this expression
volume
the entire
have:
volume
of a
\365111,the lateral surface area frustum is given by the formula
and h the altitudeof the for
spherical sector
the
equal to the product of its
area
surface
the
s of
segment
the base
find:
ball, when the
of
ball is
of a
radius.
have
we
or
of a sphericalsector,we
For
volume
third
segment
spherical
2\177rRh
or
The
2.
Corollary
of a
diameter,
its
about
result to we obtain
previous
the
Applying
both radii
ball
=
2\177rRh.
__R
=
3
altitude
h
is
a diameter
_2\177rR2h.
3
D =
2R, we
Fromthis,it is
congruent
the volume v
tThus
ball.
this
47rR 2
=
\177t
of this cylinder are respectively:
+ 27rR
2=
2 and \365v
=
2R
27rR.
volume
and the
area
of
respectively
volume
the
the cylinder circumscribed about a ball has the altitude to the diameter of the ball, and the radius of the base to the radius of the ball. Therefore the total surface area
Indeed, congruent t and
total surface area and
circumscribedabout
the cylinder
of a ball are equal
the volume
and
area
surface
The
of the
thirds
two
same ratio as
have the
or diameters.
radii
Corollary 3. to
of balls
volumes
that
evident
of their
cubes
the
3. ROUND SOLIDS
Chapter
100
\1777r_R
3,
v --
and
7rR 2
with
coincide
which
=
\337 2R
27rR 3.
the surface
respectively.
ball
the
of
67rR2,
(1) Corollary 3 was proved in the 3rd century of Syracuse. Being very fond of this result, Archimedes requested that when he was dead, it would be inscribed on his tomb. This request was honored by the Roman general Marcellus, whosesoldierkilled Archimedes during the capture of Syracuse in 212 B.C. A famous Roman statesman Cicero, who lived in the 1st century A.D., describes how he managed to locate Archimedes' forgottengrave by looking for a monument featuring a sphereand a cylinder, and was even able to read the versecarvedon the tomb. (2) As a useful exercise, we suggestthe readerprove that the surface area and the volume of a ball are equal to \177- of the total surface and the volume respectively of the circumscribed cone, whose generatrix is congruent to the diameterof the base. Combining this proposition with Corollary 3, we can write the following identity, where Q stands for either surface area or volume: 118. Remarks. Archimedes
by
B.C.
I\177cylinder
I\177ball
(3) The plane
whose
formula
120),
of the
diameter
the cylinder the
that
from
midpoint
the
of a
ball can
same
plane parallel
volume
be easily derivedfrom of radius R onto a
a ball
place
Indeed,
onto the sameplane base and altitudeare congruent
and place
its
cylinder's base.
as the
2R
of radius
(i.e.
R).
are removed,each
axis
vertex,
The denote
to
two cones
ball. To
to the planeP, and
cylinder
the
ball
the
about
cylinder,
the
of the
the cylinder's bases for have
9
that can be circumscribed
Imagine now having
(\36566).
(Figure
P
volume
the
for
principle
Cavalieri's
(\177)cone
6
4
for
its
remaining show
the
this,
distance
and
one of
solid turns
out to
draw
any
from the
section
center
101
The ball of
ball
\177r(/\177 2-
plane by d, and the radiusof the crosssectionof the area of this crosssectionis equal to \177rr2 -The cross section of the solidremaining from the cylinder
to this
ball
the
the
by
d2).
Then
r.
ring whose exterior circlehas radius /\177 and interior d (since the angle between the axisof the cone and its generatrices is 45\370). Therefore the area Ibf the ring is equal to \177r/\177 2 - \177rd2 7r(/\177 2 d2). We see that cross sectionsof both solidsby planes parallel to P have the same areas, and hencethe solids, according to Cavalieri's principle, have the samevolume. For the solid remaining from the cylinder, this volume is equalto the volume of the cylinder less twice is a
the volume of the cone:
1 \177rR \177.
which thus
2R-
2.
2
\365\177rR
\337 R
expresses the volume
Figure
(4)
simple
The
2\177rR \177-
of
the
ball
3
-2\177rR\177
of a ball can alsobe derived not quite rigorous)argument. of the ball is partitioned into very
volume
a large number of smallsolids.Each (approximately) as a pyramid with the
of
them
vertex
by
the
following
Imagine
that
small
connected
pieces, by
the
and radii
partitioned into can be considered at the center. The
to the product of the area of its base (which can be taken to be the radius R the ball). Therefore the volume v of the ball, equal to the sum of
volume of a pyramid of
R.
120
that all points on the contourof eachpieceare to the center of the ball. Thus the ball becomes
and
3
=
of radius
(although surface
entire
=
a
third
of the
is
equal
altitude
102 pyramids,
of the
volumes
the
Chapter3. is expressedas:
SOLIDS
ROUND
1 where
s denotes
But
this sum
the sum of the areas of the basesof all the pyramids. constitutes the surfaceareaofthe ball,i.e. v
Thus the
area.
3
expressed through its surface
can be
the sphere
ball:
of the
volume
3
s of
area
the
_4rRa'
2- R
can be
ball
the
of
volume
Conversely,
4rR
=
and hences =
4\177rR
the
from
found
2.
fact the above argument applies to any part of the ball a solid angle.On the surface of the ball, consider a region B boundedby a closed curve C, and connect all points of this curve with the center of the ball by radii. The resultingsurface(it is a In
(5)
known as
in
surface
conical
general
the
sense of
\36592,
directfix and the radiias generatrices) called a solid anglewith the base
with
bounds
curve
the the
part
C as of the
the ball
B and the vertex at the center. pieces and following the argument in Remark 4, we conclude that the volume V of the part of the solid angleinsidethe ballof radius R is related to the area S of the base by
V =
formula:
the
many small
into
base
the
Dividing
SR/3.
EXERCISES
190. Find the geometric locusof the
projections
of a
given point in
space to planespassingthrough another given point. 191. Find the geometric locusof the centersof the cross sections of a given ball by planes containinga given line. Consider separately the cases when the line intersects the ball, is tangent to it, or does
not intersectit.
192.
the
Find
geometric
of a ball by planes separatelythe cases outside ball. given
when
of the
locus
passing the
centers of the cross sections a given point. Consider lies inside, on the surface, or
through
point
the
193. On the surface of a given ball, find tangency points with linesdrawn from a
and
tangent
to
the
sphere.
of the
the
geometric
locus
given
point
outside the
ball
The
2.
103
b\17711
195. lineofcenters. (i.e. one Derive that
one
ball
a common
have
spheres
tangent
if
point
given
a
outside
are congruent to
each
of two balls is calledtheir are tangent to eachother
then the
extension,
(or its
of centers
point)
common
only
have
line
the
if two
that
Prove
centers spheres
the
connecting
segment
a
from
endpoint
at their
it
to
other.
The
drawn
all segments
that
Prove
19J.
given ball and tangent
tangency point lieson the other). plane at their
lies inside
tangent
tangencypoint,and this planeis perpendicular to line of centers. 196. Prove that two spheres are if and only if their line of centers is congruentto sum or difference of their radii. 197. Provethat in a tetrahedron, the three sums of pairs of opposite are congruent, then the vertices are the centersof the
tangent
the
if
edges
four
to each
tangent
pairwise
balls
198.* Provethat
tangent balls, of of these then
all
point.
sufficient
and
necessary
a
Find
the
pairs
tangency
199.
tetrahedron are centers of pairwise six common tangent planes at the points of balls pass through the same of a
vertices
if
other.
condition
for all edges of
a
same ball. 200. A sphere tangent to all faces of a polyhedron or polyhedral angle is called inscribed into it. Find the geometric locusof the centers of spheres inscribed into a given trihedral angle. 201. Find the geometric locus of points in space equidistant from to
tetrahedron
points.
three
given
202.
On a given
of ballsof a fixed given
203.*
of
Prove
a tangent
locus
are tangent
of points of tangency to the plane and to a
to it.
find the with spheres
plane, plane
that
geometric locus of the points of passing through two given points
polyhedron
205.In a
trihedral
tangent
spheres
possesses a unique inscribedand sphere. (A sphere is called circumscribed if all of its vertices lie on the sphere.) angle all of whose plane angles are right, two each other are inscribed. Computethe ratioof
any
circumscribed
about a their
which
the plane.
outside
unique
this
geometric
the
r,
radius
given
a
On
tangency
20\177.
plane, find
of radius
ball
to the
tangent
be
to
tetrahedron
radii.
206. Prove that about any regular pyramid, circumscribed, and its centerlieson the altitude. 207. Prove that into any regular pyramid, inscribed, and its center lieson the altitude.
a
a
unique
unique
ball
can be
ball
can
be
3.
Chapter
104
208. Computethe planeangle
lar
the
if
pyramid,
coincide.
209.
Prove that
210.*
of the
vertex of a regular quadranguinscribed and circumscribedballs
radius
of the
ball circumscribed
about a cube
211.* Provethat if
polyhedral angles a circle, then the
by a sphere. prism, a ball can be inscribed, then the prism is equal to two thirds of the total
a
into
of the
area
surface
can
face
be circumscribed
can
lateral
all of whose be circumscribed by
a polyhedron,
in
if
each
trihedral,
polyhedron
the
at
Im.
is
side
are
centers
the
Compute
whose
ROUND SOLIDS
surface area.
;212,.Prove that into a conicalfrustum, a sphere can be inscribed if and only if the sum of the radii of the bases is congruent to the generatrix.
that into a conicalfrustum,
;215.* Prove
and only if
of
altitude
the
a sphere
is the
geometric mean
M\177rs
is
can
be inscribed
between the diameters
bases.
the
of
diameter
The
2,1,\177.
times
the surfacearea 2,15. tall cylindrical
vessel with By how much
A
ballof
water.
with
half-filled radius
2,16. A
hollow
3
iron in
the radius of the base 6 cra is will the water level rise after a
vessel? ball of radius 15.5 cm is floating in water being How thick is the shell, if the specific gravity of
is sunk
cm
half-submersed it. iron is
volume
the
and
as small, and of Jupiter is 11 of Earth. By how many times do of Jupiter exceed those of Mars?
twice
diameter
the
than
greater,
in the
7.75 g/cvr\1773?
217.
Compute
tum,
whose
and 18cm.
218.
Given
the volume and lateral surfaceareaof a conical generatrix is 21 cm, and the radii of the bases are of radius
a ball
to a cross sectionplane,if
smaller
spherical
segment,
113 cm, the
ratio
cut
out
find
the
distance
from
frus-
27 cm
the center
of the lateral surface area of the by the plane, to the lateralsurface
cone, which has the same base as the spherical vertex at the centerof the ball,is 7: 4. 219.Along a diameter of a ball of radius 2 cm, a cylindrical radius I cm is drilled. Computethe volume of the remaining the ball. area
if
of the
segment
and the
2,20. Computethe basehas radius r =
221.
A
disk
inscribed
volume 5
cm,
into
hole
of
part
of
into a cone whose 13 cm long. an equilateral triangle. Compute the ratio of
and
ball generatrix
the
inscribed is 1 =
2.
105
ball
The
volumes of the solidsobtainedby rotating the disk and triangle about an altitude of the triangle. 222.Prove that the volume of a polyhedron circumscribed about a ballis equal to one third of the product of the radiusofthe balland the surface area of the polyhedron. 223. A ball is tangent to all edges of a regulartetrahedron.Which of the solids has greater: (a) volume, (b) surfacearea? 224\177. Compute the volume and surface area of the solidwhich consists of all those points in space whose distance from (at least one point inside 'or on the surfaceof) a given cube with the edge a does not of the
exceedr.
225.* A
of all
consists
solid
points in space
whose distancefrom
(at
one point inside or on the boundary of) a given polygon of area s and semiperimeter p doesnot exceed r. Prove that the volume V and the surfacearea \177qof the solid are given by the formulas:
least
4 V
Prove
226.
- 2rs that
\177rr2p
area
generatrix
whose
+
the surface
of the total surface
cone
+
3
S
2\177rrp
+
4\177rr \177.
area and volume volume
and
2s +
=
\365\177rr,
is congruent
of a ball are equal to 4/9 respectively of the circumscribed to the diameter of the base.
