Classic constructions geometry

John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
http://www.pitt.edu/~jdnorton

Three Impossible Constructions

In the ancient Greek geometric tradition, three construction problems were prominent.

Duplicating the cube. For a given cube, construct another cube that has twice its volume.

Trisecting an arbitrary angle. For any given angle, construct lines that divide it into three equal angles.

These apparently simple geometric constructions resisted all attempts at solution, by means of the standard procedures of ancient geometry, constructions with a straight edge and compass. It was only in the 19th century that proofs of their impossibility were provided. From the earliest times, however, it was recognized that the three constructions presented intractable problems for straight edge and compass constructions. Whether the constructions were just very difficult or whether they were impossible in principle, was less of an issue.

To sophisticated geometers, these three problems represented the boundary of straight edge and compass methods. To go beyond, required more sophisticated methods. Each construction is possible if geometers have access to curves and other devices not permitted by the use a straight edge and compass alone. Much of the ingenuity of ancient geometers was devoted to finding these methods.

Thomas Heath, A History of Greek Mathematics. Vol. 1 From Thales to Euclid. Oxford: Clarendon Press, p. 218.

Heath, in his expansive history of Greek mathematics. summarizes the role of these constructions as:

"Simultaneously with the gradual evolution of the Elements [of Euclid], the Greeks were occupying themselves with problems in higher geometry; three problems in particular, the squaring of the circle, the doubling of the cube, and the trisection of any given angle, were rallying-points for mathematicians during three centuries at least, and the whole course of Greek geometry was profoundly influenced by the character of the specialized investigations which had their origin in the attempts to solve these problems."

The importance of the problems persisted. In 1895, the leading geometer of the late 19th and early 20th century, Felix Klein, found them important enough to make them the subject of popular lectures. They were designed to inform the less academic geometers of his time of the problems and the methods that had finally shown their insolubility.

These impossibilities have the air of paradox to them. Pretty much all of basic geometry we learn in school is produced by constructions with straight edge and compass alone. One can spend a lifetime working on geometrical problems and never need to go beyond them. Hence, when problems like these three arise, it is surprising that ordinary straight edge and compass methods fail. It just seems that we need only to try a little harder and the construction will be found. Many problems appear intractable only because the solution is complicated and hard to find. But try as we might, no solution will be found. For these constructions are impossible.

The longevity of these problems is striking. We shall see that the problem of squaring the circle is the ancient formulation of the modern question of the value of π. We all learn quite early on that π is well approximated by 22/7. Its exact value, however, is a monstrous, irreducible, infinite decimal fraction, π = 3.1415926535897932384626433... There is a long history of "circle squarers" who insist that there has to be a simple, rational expression for π. From ancient times to the present, someone seeking to square the circle is a popular image of a deluded zealot, insistently and persistently trying to do the impossible.

Derivative Impossible Constructions

These three impossible constructions are just the beginning of a much larger collection of impossible construction. We arrive at them by working backwards. We posit some construction. Then we ask if succeeding in that construction would enable us to succeed in one of the three impossible constructions above. If it does enable us to succeed, we then know that the posited construction is also impossible.

For example, we might ask "can we divide an arbitrary angle into six equal angles using a straight edge and compass only?" The result would be the six-way division of the angle shown at left.

However, within that six-way division is a trisection, shown at right through the colored lines. Thus, if we could achieve the six-way division, we could also then achieve the trisection, which is impossible. Hence we conclude that a six-way division is impossible.

We might also ask if we can "square the semi-circle." That is, using a straight edge and compass only, can we construct a square with the same area as a given semi-circle:

Once again, if we could square the semi-circle, then we could find a way to square the circle. For the full circle is twice the area of the semi-circle. Once we have squared the semi-circle, all we need is to take the resulting square and construct a second square of double its area. That is, we need a square whose side is √2 times as big as the side of the first square. Since the diagonal of a square is √2 times as big as its side, all we need to do is to construct a square whose base is equal to the length of the diagonal of the original square. That is easily done in Euclidean geometry (as we shall see below):

With this construction, we have squared the circle. Since we know that is an impossibility, it follows that we cannot square the semi-circle.

