Download Constructibility and the construction of a 17-sided

CONSTRUCTIBILITY AND THE CONSTRUCTION OF A
17-SIDED REGULAR POLYGON
LAUREN HICKMAN
Advised by Dr. Natasa Macura
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LAUREN HICKMAN
1. Introduction
The ancient Greeks were able to perform a number of geometric constructions using a compass and a straightedge. They determined a number of basic constructions,
such as constructing perpendiculars, angle bisectors, and tangent lines to a circle.
However, some constructions presented a struggle for ancient Greek mathematicians.
The three most famous of these constructions are the problems of trisecting an angle,
squaring the circle, and doubling the cube. Two other construction problems are
those of dividing the circle into n equal arcs and constructing regular polygons[1]. In
1801, Carl Frederich Gauss published his book Disquisitiones Arithmeticae, part of
which addressed the problem of dividing the circle[4]. Then in the late 1800s, Felix
Klein wrote Famous Problems of Elementary Geometry, which builds on Gauss’s results to determine the constructibility of a regular 17-gon[8]. The algebra techniques
used by mathematicians such as Gauss and Klein prove that many ancient construction problems are impossible. However these same techniques have been used to show
that many of these constructions are in fact possible.
In this discussion, we will further explore this idea of constructibility. We will
prove several Lemmas which show that constructible numbers can be written in terms
of rational numbers, addition, subtraction, multiplication, division, and square roots.
In the next section, using algebra concepts such as fields, field extensions, and the
degree of field extensions, we will show that any number which is not written according
to our corollary is not constructible. Then, we will discuss a few of Carl Frederich
Gauss’s results in his proof of dividing the circle into p equal arcs. These results were
used by Felix Klein in proving that it is possible to construct a regular polygon of
17 sides. Klein’s proof of the constructibility of the regular 17-gon will be explained.
Finally, we will see one possible construction of this polygon.
In order to understand whether or not a construction is possible, we must first
know what it means for a point or a real number to be constructible.
Definition 1 ([7]). Given any line segment in the plane, choose rectangular coordinates so that its end points are (0, 0) and (1, 0).
Any point in the plane that can be constructed from this line segment using a
compass and straightedge is called a constructible point.
For the remainder of this discussion, we will refer to these starting points as
O = (0, 0) and A = (1, 0).
Definition 2 ([7]). A real number is called constructible if it occurs as one of the
coordinates of a constructible point.
Points can be constructed by performing the following operations a finite number
of times[2]:
• Draw a line through two previously constructed points.
• Draw a circle with center at a previously constructed point, and with radius
equal to the distance between two previously constructed points.
• Mark the points of intersection between two straight lines, two circles, or a
straight line and a circle.
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON
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We will show that the set of constructible points, we will call this set K, consists
of any point with coordinates that can be written in terms of rational numbers,
addition, subtraction, multiplication, division, and square roots. Any number which
is not written in this form is not constructible.
Some examples of constructible real numbers include:
√
• r7
q
p √
•
3− 7 5
Real numbers which are not constructible include:
√
• 32
• π
Once we have established which real numbers are constructible, we can examine
how to use algebra to model different constructions. A basic example of an algebraic
representation of a geometric construction would be the following problem:
Construct a square with an area of 4 units2
We can model this construction with the algebraic equation A = s2 = 4 where s
is the length of each side. This gives a solution of s = 2, a constructible real number.
Therefore we are able to construct a square with an area of 4 units2 .
We can apply this concept to more complicated algebraic equations. There exist
unique equations to represent lines and circles. A line can be written as x = ay + b
where a and b are constructible numbers. A circle is represented by (x − a)2 + (y −
b)2 −r 2 = 0 where x, y, r are constructible. These polynomials represent constructions
of specific objects. Given polynomials, we can determine whether or not the objects
that they represent are constructible. If all roots of a polynomial are constructible,
then the object that the polynomial represents is also constructible. If one or more
roots is not constructible, then that construction is impossible[7].
Polynomials with constructible roots:
• x2 + 5x − 6 has roots of x = 6, xq= −1
√
• x4 − 3x2 + 1 has roots of x = ± (3 ± 5)/2
Polynomials without constructible roots:
• x3
• x5 + 4x + 2
This use of algebraic equations can be demonstrated through problems that
the ancient Greeks struggled with. Three problems in particular were paid special
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LAUREN HICKMAN
attention by the Greeks. These are known as duplicating the cube, trisecting an
angle, and squaring of the circle[1].
The first of these problems, duplicating the cube, is also known as the Delian
problem. The problem is this:
Given a cube with volume V , construct a second cube with volume 2V .
Solution:
Suppose the first cube has side length 1. Then V1 = 13 = 1. The second cube
then has volume V2 = x3 = 2. Then √
the side length for the second cube is the root of
3
the polynomial x = 2, that is, x = 3 2 which is not constructible.
Therefore, the problem of duplicating the cube is not a possible construction.
The second of these problems, trisecting the angle, involves taking any given angle and constructing two lines which divide the angle into three equal angles. Certain
angles we are able to trisect. For example, since the angle 30◦ can be constructed,
we can obviously trisect a 90◦ angle. We will assume that our given angle is not
one of these special cases. First, cos θ = b, cos θ3 = y. By trigonometric identities,
cos θ = 4 cos3 3θ − 3 cos 3θ , which is equivalent to the cubic equation 4y 3 − 3y − b = 0.