Prove that the volume of a sphericalsegmentis equal volume of the cylinder, whose base has the radius congruent altitude of the segment, and whosealtitudeis congruent to the of the ball less a third of the altitudeof the segment, i.e. 227.*
v denotes the volume of the segment,h its radius of the ball. 228.\177Prove that the volume of a spherical frustum
where
formula: where
h
is the
altitude
= =T + + of the frustum, and rl
229. a polyhedral angle, intersect centeredat the vertex, show that sphere inside the angleis
to
proportional
\177q:
/\1772
of the polyhedral angle (or,more
the
R the by the
can
be
and
r2
are
the
radii of
the
and
The ratio
given
to
radius
it with a sphere of radius R area \177qof the part of the
Given
Remark.
is
the
h
h3
the bases.
and
altitude,
to
R 2.
considered
generally,
any
as some solid
measure
angle).
Chapter 3. ROUNDSOLIDS
106
230. 231. with
Compute
the
Compute
the
the
generatrix
polyhedral angles of the cube. measure of the solid angle at the vertexof a cone I and altitude h. measure
of the
4
Chapter
AND
VECTORS
FOUNDATIONS I Algebraicoperations
vectors
with
of vectors.
Definition
119.
distances,
volumes,
terized by
their
magnitudes,
In physics, somequantities (e.g. are completely characwith respect to a chosenunit
masses)
or
temperatures,
expressed
quantities are called scalars. Someothers (e.g. velocities, accelerations, or forces) cannot be characterizedonly by their magnitudes, because they may differ also by their directions. Such quantities are calledvectors.
by
numbers.
real
These
Figure
121
quantity geometrically,we draw an arrow space, e.g. A with B (Figure 121). We call it a directed segment with the tail A and head B, and indicate > To
this
in
a vector two points in
represent
connecting
writing
as AB.
107
Chapter 4.
108
FOUNDATIONS
AND
VECTORS
can be representedby different directed segments. By definition, two directed segments (A/} and C/\177) represent the same vector if they are obtained from each other by translation (\36572). In other words, the directed segments must have the same length, lie on the same line or two parallel lines, and point toward the same direction (out of two possible ones). When this is the case,we write AB = CD and say that the vectorsrepresented by these directed segments are equal. Note that A/} = C/\177 exactly when the quadrilateral ABDC is a parallelogram. We will also denote a vector by a single lower case letterwith an arrow over it, e.g. the vector \1777(Figure 121). 120. Addition of vectors. Given two vectors \1777and F, their sum \1777q-F is defined as follows. Pick any directed segment [4/} (Figure 122) representing the vector \177Y. Represent the vector F by the directed segment BC whose tail B coincides with the head of AB. Then the directed segment AC represents the sum \177+ F. vector
same
The
C A
122
Figure
thus defined does not dependon the choice representing the vector \1777. Indeed, if andirected segment A'B' is chosento represent \177Y,and respectively >
The of
the
other the
sum of vectors directed segment
directed
quadrilaterals the segments
B'C
segment \177 and
AA
directionas same
BB\177),
too. Thus
vector.
CC are
they
and
\177(with
\177and
ABB\177A
direction(since ogram
u+v
\177+\177
\177 are
\177are
congruent,
congruent hence
tail
the
BCC\177B
the
B \177) represents parallelograms.
F,
then the Therefore
and have the same to, and have the same ACCRA \177is a parallel-
parallel,
to, parallel
quadrilateral
the directedsegmentsA\177
and
A\177C \177represent
the
operations with
1. Algebraic on the
tail
\177iand
\276.
directed segments AB and AD
the vectors by A (Figure 123).
represent
same
the
In the planeof the
AD and DC AB, and denoteby C their intersection Then ABCD is a parallelogram, and hence DC: \177, BC = the diagonal AC of the parallelogramis a directedsegment
point. Therefore
both
representing
+
ff
\276+
\276 and
A
A \177
A
B
\177
Figure 124
Addition of vectors is more) vectors does are performed: (\177l +
+
\177
AC, and
AD
order
in
\177+
BD'
= AC
)
= ,>
AD'
vectors
all
\1777, \177Y,and
t\177 by
(or additions
three
of
sum
the
in
the
which
\1777,\177l, and
\177.
segments
directed
directed segment AA' represents of the summands or the are performed. For instance,
of the
= AB +
\276 and ,,,>
+ D'A'
AB,
A (Figure 124), and then construct all of whose edges are parallelto
the diagonal
+ BA'
AB
=
for
\177)
tail
same
additions
the
which
=
+
\276+ t\177 regardless
AA'
AA'
(ff
vectors
Then
AD.
or
AC,
sum
\276+
ABCDA'B'C'D'
parallelepiped
AB,
=
the
with
associative, i.e.
not dependon the order
the
represent
Indeed,
since
\276)
B
\177
123
Figure
\177.
\177.
B'
the
with
ABD,
triangle
lines BC
draw
the
de-
not
vectors
all
for
Indeed,
i.e. the sum does
of vectors is commutative, order of the summands:
Addition pend
109
vectors
ordering
(BD'+ D'A'):
= AD
D'A' )
=
\177.
At
the
\177i+
(\276
same
+
t\177),
time,
>
= (AB + BD') + D'A'=
(\177i +
\276)
+
VECTORS AND FOUNDATIONS
4.
Chapter
110
121.Multiplicationof vectors
apply
efficient
0
\177 \177
respect
with
of the
the vector
represents
A'B'
directed
segment
since the
triangles SAB and SA'B\177are
vector
scalar and the vector.
AB (Fig-
segment
directed
any
\1777by
a scalar
a new
form
can
one
homothety (see \365\36570-72) with to any center S. Then the
it the
to
\177,
product vector
the
represent
Namely,
ure 125) and
a vector
Given
scalars.
by
real number) c\177and denoted \177 and called the (i.e. a
directed
the
co-
words,
other
In
\1777.
similar,
the resulting
segment
to AB (or lies on the same has the samedirectionas AB when c\177is positive, and the opposite direction when c\177is negative. We will often call vectors \1777and \177 proportional and refer to
line),is
[\177[
vector
the
A\177B \177 representing
the numberc\177as
AB, and
of proportionality.
coefficient
the
parallel
is
\177
than
longer
times
A
B'
S
\177
B
A'
Figure 126
125
Figure
In the specialcaseof a any directed
segment SS
= 0,the
responding vector is calledthe
represented
by
The coris denoted by \177.
coincide.
head
and
vector
zero
is
0ff
product
and
tail
whose
Thus
0ff =
Multiplication
of vectors,i.e. for
scalars
by
all
for
\177
\1777and
any
and
centel
let S)
A'B'
/kA\177B\177C
with =
\177be
c\1777,
B'C'
ABC the
homothetic
the homothety =
every
\276 and
Indeed, let the sidesof the triangle respectively: AB the vector \177, BC \17774-\276,
with respect to addition scalar a we have:
is distributive
vectors
and
(Figure
126) represent
vector \276, and AC to /kABC (with
coefficient
c\177\276,
if.
vector
every
A'C'
\177.
=
Then
c\177(\1777 +
\276).
their
sum
respect to
Since
A'B'
A\177C \177 =
Two more
let OU
deed,
=
law
distributivity
c\1777
+ fig,
and
meaning of The
127).
\1777(Figure
line (see
number
the
with
\177)\1777=
geometric
the
from
'\177,the
follows.
properties1 of multiplication of vectors
(c\177 +
follow
+ B'C
111
vectors
with
operations
Algebraic
Book I,
(c\1773)g
=
c\177(3g)
operations with
infinite \177153)
by
scalars:
by
line OU
can be identified segment
the
taking
In-
numbers.
OU
for the unit of length and letting the points O and U representthe numbers 0 and I respectively. Then any scalar a is> represented on this number line by a unique point A such that OA = c\177OU. Furthermore, addition of vectors on the line and their multiplication by scalars is expressed as addition and multiplication of the corre: sponding numbers. For example, if B is another point on the line such that OB = \177OU, then the vector sum OA> + OB >correspondsto the number a+/\177 on the number line, i.e. aOU+\177OU > = (a+\177)OU. Similarly, multiplying by a scalar a the vectorOB (corresponding to \177 on the number line) we obtain a new vector corresponding to the product c\177/\177 of the numbers, i.e. a(/\177OU)= (a/\177)OU. Examples. (1) If a vector \1777is represented by a directed segment
A/}, then 'the oppositedirected segment BA represents (-1)\177,also denoted simply by -\1777. Note that opposite add up to 0. This is obvious from AB+BAAA-follows formally from the distributivity law:
(-1 + (2) If vectors\1777and
F are
a common
the heads
connecting
BC' =
with
AC
and
AB
1)\177:
(--1)\177,
represented
+
1\177= -g,
by the
tail, then the
the vector vectors 0\177
but
also
+ 5.
directed segments
directed
represents the difference F-\1777
segment
BC AB
(because
+
AC').
In applicationsof vector algebrato geometry, it is often convenient to represent all vectorsby directed segments with a common tail, called the origin, which can be chosen arbitrarily. Once an origin O is chosen,eachpoint A in space becomes represented by a unique vector, OA, called the radius-vector of the point A with respect XThey
to the are
associativity
origin O.
called: with
distributivity with to multiplication
respect
respect to addition by scalars.
of
scalars,
and
Chapter 4. VECTORSAND
112
the radius-vector
Compute
Problem.
122.
of a triangle ABC of its vertices.
(Figure 128),
\177
5,
Recall that the barycenteris the intersection point ans. Denote A' the midpointof the side BC, and AA \177mark the point M which divides the medianin AM' MA\177= 2' 1. We have: A/\177 = \177- 5, B\177 = 5AA
\177 =
\177,
medi-
of the
median
the
on
5
and
proportion
the
\177, and
hence
2
+ 2
AB
barycenter
the
of
radius-vectors
the
given
FOUNDATIONS
Therefore >
\177
>
the
Clearly,
a'+
\177
result
same
1
1
\177
OM-OA+AM=OA+-2AA
remains true
dians. Thus the headofthe radius-vector
other
of the
each
for
_\177
\177(a +
\276- 2\177')
(\177'+
g+
c-*)
two me-
1
,
all the three mediansand As a by-product,we have obtained lies on
theorem(BookI, point
dividing
each
\365142):
a new medians
three
the
proportion 2: 1 counting
in the
of them
with the barycenter. proof of the concurrency of a triangle meet at the
coincides
thus
from
the
vertex. c
Figure 127
123.
The
Given two
dot product.
if. F (also known as the productof the lengths
product
as
of
the
anglebetweentheirdirections.
if
Thus,
\276=
OU
. OV
vectors
product) vectors
scalar
by the directedsegmentsOU and \276.
128
Figure
OV
the
\276 and
F, their
dot
is a number defined and the cosine of the vectors are represented
(Figure
. cosXVOU.
129),
then
1. Algebraic operationswith
113
vectors
the dot product of a vector with square of the veetor'slength:
In particular,
g.ff= g2, or If
0(g,
ffl=x/-\177'ff.
the directions of two
between
angle
the
denotes
\276)
to the
is equal
itself
non-zero
vectors,then
cos0(g,3) = g 3' information about dis-
captures
operation
product
dot
the
Thus,
tanees
angles.
and
v
\1777
U
-\177
0
0
Figure 129
let us
Now
signed projection
define the
direction of a unit
vector 3, i.e.
rected segmentsAB
and
A and
through
until
line
OV with
planes P and Q perpendicular to the
B the
the number line, we
points by numbersc\177and
3,
their
does
3
difference
-
segments
representing
product:
3-
plane
c\177.
parallel planes),and if
the
ever
of
direction \177BA\177'
obtuse.
We
is find
and not
the
represent
introduce depend
positions
line
the
Identifying
of these
as
is equal to their dot through the point A
extend it to the intersection pointC with the segments of parallel lines between two
A'B \177(as
=
\177BAO )
and
therefore
0(\177,
the
that
3
-
-
c\177is >
positive
of OV, i.e. wheni.e. when the angleis
direction
otherwise,
negative
of 3
sign
The
3).
with
A\177B \177agrees
acute,
\177.
the signed projection on the choice of directed
the vectors because it 3). Indeed, draw
OU, and Then AC =
Q.
