From Euclid to Spigotry #8
Constructing the Regular Heptagon

Last modified: September 15, 2004
Reference: "A construction for a regular heptagon" by Crockett Johnson, Mathematical Gazette (1975), pp. 17-18 (attached)
Lecture topic 1: the rational root theorem
This is a generalization both of the proof, attributed to Euclid, that  is irrational and of the proof
from "Beyond Ruler and Compass" that the cubic equation that arises from trisecting a 600 angle
has no rational solutions.  This form of the proof is taken from Irrational Numbers by Ivan Niven.  

Theorem:
Suppose that x is a solution of a polynomial equation in which the coefficient of the highest power of x is 1 and all other coefficients are integers.  Then either x is an integer or x is irrational.

Proof:
Suppose that x is rational, so that x = a/b where b > 0 and a and b have no common factor except 1. 
We need to show that b = 1.

By hypothesis, x satisfies the equation

xn + cn-1xn-1 +cn-2xn-2 + ... + c1x + c0 = 0, where all the ci are integers.
Set x = a/b.
(a/b)n + cn-1(a/b)n-1 +cn-2(a/b)n-2 + ... + c1(a/b) + c0 = 0.
Multiply through by bn.

an + bcn-1an-1 +b2cn-2an-2 + ... + bn-1c1a + bnc0 = 0
Thus b(cn-1an-1 +bcn-2an-2 + ... + bn-2c1a + bn-1c0) = -an

This means that b is a factor of  -an. But now, to quote Niven,
"If b > 1, then any prime divisor p of b would divide an, and so p would divide a."
But this contradicts the assumption that a and b have no common factor except 1.
Therefore b = 1, and x = a/b is an integer.

Special cases:
If x satisfies x2 - 2 = 0, then either x is an integer or x is irrational.  But there is no integer whose square is 2, so  is irrational.

If x satisfies x3 - 3x -1 = 0 (from trying to trisect a 600 angle) , then either x is an integer or x is irrational.  But the only candidates for an integer solution are 1 and -1, and neither one works. So every solution x of this equation is irrational.

Lecture topic 2: Why Euclid could not construct a heptagon
Try the same approach, using complex roots of unity, that worked for the pentagon. For the heptagon this approach will lead not to a quadratic equation   Since solutions to an irreducible cubic cannot be constructed with compass and straightedge, the regular heptagon cannot be constructed.
Note first that
=
or 
So by constructing three so-called symmetric polynomials in y1, y2, and y3 we can build a
 cubic equation of which they are the roots.
Let R be a complex seventh root of unity.  Then

Set 
Set 
Set 
It is easy to see that .
Bold (page 59) shows that 
Finally,

Since 

Thus the equation whose roots are y1, y2, and y3 is F(y) = 0, where
     Check:

By the rational root theorem, if y is rational, it is an integer. The only candidates for linear factors of F(y) are therefore y+1 and y-1.  Since neither -1 nor +1 is a root of F(y)=0, there are no linear factors, and F(y) = 0 is an irreducible cubic that cannot be solved in terms of square roots.

Lecture topic 3: Johnson's construction of the regular heptagon (1975)
Although there is a difficult construction of the regular heptagon that is attributed by Arabic sources to a now-lost manuscript of Archimedes, the following construction, published by Crockett Johnson in 1975, is much simpler.  It is reminiscent both of Euclid's construction of the regular pentagon and of Archimedes' marked-ruler angle trisection.
The key idea is to construct an isosceles triangle in which each of the base angles is 3 times as large as the smallest angle, so that the angles are p/7, 3p/7, and 3p/7.

Start with a unit length BC.
Construct its perpendicular bisector
Construct a square BCDE of side 1.
The diagonal CE has length 

Now construct a circle of radius centered at C.

Take a ruler, with the unit length BC marked on it between point A and P. Align it so that
A is on the perpendicular bisector of BC, P is on the circle, and the ruler passes through B.

Now the angle a = 3p/7, the angle at A is p/7,
and the distance y between P and B satisfies
y3 + y2 - 2y -1 = 0 
and hence is equal to 2 cos 2p/7.

The angle QCA is equal to 3p/7 - p/7 = 2p/7.
The easiest way to construct the heptagon is
to extend CQ so that it meets the circle. 
Points V and W are now two vertices of the heptagon.  The rest can be constructed by
marking off lengths equal to WV around the circumference of the circle. 

Lecture topic 4: Johnson's proof that the angle q between either side of the isosceles triangle and the line AM is p/14.
The angle q = p/14 has the property that sin 4q = cos 3q.
But  =  and
=
Set  cos 3q - sin 4q= 0, factor out cos q, and you get
8 sin3 q - 4 sin2 q - 4 sin q + 1 = 0.

Let x = y + 1.
Since BC has length 1, 2 x sin q = 1.
By the Law of Cosines applied to triangle PAC, 1 + x2 - 2 x cos 2q = 2.
Set cos 2q = 1 - 2 sin2 q, eliminate x, and you get the equation above satisfied by sin p/14. 

Lecture topic 5: A proof that y = 2 cos 2p/7
Since BC has length 1, 2 (y+1) cos a = 1.
Apply the Law of Cosines to triangle CBP
1 + y2 - 2 y cos a = 2 or

y2 - 2 y cos a = 1.

Multiply by y + 1:
y2(y+1) - 2 y (y+1) cos a = y + 1

So y3 + y2 - y = y + 1
or y3 + y3 - 2 y - 1 = 0.

We already know that the only positive root of this equation is y = 2 cos 2p/7 
In the printed version, an excerpt from the Johnson article is here