This polynomial does not have constructible roots, and thus the trisection of the angle
is impossible, except for a few special cases of θ[6].
The third ancient problem, squaring the circle, involves taking a circle with a
given area and constructing a square with that same area. Notice that for a circle
√
of radius r = 1, the area is π. Each side of the square would then have length π.
Since π is not constructible, it is not possible to square the circle.
One other construction that the Greeks struggled with was the construction of
regular polygons. The Greeks were able to construct some regular polygons[2]:
• 2n sides
• 3 and 5 sides
• Any product a · b where a, b=2n , 3, or 5 and a, b are relatively prime
Unfortunately, this only includes a limited amount of numbers. In particular, it
includes only three primes. The question then became whether or not certain p-sided
regular polygons, p is prime, are constructible.
Before we can look at the construction of regular polygons, we must first consider
the division of a circle into p equal arcs. If we can divide a circle into p equal arcs,
then we can inscribe a regular p-gon in this circle where the vertices of the polygon
are the points on the circle that separate each arc.
Lemma 1.1 (Gauss[4]). Let p be a prime number. The division of the circle into p
equal parts by a straight-edge and compass is impossible unless p is of the form:
µ
p = 2h + 1 = 22 + 1.
where µ ∈ N
2
Note that the prime number 17 can be written as 17 = 22 +1, so we know that it
is not impossible to divide a circle into 17 equal parts. Felix Klein takes this problem
a step further and proves that is it in fact possible to perform such a construction. His
proof begins with a polynomial which represents this construction, then breaks down
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON
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the roots of the polynomial until he shows that these roots are indeed constructible
numbers.
2. Constructibility
First, we will discuss several allowable constructions. Given a constructible line
ℓ and a constructible point P (which may or may not lie on ℓ), we can draw a
line through P which is perpendicular to ℓ. We may also construct perpendicular
bisectors, angle bisectors, tangent lines to a circle, and others.
As mentioned earlier, a number can be constructed by marking the intersections
of lines and circles which have been constructed from previously constructed points.
We will state and prove two theorems which will determine algebraic equations for
the construction of circles and lines. Using these equations, we can set up systems of
equations. Solving these systems gives us constructible points.
Before proving these theorems, we will look at three lemmas. Note that in
proving these lemmas, we will also prove that a constructible number can be written
in terms of rational numbers, addition, subtraction, multiplication, and division. We
will also prove that square roots are constructible.
Lemma 2.1. If s is a constructible number, then −s is a constructible number. Moreover, the points (|s|, 0), (−|s|, 0), (0, |s|), and (0, −|s|) are constructible.
Proof. By definition of constructible real numbers, s is a coordinate of a constructible
point, call it P . Construct circle C centered at the origin with radius OP , and construct diameter P P ′, determined by OP . These are legitimate construction operations. Note that −s is a coordinate of P ′ . Therefore, −s is a constructible number.
Next, we will show that the points (|s|, 0), (−|s|, 0), (0, |s|), and (0, −|s|) are constructible.
Case 1: P lies on the x or y axis.
Then we construct a circle centered at O, with radius OP = |s|. This circle
intersects the coordinate axes at four points, (|s|, 0), (−|s|, 0), (0, |s|), and (0, −|s|),
all of which are constructible.
Case 2: P does not lie on the x or y axis.
We must first consider which coordinate of P is s. If s is the x coordinate,
construct a perpendicular from P to the x-axis. If s is the y coordinate, construct a
perpendicular from P to the y-axis. We will call the foot of the perpendicular point
Q. Next, construct a circle centered at O with radius OQ = s. This circle intersects
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LAUREN HICKMAN
the coordinate axes at (|s|, 0), (−|s|, 0), (0, |s|), and (0, −|s|), so all of these points
are constructible[7].
Lemma 2.2. If p and q are constructible numbers, then p + q is constructible.
Proof. By Lemma 2.1, (p, 0) is a constructible point. We will begin with proof of the
constructibility of p + q.
If q = 0, then p + q = p is constructible, so assume that q 6= 0. Construct a circle
with center (p, 0) and radius |q|. Then consider the intersections of this circle with
the x-axis. If q > 0, then the intersection to the right of (p, 0) is the point (p + q, 0).
If q < 0, then the intersection to the left of (p, 0) is the point (p + q, 0). Since these
are all allowable construction operations, p + q is a constructible point.
Lemma 2.3. If p and q are constructible numbers, then pq is constructible. Moreover,
if q 6= 0, then p/q is a constructible number.
If p or q are 0, then pq = 0 is constructible, so assume that p 6= 0 and q 6= 0.
Construct circle C centered at O with radius p. Next, construct a segment through
O
|p| |p|
and (1, 1) until it intersects C. This point of intersection is the point P = √2 , √2 ,
by the Pythagorean Theorem.
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON
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Construct Q = (|q|, 0), which is constructible by Lemma 2.1, and mark point
A = (1, 0). From here, draw segment AP , and construct a line parallel to AP and
through Q. Extend this parallel line until it intersects the segment OP , which may
need to be extended beyond circle C. Mark this intersection, point R.
Now, by a property of similar triangles,
OP
|p|
OR
|p||q|
=
=
=
1
|q|
OA
OQ
so OR = |p| · |q| = |pq|, pq is constructible.
Finally, we will prove the constructibility of pq . We begin by constructing 1q .