It
B
\177and
gcos0(g,
c\177=
A6'
line
the
the points A
it at
intersect
they
OV
130) represent the two vectors. B to the line OV. Forthis,
A and
points
the
1.
=
13
the Let di-
g to
vector
any
of
that
assuming
(Figure
OV
Considerprojections of
draw
130
Figure
c\177 =
AB
\337 cosXBA6'
=
\177.
3.
4.
Chapter
114
operation:
is equal
vector
product
dot
the
to the
the dot product
geometric interpretation of of any vector with a unit signed projectionof the former to the direction following
the
obtain
We
AND FOUNDATIONS
VECTORS
of the
latter.
degree
1)
124. Algebraic propertiesof the product. The dot product operation is symmetric: dot
(1)
\1777. \276 =
because
respect
It
obviously
cos
dot to either
product vector,
The
(2)
from it due to the c\1777
is
0(\1777, \276)
=
\1777. \177
i.e. for
make
c\1777
opposite
the
with
vector
opposite,and the
\276 are
(of
of \1777. Therefore in
(Figure
c\177,
so that
supplementary, remains
property
the
131). Then
\177
one follows The length of
case the
this
of negative
case
the
second
product).
with
a
scalar
any
and
the
(as
dot
the
c\177 since
directions
\276,
\1777).
all vectors3, \276,
the length.
than
equality
0(\276,
\1777and
homogeneous
of
for positive (or zero)
have
vectors
equality only
of these vectors coincide.gin and
all
cos is
symmetricity
greater
c\177] times
is obvious
for
the first
to verify
suffices
\276. \1777
directions
the
vectors
\1777
the anglesthey their cosines are
true\177
B
W
Figure
Figure 132
131
(3) The dot productis additivewith and
(\177+\177)-\177-\177.\177+\177.\177
for only
all vectors the
first
\1777, \177F, and
equality.
\177.
to
Due
We may
respect
to each
\177.(\177+\177)=\177.\177+\177.\177
symmetricity,
it suffices
assume that \177 \177 \177(because
al] three terms vanish). Due to the length of the vector \177, one
vectors:
of the
homogeneity,
reduces
the
to
verify
otherwise
dividing
each
equa]ity
to its
term
by
special
Algebraicoperations case
when
AB
=
O' the
=
tU
,
BO=
\177>
115
vectors
with
1.
Let ABO
\276,
and
(Figure 132) be a AO=
hence
projections of>the points A, =
OW
segment
directed
representthe projectionpoints
O to
and
B,
numbers
by
B',
byA',
the line of the the number
line as
this
Considering
tU.
such
triangle
Denote
\1777+\276.
and
unit line,
Then
7.
and
c\177, \177,
that
and
a) +
- a, the requiredequality holds. 125. Examples. Some applications of the dot productoperation based on the simplicity of its algebraicproperties. (1)Perpendicular vectors have zero dot product because cos 90\370= Therefore, if we denote by ff and \276the legs AB and BC of a right
Since (/S are
0.
= '7
-/S)
(7
triangle ABC, its hypotenuse its length is computedas
is
AC
then
+
ff
\276,
the
and
square
of
follows:
\1777- \276 =
since
0.
\276. \1777=
AC 2 =
Thus
AB2
+
BC 2,
and so
we have
re-provedthe Pythagoreantheoremonce
again.
given any
generally,
More
(2)
triangle ABC, put
\276 =
AB,
\276=
AC
and compute:
=
v.v_ v.4_ 4.v+
(v- 4). (v-
2=
\1777. \17774-
This is the
law
\276. \276-
2\1777.
g =
(Book
cosines
of
AB 2 4- AO 2 - 2AB. I,
AO. cos
ZBAO.
\365205).
EXERCISES
232.
Prove
that for every
closed brokenline ABODE,
+ BO
AB
+
... + DE + EA
=
6.
unit vectors is equal to 6, then of these vectorsis equalto 120 \370 . 234. Prove that if four unit vectors 15;ingin the same planeadd up to 6 then they form two pairs of oppositevectors.Doesthis remain true if the vectors do not have to lie in the same plane? 235.* Let ABODE be a regularpolygon with the center O. Prove that 233.
the
Prove that
angle
between
if
the
sum
of three
each pair
OA + OB +...
+
OE
= O.
Chapter 4. VECTORSAND
116
FOUNDATIONS
236. Along three circles lying in the sameplane,vertices are moving clockwise with the equalconstantangular out how the baryccntcr of the triangle is moving. 237.
that
Prove
if AA \177is
nn'=
a triangle Find
triangle ABC, then
in a
a median
of velocities.
+
(hA
that from segments congruentto the medians of a given another triangle can be formed. 239. Sides of one triangle are parallel to the medians of another. Prove that the medians of the latter triangle are parallel to the sides of the former one. 240. From medians of a given triangle, a new triangle is formed, and from its medians, yet another triangle is formed. Provethat the third triangleis similar to the first one, and find the coefficient of 238.
Prove
triangle,
similarity.
of AB and CD, and of BC and DE are connected segments, whosemidpointsare alsoconnected.Prove that the resulting segment is parallel to AE and congruent to \254AE. 242. Prove that a point X lieson the line AB if and only if for some scalar c\177and any origin O the radius-vectors satisfy the equation: 241.
Midpoints
by two
243.
Prove
then
I ff
244.
For
=
that
\177OA +
=
OX
vectors
if the
ff
(1
-
\177)OB.
+
\276 and
\177,
verify
\177-
\276 are
perpendicular,
\276.
vectors
arbitrary
\177 and
the theorem: the sum
the
equality:
of the diagonals of a squares of the sides. 245. Prove that for every triangle ABC and every point X in space, XA . BC + XB . CA + XC . AB = O. 246.* For four arbitrary points A, B, C, and D in space, prove that if the lines AC and BD are perpendicular, then AB 2 + CD 2 = BC2 + DA 2, and vice versa. 24 7. Given a quadrilateral with perpendicular diagonals, show that every quadrilateral, whose sides are respectively congruent to the sides of the given one, has perpendicular diagonals. 248.A regular triangle ABC is inscribed into a circleof radius R. Provethat for every point X of this circle, XA 2 +XB2+XC 9\177 = 6R 2. and derive
parallelogramis
equal
to
the
sum
of
the
of the
squares
2. Applications of
249.*
Let
A\177B\177A2B2...
Provethat the not
exceed
117
vectors to geometry A\177B\177
of
length
the
inscribed
a 2n-gon
be
vector
A\177B\177
+
+ A2B2
into a
\337 \"+
circle. does
A\177B\177
the diameter.
Consider projectionsof the verticesto any line. 250.* A polyhedron is filled with air under pressure. The pressure force to each face is the vectorperpendicular to the face, proportional to the area of the face,and directed to the exterior of the polyhedron. Prove that the sum of thesevectors equals 0. Hint' Take the dot-product with an arbitrary unit vector,and use
Hint'
Corollary2 of 2
\36565.
126.
the
If
Theorem.
is
(ABC)
the
5respectively.
radius-vectors
Then
H be the point in It
is required
the
plane
productCH-AB.
and AB = OB -
to
equal
c\177,
of the
the
vertices A, B,
of the
5 = b[ =
to show
Figure 133) of a then the radiussum of the radius-
circumcenter (O, the origin,
triangle chosen for vector of the orthocenteris vectors of the vertices. Denote
to geometry
of vectors
Applications
since
O
is the
triangle such
that H is the
orthocenter.
and C by
5,
\177, and
Let
circumcenter.
that OH =
5+
\177+
cZ
dot
the
Compute
SinceCH=OH-OC=(5+b+c\177-5=5+\177, OA = b- 5, we find:
CH.AB:(\177+\177').(b-g):\177.\177'-\177.a: \177'2_
g2=O
'
Figure 133
Vanishing of the dot product of two vectors means vectors are perpendicular (unlessone of themis zero, in
that which
these case
4. VECTORS AND
Chapter
118
FOUNDATIONS
angle is, strictly speaking, ill-defined). \177,Ve conclude that the line CH is perpendicular to AB (unless H and C coincide), i.e. (in either case)the point H lies on the altitude dropped from the vertex C to the side AB, or on the extensionof this altitude. Since the same applies to each of the other two altitudes, it follows that the three altitudes, or their extensions, pass through the point H. Corollaries. (1) We have obtained a new proof of the theorem (BookI, \177141)' Altitudes of a triangle are concurrent. (2) In every triangle, the circumcenter O, barycenter \177I, and orthocenter H are collinear. More precisely, M divides the segment OH in the proportion OM \337MH -- 1 \3372. Indeed, according to \177122\177 the
have:
we
1
1
>
+ g
= The
(1)
Remarks.
ter,
BookI,
: OH.
+
containing the circumcenter, barycenEuler\177s line of the triangle (seealso
segment is called
orthocenter
and
>
226-228).
Exercises
(2) In the previous theo.}'em,operations with formulate new results (andre-proveold ones) of plane geometry. In the nextexample, vectors
although the formulation at all. In such situations, can
interest
be
context
the
from
the origin, it
the
of
in
allow
vectors
familiar
about
does not involve any vectors to apply vector algebra, points of
problem
order
represented by their which of the given
If it is unclear
radius-vectors.
points should play
is advisable (althoughnot
to
necessary)
the
127. Problem. A
A\177B\177C \177is
respect
with
to
respect
to
the
the
center
/kABC
to
a
Given
in such
drawn
the
center
triangle
ABC
center C, C \177is centrally A, and then the triangle
from/kA\177B\177C\177
by
(Figure
a way that A \177is B, B \177is centrally
straightedge
depend
to C
symmetric
erased.
trianto
symmetric
to B
symmetric
ABC is and
not
134), a new
centrally
of
role
making point not
avoid
any artificial choice. Instead, one can chose an arbitrary mentioned in the problem -- the resultingconclusionwill on this choice.
gle
useful
to be
out
turn
one to notions
with
with
respect
Reconstruct
compass.
arbitrary point O as the origin, and denote by 5, c\177, g, of the points A, A \177, B, etc. If two points are centrally symmetricwith respect to a center\177 then the radiusvector of the center is equalto the average of the radius-vectors of the points. Therefore\177 fi'om the hypotheses of the problem, we have: Pick an
etc.
the
radius-vectors
2 (a'+a')'
2
2.
5,
2,
that
Assuming for
vectors
of
Applications
5. For
\177,and
\177,
\177o geometry
c' areffiven,
and
this, replace b in
1st, and substitute 3rd equation. We obtain: the
from
solve this system of equations the 2nd equationby its expression
the resultingexpression
5
for
into
the
Therefore,
1 75=1\177+
1\177
5--
or
g;+
1\177
+
2
4\177 \177+
>
segment
OA representing
construct
by straightedge
directed
The
to
hard
not
directed
given
B and
vertices
the last vector expression is and compass, startin>g from - OB', and \177 -- OC( The
segments a' - OA', U C of AABC' can be constructed 1\177+
and
\177+
Figure
5=
using
the
+
+
expressions:
.
134
of mass. By a material pointwe will mean a equipped with a mass, which can be any real number. Unless the oppositeis specified, we will assume all masses positive? 128.
The
The
center
in space
point
notion
following
is borrowed
from physics.
Given a system of n material points A1, A2,..., An of masses ml, m2,..., ran, their center of mass(or barycenter) is the material point whose mass m is equal to the total mass of the system:
m=ml +m2+...+mn, 2When
both
positive
pseudo-masses.
and negative
masses occur,we
refer
to the
latter ones as
.
Chapter4.
120 the position
and
+
m\177AA\177
words, the
In other equal
of
the
the
to
respect
-
of the
sum
weighted
the materialpoints with is
... 4- m\177AA\177
+
m\177AA\177
above
condition:
by the
determined
is
A
FOUNDATIONs
AND
VECTORS
6'.
radius-vectors
of mass
center
of
as the origin
to zero.