There are many more examples like these. The impossibility of duplicating the cube leads to the impossibility of duplicating any three-dimensional figure that can be fitted into a cube.

A Paper Folding Construction

Our interest here will be in the demonstrations of the impossibility of these constructions. That is the work that puts an end to millennia of uncertainty over the exact status of these problems. The intriguing question is how this sort of impossibility can be demonstrated.

The three classic problems are eventually resolved by reducing the challenge to something that can be analyzed algebraically. They turn out to the be the same as asking which numbers are constructible by a particular repertoire of algebraic manipulations. The full analysis is difficult and requires some of the advances of 19th century mathematics. These advanced require millennia of work by mathematicians after the first posing of the problems in ancient Greece.

As preparation for the account of these three classic problems, it will be helpful to consider a much simpler problem in paper folding. The problem is one of geometry and its impossibility is demonstrated by reducing it to a problem in arithmetic.

First is a specification of the tools or methods that may be used in the construction. Without such a specification, the problem is ill-defined.

Second are attempts to use those tools and methods to carry out the construction; or, as is the case here, to prove their impossibility.

The problem is about as simple as any paper folding problem can be: take a sheet of paper and fold it exactly into thirds, using only a finite number of folds:

What makes the problem challenging is that the folds employed must be drawn from a prescribed set of folds. There are three types of folds allowed, all in the same direction:

Edge-to-edge folds. The lower edge of the sheet is folded over to match the upper edge:

Edge-to-crease folds. A fold may be formed when an edge is brought to touch a crease formed in an earlier step:

Crease-to-crease folds. A fold may be formed when an existing crease is brought to touch an existing crease, where both creases were formed in earlier steps.:

Reduction to Arithmetic

If we take the height of the sheet to be of unit size, we can see that all creases from folds must be positioned at heights that bissect the sheet; or bissect the sheet edge and existing creases; or bissect two existing creases. That means that the creases must always be in the infinite set of heights:

1/2
1/4, 2/4, 3/4
1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8
1/16, 2/16, ..., 15/16
and so on
(with obvious duplications, 1/2 = 2/4 = 4/8 = ...)

If the sheet is to be folded into thirds, we need to construct creases at the 1/3 height and the 2/3 height. The possibility of the construction then reduces to the question of whether these creases arise in the creases that can be formed by the allowed folds. We will consider just the case of the crease at the 1/3 height since showing its impossibility is enough to complete the analysis.

Assume that the folding has proceed far enough that the shortest distance between neighboring creases is 1/16. We will draw all the creases formed with this shortest distance. Is there any crease that coincides with 1/3?

The crease that comes closest is the one at 5/16. If it is to be the 1/3 height crease sought, we must have the equality

Since 3x5 =15 ≠ 16, this attempt at the construction has failed. The pattern of this calculation can be extended to any possible system of folds produced by finitely many foldings. The case just considered is that of a 4-fold foldings since 16 = 2x2x2x2 = 2⁴.

Consider the case of n-fold foldings. The number of the crease at which the 1/3 fold is to be found--call it "1/3-crease-number"--must satisfy the generalization of the condition above for the 4-fold case:

The problem of folding the sheet of paper into thirds has now been reduced to a question in arithmetic. Can there be a number that satisfies this equation? We can see that, as n gets larger, we can find numbers that get closer to the desired solution:

3x5 = 15 ≠ 16 = 2x2x2x2
3x6 = 18 ≠ 16 = 2x2x2x2

3x10 = 30 ≠ 32 = 2x2x2x2x2
3x11 = 33 ≠ 32 = 2x2x2x2x2

No matter how large we choose n, we will not find a number for 1/3-crease-number that exactly satisfies the equation. Most people will find this intuitively obvious. However merely showing failure in a few cases is not a proof.

The full demonstration requires a non-trivial fact in number theory, the fundamental theorem of arithmetic. This theorem was known in its earliest forms to Euclid. The theorem says:

The number on the right hand side of the equation is 2x2x2x...x2 is factorized by its construction into a product of 2's. The theorem tells us that there is no other factorization. To solve the equation, we would need to find a factorization of 2x2x2x...x2 that includes the prime number 3. That alternative factorization would contradict the uniqueness of the prime factorization of the fundamental theorem.

