Approximate Constructions and More General Instruments

For the technical drawing expert we emphasise that we are discussing exact con-structions. There are many approximate constructions for trisecting the angle, for instance, but no exact methods. Dudley (1987) is a fascinating collection of approxi-mate methods that were thought by their inventors to be exact. Figure 10 is a typical example. To trisect angle BOA, draw line BE parallel to OA. Mark off AC and CD equal to OA, draw arc DE with centre C and radius CD. Drop a perpendicular EF to OD and draw arc FT centre O radius OF to meet BE at T. Then angle AOT approxi-mately trisects angle BOA. See Exercise 7.10.

FIGURE 10: Close—but no banana.

The Greeks were well aware that by going outside the Platonic constraints, all three classical problems can be solved. Archimedes and others knew that angles can be trisected using a marked ruler, as in Figure 11. The ruler has marked on it two points distance r apart. Given∠AOB = θ draw a circle centre O with radius r, cutting OA at X, OB at Y. Place the ruler with its edge through X and one mark on the line OY at D; slide it until the other marked point lies on the circle at E. Then∠EDO

= θ /3. For a proof, see Exercise 7.3. Exercise 7.14 shows how to duplicate the cube using a marked ruler.

Setting your compasses up against the ruler so that the pivot point and the pen-cil effectively constitute such marks also provides a trisection, but again this goes beyond the precise concept of a ‘ruler-and-compass construction’. Many other uses of ‘exotic’ instruments are catalogued in Dudley (1987), which examines the history of trisection attempts. Euclid may have limited himself to an unmarked ruler (plus compasses) because it made his axiomatic treatment more convincing. It is not en-tirely clear what conditions should apply to a marked ruler—the distance between the marks causes difficulties. Presumably it ought to be constructible, for example.

The Greeks solved all three problems using conic sections, or more recondite curves such as the conchoid of Nichomedes or the quadratrix (Klein 1962, Coolidge

FIGURE 11: Trisecting an angle with a marked ruler.

1963). Archimedes tackled the problem of Squaring the Circle in a characteristically ingenious manner, and proved a result which would now be written

3¹⁰₇₁< π < 3¹₇

This was a remarkable achievement with the limited techniques available, and refine-ments of his method can approximate π to any required degree of precision.

Such extensions of the apparatus solve the practical problem, but it is the theo-retical one that holds the most interest. What, precisely, are the limitations on ruler-and-compass constructions? With the machinery now at our disposal it is relatively simple to characterise these limitations, and thereby give a complete answer to all three problems. We use coordinate geometry to express problems in algebraic terms, and apply the theory of field extensions to the algebraic questions that arise.

7.2 Constructions in C

We begin by formalising the notion of a ruler-and-compass construction. Assume that initially we are given two distinct points in the plane. Equivalently, by Euclid’s Axiom 1, we can begin with the line segment that joins them. These points let us choose an origin and set a scale. So we can identify the Euclidean plane R²with C, and assume that these two points are 0 and 1.

Euclid dealt with finite line segments (his condition (1) above) but could make them as long as he pleased by extending the line (condition (2)). We find it more convenient to work with infinitely long lines (modelling an infinitely long ruler), which in effect combines Euclid’s conditions into just one: the possibility of drawing the (infinitely long) line that passes through two given points. From now on, ‘line’ is always used in this sense.

Constructions in C 91 If z₁, z₂∈ C and 0 ≤ r ∈ R, define

L(z₁, z₂) = the line joining z1to z₂ (z₁6= z₂) C(z₁, r) = the circle centre z1with radius r > 0 We now define constructible points, lines, and circles recursively:

Definition 7.2. For each n ∈ N define sets Pn,Ln, andCnof n-constructible points, lines, and circles, by:

P0 = {0, 1}

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}

FIGURE 12: The setP1. Figure 12 shows that

P1= {−1, 0, 1, 2,1 ± i√ 3

2 }

Lemma 7.3. For all n ∈ N,

Pn⊆Pn+1 Ln⊆Ln+1 Cn⊆Cn+1

and each is a finite set.

Proof. The inclusions are clear. Let p_nbe the number of points inPn, l_nthe number of lines inLn, and c_nthe number of circles inCn. Then

|Ln+1| ≤ ¹₂p_n(p_n+ 1)

|Cn+1| ≤ p_n¹₂p_n(p_n+ 1)

|Pn+1| ≤ ¹₂ln+1(l_n+1+ 1) + 2lncn+ c_n+1(c_n+1+ 1)

bearing in mind that a line or circle meets a distinct circle in ≤ 2 points. By induction, all three sets are finite for all n.

We formalise a Euclidean ruler-and-compass construction using these sets. The intuitive idea is that starting from 0 and 1, such a construction generates a finite sequence of points by drawing a line through two previously constructed points, or a circle whose centre is one previously constructed point and whose radius is the distance between two previously constructed points, and then defining a new point using intersections of these.

Definition 7.4. A point z ∈ C is constructible if there is a finite sequence of points z₀= 0, z1= 1, z2, z₃, . . . z_k= z (7.1) such that z_j+1lies in at least one of:

L(z_j1, z_j2) ∩ L(z_j3, z_j4) L(z_j1, z_j2) ∩C(z_j3, |z_j4− z_j5|) C(z_j1, |z_j2− z_j3|) ∩C(z_j4, |z_j5− z_j6|) where all j_i≤ j and the intersecting lines and circles are distinct.