With respect to an arbitrary originO, the center of mass can be computedin terms
4, \177,...,a\177, of
points.
the
radius-vector
of the
\177
radius-vectors
have:
\177Ve
therefore
and
1
+... +
+
This centerof
any
of n
system
m
Examples. (1) In a m\177AA\177
on
lies
A
the
segment
-
AA\177
A\177A2
negative
masses,
non-zero).
material
two
of
system
or equivalently,
= 0,
+ m2AA2
centerof mass
system is
of the
with
(even
points
m\177terial
as long as the total mass
uniqueness of the mass
and the
existence
the
establishes
formula
(Figure
points, m2 AA2. 135),
we have:
Hence the connecting m2: m\177
and divides it in the proportionA\177A: AA2: (i.e. the mass center is closer to the point of greatermass). (2)Let if, \177, and \177'b1e radius-vectors of three given material points of equal mass. Then g(ff + \177+ c-) is the radius-vector of the center of mass. Comparingwith \177122, we conclude that the center of mass coincideswith the barycenter of the triangle with vertices at the three the points,
points.
given
129. ometry
Most applications of centers of mass to geassociativity, or regrouping property.
Regrouping. rely
on their
Theorem.If a systemof material is divided into two (or more) parts, and then each part is by a \177single material point representing its center of mass, then points
replaced
the centerof mass
of
with
Say,
let
A\177,
A2
tem of five material to show that if A
and
A3, A4, A5
points \177and
A
system
resulting
the
material points coincides original system. with \177are
the
of two (or more) of mass of the
(Figure 136) be two parts of a sysm\177,..., m\177. We are required positions of the centers of mass of
masses the
center
these parts, and
with
position)
Firstly, we note that the sum m\177+
m = ral+ra2+. \337\337+ms
total mass with
to both its
mass
of
five
with
the
system
points.
material
ing
m4 + m5 are their of this pair of material
m3 +
of mass
a material point, i.e. in regard the center of mass of the whole
(as
coincides
points
and m\177'-
center
the
then
masses,
+ m2
-- ml
m'
and
respective
121
to geometry
of vectors
Applications
2.
a \177, c\177 \177, \177,...,
radius-vectors
the
to any
respect
ra
of the a5
of
coincides
\177indeed
whole system. Secondly,usthe points A \177,A \177,A\177,.. A5
origin, we find: 1
(\177---
The
J' = --
+
of this pair
of mass
center
points
material
of
has
radius-
the
vector
Thus, it coincideswith
rn
'+m '_'_ _
rn-rn
m1\177
_
A \177
Figure
A \177
rn
A4
Figure 136
135
of a given triangle with and computethe centerof mass
first. It lies
carries the mass2m. By whole system lies on the
oppositevertex
and
the
order
vertex.
rn Figure
137
each vertex
Equip
Example.
mass m (Figure 137) of one of the sides
the
mass of the
A2
A2
from
center of
of the
radius-vector
the
system.
whole
divides
Since
of grouping,
at
the
connecting
median it
in the
vertices
of that side and center of mass of the this midpoint with the
midpoint
the
theorem,
the
same
the for
proportion
2m: m counting
same we derive concurrencyof mediansonce the center
of mass is the
regardless
again.
of
AND FOUNDATIONS
VECTORS
4.
Chapter
122
130. Ceva's theorem. AA
lines
\177, BB
the vertices can B
that
if
A \177,B \177,C \177
C, C and
pairs: B and
of the
mass
only
and
if
such
masses
with
respectively,
AB
concurrent
\177 are
equipped
of
centers
become
and
BC, CA, and
CC
\177, and be
points
(Figure 138) and
ABC
triangle
the sides
on
C'
B \177, and
the
a
Given
Theorem. A',
A,
A
respectively.
A, B,
Suppose
and C are material
A
and
points,
are positions of the centersof mass of the pairs ]3 A and ]3. Then, by the regrouping property, the the whole system lies on each of the segments AA Therefore these segments are concurrent.
and C, C and A,
centerof \177,BB
of
mass
\177,and
CC
\177.
Conversely, assume that the linesAA \177,BB \177,and CC \177are concurAssign an arbitrary mass \177A = \177 to the vertex A, and then assign massesto the vertices ]3 and C so that C \177and B \177become the centers of mass of the pairs A and ]3, and A and C respectively, rent.
namely: AC
mB
AB
\177
and
C,B?,
--
\177
Then the centerof massof whole system will lie at the intersection point M of segments BB and CC On the other hand, by regrouping, it must lie on the lineconnectingthe vertex with the center of mass of the pair ]3 C. Therefore the center of mass this pair is located at the intersection point of the line AM with the
the
\177
\177.
A
and
of
A'
the side
B
C
B'
A
Figure
Corollary
(Ceva's theorem).
\177,BB
CC
\177,and
site sides,
are
if and
concurrent
AC'
C'B
vertices
with points
segments
on the oppo-
only if
BA' CB' A'C
139
In a triangle ABC, the
the
\177,connecting
C
B'
A
138
Figure
AA
B
B'A
-i.
(,)
Indeed,
concurrent, the masses:
are
lines
the
when
when rewritten in
123
geometry
of vectors to
Applications
2.
equality becomes obvious
of the
terms
mB
mc
mA
mA
mB
mc
--1.
Conversely, the relation (,) means that if one assigns masses as in the proof of the theorem,i.e. sothat mB\337 mA = AC \177' CtB and mc' mA = AB\177' BtC, then the proportion mc' mB= BAt \337ArC holds too. Therefore all three points Ct, B t, and A t are the centers of mass of the corresponding pairs of vertices. Now the concurrency property is guaranteed by the theorem.
Problem. In the lines
that
relation
the
Therefore
the
sides.
Prove
AB t =ACt, BC t = BA \177, and CA t = drawn from the vertices to the same (,) holds true, and the concurrencyfollows
circle).
corollary.
the
from
t,
segments
tangent
(as
t
of the inscribed circlewith and CC t are concurrent.
We have:
1.
Solut\361on
CB
t, BB
AA
let A', B t, and C t
139),
(Figure
ABC
triangle
a
of tangency
points
denote
Assigning masses mA -- 1/AB t -- 1/AC t, mB mc= 1/CA t= 1/CB t, we make A', B', and of mass of the corresponding pairs of vertices,and concurrency follows from the theorem.
Solut\361on 2.
1/BC
--
C
centers
t
\177the
1/BA', the
therefore
and
131. Menelaus
\177theorem.
and onlyif pseudo-masses ml, m2, and
zero
same line) if
to have ml
signs)
different
-F m2
they
Ifthe pointsare called
A3)
the
such
- 0,
-F m3
collinear,
be
A1, A2, and
points
Three
Lemma.
on the
center
can m3
(i.e.
collinear
are
A3
(they
are
lie
with non-
be equipped allowed
therefore
that m1OA1
and
then
of mass
one
-F m2OA2
can make
of the points A1
+ m3OA3
the middle and
A2
- 0. one (letit
by assigning
their masses according to the proportionm2 ' ml = A\177A3 ' A3A2. Then, for any origin O, we have: m\177OA\177+m20A2-(ml+m2)OA3 = Ij, i.e. it suffices to put m3 = -m\177 - m2. Conversely, if the required pseudo-massesexist, onemay assume (changing, if necessary, the signs of all three) that oneof them (say, m3) is negative while the other two are positive. Then m3 = -m\177 m2,
that
relation mlOA1 + m2OA2 - (m\177 + m2)OA3 = 0 means position of the center of mass of the pair of material and A2. Thus Aa lies on the segmentAiA2.
and the A3
points A1
is
the
AND FOUNDATIONS
VECTORS
4.
Chapter
124
Corollary (Menelaus'theorem.)Any points A; ure 140)lying on the sides BC, CA, and AB respectively or on their extensions, are collinear,if and only if CB'
AC'
BA'
C'B
A'C B'A
and C'
B',
(Fig-
AABC,
of
-1.
relation looks identical to (.), and it may seem puzsame relation can characterize triplesof points A', B', C' satisfying two different geometric conditions. In fact in Menelaus' theorem (see Figure 140),either one or all three of the points must lie on extensions of the sides, so that the same relation is appliedto two mutually exclusive geometric situations. Furthermore, let us identify the sides of/\177ABC with number lines by directingthemas shown on Figure 140, i.e. the side AB from A to B, BC from B to C, and CA from C to A. Then the segmentsAC', C'B, BA', etc. in the above relation can be understood as signed quantities, i.e. real numbers whose absolutevalues are equal to the lengths of the segments,and This
Remark.
how the
zling
the
determined
are
signs
BA', etc. on the respectivenumber correct
of
form
the
of the
directions
\177by the
C'B,
convention,
the
this
With
lines.
theorem is:
in Menelaus'
relation
vectors AC',
AC' BA' CB'
C'B Menelaus'
prove
To
that
always
can
we
theorem by the sigm
in Ceva's
relation
the
from
differing
thereby
(**)
-1'
-
' \177'A
A'C
a
theorem in this improved formulation,note to the vertices A, B, and C somereal
assign
a, b, and c so that C' (resp. B') becomes the center of the pair of points A and B (resp. C and A) equipped with pseudo-massesa and -b (resp. c and -a). Namely,it suffices to take:
numbers
mass of -b
\337 a
=
AC'
\337 C'B
and
-a
Thus,
have:
we
CB'
=
\337 c
means that BA' \337A'C = -c pair B and C equippedwith >
\337 b,
pseudo-masses >
+ mBOB'
3In Ceva's theorem, it is points on the extensionsof criterion
even
for
number
we
the
sides.
to
the
-\177m
putting
B
-\177inC'
-- O.
and consider (*) remains the correct
sign convention
Then the relation (or
and
equalities,
A
apply
cOC-aOA,
find:
m
+ mcOC\"= 0, also possible
:
(c-a)OB' these
b,
three lines to be concurrent (i.e. 0 or 2) of the points lie on
the
>
= aOA-OOB,
(a-O)OC'
and (b - c)OA' = bOB - cOC.Adding mA -- b - c, mB -- c - a, mc= a -
mAOA'
Then the relation (**) the center of mass of the b and >-c respectively. > >
\337 B'A.
A' is
i.e.
parallel).
When
the extensionsof
the
(*) holds,
sides.
an
2.
Applications
of vectors
125
\177o Keometry
B', and C' are collinear. for any points C' and B' in the the sides AB and CA, we can find
the points A',
Therefore Conversely,
of
extensions
previous
A' must
argument,
coincide with
BC. Thus
a
to
point
and
the
on the sidesofa triangle
or
on
for any
true
holds
three collinear
the
on the
Then,
are
the relation (**)
points
extensions.
their
B
B
C
A
B'
C
A
A'
point
relation(**)holdstrue. according points A', B', and C' collinear, i.e. the point of intersection of lines B'C'
line BC suchthat the the
interioror on
C'
Figure 140
132. The method developed
and
applied
ometry, can be explained
of barycentersdemystified. This
in
\365\365128-131
using
to
geometry
Figure
some
problems
of vectors
method,
of plane
ge-
in space.
141
the plane P in space in such a way (Figure 141) that it point O chosenfor the origin. Then, to each point A on the plane, one can associate a line in space passing through the origin, namely the line OA. When the point comes equipped with a mass (or pseudo-mass) ra, we associate to this material point on the plane the vector 5-' raOA in space. We claim that this way, the center of mass of a system of materialpoints on the plane corresponds to the sum of the vectors associatedto them in space. Indeed, if A denotes the center of mass of a system of n material points A\177, ..., An in the plane of masses m\177, ..., rr\177n, then the total mass is equal to Position
misses the
Chapter4.
126 ra =
ral +
..- +
=
raOM:
the
and
ran,
FOUNDATIONS
AND
VECTORS
vector
corresponding
--4
5
In
+
=
ra\177OA\177
regrouping property of the of the addition of vectors.
the
particular, associativity
from
+...
ralOA\177
a\177
is
in space
--'\177
... +
+
a\177.
centerof mass
follows
/?