Folding the sheet into thirds is thereby shown to be impossible, on pain of violating the fundamental theory of arithmetic.

Allowing Infinitely Many Folds

The analysis above depends essentially on requiring that there be only finitely many folds. If we allow an infinity of folds, then the construction is possible. The requisite sequence of folds is this:

First produce a crease at height 1/4.
Then a second crease 1/16 height above it.
Then a third crease 1/64 height above it.
Then a fourth crease 1/256 height above it.
Then a fourth crease 1/1024 height above it.
...

Each of these creases is possible since each can be produced by bisection of intervals between earlier creases (or edges). As the sequence continues, we come arbitrarily close to 1/3:

1/4 = 0.25
1/4 + 1/16 = 0.3215
1/4 + 1/16 + 1/64 = 0.328125
1/4 + 1/16 + 1/64 + 1/256 = 0.332031
1/4 + 1/16 + 1/64 + 1/256 + 1/1024 = 0.333008
...

Proof. Write Sum = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 + ... so that
4xSum = 1 + 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 + ... = 1+ Sum. That is, 4xSum = 1+ Sum. Solve for Sum = 1/3.

These facts about limits have an especially simple expression if we rewrite them in base-2, binary arithmetic. There we have

1/4       = 0.01₂
1/16     = 0.0001₂
1/64     = 0.000001₂
1/256   = 0.00000001₂
1/1024 = 0.0000000001₂
...

Straight Edge and Ruler Construction

With this easy example showing us the way, we can return to the harder problem of the three impossible Euclidean constructions. To make the construction problem more precise, we need a clear specification of just which geometric operations are permitted by the construction problems.

Loosely speaking, the constructions are those that can be accomplished on paper by two of the most familiar instruments of geometry: a straight edge and a compass. The straight edge allows us to draw a straight line on the paper, connecting, for example, two points.

A straight edge is NOT a ruler with distance markings that would enable us to measure distances. A compass allows us to draw circles with any center and any radius:

Images from A Text Book of Ornamental Design. Scanton: International Textbook Co., 1901. p. 8, p. 11.

For more on the history of Euclid's Elements, its importance in science overall and the transition to non-Euclidean geometry, see Euclidean Geometry
The First Great Science

These two modes of construction were codified as fundamental in Euclid's, Elements, the core exposition of geometry in the ancient world and up to the early 20th century. Euclid's masterful work commences with a long list of definitions. What is a point? What is a line? What is a triangle? What is a square? And so on. There there are axioms and common notions: things equal to the same thing are equal to one another.

None of the definitions or axioms, however, asserts the existence of anything geometrical. We can give a definition of a unicorn, for example, without asserting that there is such a beast. That crucial work of defining what geometrical entities exist is taken over by Euclid's five postulates. The first three codify in abstract terms what can be drawn with a straight edge and compass.

Here are the pages showing these postulates from Billingsley's first English edition of the Elements of 1582:

In more modern language (Heath's translation of 1908, Cambridge University Press) the three postulates are:

In giving these three postulates, Euclid is restricting the sorts of things we might naturally be inclined to do with real instruments. For example, we may be inclined to identify a point on a circle by sliding our straight edge towards the circle and waiting until it makes first contact with the circle. This manipulation is not authorized by Euclid's postulates and hence may not be used.

What is left tacit is how the points present in these postulates are identified. There are two modes.

To see why this matters, without it, we could trisect an angle of square a circle just by assuming that we have placed a point that happens to have exactly the property needed for the construction.

The second is less direct. A point can be chosen anywhere and then connected according to the postulates. However the selection must be generic in the sense that the result sought must not depend on a fortuitious choice of just the one or few points that would deliver the result. The result must be recoverable no matter where the point happens to be placed.

For example, we can construct a right angle by selecting any point on a semicircle and connect it with the ends of a diameter AB. It does not matter which point C, C', C'', ... is chosen. The angles ACB, AC'B, AC''B, ... are all right angles.

