Download Constructions with ruler and compass

Chapter 1
Constructions with ruler and
compass
1.1
Constructibility
An important part in ancient Greek mathematics was played by the constructions with ruler and compass. That is the art to construct certain figures in
plane geometry using only ruler and compass starting from a given geometric
configuration. For example, given a line l and a point P not on l, construct
the line through P perpendicular to l. Or, given two lines l, m, intersecting
in a point, construct an angle bisector. There are a number of construction
problems that could not be solved in Greek times. The most notorious ones
are:
1. (squaring the circle) Given a circle, construct a square having the same
surface area.
2. (doubling the cube) Given the edge of a cube, construct an new edge such
that the volume of the corresponding cube is twice the original one. Also
known as the Delian problem.
3. (trisection of an angle) Given two lines intersecting through P , construct
a line through P which trisects the angle.
Nowadays we know that such constructions are impossible. This follows from
developments in algebra from the 19-th century. For example, the impossibility
of the Delian problem and the trisection of an (arbitrary) angle was proven
by Wantzel in 1837. In this chapter we show how that can be done. In 1881
Lindemann showed that the quadrature of the circle is impossible, and this was
done by showing that π is transcendental over Q. This requires non-algebraic
techniques which we will not go into.
To understand Wantzel’s results we must formalize the concept of construction
first. The general idea is that a construction starts with a finite set of points
in the plane, which we consider constructed. The basic construction steps are
the following,
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CHAPTER 1. CONSTRUCTIONS WITH RULER AND COMPASS
1. Given any two constructed points A, B we can draw a straight line through
A and B, or a circle with center in A and passing through B. We call
such circles and lines constructible.
2. Given any pair C1 , C2 where Ci can either be a constructed line or circle,
we can determine their points of intersection. We add these new points
to our constructed set of points.
A few warnings:
1. Not all points on a constructible line or circle are constructible.
2. The ruler is unmarked, so you cannot measure distances with it.
Here are a few basic constructions. The numbering in the figures depicts the
order of the operations.
Construction 1.1.1 Given a line l and a point p, one can construct a line
through p perpendicular to l, cf. figuur 1.
Figure 1.1: Construction: line through p, perpendicular to l, left p 6∈ l and right
p ∈ l.
Construction 1.1.2 Given a line l and a point p not on l, one can construct
a line through p, parallel to l. cf. figure 2 (we use construction 0.1.1).
Construction 1.1.3 Given a line l, a point p ∈ l and two other (constructible)
points q, r. One can construct a point s ∈ l such that the distance between p
and s equals the distance between q, r, cf. figure 2 (we use construction 0.1.2).
We now come to the classical construction problems mentioned above. In all
three we can start with the simplest basis configuration, namely two points.
This is obvious for the quadrature of the circle, we only need to give the center
1.1. CONSTRUCTIBILITY
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Figure 1.2: Construction of parallel
of the circle and a point through which the circle passes. For the Delian problem
we can take two points and the distance between them will be the side of the
initial cube. For the trisection of an angle it suffices to take the angle of 600 . In
that case one can start with two points and form an equilateral triangle having
the distance between them as sides.
We may as well assume that in a basis configuration the distance between the
points is 1.
Definition 1.1.4 A real number a is said to be constructible if |a| occurs as the
distance between two points of a construction whose base configuration consists
of two points with distance 1.
Using the constructions 0.1.1, 0.1.2, 0.1.3, we can introduce a system of coordinate axes in which the two basic starting points read (0, 0) and (1, 0). The
following theorem is then more or less obvious.
Theorem 1.1.5 Consider constructions with a basis configuration of two points
(0, 0) and (1, 0) (so we assume we also have a coordinate system). Then a point
punt p = (a, b) can be constructed if and only if a, b are constructible numbers.
Proof: Given a constructed point (a, b) we can project it on the x- and y-axes
using construction 0.1.1 to get (a, 0) and (0, b). Then |a| is the distance between
the constructed points (a, 0) and (0, 0). Similarly for |b|.
Conversely, given any constructible number a, we can use it to construct the
point (a, 0) via construction 0.1.3. Similarly we can construct (0, b). One then
finds (a, b) by intersection of parallels via construction 0.1.2.
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Theorem 1.1.6 The constructible numbers form a subfield of R containing Q.
We shall denote the field of constructible numbers by Qconstr .
Proof: We show that for every two constructible a, b ∈ R with a, b ≥ 0 the
sum a + b, the difference a − b, the product ab and, if a 6= 0, the inverse a−1
are constructible. Our assertion then follows by definition of a subfield of R.
The sum and difference can be constructed via construction 0.1.3.
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CHAPTER 1. CONSTRUCTIONS WITH RULER AND COMPASS
Figure 1.3: Construction of the product
For the product we use the following geometric picture of proportional triangles
which can easily be constructed, see 3.
Suppose the left hand triangle is given and one side of the right hand triangle. Then the second triangle cvan be completed by drawing parallel lines via
construction 0.1.2.
To construct ab te construeren, we take r = 1, s = a and r0 = b. From r/s =
r0 /s0 it follows that s0 = ab. To construct a−1 we choose r = a, s = 1 en r0 = 1.
Then s0 = a−1 .
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Theorem 1.1.7 If a ∈ R, a > 0 is constructible, then so is
√
a.
Proof: We use the construction in figure 4. The left hand triangle has side
r = a and the right hand (gelijkvormige) driehoek heeft zijden r0 = s en s0 = 1.
