Ruler-and-Compass Constructions
7.5 Construction From a Given Set of Points
There is a ‘relative’ version of the theory of this chapter, in which we start not with {0, 1} but some finite subset P ⊆ C, satisfying some simple technical condi-tions. This set-up is more appropriate for discussing constructions such as ‘given an angle, bisect it’, without assuming that the original angle is itself constructible. In this context, Definition 7.4 is modified to:
Definition 7.17. Let P be a finite subset of C containing at least two distinct el-ements, with 0, 1 ∈ P (to identify the plane with C). For each n ∈ N define sets Pn,Ln, andCnof points, lines, and circles that are n-constructible from P by:
P0 = P L0 = /0 C0 = /0
Ln+1 = {L(z1, z2) : z1, z2∈Pn}
Cn+1 = {C(z1, |z2− z3|) : z1, z2, z3∈Pn}
Pn+1 = {z ∈ C : z lies on two distinct lines in Ln+1} ∪
{z ∈ C : z lies on a line in Ln+1and a circle inCn+1} ∪ {z ∈ C : z lies on two distinct circles in Ln+1}
A point is constructible from P if it is n-constructible from P for some n.
The entire theory then goes through, with essentially the same proofs, except that the ground field Q must be replaced by Q(P) throughout. The constructible points are precisely those in Q(P)py, defined in the obvious way, and they are characterised by the existence of a tower of subfields of C starting from Q(P) such that each successive extension has degree 2. More precisely, Theorem 7.11 becomes
Theorem 7.18. A complex number α is an element of Q(P)pyif and only if there is a tower of field extensions
Q(P) = K0⊆ K1⊆ . . . ⊆ Kn= Q(α) such that
[Kj+1: Kj] = 2 for0 ≤ j ≤ n − 1.
The proof is the same.
EXERCISES
7.1 Express in the language of this chapter methods of constructing, by ruler and compasses:
(a) The perpendicular bisector of a line.
(b) The points trisecting a line.
(c) Division of a line into n equal parts.
(d) The tangent to a circle at a given point.
(e) Common tangents to two circles.
7.2 Estimate the degrees of the field extensions corresponding to the constructions in Exercise 7.1, by giving reasonably good upper bounds.
7.3 Prove using Euclidean geometry that the ‘marked ruler’ construction of Fig-ure 11 does indeed trisect the given angle AOB.
7.4 Can the angle 2π/5 be trisected using ruler and compasses?
7.5 Show that it is impossible to construct a regular 9-gon using ruler and com-passes.
7.6 By considering a formula for cos 5θ find a construction for the regular pen-tagon.
7.7 Prove that the angle θ can be trisected by ruler and compasses if and only if the polynomial
4t3− 3t − cos θ is reducible over Q(cos θ ).
7.8 Verify the following approximate construction for π due to Ramanujan (1962, p. 35), see Figure 16. Let AB be the diameter of a circle centre O. Bisect AO at M, trisect OB at T. Draw TP perpendicular to AB meeting the circle at P.
Draw BQ = PT, and join AQ. Draw OS, TR parallel to BQ. Draw AD = AS, and AC = RS tangential 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.
(You will need to know that π is approximately355113. This approximation is first found in the works of the Chinese astronomer Zu Chongzhi in aboutAD450.) 7.9 Prove that the construction in Figure 10 is correct if and only if the identity
sinθ
3 = sin θ 2 + cos θ
holds. Disprove the identity and estimate the error in the construction.
Exercises 103
FIGURE 16: Srinivasa Ramanujan’s approximate squaring of the circle.
7.10 Show that the ‘compasses’ operation can be replaced by ‘draw a circle centre P0 and passing through some point other than P0’ without altering the set of constructible points.
7.11 Find a construction with infinitely many steps that trisects any given angle θ , in the sense that the angle φnobtained by stopping the construction after n steps converges to φ = θ /3 when n tends to infinity. (Hint: consider the infinite series
1 4+ 1
16+ 1 64+ · · · which converges to 13.)
7.12 A race of alien creatures living in n-dimensional hyperspace Rnwishes to du-plicate the hypercube by ruler-and-compass construction. For which n can they succeed?
7.13 Figure 17 shows a regular hexagon of side AB = 1 and some related lines. If XY = 1, show that YB =√3
2. Deduce that the cube can be duplicated using a marked ruler.
7.14 Since the anglesθ3,θ3+2π3,θ3+4π3 are all distinct, but equal θ when multiplied by 3, it can be argued that every angle has three distinct trisections. Show that Archimedes’s construction with a marked ruler (Figure 11) can find them all.
7.15 Prove that the regular 11-gon cannot be constructed with ruler and compass.
[Hint: Let ζ = e2πi/11and mimic the proof for a heptagon.]
7.16 Prove that the regular 13-gon cannot be constructed with ruler and compass.
[Hint: Let ζ = e2πi/13and mimic the proof for a heptagon.]
7.17 The regular 15-gon and 16-gon can be constructed with ruler and compass. So the next regular polygon to consider is the 17-gon.
FIGURE 17: Duplicating the cube using a marked ruler.
Why does the method used in the previous questions fail for the 17-gon?
7.18* Prove that an angle (which you must specify and which must itself be con-structible) cannot be divided into five equal pieces with ruler and compass.
[Hint: Do not start with 2π/3 or π/2, both of which can be divided into five equal pieces with ruler and compass (why?).]
7.19 If α ∈ Q, prove that the angle θ such that tan θ = α is constructible.
7.20* Let θ be such that tan θ = a/b where a, b ∈ Z are coprime and b 6= 0. Prove the following:
(a) If a + b is odd, then θ can be trisected using ruler and compass if and only if a2+ b2is a perfect cube.
(b) If a + b is even, then θ can be trisected using ruler and compass if and only if (a2+ b2)/2 is a perfect cube.
(c) The angles tan−12/11 and tan−19/13 can be trisected using ruler and com-pass.
[Hint: Use the fact that the ring of Gaussian integers Z[i] = {p + iq : p, q ∈ Z} has the property of unique prime factorisation, together with the standard formula for tan 3θ in terms of tan θ .]
This Exercise is based on Chang and Gordon (2014).
7.21 Mark the following true or false.
(a) There exist ruler-and-compass constructions trisecting the angle to an arbitrary degree of approximation.
(b) Such constructions are sufficient for practical purposes but insufficient for mathematical ones.
(c) A point is constructible if it lies in a subfield of C whose degree over Q is a power of 2.
(d) The angle π cannot be trisected using ruler and compass.
(e) A line of length π cannot be constructed using ruler and compass.
Exercises 105 (f) It is impossible to triplicate the cube (that is, construct one with three
times the volume of a given cube) by ruler and compass.
(g) The real number π is transcendental over Q.
(h) The real number π is transcendental over R.
(i) If α cannot be constructed by ruler and compass, then α is transcendental over Q.