Construct
circle
C centered at O with radius 1, and construct segment OB, where
1
1
B = √2 , √2 . This is the intersection of the line through O and (1, 1) and circle C,
which is a valid construction operation.
Construct Q = (|q|, 0) and segment QB. Then construct a segment through A
and parallel to QB until it intersects segment OB, which we can extend if necessary.
Call the intersection of this parallel and OB point R.
By properties of similar triangles, R =
1
.
q
By multiplication, the real number p ·
1
q
1
, 1
|q| |q|
, and we can construct the point
is constructible, so
Theorem 2.4. If c is a constructible number, then
p
p
q
is constructible[7].
|c| is also constructible.
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LAUREN HICKMAN
Proof. Consider the semicircle below with diameter 1+|c|. Construct the perpendicular through the diameter, where the perpendicular divides the diameter into segments
of length 1, and |c|. Now, draw segments from where this perpendicular intersects
the circle to the endpoints of the diameter. Notice that this forms a large triangle
and two smaller triangles. All three of these triangles are similar triangles.
Notice that
tan α = d =
c
d
Then
d2 =√c
d= c
Since d is constructible,
√
c is also constructible.
Once the result of the previous Lemma is established, we know that any real
number
which can be written using nested square roots is also constructible. That is,
1
a 2b where b is a nonnegative integer and a is constructible according to the previous
three Lemmas. We can combine the results from Lemmas 2.1, 2.2, 2.3, and Theorem
2.4 to give us the following corollary:
Corollary 2.5. If a number is written in terms of addition, subtraction, multiplication, division, and square roots of rational numbers, then it is a constructible real
number.
Once this corollary for constructible points has been established, it can also be
used to examine constructible lines and circles.
Definition 3 ([7]). Consider a line, determined by constructible points X and Y . If
P and Q are constructible points which lie on line XY , then the segment P Q is a
constructible segment.
Definition 4 ([7]). A circle C is a constructible circle if its center is a constructible point and its radius r is a constructible positive real number.
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON
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These definitions of constructible segment and constructible circle are further
explored in the following theorems. These theorems establish equations for lines and
circles, which can then be used to determine constructible points.
−→
Theorem 2.6. Consider the line OA
For each constructible line l which is not parallel to OA, there corresponds a
unique pair of constructible numbers a and b such that
x = ay + b
is an equation for l.
Additionally, for each constructible line l which is parallel to OA, there is a
unique constructible number c such that
y=c
is an equation for l.
Proof. Let constructible line l, not parallel to OA be given, and suppose P Q is a
constructible segment on l. Let the coordinates of P and Q be (s, t) and (u, v),
respectively. Consider
u−s
m=
.
t−v
Also consider the triangles formed by constructing the perpendiculars from P
and Q to OA (we will call the feet of these perpendiculars P ′ and Q′ , respectively),
forming triangles △P ′ OP and △Q′ OQ. These triangles are similar and thus any
corresponding sides will be in the same ratio. From this, we know that the value m
will be the same for any pair P and Q, constructible points on l, with P 6= Q. Since
v 6= t, m is a constructible number by Lemma 2.12. The equation
x−s
m=
y−t
is therefore an equation of l, where (x, y) is any point on l which is not P . We
may rewrite this equation as
x = ay + b, where a = m and b = s − tm.
Since for any two constructible points on l, m remains the same, a is uniquely
determined by l. Additionally, b is uniquely determined by l, since the point (0, t−sm)
is the point at which l crosses the y − axis. Finally, a and b are both constructible
by Lemma 2.12.
Next, consider the case where AB is parallel to OA. Then
y−t
=0
x−s
is an equation for l, where P = (s, t) is a point on l. The point (0, t) is the
intersection of l and the y-axis, and is uniquely determined by l. We can write this
equation for l in the form
y = c where c = t[7].
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LAUREN HICKMAN
Theorem 2.7. For each constructible circle C, there is a unique equation
(x − a)2 + (y − b)2 − r 2 = 0
where a, b, and r, (r > 0) are constructible real numbers.
Proof. Let C be a constructible circle. By definition, its center P is a constructible
point, and its radius is a positive constructible real number, r. Let P have cartesian
coordinates, (a, b). Then for any point (x, y) on C, C corresponds to the equation:
(x − a)2 + (y − b)2 − r 2 = 0.
Since a, b, and r are uniquely determined by C, this correspondence is unique[7].
Using these equations for constructible lines and circles, we consider three types
of systems of equations. Solving these systems gives us constructible points.
(a) A system of two linear equations:
A1 x + B1 y = C1
A2 x + B2 y = C2
(b) A system of one linear equation and one quadratic equation:
Ax + By = C
(x − a)2 + (y − b)2 = r 2
(c) A system of two quadratic equations:
(x − a1 )2 + (y − b1 )2 = (r1 )2
(x − a2 )2 + (y − b2 )2 = (r2 )2
To demonstrate this, let us consider the system of equations:
(1) x = 2x − 3
(2) x2 + (y − 1)2 = 4
Substituting x into (2), we get
(2y − 3)2 + (y − 1)2 = 4
or
5y 2 − 14y + 6 = 0
Since the solution of this system of equations is expressed in accordance with
our constructibility corollary, (x, y) is a constructible point.
Notice that substitution gives us a single equation of degree 2. This will be the
case for any system consisting of one line and one circle. It can also be shown that
breaking down system (a) (either by substitution of elimination), will yield a linear
equation, which will obviously have constructible roots. For system (c), we get an
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 11
equation of degree 4, which we can also show has constructible roots, as equations of
degree 2k can be solved by square roots.