Figure
to points
origin
in
projective
method of associating linespassingthrough plane P turns out to be fruitful and leads
The above
Remark. the
142
of the
to the so-called projectire geometry.In projectire geometry, beside ordinary points of the plane P, there exist \"points at infinity.\" They correspond to lines pgssingthrough the origin and parallel to P (e.g. EF on Figure142).Moreover, lines on the plane P (e.g. AB or CD) correspond to planes ((2 or _R) passing through the origin. When A/\177IICD, the lines do not intersect on the plane P, but they
geometry to
corresponding
\"point\"
the
intersect
namely
infinity,\"
\"at
at
the
of intersection of the planes(\177
line EF
R. Thus, the optical illusion that track meet at the line of the horizon and
two
parallel
becomes
rails reality
of a railroad in projectlye
geometry.
EXERCISES > > 251. In the plane,let A,\177B, C, D,> E be> arbitrary points. Construct point 0 such that OA + OB + OC= OD + OE. 252.* In a circle, threenon-intersecting chordsAB,CD,and EF are given, each congruent to the radiusof the circle,and the midpoints of the segments BC, DE, and FA are connected.Prove that the
the
resulting are
25s4.
is equilateral.
triangle
253. Provethat
if
Prove
that the
positeedgesofa 255. sions
a polygon
has several
axes of symmetry, then they
concurrent.
three segmentsconnectingthe midpointsof op-
tetrahedron
bisect
each
other.
Prove that bisectors of exterioranglesof a triangle of the opposite sides at collinearpoints.
meet
exten-
2.
127
to geometry
vectors
of
Applications
25(7. Formulate and prove an analogueof the previous problem for of one exterior and two interior angles of a triangle. 257. Prove that if vertices of a triangle are equipped with masses proportionalto the opposite sides, then the center of mass coincides bisectors
incenter.
the
with
258.* Prove that tangents to a circleat the vertices of an inscribed triangle intersect extensionsof the oppositesidesat collinear points. 259. In the plane, three circles of different radii are given outside each other, and for each pair, the external common tangents are drawn up to their intersection point. Prove that the three intersection
collinear.
are
points
the plane, three pairwise each other, and for each pair, the 2(70. In
mon
is
tangents
disjoint circles are given outside intersection point of internal comProve that the three lines, connecting the center of the remaining circle, are
constructed.
each
with
point
intersection
concurrent.
Prove
2(71.
reformulation of Ceva's theorem: On the (Figure 138), three points A/, \177are chosen. Prove that the lines AA \177,BB \177,and CC \177are if and only if
B
C
\177,and
concurrent
following
the
sidesBC,
262.
Give a
263.
Two
of /kABC
AB
and
CA,
sin ZACC'
sin
ZBAA
sin ZC'CB
sin
\177A\177AC
\177B\177BA
\177
=1.
and
ABC
triangles
sidesof the other are
\177C'BB
sin
similar reformulation of Menelaus'theorem.
through the verticesof each
triples
\177sin
A\177B\177C\177
drawn.
are
lines of
the
then
concurrent,
are
lines that
them, Prove
of
given in the plane, and parallel to the respective if the lines of one of the
the other triple areconcurrent
too.
26\177.* Prove
and
lines
AB
\177and
theorem:
Pappus'
A \177, B \177, C \177on
BA
then
another, \177,BC
\177and
CB
If points A, B, C lie on one line, intersection points of the
the three \177, AC
\177and
CA
\177,are
collinear.
case of parallellinesusing projective geometry, i.e. by restating the problem about points and linesin the plane in terms of corresponding lines and planesin space. 265.* Prove Desargues' theorem: In the plane, if the lines AA \177, BB\177, and CC \177connecting vertices of two triangles ABC and A\177B\177C \177 are concurrent, then the three intersection points of each pair of tt\361nt: Reduce
extended
and H\361nt:
to the
C\177A \177) are
Represent
AB and A\177B \177,BC and B\177C and vice versa. diagram as a projection from space.
sides
respective
collinear, the
(i.e.
\177,and
CA
128
133.
AND FOUNDATIONS
of geometry
Foundations
3
VECTORS
4.
Chapter
theorems of geometry can but also confirmedby direct observation. This intuitive, visual nature of geometry allows one to discover many geometric facts long before they can be rigorously justified. The ancient Egyptians (at about 2000 B.C.) were using this empirical method of establishing the simplest geometricresults neededfor practical purposes. However, self-evidence of conclusions, derived by observing diagrams, can be deceptive, especially when the Many
\"Elements.\"
Euclid\177s
be not only
by
proved
way
of reasoning,
diagramsbecome
complicated.
The
from
ture
elements
who inherited
Greeks,
ancient
of
cul-
mathematical
generalized their observationsand devel-
the Egyptians,
more reliable forms of reasoning. All geometric results were now confirmed by flawless logicalarguments relying only on explicitly made assumptions about diagrams,thus rendering the conclusions independent of accidental detailsof specific diagrams. Around 300 B.C., a Greek geometer Euclid gave a systematic exposition of basic geometric knowledge\177of his time in a seriesof 13 under the common title Elements. This work laid down foundations of the mathematical method and remains, even by the standards of modern oped
books
science, a
quite satisfactory accountof elementary geometry.
common notions (also of the 23 (in the
as
known
definitions
1.
A
point
2.
A
line
that
is is
which has
a breadthless
axioms).
translation
English
and
postulates,
definitions,
with
begins
exposition
Euclid's
Here
are the first
seven
[1]):
no part.
length.
$. The extremities of a line are points. A straight line is a line that lies evenly with the points on itself. 5. A surface is that which has length and breadth only. 6: The extremities of a surface are lines. 7. A plane surface is a surface which lies evenly with the straight lines on itself. Then there follow definitions of angles, circles, polygons, triangles, 4-
quadrilaterals, paralleland perpendicular lines,etc. There
1.
Things
are
five
which
%ommon are
equal
notions\" to the
about arbitrary
quantities:
same thing are also equal
to
one
another.
be added to equals, the wholes are equal. be subtractedfrom equals, then the remaindersare equal. 4. Things which coincide with one another are equal to one another. 5. The whole is greater than the part. 2. If equals $. If equals
3. \337Foundations
bear specificallygeometriccontent:
The postulates Let
the
129
of geometry
be postulated:
following
i. To draw a straightlinefrom any point to any point. 2. To produce a finite straight line continuously in a straight line. $. To describe a circle with any center and distance. 4. That all right angles are equal I to one another. 5. That, if a straight line falling onto two straight lines make the interior angles on the same side lessthan two right angles, the two straight lines, if producedindefinitely, meet on that side on which are the angles less than the two right angles.
Then Euclid proceedsto stating geometric propositions, and deriving them logically, using the definitions, ioms, postulates and previously proved propositions (pretty the way we do it in this book).5 after another,
134.Non-Euclidean geometry.
the
From
view, logical foundations inherited them from Euclid,
of elementarygeometry, free of
not
are
One is the to acknowledge assumptionsused our arguments. congruent figures assumes ures in space as solidobjects.
form
the
of we
defects.
a number of implicit instance, the definition of of moving geometric fig-
explicitly
failure
For
in
the
axmuch
point
modern in
one
possibility
vague character of basic definitions,which do what points, lines,and planesare, but merely explain in what direction the everyday meaning of these words becomes idealized in mathematics.6 really
is the
one
Another
not
tell us
the
Historically, etry
emerged
of
non-Euclidean
as
need
solid logicalfoundations of geomdiscoveryin the early 19thcentury i.e. consistent geometric theories
for more
a result of the geometries,
where Euclid's 5th postulate doesnot hold cussed in Book I, \365\36575-78, the 5th postulate
parallel postulate: line
parallel
to the
Through
given one
can be drawn, and such a line
4 congruent in our terminology 5A notable distinction however is that than ours: he does not allow himself even not shown beforehand how to construct
scopeof
point
every
As it was disis equivalent to the not lying on a given line, a true.
discourse
Euclid's
to talk about this
is more
is
unique.
conservative
a certain diagram
if it was explains the role and decree that it is possibleto draw and to draw a circle of arbitrary
diagram.
This
the first three postulates, which merely through two points, to extenda segment, and radius. 6Historians still argue about the role attributed by Euclid to his first definitions. E.g. according to [3],they originally formed an unstructured successionof sentences,which became separated and numbered only in later translations. a line center
In
4. VECTORS AND
Chapter
130
Bolyai, showed where the parallel
Janos theories
ometric
are just as rich in
and which
tion,
Lobachevsky
geometers: Russian Nikolai
1830, two
1820
Hungarian
FOUNDATIONS
and
that there exist ge\177
independently
postulate is replacedby as
content
the
its
nega-
classical Euclidean (as did many others
geometry. They arrived at this resultby trying before them) to derivethe 5th postulatefrom the remaining ones, but were the first to realizethat such derivation is impossible. 135. Hilbert's axioms. In 1899,David Hilbert gave the first fully rigorous account [2] of foundations of elementary geometry,both
Euclideanand non-Euclidean.
It is cepts
attempting
defined
of previously
to accurately ones, and those
or later
one will sooner
concepts,
defined
earlier
by
that
inevitable,
terms
in
define all con-
in terms of
run
out
of
even
those
defined,\" and end up with a collection of undefinable whose meaning can be conveyed only intuitively. Hilbert chooses points, lines, and planes to be the undefinable notions of geometry. Furthermore, he assumes that thesegeometric objects can be (or not) in ?train relations with each other, namely: a point can lie on a line, a point can lie on a plane, a line can lie on a plane, a point can lie between two other points lying on a given line, (note that using these relations, one can formulate definitions of segments, angles,etc.),and two given segments (or angles) can be ;'previously
notions,
One can
intuitive
other.
each
to
congruent
try to interpretthese the
but
way,
and
notions
that interpretations are considered propertiesofthe notionsand relations such
are
of axioms.
axioms
in the
by a
established
are obtained
propositions
geometric
further
All
relations
usual
axiomatic approachis irrelevant. The meaning and
of Hilbert's
point
whole
certain
list
from the
which in principle can be doneformally, nature of the objectsinvolved. By the expression of Hilbert himself, \"it must be possible to replace in all geometric statements the words point, line, plane by table, chair, mug.\" To illustrate the character of Hilbert's axioms, we list the threeaxioms logical
by
not
derivation,
to the
appealing
of order: (i)
If
the
points
A, B,
then B lies alsobetween (it) If
and
A
point B C
lies
(iii) Of
lying
C
are
any
C
and
two points
between
between A and
one which
C lie on a line, and
A
D.
and
B
lies
between
A and
C,
A.
of a line, then there exists at least one at least onepoint D such that
C, and
three points lying on a line, there is always lies betweenthe other two.
one
and
only
3.
Foundations
131
o\234 geometry
going to present herethe whole let us mentiononly that it includes
are not
We
20 axioms;
of Hilbert's
system
postulate
the parallel
(or a negationof it), Archimedes' axiom(seeBookI, \365145), and last item in the list, the axiom of completeness: To a system of points, lines, and planes, it is impossible to add other elements in such a manner that the system thus generalized formsa new geometry obeying all of the previous axioms.
as the
136.The set-theoreticapproach. Hilbert's new area of mathematics: mathematical logic,
laid
work
While
down a
modern
the
approach to geometry relies on a different foundation, which better balance between rigorand intuition. It was proposed by Hermann Weyl and is basedon the algebraic notion of
strikes a in 1916 a vector
space.
makes Weyl's approachparticularly attractiveis that it very general undefinable notions, such as setsand elements, which permeate all of mathematics, and not only geometry. Set theory first emerged at the end of the 19thcentury in the work of a German mathematician Georg Cantor as the theory comparing various infinite sets, but soon becamethe universal language of What
only
uses
modern
mathematics.
Sets are thoughtof ascollections ofobjectsofany nature. It may to express this moreformally than by saying that a set is consideredgiven if for every object it is specified whether it is an element of this set or not. All further notions are introduced on the basis of this relationbetween elements and sets. For example, one set is calleda subsetofanother, if every element of the first set be impossible
is alsoan elementof the
second.