The remaining two postulates are important, but raise issues not pertinent to the present concerns. The fourth postulate asserts the equality of all right angles. The fifth postulate concerns parallel lines. It became the subject of massive scrutiny and its denial figured prominently in the discovery of non-Euclidean geometries in the nineteenth century. This episode is described in detail in Euclid's Fifth Postulate and subsequent chapters.

Constructions

We can do a lot of geometry using the resources that Euclid provides. Here are some simple constructions that will be useful in the ones following.

We can always find the midpoint of some interval AB by drawing two circles of radius AB, one centered on A and the other centered on B. (Allowed by postulate 3.) The two circles intersect at C and D. We join C and D with a straight line. (Allowed by postulate 1.) Where CD intersects AB at E is the midpoint of AB; and the line CED is the perpendicular erected at E to AB.

Duplicate a square

The duplication of a square is carried out as follows. Assume that we have a square of unit side and hence unit area. Its diagonal AB is √2. Hence, if we draw a square using the diagonal as a base, the new square has sides √2 and area 2 = √2 x √2. We can draw the square by erecting perpendiculars at the ends of the diagonal AB. We then use circles centered on A and B to locate the remaining two points in the square.

If duplicating a square is so easy, can we do the same thing with a cube? The analogous construction fails. The diagonal of a cube with unit sides is √3. If we draw a cube using this diagonal as a side, we end up a volume √3 x √3 x √3 = 3√3 ≠ 2.

Bisecting an angle

To bisect an angle AOB, we draw a circle with the center at O. The circle cuts AO at A' and BO at B'. We connect A' and B' and find its midpoint C. The line OC bisects the angle AOB.

At first it seems that the analogous construction could also be used to trisect the angle AOB. Instead of bisecting A'B', we trisect it. (Trisection of an interval is possible using a straight edge and compass.)

Why is this the correct trisection? The magnitude of an angle is defined by the ratio of the circle arc it subtends to the radius of the circle. Therefore trisecting the circle arc divides the angle into three equal parts from the definition.

The construction fails to trisect the angle AOB. The correct trisection would trisect the segment of the arc of the circle connecting A' and B'. That correct trisection, shown below, leads to angles different from the first construction. (The differences are too small to be apparent in these figures. Small or not, they mean that an exact trisection has not been constructed.)

Alas, there is no straight edge and compass construction that allows us to trisect an arbitrary arc of a circle. The construction is equivalent to the trisection of an angle.

If we repeat the two constructions for a larger angle, the difference between them becomes visible in the figure. The construction on the left shows the division of the angle that results from trisection of the straight line. That division differs from the correct angle trisection shown on the right that results from trisection of the arc.

Rectifying a circle

The method of exhaustion is a method of striking power in ancient geometry that employs a version of what will eventually be generalized in the 17th century as the integral calculus. Here it is used to relate the area of circle to its circumference.

To use the method, a circle is filled with many, narrow triangles. Their apexes are at the center of the circle and their base points lie on the circumference.

The area of each triangle is (1/2) height x base. The height is approximated by the circle radius and the base is approximated by a segment of the circumference. That is we have

Each triangle is equal in area to the rectangle shown with the same base but half the height of the triangle.

We approximate the area of the circle by adding the areas of all these triangles:

Area circle ≈ (1/2) radius x (circumference segment)
                       + (1/2) radius
                              x (next circumference segment
                                    + ...

That is, the area of the circle is well approximated by a sum of all the rectangles. Their combination produces a new rectangle with a height of approximately half the radius and a base equal in length approximately to the circumference of the circle.

This result is easily affirmed by combining two standard results in geometry for a circle:
   area = pi x radius²
   circumference = 2 x pi x radius
Eliminating pi from these two equations gives us
   area = (1/2) radius x circumference

Now comes the interesting move. As we increase the number of triangles without limit, making each of them arbitrarily narrow, the heights of the triangles approach the radius of the circle and the sum of the circumference segments approaches the circumference of the circle arbitrarily well. Hence we can drop the "approximately equals" and recover the main result:

This result looks close to the much-sought squaring of the circle. The rectangle shown is equal in area to that of the circle. It is exactly half the circle radius in height and exactly the circle circumference in length.

