Download compasses

CHAPTER FIVE
compasses Ruler and
According to Plato the only 'perfect' geometrical figures are the straight ·lin~ and the circle. In ancient Greek
metry this beliefhad.the effect of restricting the instruments available for performing geometrical ·-geo
constructions to two: the ruler and the compasses. The ruler, furthermore, was a single unmarked
.straight'edge
With these instrl1;ments alone it is possible to perform a wide range of constructions. Lines' can be divided
into arbitrarily "many equal segments, angles can be bisected, parallel lines drawn. Given any polygop it is
However, there are many .possible' to ' construct a square of equal area, or twice the area. And so on
ought to be constructible, for which the tools of intuitively geometrical concepts which
ruler and compasses are inadequate. There are three famous constructions which the Greeks could not. :perform
These ask respectively for a cube .tureo! the circleandQua~ra ,Duplication of the cube, Trisection" of the angle
twice the volume ofa givenJcube, an angle one third the size of a give~ angle, and a square of area equ~l to. a
.givenciic1e
It is not surprlsing that the Greeks found these con":, structionsso difficult..They ate impossible. But the
sus.picion. that solutions. did the ,Greeks had ,neither the methods to prove the impossibiliti nor, it, appears
not exist.·,Jri
GALOIS THEORY
.. consequence they expended considerable ingenuity in a fruitless search for solutions
By going outside the ~hltonic constraints all three problems can be solved, and the Greeks found several constructions'
-involving conic sections or more recondite curves sUGh as th~ conchoid of Nichomedes or the qu adra
-Coolidge [35J). Archimedes, tackling . the problem of quadrature of the circle in acharacteristi .)42[ trix (see Klein .
,cally ingenious mannerproved a result which would now be 'written
~ 3? < n < .,3
.able achievement with the limited techniques available \a remarl
With the machinery now at our .disposal it is relatively simple to. give a complete answer to all three
problems. We use coordiqate geometry to .express them ill algebrai<; terms, and apply our theor y of field
_ .extensions to the algebraic problems which arise
formulation !(Algebrai
of 'compass construction. Assume that a set Po-andOur first step is to formalize the intuitive idea of a ruler
points in the Euclidean plane' R2 :is given, and consider operations of the following two kinds
Operation1 raw a Pod (Ruler): Through any two pQints of
.straight line
'
and whose radius is equal to the ,Po (Compasses): Draw a circle, whose centre is gl poin't of 2 Operation
distance
.Ro between some pair of points in'
The Greeks preferred a restricted version of operation 2, namely: draw ('a ,Po circle; centre sorne point of
Our operation 2 can be performed by a sequence·gF such ;Po and passing through some other point of
matel" mak~s no differ'encewhich~ version we uSe. Tl1e one given above is· more operations, so itl1lti
).fo'r: our purposes -convenient
Definition. The points of intersectlQn~f "any two distinct
~RULER AND COMPASSE
lines or circles; drawn' using operatiQns1 ~Po constructible, in one' stl~pfrom be o:r~ 2, are. said to
A pointrE R2 is:const,ructiblefrom Po: if there' isa finite seq uence
,
'
ofpoh~ts ofR the point n , ... ,1 = i such that for each h
'~is cOLstnictible in one step fr.9~ the se
'
rl'"'' r{ u Poi ·d point of a given line can be realized ,within our' -shan show how the standard construction of the mid Example. We
formal framework. Suppose we are given two points,
2
.
2 ' '
,'Ph P2 .R (Fig~ 2) E
.Fig2
Po Let-.,= '}P2 'PI{
(i). Draw the.linePIP2 ~operation(1) .
of radius Dl'aw the Circle centre P'l of radius PIP2 (operation 2). , (3) Draw the Cir,cle centre P3-)2( .
be the points of intersection of these circles. (5) ,Draw the'line r2 PIP2 (operation 2). '_ (4) Let rland'Ir2
operation(1).
be the intersection of the lines Letr3 ,)6(PfP2 andr1r2• Th'e!1 the seq ueneer r ~ defi~es a '2 r 'l
• ;point of the. line PIP2-construction of the,' mid
59
60
GALOIS THEORY
by two points lying on it, and a circle SInce a line is always specified '.