3. Fields
We have seen how to use equations for constructible lines and circles in order
to find constructible points. All systems of constructible lines and circles will yield
constructible roots, since we know that these objects are already constructible. What
if we begin with an equation, but we are not given that it represents a possible
construction? How do we determine constructibility without that information? We
start with some algebraic definitions.
Definition 5 ([3]). A ring is a set R with two binary operations, + and · on R,
satisfying the following:
(i) (a + b) + c = a + (b + c).
(ii) a + b = b + a.
(iii) Exists 0 ∈ R, such that a + 0 = a ∀a ∈ R.
(iv) For each a ∈ R, exists (−a) ∈ R such that a + (−a) = 0.
(v) (a · b) · c = a · (b · c).
(vi) a · (b + c) = a · b + a · c and (b + c) · a = b · a + c · a.
An example of a ring is Z, as it possesses these properties.
Definition 6 ([3]). A ring R is called a commutative ring if for every a, b ∈ R,
a · b = b · a.
Not only is Z a ring, but it is also a commutative ring. For any two integers, a, b ∈ Z,
ab = ba.
Definition 7 ([3]). Let R be a ring. A unity in R is a nonzero element that is
an identity under multiplication. If an element a ∈ R has a multiplicative inverse,
a−1 ∈ R, then a is called a unit.
For example, consider the ring Q, which has a unity, 1. We know that
Q. Since 21 · 2 = 1, 12 and 2 would be units.
1
2
and 2 are in
Definition 8 ([3]). A field is a commutative ring with unity in which every nonzero
element is a unit.
In our case of constructible numbers, we will focus primarily on the field Q.
Definition 9 ([3]). A field E is an extension field of a field F if F ⊆ E and the
operations of F are those of E.
Many times, extension fields are created
√ by joining an existing field with additional
√
elements. For example, the set {a + b 2|a, b ∈ Q}. We refer to this field as Q( 2).
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LAUREN HICKMAN
Theorem 3.1 ([5]). K is an extension field of Q.
Proof. Since 1 ∈ K and sums and differences of constructible numbers are constructible, Z ⊆ K and since quotients of constructible numbers are also constructible,
Q ⊆ K. Therefore, since the operations of Q and K are the same, K is an extension
field of Q.
Definition 10 ([3]). Let E be an extension field of a field F . E has degree n over
F and write [E : F ] = n if E has dimension n as a vector space over F .
√ √ For example, [Q (2) : Q] = 1, Q 2 : Q = 2, Q 3 2 : Q = 3. Regarding
√
√
3
3
E
as
a
vector
space,
consider
Q
2
,
an
extension
of
Q.
Then
Q
2
has basis
n √ √ o
√
3
3
3
1, 2, 22 , so the dimension of the vector space is 3 and we write Q 2 : Q =
3.
Theorem 3.2 ([3]). Let K be a finite extension field of the field K and let E be
a finite extension field of a field F . Then K is a finite extension field of F and
[K : F ] = [K : E][E : F ].
Essentially, this theorem tells us that for any series of extension fields, K0 ⊆ K1 ⊆...⊆
Kn−1 ⊆ Kn , we can find [Kn : K0 ] by taking the product of [Ki : Ki−1 ] for 1 ≤ i ≤ n.
That is,
[Kn : K0 ] = [Kn : Kn−1 ][Kn−1 : Kn−2 ]...[K2 : K1 ][K1 : K0 ].
Theorem 3.3 ([5]). The number α is constructible if and only if there exists a sequence of real fields K0 , K1 , ..., Kn such that
α ∈ Kn ⊇ Kn−1 ⊇ ... ⊇ K0 = Q
and
[Ki : Ki−1 ] = 2 for 1 ≤ i ≤ n.
We can see that this theorem is true by examining some of our previous results. From the section on constructibility, all constructible numbers are found by
solving systems of linear and quadratic equations. Suppose α represents a constructible number, found by solving one of these systems. Since α is a root of either a linear equation or a quadratic equation, [Q (α) : Q] = 1 or [Q (α) : Q] = 2.
Then α can be used to construct more lines and circles, so α becomes a coefficient of new systems of equations. Suppose that [Q (α) : Q] = 2 and β is a constructible number found by solving a new system of equations which is found using
α. Then [Q (β) : Q (α)] = 1 or 2. If [Q (β) : Q (α)] = 2, then using Theorem 3.2,
[Q (β) : Q] = [Q (β) : Q (α)] [Q (α) : Q] = 4. This process can continue, yielding more
field extensions of degree 2k for some k.
Corollary 3.4 ([5]). If α is constructible, then [Q(α) : Q] = 2r for some r ≥ 0.
Proof. By the above theorem, if α is constructible, then α ∈ Kn ⊇ Kn−1 ⊇ ... ⊇
K0 = Q where Ki is an extension field of degree 2 over Ki−1 , that is
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 13
[Ki : Ki−1 ].
By theorem 3.1
[Kn : Q (α)] [Q (α) : Q] = [Kn : Q] = [Kn : Kn−1 ] [Kn−1 : Kn−2 ] ... [K1 : Q] = 2n .
Therefore, [Q (α) : Q] divides 2n , so [Q (α) : Q] = 2r for some 0 ≤ r ≤ n.