All that remains to square the circle is to find a way of constructing a straight line whose length equals the circumference of a circle. While the figure has been drawn confidently above, the width of the rectangle was not determined by a straight edge and compass construction. As before, alas, there is no way to construct this length with a straight edge and compass. To do so would be equivalent to squaring the circle.

Construction of Lengths

That all efforts to carry out the three constructions have failed gives us some reason to believe that they may be impossible in principle. However we cannot be sure of the impossibility until a definitive proof is provided. That proof eventually appeared in the nineteenth century. It was based on a reformulation of the construction problems. Each could be seen to be equivalent to the task of constructing a prescribed length. This was the first step.

For the case of the duplication of the cube, the construction is equivalent to forming a length that is a cube root. That is, if the original cube has side one, then the cube sought as a side ³√2. For then the volume of the cube is ³√2 x ³√2 x ³√2 = 2.

The squaring of the circle amounts to the task of constructing a length equal to π. For if the original circle has a radius of unity, then its area is π. The corresponding square has a side of length √π, since then its area is √π x √π = π. As we shall see, we can construct the square root of a length with a compass and straight edge. So, if we can construct a length equal to π, the rest of the construction is manageable.

The trisection of an arbitrary angle can be reduced again to the construction of a length. To do so, we represent the original angle and the trisected angle by their heights in the figure below, where the circle has unit radius. They are y for the original angle and x for the trisected angle.

If the original angle is small, then, to very good approximation, the height for the trisected angle is one third that of the original angle, as shown on the right of the figure:

If that were all there is to the problem, it could be solved quickly. As we shall see, dividing a length into three equal parts is achievable by a straight edge and compass. However, this simple relation fails when the angle to be divided is not small. Then the linear relation needs a correction in the cube of x:

That is, the problem reduces to this: Given a length y, construct a length x that satisfies this equation.

For those who want the more technical details: The above formulae derive from the trigonometry of trisecting angles. If the original angle is θ, then y=sin(θ) and x=sin(θ/3). The trigonometric relation between them is
sin(θ) = 3sin(θ/3) - 4sin³(θ/3).
When θ/3 is small, sin(θ/3)≈θ/3 and the cubic term becomes negligible. This exposition differs from the more usual exposition, which is given in terms of cos(θ). The governing relation is:
cos(θ) = 4cos³(θ/3) - 3cos(θ/3)
Expressing the dependency between the angles in terms of sin(θ) makes it easier to see how the relation arises. It is merely a large angle correction to the the trisection of the chord enclosing the angle.

What Can be Constructed

With this simpler characterization of the three classic construction problems, we can now ask if they are within the reach of straight edge and compass constructions. The general result is that straight edge and compass construction allows us to carry out five operations on the numerical values of lengths in a Euclidean geometric space:

To see how these operations are realized, assign numerical lengths a, b, c, ... to the intervals in the figures below.

To multiply and divide, we need to introduce a fixed unit of length. To multiply lengths that measure a and b using that unit, we proceed as follows. On intersecting straight lines, we mark out adjacent distances 1 and a on one line and b on the other.

We connect the ends of the interval b and 1 with a straight line. We draw a second line parallel to this line through the end of a. Where it intersects the second line marks the end of the interval of length axb.

For division, we employ essentially the same procedure, but now in the reversed direction.

Extracting a square root requires a slightly more complicated construction. To extract the square root of a, we form an interval ABC of length a+1, with AB=a and BC=1. We draw a circle with AC as its diameter. A perpendicular to AB at B intersects the circle at D. It is easy to prove using the Pythagorean theorem that BD = √a.

Proof: For the right angle triangle ABD, we have from Pythagoras' theorem that
    AD² = AB² + BD²
For the right angle triangle CBD, we have
    CD² = BC² + BD²
Adding we find
    AD² + CD² = AB² + BC² + 2BD²
Triangle ADC is also a right angle triangle. For it we have
    AC² = AD² + CD²
Combining the last two equations, we have
    AC² = AB² + BC² + 2BD²
Substituting AC = a+1, AB=a and BC=1, we have
    (a+1)² = a² + 2a +1 = a² + 1 + 2BD²
Eliminating equals and dividing by two we have
    BD² = a
or BD = √a.