by its centre and a point on its circumference, all the traditional
geometrical constructions of Euclidean . geometry JaIl within the scope of
.our formal definition
Field theory enters in a fairly natural way. With each stage in the
construction we associate the subfield of R . generated by the coordinates
be the subfield of R _ generated by .Ko of the points constructed. Thus let
coordinates of the point inY and -x thePD' 'Xi( has coordinates If riYi) then induct.ively ,
we defineKi ,that is to say ,Yi and Xi by adjoining Ki~l to be the field obtained from
Ki= X(Ki-1i .yJ ,
Clearly
,With the above notation .Lemma 5.1Xi andYi K In are zerosjof quadratic polynomials over Ki- 1 •
There are three cases to. consider: line meets line, line meets circle, and circle meets circle. Each case .Proof
.'geometry; as anexampJewe shall do the case 'line meets circle-is handled by coordinate
Fig. 3
RULER AND COMPASSES
61
L~t A, B, C be points whose coordinates(p, q), (r,s), ,t(u) Draw the line AB and the circle ,lie in Ki-1
since l',-Ki lies in w2 as in Fig. 3. (Note that ,Ki-1 E w2 where ,w centre C, radiusw is the distance
-between two points whose coordi
nates are inKi-1 Use Pythagoras.) The "equation of the line AB is •
x-p= q-y
(1)
q-p s-r
and the equation of the circle is
=UYz-t)2+(y-(Xw2•
(2)
Solving(1) and (2) we obtain
t)2- x(
(r - p)
+(
w2
= Uf-x_ p)+q\q-s'(
coordinates of the, intersectioi1 points X and Yare zeros of a quadratic polynomial -so the x
over'Ki The same holds,.1...::
.coor~inate's-for the y
'
,
'
....
,
'
"
"
,"
,
",
"
. aic consequence of the-deduce analgebi-We 'may now
.existence of a construction fo~ a ,given point
•.•.•.. '.
"
"
b
,x( = Theorem. If r 5.2y) is constructible from a subset.Po
ofR2, and if nates ,of the point's of Po, then :the degrees "Ko is the subfield afR generated by the coordi
'ahd ' Ko(x):KoJ[[Ko()~):,K\)J '
.are powers of2 ,
ing. By 5.1 and 4.3 we haveWe use the notation which we have been accumuiat .Ploof ,
"[Ki-1(Xi): Ki- tJ '=1 or2.,
,
.
"
....
".
~overKi: Th~ value 2 occurs if the quadratic polynomIal(1
Similarly ) .1 is a zero is irreducible;' otherwise the value is xi of which
62
GALOIS THEORY
]l-Ki-1(Xi, Yi):~i[ ,Therefore
.4 or ,2 ,1 = , ]l-l(XJ:Ki-l(Xi)][Ki-Yi):Ki ,Xi(Ki-1[ ==
for our argument.) , Hence -This observation is not required .~T~alue 4 never arises, aUbe reader may care toprov(
Ki:Ki-1J[,L::) .a power of 2. By induction (cf
is ]Kn:Ko[ we see that )4:5 Exercisea
'power of2.But since
Kn :Ko(x)l[K~(x): Ko]['= ]K~ :Ko[
.isa power of 2 ]Ko(y):Ko[ , is a power 'Qf. 2. Similarly ]Ko(x): Ko[ it follows that
Impossibility proofs
We shall now apply'the above theory to prove that theredo compass constructions -and-not exist ruler
the introduction to this chapter. (For the ·techl1ica. drawing -for the three classical problems menti 1ed in
expert'we approximate onstructions. There'are many" exact emphasiz~ that we are.discussing
).constructions· for. trisecting the 'angle, for instance;. but rio exact methods
.. compass .consp:uctions ' -and-The cube cannot be duplicated' using ruler .S.3 Theorem
We are given a cube, and hence a side of the cube, . which we may take to be the unit interval .Proof .,
. on the x~axis
. 'O)} s~ that ,1( ,)O, 0({ = Po Therefore we may assume'that .
'Ko=Q. could' construct the point_\\If we could duplicate the cube then, (a, where )0a3 = Therefore by .2
:and this is irreducible overQ by-over Q, 2 - t3 [Q(a):Q] would be a power of 2. But ais a zero of the polynomial 5,2
it This' is .3 = ](a):Q~[ by4,3 nomial ora overQ, aI1d ·-2is the miI).imuin polY - t3 Eisenstein's crit~rioi1. Hence
' .~contradic~ion. 'Therefore the cube cannot be duplicatc
RULER AND COMPASSES
.compass constructions-andThe angle n/3 cannot be trisected using ruler .Theorem 5.4
,is equivalent to constructing the point (n n/3 To cOI)struct an angle trisecting .Proof0) given(0, 0) ,and (10),
eX where=, cos(n/9). ).See Fig. 4(
Fig. 4
From this we could coflstruct ({3, 0) where {3= cos(nj9). From elementary trigonometry we recall the .2
formula
)38(cos== .)co~8 3 - )8(cos3 4
e we put. If= nj9'
)38(then cos= !, and we find that,
1-'3{3 - 33{,= .O
:; )J{,t Now£3 - )1 +sincef(t ,is irreducihle overQ 3t-1= Eisenstein's cr~terjon. by irreJucible is-
.a contradiction ,3
=
3 -3t2i + t,3
Ql :)As in the previous theorem we have [Q({3
"
-and-circle cannot be squared using ruler rh'e-Theorein..5.5.
.compass constructions
,Q( uch acoQstruction is equivalent to one of the point~ .Proof.J'7r) From
thiswecan easily consfru<;t
64
GALOIS THEQR Y .