The contrapositive assures that
Corollary 3.5. If [Q(α) : Q] 6= 2r for some r ≥ 0, then α is not constructible.
This last corollary is crucial to our study of√constructibility. To
√
demonstrate
the
3
3
use of this corollary, consider the real number 2. We have that Q 2 : Q = 3,
√
and since 3 6= 2h for some h, then 3 2 is not constructible. This tells us that the
inverse of corollary 2.5 holds true.
Corollary 3.6. If a number is not written in terms of addition, subtraction, multiplication, division, and square roots, then it is not constructible.
4. Division of the Circle Into p Equal Arcs
The problem of dividing the circle into n equal arcs is one that has been around
since ancient times. The ancient Greeks had shown the possibility of dividing the
circle into 2n , 3, and 5 equal arcs. They could also construct divisions of any product of
these numbers, ab where a and b are relatively prime. But what about other numbers?
More specifically, what about other primes? As it turns out, only certain primes
can be constructed. Carl Frederich Gauss proved this theorem in his Disquisitiones
Arithmeticae:
Theorem 4.1 (Gauss[4]). If p is a prime number, the division of the circle into p
equal parts is impossible unless p is of the form
µ
p = 22 + 1.
Gauss’s proof is very technical and includes several cases which are not applicable to our discussion. Because of this, his full proof will be excluded. However,
several pieces of his proof will be discussed, as they are essential to Klein’s proof of
constructing a regular 17-gon.
To begin, we look at the complex plane. First, note that complex numbers are
denoted z = x + iy for x, y p
∈ R. They can also be written
in the form x + iy =
y
2
2
r (cos θ + i sin θ) where r = x + y and θ = arctan x . This form of complex
numbers is another important ingredient of Klein’s proof.
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LAUREN HICKMAN
Once we have established these different forms of complex numbers, we can use
the complex plane to assist in the problem of dividing the circle. Construct a circle
of radius 1, centered at the origin in the complex plane. If we divide this circle into
n equal parts, this is the same as finding the roots of the equation z n = 1 in C.
Trivially, z = 1 is a root of this equation. Geometrically, this root represents the first
point of the division of the circle. We can disregard this point and look instead at
the other roots of the equation. Algebraically, this means we divide z n − 1 = 0 by
z − 1 to get the equation[8]:
z n−1 + z n−2 + ... + z 2 + z + 1 = 0.
Definition 11. The equation z n−1 +z n−2 +...+z 2 +z+1 = 0 is called the cyclotomic
equation.
Notice that for n = 17, the cyclotomic equation z 16 + z 15 + ... + z 2 + z + 1 = 0
represents the division of the circle into 17 equal parts. This equation is the starting
point for Klein’s proof of the construction of a regular 17-gon.
Another ingredient of Gauss’s proof involves some number theory. Specifically,
he uses Fermat’s theorem[4].
Theorem 4.2 (Fermat’s). Let p be a prime number and a be an integer which is not
divisible by p. These numbers satisfy the congruence
ap−1 ≡ 1(modp).
Now, let s be the lowest exponent such that as satisfies the above congruence.
We can show that s is a divisor of p − 1, but more specifically, we consider the case
when s = p − 1.
Definition 12 ([4]). If s is the smallest exponent such that as ≡ 1(mod p), and if
s = p − 1, we call s a primitive root of p.
To demonstrate, consider a number as (mod 17). The number 3 is a primitive
root, since for any s < 16, 3s is not congruent to 1(mod 17):
31 ≡ 3(mod 17) 35 ≡ 5(mod 17) 39 ≡ 14(mod 17) 313 ≡ 12(mod 17)
32 ≡ 9(mod 17) 36 ≡ 15(mod 17) 310 ≡ 8(mod 17) 314 ≡ 2(mod 17)
33 ≡ 10(mod 17) 37 ≡ 11(mod 17) 311 ≡ 7(mod 17) 315 ≡ 6(mod 17)
34 ≡ 13(mod 17) 38 ≡ 16(mod 17) 312 ≡ 4(mod 17) 316 ≡ 1(mod 17)
5. Klein’s proof of The Construction of the Regular Polygon of 17
Sides
Gauss proved that it is impossible to divide the circle into p equal parts unless
µ
it is of the form p = 22 + 1. However, this does not show that it is possible to
2
divide a circle into p = 17 = 22 + 1 parts; it only shows that it is not impossible.
Klein takes this a step further by showing that it is in fact possible to perform
such a construction. It turns out that the division of the circle into p equal parts is
equivalent to constructing a regular p-gon. In this section, we present Felix Klein’s
algebraic proof for this construction with additional explanations and commentary.
Klein’s method involves breaking down the roots of the cyclotomic equation of degree
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 15
16. He shows that these roots can be arranged in a particular way, so that they
also represent roots of quadratic equations. Because the roots of quadratic equations
are written in terms of addition, subtraction, multiplication, division, and square
roots, we know that they are constructible. Thus, in showing that these roots are
constructible, Klein shows that it is possible to construct a regular 17-gon[8]. To
begin this proof, we consider the roots of the cyclotomic equation
xp − 1
= x16 + x15 + ... + x2 + x + 1 = 0.
x−1
The roots of this equation, denoted ǫk for k = 1, 2, ..., 16, can be put into polar
form with r = 1 and θ = k 2π
for each k. (Note: This θ comes from the division of
17
the circle into 17 equal parts. The angle for k of these equal arcs will be k times the
):
measure of one arc, or 2π
17
ǫk = cos
2kπ
2kπ
+ i sin
, k = 1, 2, ...16.