These five operations exhaust our constructive powers for lengths by means of straight edge and compass. We can, however, combine them as many times as we like, in all combinations, as long as we employ only finitely many combinations.

The inclusion of a unit length may seem to introduce an unnecessary specialization that might compromise the generality of the result. Many numerical results differ according to the unit chosen. For example, if the unit assigns 1 to some interval, then its square root is the same number √1 =1. If we change the unit so that this same interval is now assigned the number 4, then extracting the square root halves the number, √4 =2.

We can see that introducing a unit length does not harm matters, in so far as it does not affect any of the results of interest to us.

For example,the Pythagorean theorem may be expressed for the sides of some right angle triangle with lengths a, b and c as
    a² + b² = c².
If we change our unit of length so that these same sides have lengths 5a, 5b and 5c, then Pythagoras' theorem becomes
    (5a)² + (5b)² = (5c)²
which is
    25a² + 25b² = 25c²
dividing by 25 returns the original formula. The theorem holds with one selection of a unit if and only if it holds with the other.

If we want to duplicate a unit cube, we need to construct an interval of length ³√2. If that cube has a side length 5 and volume 125 with some new unit, then to duplicate it, we need to construct a cube with a side ³√(2x125) = 5³√2. Again, the problem reduces to extracting the same cube root, independently of the units used.

Field of Constructible Numbers

When we combine these five operations, as many times as we like, in all conceivable combinations, the resulting huge set of numbers is the field of constructible numbers. It has the key property of closure. That is, these five operations, acting on anything within the field merely leads us to other numbers in the field.

This closure property allows us to restate the question of the possibility of the three construction problems as a problem to do with these numbers:

Is the numerical value of the length required within the field of constructible numbers?

For the numbers in the field exhaust the constructive powers of straight edge and compass.

Duplicating the cube

Extracting a cube root is required for the duplication. An important fact about the field of constructible numbers is that it does not include cube roots.

An accessible account of these results, giving much of the mathematical steps required, is provided by Richard Courant and Herbert Robbins, What is Mathematics? 2nd ed. Oxford: Oxford Univ. Press, 1996, Ch. 3 and Benjamin Bold, Famous Problems in Geometry. New York: Dover, 1982.

Here, all I can do is state this key result. It is not obvious that no combination of the five allowed operations must fail to extract a cube root. It is one of the earliest result of 19th century group theory.

In the paper folding problem, solving an apparently simple construction problem ended up depending on a basic fact of arithmetic, the fundamental theory of arithmetic. Group theory and the closure of the field of constructible numbers plays the same role for the geometric problems.

Trisecting an arbitrary angle

The impossibility of trisecting an arbitrary angle follows from a more complicated version of this last problem. The challenge is not merely to extract a cube. It is to solve the cubic equation shown in the figure for some fixed value of y:

Once again, general solutions to this equation, for fixed values of y, are not in the field of constructible numbers.

One might imagine that this more complicated equation makes matters all the more intractible. Curiously, the situation is different. If we choose the value of y carefully, it turns out that a solution of this last equation is in the field of constructible numbers.

This case arises when we seek to trisect a right angle. Then we have y=1 and x=1/2. (From sin(90) = 1 and sin(30)=1/2.) Putting these values into the equation above we have

The trisection is then easy, as the figure shows. All that is needed is to bisect the unit height associated with the 90 degree angle. We then transport that midpoint to the enclosing circle with a line parallel to the base to identify the 30 degree trisection.

Squaring the Circle

The impossibility of squaring the circle now also follows by analogous reasoning. To succeed, we need to construct a length equal to π. This number π is not in the field of constructible numbers.

This particular case turns out to be more intractible than the other two constructions. These other constructions required only the extraction of a cube root or the solution of a cubic equation. A major result of 19th century mathematics is that π is a "transcendental number." That means that π cannot be recovered as the solution of a cubic equation, like the one above, or indeed of any analogous equation that might employ higher powers of x, as long as only finitely many powers are employed.

Other methods

Beyond straight edge and compass

If we limit ourselves to straight edge and compass constructions, nos solution is possible for these three classic construction problems. However, we need only a slight increase in the methods admissible to make the constructions possible. One approach is to give us access to curves other than arcs of circles.