Euclid, we considered linesand planesas sepamerely as sets of points lying on them. 7 One simplifying distinction of the set-theoreticapproachto geometry is that only the set of points and its properties need to be postulated, while lines and planes are defined simply as certain subsets of the set of points, so that all their properties become theorems. Another ingredient of Weyl's approach to geometry is the set R of (positive, zero, and negative) endowed with the So
far,
entities,
rate
following and
not
real numbers
usual operationsof addition,subtraction, multiplication, sion by non-zero numbers. To define real numbers and the operations, it suffices to introduce them in terms of signed of decimal numerals (as outlinedin Book I, \365\365151-153). same line couldbe represented
7In particular, the fits a
diagram), and
it
made
sense
to talk about
by
any finite
%xtending\"
the
and
divi-
arithmetic
sequences
piece of it line.
(that
VECTORS AND FOUNDATIONS
4.
Chapter
132
137.Axiomsofa vectorspace.
space
a vector
definition,
By
set, equippedwith operations of addition and multiplication by scalars which are requiredto satisfy certain axioms. The set will be denoted here by )2, and its elements will be referred to as vectors. Thescalars form the set R of real numbers, which includes thereforethe numbers 0 and 1. To every vector \177 and scalar c\177 (i.e. is a
associates a new \177,
axioms
The
following
for all
require that holds true:
(i)
(\177
+
vectors \177, \177,\177 and
\177)
+
\177 +
(ii)
unique vector
is a
There
(iii)
=
\177+
(\177
\177=
\177+
\177,
\177
(v)
\177(\177 +
(\177i)
(\177
+
\177)
\177)4
,
(viii)
=
\177 +
=
\1774
6,
=
4.
of
c\177,/3
the
\177),
6 =
\177+
+
\1777for
all
\1777,
\177<
In words, the axioms express: associativity addition of vectors, existence and uniqueness scalars,
scalars
all
\177.
= (\1773)g
04 = \1774
sociativity of multiplication by to addition of vectors,to addition
+
6 such that
\177(3g)
(iv)
\275ii)
a new vector denoted\1777+
associates
of addition
pair
every
To
c\177.
of multiplication of vectors \1777and
operation
the
R),
and
)2
denoted
vector,
operation
the
sets
of the
elements
any
of
commutativity
and
zero
the
of
its
distributivity
scalars,
and
vector, aswith respect
the way
the multi-
plication by the scalars 0 and i acts. Here is an exampleof explicit derivation from the axioms. For every vector \177 there exists a unique opposite vector, i.e. a
vector
\1777such
that
we
have:
Indeed, \177=
(\177 +
0\177:
\177 +
(-\177)\177+
=
\177)\177+
(-\177
+
is
\177=
((-\177)\177
=\276+5:
\1777+
(-1)if,
\276,
\177+
\177,
\177=o\177+
We will denote \177-
also
write
sums of
of opposite
definition
each
adding
(iii) and
\177
\177=5+
of
\177
vectors
(vi),-1 + 1: (ii) and
+
side
(-\177). Due
to (-1)\276
(i)
0,
and
(vii)
(iii)
the vector oppositeto a vector
\1777instead
1 + (-1)
(viii)
and
definition of oppositevectors
\177),
\276: (-1)\276.
i.e.
write
to
.then
(\177+
+ \177)+
and
(vi)
opposite
(-\177)\177+
(-\177)\177 =
of:
(vii), and 0--
(-\177))\177
if
5:
\177.
Because
: D7+ (-1)ff = and hence (-1)\1777 Conversely,
\177'=
to the
ff
simply
by
axiom (i) we
several vectors without parentheses,e.g.
-\276
can
3.
\177ound\177tions
are
called
all equal to 0,
exist scalars c\177,/3,...,\"/, not
if there
dependent
linearly
Several vectors ff,\177Y,...,\177
dimension.
and
Subspaces
138.
133
o\234 geometry
that
such
-.' +
+
o\177u
7\177
=
O.
are called linearly independent otherwise, i.e. if this possible only when c\177 = /3 = ... = 7 = 0. Clearly, including new elements in a linearly dependent set of vectorsleaves it linearly dependent. A vector space is said to be of dimension/c if it contains a set of/c linearly independentvectors,but every/c + 1 vectors are linearly dependent.
The vectors equality is
form
of the
expression
An
/3\276
is
called
cients
combination Given a set
a linear c\177,/3,...,7.
+
+...
of vectors of vectors in
linear combinations of these vectorsform
set of 1/whichis a vector
on
space
operations.Indeed,
scalar
and
sums
7\177
1/, all a sub-
space 14/, i.e.
vector
a
coeffi-
with
\1777,\177Y,...,\177
a subspace
respect to the same of linear combinations
its own with multiples
linear combinations of the samevectors.The same applies to the opposites of such linear combinations, and to the vector 6 = 0\1777 + 0\177Y+ ... + 0\177, which therefore lie in 14/. Thus the operationswith vectors are well-defined in 14/, and satisfy axioms(i)-(viii) because the axioms hold true in the ambient vector
of
themselves
are
vectors
given
space
of a
multiples
scalar
All
(1)
Theorem.
vector form a subspaceof dimension 1. All
(1)
Indeed,
of two
combinations
linear
(2)
single non-zero independent
linearly
vectors form a subspaceof dimension 2. consisting of
\177=/3ff
ff
vector
a nonzero
forms
a linearly
0, then lff\177- (c\177//3)\177 = 6. (2) To prove that all linearcombinations pendent vectors ff and \276 form a subspace and
if
\177
-
6, then
and
5 =
05 q- 0\177 q-
Suppose \"/\177, \"/2
such
three
any
/31 \177 q-/327\177,
dependent. cients
set
if/3 :/:
to show that \177--
independent
scalar multiples ff\177= c\177ff and Indeed, if/3 = 0, then 0ff\177+l\177 =
dependent.
linearly
are
ff
any two
while
element,
one
of
is non-zero.
that
\"/1\177
linear
of
two
linearly
of dimension combinations:
q- \"/27\177 are
linearly
dependent.
inde-
2, it suffices Indeed,
hence the vectors are linearly at least one of the coeffiFor certainty, let \"/2 \177 0. Then the vectors 15 =
6, and
\177 \177
6.
Then
Chapter 4. VECTORSAND
134 5'
= 5-(c\1772/%)5,
vector, \1775'
=
+/3\177t
vectors
scalars
some
find
are scalar
\177-(/32/72)5'
\177\"=
the
Therefore
\276.
one can
and
5' and
to 0,
the vectors 5, \177, and 5 gemark. One can show that
and hence
5/2
are all
/
There exist
independent
linearly
three
dependent.
linearly
of three 3, etc. of points of solid V of dimension 3. axiom, the axiom
combinations of dimension
linear
linearly independentvectorsform a subspace 139. Points, lines and planes. The Euclidean geometry is defined as a vector The latter conditioncanbe considered as an of dimension:
set space
extra
vectors,
four vectors
but any
dependent.
linearly
set
of points
comes equipped thereforewith
a
distinguished
O, corresponding to the vector 6. To an a vector if, called its radius-vector. will continue using the notationUV for the vector \276if. the origin
point, point
such that
+/3 5/2
The
i.e.
dependent,
equal
both
6. Thus
og+/3O-
are
multiples of a single
linearly
\177' are
not
c\177and/3,
FOUNDATIONS
U, there
arbitrary
corresponds
v
Figure
We
T
Figure
143
144
Subspaces of dimension 1 and 2 are examples of lines and planes of geometry, namely those lines and planes that passthrough the origin O. By definition, an arbitrary line (or plane),not necessarily passing through the origin, is obtained from a subspace of dimension ,-> solid
1 (respectively
Figures a subset
translation, i.e. by addinga fixed vector 144) to all vectorsofthat subspace.In other
2) by
143 and of ]2 is
called a line (respectively
plane)
if the
(OU, words,
differences
3. Foundations
(OV-OU: The
145) of radius-vectors of its points I (respectively 2). how Euclid's postulates about lines under the vector approach to foundaand
143
Figures
UV,
of dimension
subspace
a
form
135
of geometry
shows
result
following
and planes becometheorems tions of solidgeometry. For
(1)
Theorem.
unique line passing
them.
through
three points' not
every
For
(2)
there is a
points,
distinct
two
every
line,
same
the
on
lying
there emists a unique plane passingthrough them. (3) If two points of a given line lie in a given plane, then every point of the line lies in that plane. (4) If two distinct planes have a common point, then they intersect in a line passing through that point. (5) Given a line, and a point not lying on it, there emists a unique line passing through this pointand parallel to the line.
given
two distinct points U and
(1) Given
V
set of
the
143),
(Figure
points with the radius-vectors OU + tUV, where t is an arbitrary real number, containsU (for l = 0), V (for l = 1), and forms a line, since all differences of such vectorsare scalar multiples of the vector UV. Conversely, if V, then the vector for
T lies
on the line previously If three
(2)
line passin>g to UV,
proportional
be
must
t. Therefore
number
real
some
on any
T lies
point
a
UT
OU+
OT -
tUV,
U and
throug>h
i.e. UT
- tUV
the
i.e.
point
described.
U, V, and
given points
(Figure
W
144)
are
non-
collinear, then the vectorsUV and UW are linearly indep>endent. The set of points with the radius-vectors OU + xUV 4- yUW, where x and y are arbitrary real numbers, contains the three given points (take the pair (x,y) to be (0,0), (1,0)and (0, 1)) and forms > a plane. Indeed, differences of such radius-vectors have the form ceUV +/\177UW and thus form a subspaceof dimension2. Conversely, if T is any point V, and W, then the three vectors the same subspaceof. dimension 2. they are linearly dependent, and henceUT-- xUV+y>UW real numbers x and y. Thus OT - OU + xUV + yUW, i.e.
on ) a plane passin\177 > UT,
UV,
Therefore for
some
and
must lie in
the point T lieson the (3) Let and V
79
(Figure
be distinct
U,
through
UW
previously
plane
145)
be a
points on a plane
Then the vector UV =
OV
-
described.
subspace of
dimension 2, and
P obtained
OU lies in 79,
from 79 by
let
U
translation.
as well as gUV where
136
4.
Chapter
AND
VECTORS
FOUNDATIONS
any real number. Thereforeall pointswith the radius-vectors 4- tUV lie on the plane P. (4) Suppose that two planes P and P' (Figure 146) intersect a point U. In the plane P (respectivelyP'), pick two points and W (respectively V' and I/V \177) not collinear with U. Then
t is> OU
at
V
the vectors UV and
UV
(respectively
UI/V
\177and
UI/V
\177) are
linearly
are linearly dependent. Thereforethere exist real numbers \177, fi, \177, fi\177 not all equal to 0 such that \177UV + fiUW = \177UV \177+ fi\177UW \177. Denote by 5 the vector represented by either of the equal expressions. The point A with the radius-vector OA = OU + 5 lies in both planes. It is different from U, becauseotherwisewe would have 5 = \177 in contradiction with linear independence of each pair of vectors.By (3), the planes intersect along the whole line UA. Any common point B of the planes P and P\177 must lie on this line. Indeed, otherwise we would have three non-collinear pointsU, A and B lying in two distinct planes, which contradicts(2). Since the
independent.
Figure
(5) Let
on it. The whose
y =
\276 has
dimension
four vectors
146
and W a point not lying and W consists of all points have the form OU + xUV + yUW. Taking the planecontains the line formed by points with
passing
see that
3, the
Figure
145
UV (Figure 147)bea given plane
radius-vectors
I we
space
through
line
U, V,
the radius-vectorsOW + xUV. It passes through W (when x = and does not intersect the line UV. For if it did, the intersection point would have the radius-vectorOW 4- xUV = OU + tUV. This would imply: UW - (t - x)UV, i.e. that the point W lies on the line UV, which is impossible. Thus, we have found a line passing
3.
To
,>
3
UW,
UV,
=
+ ?WZ
3UW
and W > lie in the and WZ must be
V, >Z,
U,
vectors
any
consider
uniqueness,
prove
points
the
parallel to the line UV.