17
17
For k = 1,
ǫ1 = cos
2π
2π
+ i sin .
17
17
Notice that using trigonometric identities, we get
2π
2π
2π
2π
4π
4π
ǫ21 = cos2
− sin2
+ 2i sin
cos
= cos
+ i sin
= ǫ2 .
17
17
17
17
17
17
This property holds for all k, that is
ǫk1 = ǫk .
Geometrically, the roots of this cyclotomic equation are actually points on the
unit circle in the complex plane. More specifically, these points are the vertices of the
regular 17-gon inscribed in this circle.
Our next step is to arrange the roots of this equation in a set order. We will
use a primitive root to aid us in this process. By definition, a is a primitive root
of 17 when the least solution of as ≡ 1(mod17) is s = 17 − 1 = 16. As mentioned
previously, 3 is a primitive root of 17, since s = 17 − 1 = 16 is the smallest value such
that 3s ≡ 1(mod17):
31 ≡ 3(mod 17) 35 ≡ 5(mod 17)
32 ≡ 9(mod 17) 36 ≡ 15(mod 17)
33 ≡ 10(mod 17) 37 ≡ 11(mod 17)
34 ≡ 13(mod 17) 38 ≡ 16(mod 17)
39 ≡ 14(mod 17) 313 ≡ 12(mod 17)
310 ≡ 8(mod 17) 314 ≡ 2(mod 17)
311 ≡ 7(mod 17) 315 ≡ 6(mod 17)
312 ≡ 4(mod 17) 316 ≡ 1(mod 17)
We then arrange our roots, ǫk , so that the indices are the above remainders in
order. Our list is as follows:
ǫ3 , ǫ9 , ǫ10 , ǫ5 , ǫ15 , ǫ11 , ǫ16 , ǫ14 , ǫ8 , ǫ7 , ǫ4 , ǫ12 , ǫ2 , ǫ6 , ǫ1
16
LAUREN HICKMAN
Notice that if 3k ≡ r(mod17), then
k
ǫr = ǫr1 = ǫ31 .
Consider the next remainder in our list, and call it r ′ . For example, if r = 3,
consider r = 9. We have that r ′ ≡ 3k+1 (mod 17). Now,
′
k+1
ǫr′ = ǫr1 = ǫ31
3
k
= ǫ31
= ǫ3r .
Therefore, each root in our list, ǫr is the cube of the preceding root. This gives
us the following cycle:
We will decompose this cycle into sums, each of which is called a period. In our
case of 16 roots, we will have periods containing 8, 4, 2, and 1 roots, as these are the
divisors of 16. The list above lists all of the periods containing 1 root. To address
the cases for periods containing 8, 4, and 2 roots, we complete the following process.
Begin by forming two periods of 8 roots, denoted by x1 and x2 , where x1 includes
the roots in even positions in our list and x2 includes the roots in odd positions.
x1 = ǫ9 + ǫ13 + ǫ15 + ǫ16 + ǫ8 + ǫ4 + ǫ2 + ǫ1
x2 = ǫ3 + ǫ10 + ǫ5 + ǫ11 + ǫ14 + ǫ7 + ǫ12 + ǫ6
We can further break down each of these x periods into four periods with four
terms each. Again, we will take the roots in ”even” and ”odd” positions within each
x, and form the following sums:
y1 = ǫ13 + ǫ16 + ǫ4 + ǫ1
y2 = ǫ9 + ǫ15 + ǫ8 + ǫ2
y3 = ǫ10 + ǫ11 + ǫ7 + ǫ6
y4 = ǫ3 + ǫ5 + ǫ14 + ǫ12
Again, operating in the same way on
terms each:
z1 = ǫ16 + ǫ1
z2 = ǫ13 + ǫ4
z3 = ǫ15 + ǫ2
z4 = ǫ9 + ǫ8
y1 , y2 , y3 , y4 , we get eight periods with 2
z5
z6
z7
z8
= ǫ11 + ǫ6
= ǫ10 + ǫ7
= ǫ5 + ǫ12
= ǫ3 + ǫ14
Notice that the remainders corresponding to the roots forming a period zi for
i = 1, 2, ..., 8 always sum to equal 17. For example, z1 = ǫ16 + ǫ1 , 16 + 1 = 17. For
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 17
each zi , these roots can be written generally as ǫr and ǫ17−r . Using the polar form for
ǫk ,
2π
2π
ǫr = cos r
+ i sin r , and
17
17
ǫ17−r = cos (17 − r)
= cos 2π −
2π
2π
+ i sin (17 − r)
17
17
2π
2π
+ i sin 2π −
17
17
2π
2π
− i sin r
17
17
by the formulas cos (u − v) = cos u cos v + sin u sin v and sin (u − v) = sin u cos v −
cos u sin v.
Therefore, when we add these roots together, the imaginary portions cancel. We
are left with a real number,
2π
ǫr + ǫ17−r = 2 cos r .