√
Dus geldt a = rs0 = r0 s = s2 , dus s = a.
Figure 1.4: Constructie van de vierkantswortel
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Theorem 1.1.8 Given four points pi = (ai , bi ), i = 1, 2, 3, 4 having coordinates
in a subfield K of R. Let A, B be straight lines or circles that can be constructed
from p1 , p2 and p3 , p4 respectively. Then the coordinates of the points of inter√
section between A en B are contained in K or in a quadratic extension K( r)
of K, where r ∈ K, r > 0.
Proof: A straight line through the points p = (a, b) en q = (c, d) is determined
by the linear equation
(c − a)(y − b) = (d − b)(x − a).
1.1. CONSTRUCTIBILITY
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The circle with center p and passing through q is determined by the quadratic
equation
(x − a)2 + (y − b)2 = (c − a)2 + (d − b)2 .
The point of intersection between two lines through the given points is determined by a system of 2 equations with coefficients in K. Hence the coordinates
of this point of intersection lie in K.
The coordinates of the points of intersection between a circle and a line can
be found by elimination of y (or x) from the equations of the line and the
circle. We obtain a√quadratic equation with coefficients in K for x (or y). The
solutions lie in K( D) where D is the discriminant of the quadratic equation.
When D = 0, there is only solution, corresponding to one point of intersection,
and its coordinates lie in K. When D > 0 then we have two real solutions,
√
corresponding to two points of intersection, and their coordinates lie in K( D).
The case D < 0 corresponds to the case when the line and circle do not intersect.
Finally, to obtain the intersection points between two circles we need to solve a
system of two quadratic equations. However, both equations have the quadratic
term x2 + y 2 . After substraction of these two equations we are left with a linear
and a quadratic equation. We now proceed as in the previous case.
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Theorem 1.1.9 Let a be a constructible number. Then there exists a tower of
extensions
Q = K0 ⊂ K1 ⊂ · · · ⊂ Kn = K
such that:
1. K is a subfield of R,
2. a ∈ K,
3. for i = 0, . . . , n − 1 there exists 0 < ri ∈ Ki such that
√
Ki+1 = Ki ( ri )
√
and ri 6∈ Ki .
Conversely: Let
Q = K0 ⊂ K1 ⊂ · · · ⊂ Kn = K
be a chain of extensions satisfying the above conditions, then every element of
K is contructible.
Proof: Choose the coordinate system as before such that the starting points
are (0, 0) and (1, 0). The construction of a consists of several steps. In each step
one determines the points of intersection of constructible lines and circles. By
adjoining the coordinates of these points to Q we build a sequence of quadratic
extension in virtue of Theorem 0.1.8.
Conversely, by Theorems 0.1.7 and 0.1.6 every element in K is constructible.
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CHAPTER 1. CONSTRUCTIONS WITH RULER AND COMPASS
Corollary 1.1.10 Let a be a constructible number, then a is algebraic over Q
of degree 2n .
Proof: Let K be the highest degree field in a tower given Theorem 0.1.9, and
a ∈ K. Then [K : Q] = 2m because of the tower formula. The tower formula
also implies that [Q(a) : Q]|2m , since Q(a) ⊂ K.
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Corollary 0.1.10 implies the “impossibility” of the classical contruction problems
we mentioned in the beginning.
Theorem 1.1.11
1. Quadrature of the circle is impossible.
2. The cube cannot be doubled in volume.
3. The angle of 60o cannot be trisected.
Proof: 1) Take a circle with radius 1. A square with the same area would have
side π. Since π is transcendental 1 it is not algebraic,
hence not constructible.
√
3
2) Take a cube with √
side 1. The doubled cube has 2, but that number is not
constructible, since 3 2 has minimal equation X 3 − 2 which has degree 3 and
not a power of two.
3) Notice that trisection of 60o would imply the contructibility of a = cos 20o .
Using the formula cos 3α = 4 cos3 α−3 cos α with α = 20o we get 8a3 −6a−1 = 0.
So 2a satisfies X 3 −3X −1 = 0. But this polynomial is irreducible in Q[X], since
reducibility would imply that there is an integer zero which divides −1. This is
clearly not the case, so a is a cubic number, and therefore not constructible.
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The converse of Corollary 0.1.10 does not hold. For example, there are extensions Q(α) of degree of 4 with α ∈ R not constructible. In general it is
harder to derive positive construction results using Theorem 0.1.9. For this
we require more information concerning the subfields of a field extension. The
main theorem of Galois theory gives us precisely such information. One of the
main applications will be the construction of the regular 17-gon as discovered
by Gauss.
1.2
Exercises
2πi
1. Let ζ = e 5 . Show that ζ has minimal polynomial X 4 + X 3 + X 2 + X + 1
over Q. Show that ξ := ζ + ζ −1 = cos 720 and ξ satisfies ξ 2 + ξ − 1 = 0.
Use this to contruct the regular 5-gon.
2. Can the angles 900 and 1200 be trisected with ruler and compass? Prove
or disprove.
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voor een bewijs, zie bv. L. Berggren; J. Borwein, P. Borwein, Pi: a source book. SpringerVerlag, New York, 2000.
1.2. EXERCISES
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3. We are given two points P, Q with distance 1. Is it possible to construct
with ruler and compass a point R on the line P Q such that P R · QR2 =1?