A simple extension, attributed to Archimedes, is called neusis. It allows us to maneuver about with a straight edge with distances marked on it. It can be used in a simple trisection of an arbitrary angle. The figure shows the angle to be trisected.

A straight edge is used to measure the radius of circle enclosing the angle. Two marks are made on the straight edge to indicate the radial distance.

That straight edge is then shifted and rotated until the configuration shown below is achieved. That is, the straight edge connects point B with a point D on the extended diameter of the circle, such that the length CD equals that of the radius of the circle.

The resulting angle ADB trisects the original angle AOB. Since the result is a trisection of the angle, we know that this manipulation cannot be simulated by straight edge and compass construction.

Proof: Triangle DCO is an isosceles triangle, so its two base angles, CDO and COD, are equal. Call their angle size α. By the exterior angle theorem, angle OCB is equal to their sum, that is, 2α. Triangle OCB is an isosceles triangle, so its base angles, OCB and OBC are equal, and equal to 2α. Applying the exterior angle theorem to triangle ODB, we have that the original angle AOB is equal to the sum of the angles ODB and OBD, that is a + 2α, which is 3a. Hence ADB = a trisects AOB = 3a.

Infinitely many steps

The ancient geometers were certainly unwilling to entertain seriously a construction that requires us to complete and infinity of steps. As far as practically achievable geometry is concerned, the restriction is apt and not one we would not want to rescind.

However, as an abstract matter of theory, allowing infinitely many steps makes all the impossible constructions possible, even with just a straight edge and compass.

An easy case is the trisection an angle. We saw above in the paper folding problem that an infinite sequence of bisections can yield a third. The relations was summarized in the infinite series:

This series gives us a recipe for trisecting an angle using an infinity of bisections, each individually achievable by straight edge and compass construction:

Two bisections take us to the "quarter" angle, here labeled "1/4."
We add to that angle the tiny angle 1/16 produced by two further bisections of 1/4.
We add to that angle the tiny angle 1/64 produced by two further bisections of 1/16.
etc.

Constructing a length equal to π also is within reach of straight edge and compass if infinitely many steps are allowed. The recipe for them comes from the very many infinite series expressions for π that have been discovered by mathematicians over the centuries.

These expressions require only the four operations addition, subtraction, multiplication and division. Completing the infinitely many in the series would yield a length of π.

I have added some shading of like terms to aid in untangling the nestling of the square roots.

Here is a formula that employs square roots, so that its individual steps are still accessible to straight edge and compass construction:

2/π = √(1/2)
. √(1/2 + (1/2)√(1/2))
. √(1/2 . 1/2√(1/2 + (1/2)√(1/2))). ...

As a practical matter, however, these simpler series are poorly adapted to the real computation of π since many terms must be computed in order to achieve a good approximation to π.

To Ponder

Show the set of all rational numbers form a field. That is, the set is closed under addition, subtraction, multiplication and division (exclude division by zero).

Show that if we employ only the operations of addition, subtraction, multiplication and division of rational numbers that we cannot form √2.

Show that the set of numbers of the form a + b√2, where a and b are any rational numbers, form a field closed under addition, subtraction, multiplication and division (exclude division by zero) and square roots of positive numbers. (This needs a lot of algebra, but is manageable. Full proof in Benjamin Bold, Famous Problems in Geometry. pp. 11-12.)

How can we be sure addition, subtraction, multiplication, division and the extraction of square roots exhaust the repertoire of independent constructions possible with straight edge and compass?

We have seen that allowing infinitely many steps in the construction lets us trisect an angle. The various infinite series for π open the same possibility for squaring the circle. Can you see how to convert one into the squaring of the circle?

Can you find a construction with infinitely many steps that allows for the duplication of the cube?

We can enhance our constructive powers by employing curves that cannot be drawn by a straight edge and compass alone; and again by allowing infinitely many steps in the construction. Is there a repertoire of constructive procedures that can do all constructions? Or will every repertoire have a "nemesis," akin to a diagonalization, that lies beyond its constructive powers.