.})O ,1( )0 ,O({ from
(O,ir). So·[Q(n): Q] .algebraic over If iK and iil particular ,2 is a power ofQ. On the other hand there is
a. famous theorem of Lindemann which asserts that7C algebraic not is
..
overQ. .The theorem follows
.We shall prove Lindemann's theorem in the next chapter
The proof involves ideas off the mainrtack: of the book~'and . hastherefor~ b,een segregated into a
he wishes; the if . separate chapter. The reader ,Who is willing to take it on trust can skip th~ proof
.. results are not used anywhere else in the . book
may consider other possible methods of. ·Instead of the traditional ruler and c.ompasses, we
rurer and Mohr showed that every· construction performable·' with. 1672 construction. Thus. in
compasses can be performed with compflsses alone (provided we agree'that a line is constructed when
buted to Mascheroni (1797)~ In 1818 ' -given); this result is more commonlvattri two points. on itare
a Brianchon considered constructions with ruler 9nly.Poncelet suggested that instead of compasses
Steiner idea was' taken up by should suffice; and this ,centre __ its _J~th~r~~jJb{to ,singl~fixed circle
in 1833 .. Ifthe centre of the circle is not known~ then fewer constructions are possible. Hilbert asked,
how many circles must b~ given in .order for the.centr:~of one of them
showed that for two general circles this is ,1912 to be constructed using ruler alone. C~uer, in
trine Grossmann At about the same .are concentric or ,iIppossible,but can be done if they cut, touch
. that three . linearly i·n.dependent circ1essuffice for ge'ometric constructions with only a ruler ~sho
and~compass constructions can be performed using a tw<?~edged -It. is also known that all ruler .: .
int(Referencesfor all these-po-a-ruler, whose 'edge~ are either parallel or meet at
).]results may be found in Kleiri,[ 42
.
With additional instruments, more constructions aTe pps~ sible .. Thus·.with· a marked ruler the angle
can be trisected (see Exercise 5.3). A device fdr dividing th~ angle into . arbitrarily many equal parts can
.. .]9[ he found. in Cundy and RolIett
Exercises COMPASSES RULER AND
:~Express in the language of this chapter methods of constructing, by ruler and compasse 5.1
.(a) The perpendicular bisector of a line. (b) The points tris~ctinga line· .
.equal parts n (c) Division of a line into
.(d) The tangent to a circle at a given point. (e) Common tangents to two circles
65
sions by giving reasonably good upper Estimate the degrees of the corresponding field exten 5.2
.bounds
The ruler .)5 .Verify the following construction for trisecting an angle using a mark ed ruler (Fig 5.3
has marked on it two points distancer .apart
A
Fig. 5
Given .L cutting ,r ., draw a circle centre 0 with radius )( = AOBOA atX, OB at Y. Place the ruler
with its edge through X and one mark: on the lineOY at D; slide it until the other marked point lies
E. Then at . on the circleL EDO= .8/3
?be trisected usirig ruler and compasses 2n/5 Can the angle 5A
.gon using ruler and compasses-Show that itisiinpossible to construct a regular 9:
s.~
66
GALOIS THEORY
.tion for the regular pentagonBy considering a formula for cos(Se) find a construc 5.6
Prove that the angle () can be trisected by ruler and compasses if and only if the polynomial 5.7
)8(cos - 3t - 4t3
.)is reducible over Q(cos«())
Discuss the quinquisection (cFvision into 'S.5S )equal partsof :angles
.p. 35 (Fig. 6) 24J[ due to Ramanujan n Verify the following approxima,e construction for 5.9
B
' Fig. 6
Let AB"be the diameter of a circle,centre O. Bisect AO at M~trisect OB at T. Draw TP perpendicular to
.and join AQ ,PT ='·AB meeting the circle atP; J:?raw. BQ
RULER AND COMPASSES
67
Drawos, TR parallel to BQ. Draw AD= AS, and
AC -= RS tangenti~l to the circle at A. Join BC, BD, CD. Make BE = BM. Draw EX parallel to CD. Then the
.square on BX has approximately the same area as the circle
to know that-You will need (n is approximately 355/113. This approximation is first found in the works of the
).A.D Chinese astronomer Tsu' Ch'ung Ching in about 450
/Mark the following t~}le or false 5.10
. approximation (a) There exist constructions trisecting the angle to an aibitniry degree of .poses but insufficient for mathematical ones _ -(b) Such constructions are sufficient for practical pur .
whose degree over the subfield generated -(c) The coordinates ora constructible point lie in a subfield of R
.by the coordinates of the given points )s a power of 2
.cannot be trisected using ruler and compasses n (d) The angle.using ruler and compasses On ,1( ,)0 ,O({ cannot be constructed from n (e) A line of length
(f) It is impossible to triplicate the cube by ruler and
.compasses
is transcendental over n (g) The real numberQ. real number -(h) Then .. is transcendental over R
,(i) If (0a) is transcendental over a by ruler and compasses then })O ,cannot be constructed from {(O, 0), (1
Q.
~(j) Geometrical problems ·cannot always be solved geometrically