Given an initial set of points in the plane, an image can be "constructed" if it can be produced by a finite number of successive drawing operations using a straightedge and compass. The straightedge allows one to draw a line through any two designated points. The compass allows one to draw a circle of a given radius around a designated point. The radius can be set to some random value, or to the distance between two designated points. Additional points are determined by the intersections of lines and circles.
The ancient greeks developed this concept, and established procedures for constructing many common shapes. For example, given two points one unit distance apart (setting the scale), a unit equilateral triangle can be constructed. Let the segment determined by the two initial points be the base of the triangle, draw a circle of radius 1 about each endpoint, and let the intersection of these two circles be the apex of the triangle.
Actually the circles intersect in two points, one above the base and one below. So you can draw an equilateral triangle pointing up and another one pointing down, both sharing a common base. Draw a line through the two apexes and you have done two things: cut the base in half, and drawn a line perpendicular to the base. These are both useful operations. For instance, you can draw two perpendicular walls one unit apart, then a ceiling, to make a unit square.
[ triangles pointing up and down, two circular arcs intersecting in the apexes, perpendicular bisector from one apex to the other ]
But other shapes, such as the heptagon, eluded the best mathematical minds for 2,000 years. And for good reason; most of these shapes cannot be constructed. It took the mind of Gauss to determine what can be constructed and what cannot.
Given two points 1 unit apart, draw an infinite line through these points. If these points are denoted 0 and 1, draw a circle of radius 1 about 1, and mark its rightmost intersection with the line. Call this point 2. Repeat this process, marking all the integers along the x axis.
To draw the y axis, draw circles of radius 2 about 1 and -1. They intersect in two points on the traditional y axis, at ±sqrt(3). A line drawn through these points becomes the y axis. Establish the integers along the y axis, just as you did for the x axis. It is now possible to reach any lattice point in the xy plane.
If you want to cut a line segment in half, draw the perpendicular bisector, as shown above. Repeat this to cut the segment into quarters, then eighths, and so on. Thus any distance n/2m can be constructed. At this point any figure can be approximated to arbitrary precision, and the engineers are satisfied, but the mathematicians press on.
The unit segment can be divided into 2n equal parts by drawing circles of radius n around both endpoints, and treating their intersection as the apex of an isosceles triangle. Mark off integer lengths along the sides of this triangle, and project these points down into the base, thus cutting the base into 2n equal parts. (This is accomplished by connecting points above with their mirror images below, one isosceles triangle pointing up and one pointing down.) Now all rational coordinates are accessible.
[ tall isosceles triangle, 7 points marked along each side, perpendiculars drawn down to the base ]
If you have a segment of length l, set the compass to this length and draw a circle of radius l centered at the origin. This gives x and y coordinates of l. Conversely, any point with an x coordinate of l admits a perpendicular down to the x axis, and the segment from there to the origin has length l. Therefore coordinates, x or y, and lengths, are interchangeable.
Assume you are given, or you have derived, two segments whose lengths are x and y. On another line, somewhere in the plane, concatenate the distances x and y to produce the distance x+y. You can also find x-y just as easily.
Although it is not immediately obvious, you can derive the geometric mean of x and y, thanks to similar triangles. Draw a horizontal segment ab of length y. Place the point c between a and b, such that ac has length x. Let m be the midpoint of ab and draw a circle, centered at m, whose diameter is ab. Draw the vertical line through c and let it intersect the circle at d. Now ∠ adb is an inscribed angle, and is half the central angle (which happens to be a straight line). Therefore adb forms a right triangle. At the same time, acd forms a right triangle, and both triangles share a common angle at a, and a common side ad. They are similar triangles, and the distance ad = sqrt(xy).
[ picture for the above ]
Apply this procedure to the distances 1 and y and find the square root of y. Irrational numbers have entered our world! (We saw this earlier when we constructed the square root of 3.) Of course most of the real numbers will always be inaccessible, since they are uncountable, while sequences of construction operations are countable.
Deriving x2 is not hard. Let the horizontal segment ac have length 1. Draw a circle of radius x about a, and draw a vertical line through c. The circle and vertical line intersect in d. Draw a line through d, perpendicular to ad. Let this line intersect the line ac in the point b. Once again triangles acd and adb are similar. The segment ab has length x2.
[ picture for the above ]
This only works for x > 1. If x is smaller than 1, it is certainly larger than 1/n for some n. Set the length of ac to 1/n and follow the above procedure. The resulting segment ab has length nx2. Divide this by n to get x2.
Multiplication is an immediate corollary. Take the geometric mean of x and y and square it to get xy.
Now find the inverse of x. Reverse the roles of 1 and x in the previous construction. Thus ac has length x, which we assume is less than 1, and a circle of radius 1 is drawn about a. If y is the length of ab, we have x is to 1 as 1 is to y, hence y = 1/x.
[ picture for the above ]
It is interesting to see how this construction leaves 1/0 undefined. When x is 0 c and a coincide. The line through d runs parallel to b, and there is no way to complete the diagram.