W and
through
geometry
of
Foundations
\177 for
scalars
some
line WZ parallel to UV. Since sameplane, three difference linearly dependent,i.e. + the
?
c\177,3,
we find a
+ (\177/3)WZ = -(\177/3)UV. Adding point with the radius-vector OW
OU-
(\177/3)UV,
which
UW
then
0,
OU
+
both
to
(\177/3)WZ
=
both parallel lines WZ and that ,8 = 0. This implies multiples of each other: i.e. the line WZ parallel to
lies on
therefore
to zero. If
all equal
not
sides,
\177
>
o\177UV
UV. Sincethis isimpossible, conclude that the vectors and WZ are scalar we
UV
=
WZ
\177UV. Thus
UV coincideswith
+
OZ = OW line
the
\177UV,
constructed.
previously
Figure 147
140. Inner products. The concept the axiom of dimension)is sufficient
of to
a vector
describe
space (satisfying set of points,
the
as well as lines and planes,of Euclidean geometry. To introduce measures of lengths and angles,an extra datum is required. An inner product on a vector space\276 is an operation that, to every pair of vectors3 and \276, assigns a real number, denoted 3. \276, in such a way that the properties of symmetricity and bilinearity are
for
satisfied:
3,
all vectors
\337 V--V'
+ An
product
inner
vector
\1777.
A
vector
.
scalars
\177,t\177 and
,
+
=
is called Euclidean space endowed with
calleda Euclideanvectorspace. In
a
Euclidean
vector
space,
c\177,\177
if 3-ff > 0 for a Euclidean
lengths of
every
non-zero
inner product is
vectors are defined
as
This expression root is non-negative.
then defined as the length of their radius-vectors.
of
\177-\177
0 \370to
from
angles
of
measure
distance
The
is
Fora
measure can be taken.Thecosine
vectors 4 and
by the
guaranteed
For
Lemma.
value of the between -1 and
following Cauchy-Schwarz all vectors 4 and \177 of
<
(4.
vector
Euclidean
a
Put
= 4.4 = [\1777\177,B -- 4. expression of dcgrcc _<
A
(t4 + vectors
non-zero
(t4
\276).
If t is a solutiontothe
4B 2 -
discriminant
+
\276,
\276)
mul-
q- 2Bt
to a
the
examine
[\177\177,and
real number t:
+ U.
2Bt q- C = O, then \1774 + g = (\177 inner squares), and henceoneof
the other. If, alternatively, then (as it is known B2 <
i.e.
141. Congruence. geometric isometry if it
the
algebra)
AC as required. 8 (of
the plane) between
distances
pairwise
preserves
quadratic
the from
transformation
A
is called an
respect
2 +
negative,
is
4AC
\1777. \1777--
At 2
=
At
equation
=
C
2 with
have positive
is a multiple of has no solutions,
vectors
equation
vectors are scalar
other.
each
of
following
the
is
property
inequality.
141
only when the
holds
equality
the
tiples
(since
cosine,the expres-
1. This
we have:
space, and
formula:
=
to lie
needs
R.H.S.
the
between
a legitimate
represent
to
order
angular two non-zero
of the
cosine
\370,the
angle
of the
cos
In sionon
180 the
of
by means
defined
\177Y is
number under the squarebetween two points U and V the difference U\177: Ok the
because
sense
makes
sign
AND FOUNDATIONS
VECTORS
4.
Chapter
138
points.
One can
show (seethe last section)that isometriesof the
plane are: translations (by any vector), angle, about any center), reflections (about any
clidean any
sitions obtained
of these by
their
consecutive
This
SThat
is,
any
rule
or compo-
application.
Respectively, one can calltwo of them can be obtained plane.
line),
Eu-
(through
transformations, i.e. the geometric transformations
if one the
rotations
definition assigning
plane
geometric
from
the
grounds to every
point
figures
congruent
by an
isometry
other
the concept of F
another
of
congruence
point of F.
of
on
139
3. Foundations of geometry
cept of
congruence slightly
symmetric can be obtained
are
also reflections
but
translations,
and
axis
any
center,
any
result, geometricfigures (but not necessarily congruent, see for instance from each other by isometries. As a
plane.
any
or
a con-
of
space
Euclidean
in
result
the one used in this book. dimension 3 include not
from
different
Indeed, isometriesof a only rotations about about
definition would
figures, this
to space
applied
If
through
figures
\"moving
precise the intuitive idea change\" (see BookI, \365l).
it makes space without
Namely,
foundations.
satisfactory of
which \36549)
EXERCISES
the 5th postulate, derivethat lines makes interiorangles on the
From
266.
onto two
two right angles,then the two meet on the othersideto which
straight
Does this
of order.
axioms
Hilbert's
parallel postulate. form a set of
numbers
rational
line,
number
same
of the
statement
the
set
satisfy
the
of completeness?
axiom
268. Show
that the requirement in
that the element
\177 is
the
the
are
angles.Deduce uniqueness 267. Show that on the points satisfying
line falling side greater than lines, if produced indefinitely, angles greater than two right straight
a
if
269.
that
Prove
that
Prove
270.
for
every
vector R/C
\177of
of
all
real numberssatisfiesthe axioms respect to the following component-wise by scalars and addition:
271.
Can
a
set
of vectors
of a
axiom, and the axiom(ii).
of this
the set
vector space, namely, it follows from (iii)
axiom
is redundant;
unique,
statement
existence
the
a vector
space,
\177+\177+\177+\177:
4\177Y.
k-tuples (Xl,...,x/c) of of a vector space with operations of multiplication
ordered (i)-(viii)
.
be linearly
dependent
if
it
contains
only
one element?
272. Prove that every subset is linearly independent.
273. are
Prove
linearly
274\177.
Prove
that
in
R/C, the
of a
independent
set of vectors
(1,0,...,0),...,
(0,...,0, 1)
linearly
k elements
independent. that
respectively.
R 1
and R 2
are
vector
spaces
of dimension
1 and 2
140
4.
Chapter
275.* Give an containing
one wishes.
FOUNDATIONS
AND
VECTORS
example of a vector spaceof infinite linearly independent vectorswith
276.*
Which
vector
spaces of
i.e.
dimension,
sets of k
k
as
as
large
five parts of the theorem in \365139 remain true in dimension greater than 3? 277. Prove that planes in space obtained from each other by translation either coincide or do not intersect (i.e. are parallel), and conversely, parallelplanesare obtained from each other by translation. 278. Provethat two lines in space obtained from each other by translation either coincide or are parallel (i.e. do not intersectand lie in a plane), and conversely, parallel lines areobtainedfrom each other the
of
by translation.
that the vectorspaceR\177,equipped
279. Check
uct
(x\177,...,x\177)
..., (0,... ,0,
280.
are
in a
2)
Euclidean two
space:
inner
the triangle inequality: for vector
a Euclidean
\177-\1772
285.
Prove
gles
between
vectors
to zero.
space, every set e\177,..., \177'\177 of vectors is linearly independent. product of their linear combination with
d-IBd
2
radius-vectors
arbitrary
space,
Prove
284.
is equal
perpendicular
Use the Cauchy-Schwarz the law of cosinesfor
Hint'
non-zero
product
inner
vectors.
283.* Prove \1777,\177,t\177 in
their
vector
Euclidean
a
in
the
Compute
of the
and
plane (i.e. a Euclidean vectorspace perpendicular vectors of unit length.
every
pairwise
Hint' each
if
o\177ly
contains
282. Provethat non-zero
prod-
inner
Euclidean,
vectors(1,0,...,0),
Euclidean vector space, two
if and
that
Prove
dimension
is
x\177y\177,
\177).
Prove that perpendicular
281.* of
the
with
d-
x\177y\177 d-...
and pairwise anglesof the k
lengths
compute
=
(y\177,...,y\177)
-
- 2IA/}
inequality. triangles
ABe'
in a
Bd cos ZABC:
Euclidean vector
\177-\177 2.
that an isometry of a Euclideanvectorspacepreserves anany segments, i.e. if it transforms/NABC into/NA'B'C',
then ZABC=
286.
\276
(i.e.
\177 +
is
an
\276)
can
about
the
transformation
of a
Euclidean vector spaceby assigning to a point \177 the
a
fixed point
isometry.
Prove that every geometric transformationof the planethat obtained by composing translations, rotations, or reflections, also be described as the rotation about the origin or reflection a line passing through the origin, possibly followed by a trans-
287.* can
a translation
that
Prove
vector
be
lation.
to non-Euclidean geometry
Introduction
Let ]? be a Euclidean vector space i.e. any vector of unit length
in space.
Coordinates
142.
3. Let
of dimension
vector,
a unit
be
\177
be found by choosing any non-zero vectorand dividing length). Take any vector \1777linearly independent of e\177 and from it a scalar multiple of \177*\177 so that the resulting vector is perpendicular to e\177. For this, put s = \1777- \177'\177 so that
can
(which
its
by
it
to non-Euclidean geometry
Introduction
4
subtract s\177
\177-
(\177
from
tract
=
s\177
\177'
\276-
vectors
of the
-
s2e-\177
e-\177.
Since
s\177'\177
= F.
\177, s2
of
independent
combination
linear
vector
resulting
put
\276a
vector
\177'\177 and
\177'2.
space ]2 is of dimension>
the vector
because
exist
0.
- s =
s
length we obtain a unit
by its
vector \276linearly
take any
Next, vectors
=
\1772
eq.
to
perpendicular
s eq \337\177
-
\177. \177
vector
resulting
the
Dividing
eq =
- s\177).
\177'\177 and
\177 \337 \177*\177 = \1772 \337 \177'2 =
1 and
Sub-
the
that
For this,
to them.
perpendicular
is
\1772 so
(Such
2.)
\177'\177 \337 \1772
=
O,
we have:
Dividing the resultingvectorby to
perpendicular
e\177 and
coordinate system Let
\177 be
any
dimension 3 are not all equal to so
\177
0,
since
space.
0
vectors
-*
--
-\177 s\177e\177 +
x\177,
nates
of the
x2,
s3
Dividing
so, we find: Xl
e-*l
q- x2\177'2
q- x
3 e\1773,
x3
6--therefore
of
space
s0, s\177, s2, S3 \177, 3- Moreover,
independent.
linearly
\177'\177, e\1772,\1773 are
$2 \177, 2+
vector
are some real numbers. They are calledthe vector \177 with respect \177to this coordinate system. The coordinates are uniquely determinedby the vector. ' \177 + x' 3 \177. \177 = x\177'\177 + x2e2 3, then
where
and
sox
that
such
3\177--
if
a unit vector e-\177 constructed a Cartesian
we obtain
we have
4th vector. Since any 4 vectorsin a linearly dependent,thereexist scalars
the
this equality by
in
length
its
fact
In
\177'2.
coordi-
Indeed,
(331
x\177
=
'x\177,
x2
=
x\177,
dence of the vectorse\177, e\1772,\1773.
333
---
33\177 due
to
the
linear
indepen-
4.
Chapter
142
Furthermore, for
every
OZ\177 ---
the
i.e.
multiplying
If
(OZ331)\177*
c\177we
AND FOUNDATIONS
have:
1 -+- (OZ332)\177 2 -+- (OZ
3)\1773,
coordinates of a scalar multipleof a vector are obtained the coordinates of the vector by this scalar.
\1777=
Y292 + Y393
y\177g\177+
\177-+-
\177: (Xl
is another
-+- \177/1)\177'1-+- (\177U 2
by
vector, then
-+- \177/2)\177,2-+- (\177U 3 -+- \177/3)\177I,3,
sum of two vectors are obtainedby adding of the vectors. Since pairwiseinner prodthe vectors g\177, e-\177, e-\177 are equal to 0, and their inner squares coordinates of the coordinates
the
i.e.
scalar
VECTORS
corresponding
of
ucts
are
1, we
to
equal
find: :\177' \177----
These
formulas
ordered k-tuples operations by
can be
Xly
generalized. Denote by
(x\177,..., z\177)
of multiplication
1 \276- X2y 2 \276- X3y 3.
of
real
numbers.
by scalars,
R
set
\177the
Introduce
of all
in R \177 the
addition, and inner product
formulas:
the
+
(w,...,w):
+
+
,:%). (w,... ,w) = :aw +-'-+ hard to verify that the axioms (i)-(viii) as the propertiesof symmetricity and ner product, are satisfiedin R \177. In particular, It
is
not
as well
of
a vector
bilinearity
space,
of an
in-
k-tuple (0,..., 0) plays the role of the vector iJ, and all other k-tuples have positive inner squares: \177c\177 +.. \337 + \177c} > 0. Thus R \177equipped with these operations is a Euclidean vector space(of dimension k). It is called the coordinate Euclidean space. Thus our previous construction of a Cartesian coordinate system in 12 establishes the following theorem. the
Any Euclidean vector space of dimension 3 can be identified with the coordinate Euclidean space 1% 3 by associating to each radius-vector the ordered triple of its coordinates with respect to a Cartesian coordinate system. Remarks. (1) The same result holds true for a Euclidean space of any finite dimension k, for the plane in particular, where k = 2. (2) The identification of 12 with the coordinate Euclidean space is not unique, but depends on the choice of a Cartesiancoordinate Theorem.