In the first case, the lines must not be parallel in order to have non-empty in-tersection; in the other cases, the line must meet the circle and the two circles must meet. These technical conditions can be expressed as algebraic properties of the z_j.

We can now prove:

Theorem 7.5. A point z ∈ C is constructible if and only if z ∈ Pnfor some n∈ N.

Proof. Let z ∈ C be constructible, using the sequence (7.1). Inductively, it is clear that z = z_k∈Pk.

Conversely, let z ∈Pk. Then we can find a sequence z_j∈Pj, where 0 ≤ j ≤ k, satisfying (7.1).

To characterise constructible points, we need:

Definition 7.6. The Pythagorean closure Q^pyof Q is the smallest subfield K ⊆ C with the property:

z∈ K =⇒ ±√

z∈ K (7.2)

The Pythagorean closure of Q exists because every subfield of C contains Q, so Q^py is the intersection of all subfields of C satisfying (7.2).

Constructions in C 93 The main theorem of this section is:

Theorem 7.7. A point z ∈ C is constructible if and only if z ∈ Q^py. Equivalently,

∞ [

n=0

Pn= Q^py (7.3)

Pre-proof Discussion.

We can summarise the main idea succinctly. Coordinate geometry in C shows that each step in a ruler-and-compass construction leads to points that can be expressed using rational functions of the previously constructed points together with the square root of a rational function of those points. Conversely, all rational functions of given points can be constructed, and so can square roots of given points. Therefore anything that can be constructed lies in Q^py, and anything in Q^pycan be constructed.

The details require some algebraic computations in C and some basic Euclidean geometry. We prove Theorem 7.7 in two stages. In this section we show that

(A) Pn⊆ Q^pyfor all n ∈ N.

In the next section, after describing some basic constructions for arithmetical opera-tions and square roots, we complete the proof by establishing

(B) If z ∈ Q^pythen z ∈Pnfor some n ∈ N.

Equation (7.3) is an immediate consequence of (A) and (B).

Proof of Part (A). Part (A) follows by coordinate geometry in C ≡ R². The details are tedious, but we give them for completeness. Use induction on n. Since P0= {0, 1} ⊆ Q, we have P0∈ z. Suppose inductively thatPn⊆ Q^py, and let z ∈Pn+1. We have to prove that z ∈ Q^py.

There are three cases: line meets line, line meets circle, circle meets circle.

Case 1: Line meets line. Here {z} = L(z1, z₂) ∩ L(z₃, z₄) where the z_j∈Pn⊆ Q^py (induction hypothesis) and the lines are distinct. Therefore there exist real α, β such that

z= αz1+ (1 − α)z2

z= β z₃+ (1 − β )z₄ Therefore

α =β (z3− z₄) + z₄− z₂ z₁− z₂ Since α, β ∈ R, we also have

α =β (z₃− z₄) + z₄− z₂ z₁− z₂

where the bar is complex conjugate. These two equations have a unique solution for

α , β because we are assuming that the lines meet at a unique point z, and the solution is:

α = z₂(z₄− z₃) + z₂(z₃− z₄) − z₃z₄+ z₄z₃ (z₁− z₂)(z₃− z₄) + (z₄− z₃)(z₁− z₂) β = z3(z₁− z2) + z₃(z₂− z1) − z₂z1+ z₁z2

(z₄− z₃)(z₂− z₁) + (z₁− z₂)(z₄− z₃) so α, β ∈ Q^py. Then z = αz1+ (1 − α)z2∈ Q^py.

Case 2: Line meets circle. Here z ∈ L(z₁, z₂) ∩ C(z₃, |z₄− z₅|) where the z_j ∈ Pn⊆ Q^py(induction hypothesis). Let r = |z₄− z₅|. There exist α, θ ∈ R such that

z= αz1+ (1 − αz2) z= z₃+ re^iθ Therefore

α (z₁− z₂) + z₂ = z₃+ re^iθ α (z1− z₂) + z₂ = z₃+ re^−iθ

where we take the complex conjugate to get the second equation. We can eliminate θ to get

(α(z1− z₂) + z₂− z₃)(α(z1− z₂) + z₂− z₃) = re^iθ.re^−iθ = r²= (z₄− z₅)(z₄− z₅) which is a quadratic equation for α with coefficients in Q^py. Since the quadratic formula involves only rational functions of the coefficients and a square root, α ∈ Q^py. Therefore z ∈ Q^py.

Case 3: Circle meets circle. Here z ∈ C(z₁, |z₂− z₃|) ∩ C(z₄, |z₅− z₆|) where the z_j ∈Pn⊆ Q^py (induction hypothesis). Let r = |z₂− z₃|, s = |z₅− z₆|. There exist θ , φ ∈ R such that

z = z₁+ re^iθ z = z₄+ se^iφ Take conjugates and eliminate θ , φ as above to get

(z − z₁)(z − z₁) = r² (z − z₄)(z − z₄) = s²

Solving for z and z (left as an exercise) we find that z satisifies a quadratic equation with coefficients in Q^py. Therefore z ∈ Q^py.