If x is larger than 1 it is certainly less than n for some n. Draw a circle of radius n, instead of radius 1. This gives: x is to n as n is to y. So y = n2/x. Divide by n2 to find the inverse of x.
The standard mathematical operators have been implemented, and the set of constructible distances forms a field, with the rationals as subfield. If an additional distance, or coordinate, is brought in, call it w, all the distances in the field extension Q(w) are accessible. This includes 3+w2, 17/w, etc. Each new distance establishes a larger field extension, building a tower of fields within the reals.
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Given a distance r, we know how to take the square root of r.
Therefore all points in the field Q(sqrt(r)) are constructible,
for any positive value of r.
Take any real number s from this field and adjoin its square root. This creates another field extension of dimension 2. Continue this process indefinitely, building a tower of quadratic extensions. |
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With square roots at our disposal, any quadratic extension is fair game. Adjoin u, where u is the root of an irreducible quadratic polynomial over a prior field. Take the square root of the discriminant, then employ standard mathematical operations to obtain u. Therefore, any algebraic number that lives in a tower of quadratic extensions is constructible.
Now consider the converse. Each construction operation locates another point in the established field of points, or it extends the field. If the extension is always quadratic, we're done.
Consider the available techniques for identifying an additional point. A new point might be the intersection of two lines derived from two pairs of previously designated points. Find the intersection by solving two simultaneous linear equations in the plane. This incorporates only field operations, so the new point lies in the preexisting field. This does not create a field extension at all.
Another possibility is the intersection of a circle and a line. Write the equation of the circle as:
(x-x0)2 + (y-y0)2 = r2
Replace y with a linear function of x. The result is a quadratic equation in x. This can only create a quadratic field extension. The new point x, identified by construction, generates a field extension of dimension two. Since y is linear in x, it is part of the new extension as well.
The last possibility is the intersection of two circles. Write the two equations and subtract one from the other. The squared terms cancel, leaving a linear equation. This is, in fact, the equation of the line that passes through the two points of intersection. Intersect this line with either of the two circles, using the procedure described above. Once again the new points generate a field extension of dimension 2.
In summary, u is constructible iff it belongs to a tower of quadratic field extensions.
Don't assume every extension of dimension 2n is constructible. There are quartic extensions (i.e. extensions of dimension 4) that are not the composition of two quadratic extensions.
Given an angle in the plane, can you bisect the angle using straightedge and compass?
If v is the vertex, draw a circle about v, of any radius. Let it intersect the two rays in the points s and t. Draw two more circles, having the same radius, about s and t. These intersect in v, but they also intersect in u, inside the angle. Now the ray from v through u bisects the angle.
[ picture for the above ]
The Greeks wondered if an arbitrary angle could be trisected, i.e. cut into three equal pieces. Mathematicians worked on this problem for 2,000 years without success. Now we know why; it can't be done.
As a warmup, show the cube root of 2 is not constructible. If it were, it would belong to a field extension of dimension 2m. Yet x3-2 has no rational roots, and is irreducible. Hence the cube root of 2 belongs to a field extension of dimension 3. Since 3 does not divide 2m, the cube root of 2 is not constructible. Similar reasoning holds for the 5th root of 7, and the 6th root of 38, and so on.
Now return to angle trisection. If 60 degrees can be trisected, then the coordinate x = cos(20) is constructible. Let c be the cosine of 20 degrees and let s be the sin, and refer to the triple angle formula.
cos(3θ) = cos3(θ) - 3cos(θ)sin2(θ)
The cosine of 60 is ½. Replace sin2 with 1-c2 and get this.
½ = c3 - 3c(1-c2)
1 = 8c3 - 6c
8c3 - 6c - 1 = 0
Try the rational roots with numerator ±1 and denominator 1 2 4 or 8. None of these work, hence the cubic is irreducible. The point c generates an extension of dimension 3, and is not constructible.
Another vexing problem was called "squaring the circle". Given a circle, find a square of equal area. If the circle has radius 1, you must construct a square whose side is sqrt(π), which implies π, which is not even algebraic! (Not proved here.) Transcendental lengths cannot be constructed, hence it is not possible to square the circle.
[ picture of circle and square of equal area ]
When is a regular n-gon constructible?
Given a unit length, we know how to build an equilateral triangle and a square. A hexagon is easy; put six triangles together.
Any angle can be bisected, as shown in the last section. Use this to turn an n-gon into a 2n-gon. For example, a square can become an octagon.
If m is a proper factor of n, an n-gon implies an m-gon. For instance, connect every third point in a 15-gon to get a pentagon. Both shapes have the same radius, which could be set to 1. However, you may want your shapes to have sides of length 1, whence you will need to rescale the resulting pentagon. This is not hard to do. Find the center of the oversized pentagon and draw all 5 radii from the center to the corners. At one corner v, travel 1 unit along the side to a point u. This segment will become the base of an isosceles triangle, with the radius as one of its sides. Draw a line through u, parallel to the second radius, until it intersects the first radius at c. c becomes the center of the new pentagon. Place 5 copies of this isosceles triangle around c, and the resulting pentagon has sides of length 1.