17
= cos r
Then, for each of the periods zi , we now know that zi is a real number, and we
can write:
z1 = 2 cos
2π
17
z5 = 2 cos 4
2π
17
z2 = 2 cos 4
2π
2π
z6 = 2 cos 4
17
17
z3 = 2 cos 4
2π
2π
z7 = 2 cos 4
17
17
z4 = 2 cos 4
2π
2π
z8 = 2 cos 4
17
17
Additionally,
y1
y2
y3
y4
= z1 + z2
= z3 + z4
= z5 + z6
= z7 + z8
And
x1 = z1 + z2 + z3 + z4
x2 = z5 + z6 + z7 + z8
Now that we have established exactly what our periods are, we will determine
the magnitude of each one (specifically, each zi ) relative to others. To do this, first
consider a semicircle of radius 1. Divide this semi-circle into 17 equal parts, and label
the points of division A1 , A2 , ..., A17 , where the segment between our starting point,
O, and A17 is the diameter. (Note: this is not an actual construction, but rather a
18
LAUREN HICKMAN
method used for determining magnitude.) Next, label the distances between O and
each Al , as S1 , S2 , ..., S17 , corresponding to A1 , A2 , ..., A17 respectively.
Consider ∠Al A17 O, labeled as ∠X in the diagram below. Notice that m∠X is
equal to half the measure of the arc Al O. This arc has measure 2lπ
, since the measure
34
1
of this arc is l times the measure of arc OA1 , and arc OA1 measures 34
of the circle.
(17−l)π)
lπ
Thus m∠X = 34 . Also notice that for ∠Y = ∠Al OA17 , m∠Y = 34 . Now,
sin
Sl = 2 sin
lπ
Sl
=
34
2
lπ
17 − lπ
= 2 cos
34
34
for some h. Therefore, we must have
Notice that each z = 2 cos h 2π
17
(17 − l) π
2π
=h
34
17
4h = 17 − l
l = 17 − 4h
From here, we plug in the values of 1, 2, 3, ..., 8 for h. This corresponds to the h
values in each z period. That is, for z1 , h = 1, for z2 , h = 4, for z3 , h = 2, etc. To
demonstrate, let h = 1.
l = 17 − 4 = 13
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 19
S13 = 2 cos
S13 = 2 cos
(17 − 13) π
34
2π
4π
= 2 cos
= z1
34
17
So z1 has relative magnitude S13 , which corresponds to the length from O to
A13 . Once this is completed for all of h = 1, 2, ..., 8, we get corresponding values
l = 13, 9, 5, 1, −3, −7, −11, −15. Therefore
z1 = S13
z5 = −S7
z2 = S1
z6 = −S11
z3 = S9
z7 = −S3
z4 = −S15 z8 = S5
Since Sl increases as i increases:
z4 < z6 < z5 < z7 < z2 < z8 < z3 < z1 .
The relative magnitudes of each z period can be used to determine the relative
magnitude of each y period. Consider the image below.
Using the triangle inequality, we know that
Sk+p < Sk + Sp .
Additionally, the vertex at point Ak can be shifted along the circle. This is
accounted for algebraically by shifting magnitudes Sk to Sk+r and Sp to Sp+(17−r) .
This expands our inequality, and
Sk+p < Sk+r + Sp+(17−r) .
To determine the relative magnitude of y1 , y2 , y3 , and y4 , we can take their
differences. To demonstrate, consider y1 − y2 = S13 + S1 − S9 + S15 . Using the
inequalities above, we know that S13 + S1 + S15 > S13+1+15 Additionally, when we
consider (13 + 1 + 15) mod 17, we get S13+1+15 = S12 > S9 , so y1 − y2 = S13 + S1 −
S9 + S15 > 0, and y1 > y2 . Taking the differences between the rest of our y periods
gives us the following:
20
LAUREN HICKMAN
y1 − y2 = S13 + S1 − S9 + S15 > 0
y1 − y3 = S13 + S1 + S7 + S11 > 0
y1 − y4 = S13 + S1 + S3 − S5 > 0
y2 − y3 = S9 − S15 + S7 + S11 > 0
y2 − y4 = S9 − S15 + S3 − S5 < 0
y3 − y4 = −S7 − S11 − S3 − S5 < 0
Therefore,
y3 < y2 < y4 < y1 .
Using this same process with x1 and x2 , we determine
x2 < x1 .
Next, we will find quadratic equations for which each of our periods is a root.
To begin, consider the periods z1 and z2 . First, we have that
z1 + z2 = y1 .
Our next step is to determine z1 z2 :
z1 z2 = (ǫ16 + ǫ1 ) (ǫ13 + ǫ4 ).
Notice that:
Then
ǫk · ǫp = ǫk1 · ǫp1 = ǫk+p
= ǫk+p .
1
z1 z2 = ǫ16 ǫ13 + ǫ16 ǫ4 + ǫ1 ǫ13 + ǫ1 ǫ4
z1 z2 = ǫ16+13 + ǫ16+4 + ǫ1+13 + ǫ1+4
z1 z2 = ǫ12 + ǫ3 + ǫ14 + ǫ5
z1 z2 = y4 .
Therefore we may construct the quadratic equation with roots z1 and z2 :
(z − z1 )(z − z2 ) = z 2 − (z1 + z2 )z + (z1 z2 ) = z 2 − y1 z + y4 = 0.
Since z1 > z2 , these roots may also be written as:
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 21
z1 =
y1 +
p
y1 −
y12 − 4y4
, z2 =
2
p
y12 − 4y4
.
2
We must determine y1 and y4 . First, consider y1 and y2 which form the period
x1 . That is,
y1 + y2 = x1 .