4[.
to
Introduction
In other words,
system.
if
143
geometry
non-Euclidean
triple
different
a
of
e\177,\1772,\1773
pairwise
chosen, the samevector \177 will have a different triple of coordinates with respect to it, and will therefore be assigned a different element of R s. (3)We see that within the vector approach to geometry,not only can the foundations be described by a conciseand unambiguous set of axioms, but also a model satisfying all the axioms can be explicitly constructed. The only datmn needed for this is the set R of real unit
perpendicular
with
equipped
numbers
vectors is
arithmetic
ordinary
according to the theorem,any
such
can
model
operations. Moreover, be identified with the
coordinateone,R3. Inthissense,one say geometry obeying the requiredaxiomsexists can
solid
that
and
is
Euclidean
unique.
143. The Klein model. A simple variant of plane geometry that satisfies most of Euclid's(or Hilbert's) axioms but disobeys the parallel postulate was proposedin 1868 by an Italian mathematician Eugenio Beltrami and then improved on by the Englishman Arthur Cayley and the GermanFelixKlein.
Figure
On
and
given a through
plane, take any disk, and declarethe set of this disk to be the set of allpointsof the Klein of the disk to be linesofthe Klein model. 9 Then,
Euclidean
the
of
points
interior
model,
148
chords
line AB (Figure148)and a pointC outside it, one can draw C as many lines as onewishes that are parallel to AB in the
Klein model (i.e.donot intersect AB inside the In fact, the Kleinmodeldoesnot yet qualify
Euclidean late, also
because
geometry,
but
9This Hilbert's
some
proposal metaphor)
other
disk): for the
role of a non-
not only the parallel postuof Euclidean geometry. Namely, if
it disobeys
axioms
goes therefore, although of table, chair, mug
not taking
too far, in the direction (using on the role of point, line, plane.
Chapter
144
4. VECTORS
AND FOUNDATIONS
in the Klein model is measuredin the Euclid's 4th postulate (about existence of circles of arbitrary radius with arbitrary centers) no longerholds true. However from Hilbert's point of view, it is the axiomof completeness that fails in the Klein model. Namely, one can enclose the disk of all points of the Klein modelinto a larger disk and thus get a new Klein modelwhich, compared to the original one, has extra pointsand lines.We will see later that these defects can be corrected by changing the concept of distance (in such a way that the points of the boundary circle becomeinfinitely far from interior points). The significance of the Klein modelbecomes more clear in the context of numerous attempts, known in the history of mathematics, to derivethe parallelpostulatefrom the others. Following the method of reductio ad absurdurn, one would start with the negation of the parallelpostulateand try to reach a contradiction. The Klein modelshows that one does not come to a contradiction by merely rejecting the parallel postulate. Moreover, if any argument that makes sense in the Klein model couldlead to a contradiction, this would mean that a contradiction is found in the classical plane geometry. This is becausethe Klein modelis described in terms of the classical plane geometry Thus the plane geometry, where the parallel postulate holds true, cannot be consistent(i.e. free of logical contradictions) unless the Klein model, wherethe parallel we
usual
distance way, then
the
that
assume
Euclidean
postulate 144. the
also
is
fails,
Spherical
It consistsof
therefore this mation to
all
geometry. In the Euclidean spaceRs, consider coordinate equation
by the
given
surface
consistent.
points
whose
radius-vectors
have the
length R. It is
R centered at the origin. Geometry on geometry. It provides an approxigeometry of the Globe, and a model of non-Euclidean geometry. Namely (Figure 149), one defines lines of spherical geometry as great circles of the sphere. Every great circle is obtainedby intersecting the sphere by a plane passingthrough the center. Any two such planes intersect alonga line in space passing through the center (O). Thislinecuts through the sphere at two diametrically opposite points (C and C\177), which therefore lie on both great circles. This shows that in spherical geometry, every two lines intersect. Thus, the Euclideanparallel postulate fails in spherical geometry, because them arc no parallel lines there at all. sphere
the
surface
is
called
of radius
spherical
145
4. Introduction to non-Euclideangeometry geometry
of
length
The
segment (e.g.
a line
AB, Figure 149)in
is taken to be the Euclideanarclength the great circle,i.e. \177-R(a/2d), where
of
spherical
corresponding
the
measures of the straight angleand centralangle A\270B, respectively. Areas of regions in spherical geometryare also defined in the natural way, i.e. as Euclidean areasof the corresponding parts of the sphere (e.g. 4\177R 2 for the total area of the sphere). To measure angles on the surface of the sphere, one measuresthe Euclidean angle (A\177C\177B \177) between the rays tangent to the corresponding great circlesat a point arc of
(\177)
of
2d
and
a are
intersection.
c
Figure 149
The
Theorem.
angle
(ABC,
angles of a sphericaltrithan 2d. More precisely,
of interior
sum
Figure
greater
is
150)
the ratio ofthesumofthe angles(a,/3,
and
angle
2d exceeds I
ratio
the
by
triangle to the area
150
Figure
great
the
of
of
if)
spherical
the
the
to
straight
area S of
the
disk:
-14--- S
a+/3+s
\177R \177'
\177,d
lune (CAC'B, shaded on Figure The lune is swept by one of the semicircles rotated about the axis CC\177through the angle \177 -\177AOB. It is clear (see also Remarkin \365112) that the area of the lune is proportionalto the angle of rotation. The area is therefore examine
Firstly,
a spherical
149)enclosed
two
between
equal to
4\177R\177(-/4\177)
=
great
\177R\177(-/\177).
semicircles.
The
sides of the lune at the
vertexC' isa
angle
plane
ACC\177B
(since
the
A\177C\177B \177 is
angle linear
tangent
A'CB' angle
of
to the
between the dihedral
the
sphere at
the
Chapter4.
146 Ct
point \177
the
Note that
t.
Thus
CAC
circles
great
two
radius OCt).
to the
perpendicular
therefore
is
and
ZAtCtB
=
FOUNDATIONS
AND
VECTORS
of which is symmetric to the sphereand hasthe same We conclude that a pair of centrally symmetric with interior angles a at the vertices have the total Consider now any three great circles (Figure 180). divide the sphere into four pairs of centrally symmetricspherical Let/kABC and/kAtBtC be one of the pairs. The interior spherical anglesof these and the angles vertical to them are at the same interior angles of three pairs of centrally symmetric
the sphere
divide
into the
about
CACtB
lune
lunes,
four
one
O of the
center
area.
lunes
area
They
triangles.
t
triangles,
time
lunes
ABAtC and ACtAtB t of lunes with the vertices A and At). The total area of these three pairsof lunes is equal to 27r/i\1772(\177 +/\177 + 7)/d, where \177,/\177, '7 are the angles of/NABC. On the other hand, these three pairs of lunes cover each of the spherical triangles ABC and AtB\177Ct three times, and the rest of the sphere the
(e.g.
pair
once.
we
Thus
have:
27rR 2 oz
\177-/\177+
'7 =
47rR2 +
4\177q.
d
by the
Dividing
As a the points
total area of the sphere,we
required
the
obtain
model of non-Euclideanplane,spherical
geometry -of points is declare that represents a
geometry
pair of
following obvious flaw: through a of the sphere, pass infinitely many
great
Euclideanor not -one allowed). This defectis to each pair of syrrzmetric single elementof set of points. exactly
easy
centrally
the
result. suffers
symmetric
centrally
circles (while in plane through every pair
line
correct: points
it suffices to of the sphere
This way
one obtains
the spherical modelof geometry, which obeys all but one axiom expectedof a plane, namely the parallel postulate, which is replaced the property that every two lines meet at exactly one point. Sincethe spherical is constructed entirely in terms of the coordinateEuclidean spaceRs, this of non-Euclidean geometry turns out to be at least as consistent non-Euclidean
geometric
with
model
version
as
solid
Euclidean
geometry.
It is useful to outlined in \365132, projective
plane
comparethe spherical
model
of
the
can be
projectire
plane.
with
the
The set of
defined as the setof 1-dimensional
construction,
points of
the
subspaces
s (i.e. lines passing throughthe origin).If we pick in R s a plane P (Figure 151)not passing through the origin, those 1-dimensional subspaces (e.g. a) which are not parallel to P intersect P at onepoint each, and thus can be identified with these points of the plane P. in R
4.
to
Introduction
147
geometry
non-Euclidean
The projective plane alsocontains 1-dimensional subspaces parallel P (e.g. b), which are thereforenot represented by points of P (but can be interpreted as its \"points at infinity\.") Note that each 1-dimensional subspace in R 3 intersects a sphere centered at the origin at a pair of centrally symmetric points (A and A', or B and B'). Thus, the set of points of spherical geometry is identified with the set of points (all-- finite or infinite) of the projective plane. This way, the projective plane becomes endowed with notions of distance betweenpoints inheritedfrom the sphere, i.e. determined by the
to
angie (AOB)betweenthe 1-dimensional subspaces.
145.
The
introduce
Minkowski the
It is
space. In the coordinatevector
straightforward to verify
Y2,
Y3) --
that
space
product'
inner
Minkowski
(X0,:81,:82)' (Y0,
152
Figure
151
Figure
4.
--XOYO
the
XlYl
4. x2Y2.
and bilinearity
symmetricity
properties, required of an inner product,are satisfied. However this inner product is not Euclidean,becausethere exist non-zero vectors 3 = (x0, xl, x2) whose inner square 3.3 is zero or even negative.The set of points satisfying 3.3 = 0 is given by the equation x0\177-- x\177 4- x22. Fixing a norazero value of x0 we obtain a circle on a plane, parallel 'to the coordinate (xl, x2)-plane, with the center lying on the x0-axis, and radiusequalto x0 \337On the other hand, if a point 3 liesonthis surface, then all points with the radius-vectors proportional to 3 do too. Thus, 3.3 = 0 is a conical surface(Figure152)with the vertex at the origin, obtained by rotating a line (e.g. OA) about the xo-axis. The spaceR 3 equipped with the above inner product is called Minkowski space (as opposed to Euclidean space), after a German
mathematician HermannMinkowski
who
connection
with Einstein's theory' of
relativity.
introduced Following
1908 in
it in a
physi-
4. VECTORS AND
Chapter
148
FOUNDATIONS
the surface \177. \177 = 0 the light cone. l\370 It is formed by light-like vectors (e.g. OA),and separatesspace-like vectors (e.g. OB), which satisfy \177. \177 > 0 and lie between the two halves of the cone,from time-like vectors (e.g. OC or OD), which satisfy \177. \177 < 0 and fill the two interior regions of the cone. we call
terminology,
cist's
In the Minkowski space,subspaces are of dimension 1, and this property other inner product spaces. Note that the distance containing
vectors
vector
the
when
whose
vectors
Two
that
Assume
\276 is
time-like vector
if,
vector, are space-like.
if.
ff
subspace
the \276
2, contains
and
0, if.
\276--
+
=
\337 (oe
of
<
light-like vectororthogonalto a
or
a time-like
i.e.
or-
vectors
non-zero
space,
Minkowski
to a time-like
thogonal
from
Minkowskispaceis well-defined AB (Figure 152) is space-like. inner product vanishes are called orthogonal.
Theorem.In the
i.e.
it
B in the
A and
points
two
between
only
space-like
no
distinguishes
0,
and
\276.
Then
2,
