Chapter 1 Constructions with ruler and compass 1.1 Constructibility An important part in ancient Greek mathematics was played by the construc- tions 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, 1 2 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 62 l and right p 2 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 2 l and two other (constructible) points q; r. One can construct a point s 2 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 3 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 jaj 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 coor- dinate 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 jaj is the distance between the constructed points (a; 0) and (0; 0). Similarly for jbj. 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. 2 Theorem 1.1.6 The constructible numbers form a subfield of R containing Q. constr We shall denote the field of constructible numbers by Q . Proof: We show that for every two constructible a; b 2 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. 4 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 trian- gle. 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. 2 p Theorem 1.1.7 If a 2 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. p Dus geldt a = rs0 = r0s = s2, dus s = a. Figure 1.4: Constructie van de vierkantswortel 2 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 p ; p and p ; p respectively. Then the coordinates of the points of inter- 1 2 3 4 p section between A en B are contained in K or in a quadratic extension K( r) of K, where r 2 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 5 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 deter- mined 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 ap 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,p 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 +y2. After substraction of these two equations we are left with a linear and a quadratic equation. We now proceed as in the previous case. 2 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 2 K, 3. for i = 0; : : : ; n − 1 there exists 0 < ri 2 Ki such that p Ki+1 = Ki( ri) p and ri 62 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.