Now consider an n-gon and an m-gon, where m and n are coprime. Work with central angles, rather than interior angles. Thus the central angle of the n-gon is c/n, where c is a special code for the entire circle, e.g. 360° Add the two central angles together to get c(m+n)/mn. Note that m+n is coprime to both m and n, hence coprime to mn. There is some k such that k(m+n) = 1 mod mn. Copy the angle k times over, wrapping around the circle, and find an angle that is equal to c/mn. This is the central angle for an mn-gon.
In summary, an mn-gon is constructible iff an m-gon and an n-gon are constructible, for m and n coprime. It is enough to investigate n-gons where n is a prime power.
Central angles can be bisected, and the square is constructible, so all powers of 2 are fair game. This includes a power of 2 times any other constructible shape, such as a 48-gon, which is a triangle enhanced by a factor of 16. So assume n is an odd prime power.
An n-gon defines, and is defined by, the radii from the center to the corners. Thus an n-gon can be constructed iff the corresponding cyclotomic extension can be constructed.
At this point I'm going to switch to complex numbers, wherein the xy plane becomes the complex plane. This supports the construction of cyclotomic extensions.
The base field is Q[i], with rational x and y coordinates. Thus far, every point z implies its rectangular coordinates. If x+yi is in the field, then x and y are also in the field. I will call this a rectified field.
Let E be rectified, and adjoin the real algebraic number u to build F. Now F consists of linear combinations of powers of u with coefficients in E. Separate these coefficients into real and imaginary parts, and verify that F is rectified.
Add, subtract, multiply, or divide complex numbers in a rectified field using real math followed by some bookkeeping. Even division works, because the norm, in the denominator, is x2+y2, which lies within the field.
Take the square root of a complex number z by taking the square root of the radius and bisecting the angle. This adjoins two real numbers in succession, the square root of the radius and an implicit square root to bisect the angle, but the result is a rectified field containing sqrt(z). Each step adjoins the square root of a real number, as we have done before.
Let L be a quadratic complex extension of a rectified field E. Take any z in L-E and write a quadratic equation in z with coefficients in E. Solve using the quadratic formula, which entails the square root of the discriminant. If the discriminant is d, a complex number, take the square root of d as described above. From there it's just field operations using elements of E. This is one or two quadratic extensions, and creates a rectified field.
Given a tower of complex quadratic extensions, convert each into two real quadratic extensions, so that fields are rectified as you go. Any point z that lives in this tower has x and y coordinates in the tower. These are constructible, and so is z.
Revisit the intersection of circles and lines. Each brings in the square root of a real number, and builds a tower of rectified fields, where each extension is quadratic. Therefore the constructible points are precisely those that live in a tower of real or complex quadratic extensions.
Suppose a p2-gon is constructible. The primitive root of 1 creates an extension whose dimension is p×(p-1), hence divisible by p. For p odd, this is impossible. And a pk-gon is impossible, since that would imply a p2-gon. So restrict attention to a p-gon, where p is an odd prime.
The dimension of the cyclotomic extension is p-1, and this has to be a power of 2. That rules out the heptagon, for example. In fact, p has to be a fermat prime.
Just because z generates an extension of dimension 2k, does not mean it meets our tower requirements. In this case it does, because a cyclotomic extension is galois, with a cyclic galois group generated by z → zb, where b is primitive mod p. A descending chain of subgroups corresponds to an ascending tower of fields, and each extension is quadratic. So the p-gon is indeed constructible.
In summary, an n-gon is constructible iff n is a power of 2 times a (possibly empty) product of distinct fermat primes.
As a proof of concept, let's construct the pentagon. The points of the pentagon are the fifth roots of 1, which lie in an extension of dimension 4. We need to climb the tower, constructing a point of dimension 2, then using this field to find the points of dimension 4.
If you're looking for an intermediate field of dimension 2, try the field fixed by complex conjugation. The subfield fixed by this automorphism lies on the real line, i.e. the x axis. If r is the fifth root of 1, let s = r+r4. This lies on the real line, yet its value, 2×cos(72), is not rational. Square s and see what happens.
s2 = r2+2+r8 = r2+r3+2 = 2-r-r4-1 = 1-s
Use the quadratic equation and the construction techniques to solve s2+s-1 = 0. Then mark this point on the x axis.
Use the segment from the origin to s as a base and draw the perpendicular bisector. This line intersects the unit circle in r and r4. The pentagon follows from there.
Gauss patiently developed a procedure for constructing the 17-gon, corresponding to the next fermat prime. Computers have since constructed the 257-gon and even the 65537-gon. If there are no other fermat primes, as is conjectured, then there are no other constructible n-gons.