Additionally,
y1 y2 = (ǫ13 + ǫ + 16 + ǫ4 + ǫ1 ) (ǫ9 + ǫ15 + ǫ8 + ǫ2 ).
y1 y2 =
16
X
ǫk
k=1
y1 y2 = −1.
Therefore, y1 and y2 are the roots of the equation:
(y − y1 ) (y − y2 ) = y 2 − x1 y − 1 = 0.
And since y1 > y2 ,
y1 =
x1 +
p
x21 + 4
2
, y2 =
x1 −
p
x21 + 4
p
x22 + 4
2
.
Similarly, for y3 and y4 ,
y3 + y4 = x2
and
y3 y4 = −1.
Therefore, y3 and y4 are roots of the equation
y 2 − x2 y − 1 = 0.
And since y4 > y3 ,
y3 =
x2 +
p
x22 + 4
2
, y4 =
x2 −
We will determine x1 and x2 . First,
x1 + x2 =
16
X
k=1
And
ǫk = −1.
2
.
22
LAUREN HICKMAN
x1 x2 =
(ǫ13 + ǫ16 + ǫ4 + ǫ1 + ǫ9 + ǫ15 + ǫ8 + ǫ2 ) (ǫ10 + ǫ11 + ǫ7 + ǫ6 + ǫ3 + ǫ5 + ǫ14 + ǫ12 ).
Expanding this gives us:
x1 x2 = 4
16
X
k=1
ǫk = −4.
Therefore, x1 and x2 are the roots of the equation
x2 − x − 4.
And since x1 > x2
√
√
−1 − 17
−1 + 17
x1 =
, x2 =
.
2
2
We can work back through x1 , y1 , and y4 to find an expression for z1 . First
√
−1 + 17
x1 =
,
2
r
y1 =
y4 =
√
−1+ 17
2
√
−1− 17
2
+
+
√ 2
−1+ 17
2
2
r
√ 2
−1− 17
2
+4
=
−1 +
=
−1 −
+4
2
√
p
√
17 + 34 + 2 17
,
4
√
p
√
17 + 34 − 2 17
.
4
Then, plugging into z1 , we get
√
−1+ 17+
√
√
34+2 17
4
z1 =
+
s
√
√
√
−1− 17+ 34−2 17
4
4
2
−4
√
√
√
−1− 17+ 34−2 17
4
2
.
Finally z1 reduces to
z1 =
p
√
17 + 34 + 2 17
+
8
q
p
√
√
√ p
√
68 + 12 17 − 16 34 + 2 17 − 2 1 − 17
34 − 2 17
−1 +
√
8
.
Notice that since z1 consists of only addition, subtraction, multiplication, division, and square roots, z1 is a constructible number. Also, z1 = ǫ16 + ǫ1 , and dividing
z1 by 2 gives us the real portion of both ǫ16 and ǫ1 . This is also a real number. We
can now construct a perpendicular through this point, which intersects the circle at
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 23
ǫ16 and ǫ1 . Therefore, these are constructible. This is one side of our polygon, so
the other sides can be constructed from this. The construction of a regular 17-gon is
possible[8].
6. Constructing the regular 17-gon
Dozens of different constructions for the regular 17-gon have been given, some before Klein’s algebraic proof, some after. Gauss and Klein both gave possible constructions, both of which were fairly complex. The following construction demonstrates a
more modern (and much more simple) construction of the polygon.
First, construct a circle with center O and diameter BA. Construct the
perpendicular bisector of BA, segment OC.
Construct OD = OC/4. Then construct segment DA.
24
LAUREN HICKMAN
1
Bisect angle ODA twice to construct angle ODE, where the measure of ODE is
4
that of angle ODA.
Construct angle EDF = 45 degrees. Then construct a semi-circle with diameter
AF , intersecting OC at G.
Construct a semicircle with center E and radius EG, which intersects AB at H and
K. Construct HL and KM, both perpendicular to AB.
CONSTRUCTIBILITY AND THE CONSTRUCTION OF A 17-SIDED REGULAR POLYGON 25
360
720
Measure of angle LOM =
, so bisecting this angle gives us
. We have now
17
17
divided the circle into 17 equal parts. Connecting these points of division gives us
our 17-sided regular polygon[2].
7. Conclusion
The results from Gauss and Klein show a beautiful relationship between algebra
and geometry. The connection between these subject areas is what drew me to the
topic of geometric constructibility. This geometric application of algebra is very
tangible, and it shows that algebra is essential to the understanding of geometry.
Similarly, geometry can aid in the understanding of algebraic concepts. By studying
the relationships between topics in mathematics, we can understand much more about
these different subject areas.
26
LAUREN HICKMAN
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
Greek Mathematical Works, volume I. Thales to Euclid.
Benjamin Bold. Famous Problems of Mathematics. Van Nostrand Reinhold Company, 1969.
Joseph A. Gallian. Contemporary Abstract Algebra. Brooks Cole, 2004.
Carl Frederich Gauss. Disquisitiones Arithmeticae. Yale University Press, 1966.
William J. Gilbert. Modern Algebra with Applications. John Wiley and Sons, Inc., 1976.
Hilda P. Hudson. Ruler and Compasses. Longmans, Green and Co., 1916.
Nicholas D. Kazarinoff. Ruler and the Round. Prindle, Weber and Schmidt, Inc., 1970.
Felix Klein. Famous Problems of Elementary Geometry. Ginn and Company, 1897.