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### Construction of a Regular Heptadecagon

```Date: 12/27/2009 at 04:58:01
From: David
Subject: Construction of a regular Heptadecagon

Gauss derived a finite algebraic expression for sin(pi/17) which lead
to an algorithm for the construction of the regular 17-gon.  The
expression was:

1/8 * sqrt(34-2sqrt(17)-2sqrt(2)sqrt(17-sqrt(17))-
2sqrt(68+12sqrt(17)+2sqrt(2)(sqrt(17)-1)sqrt(17-sqrt(17))-
16sqrt(2)sqrt(17+sqrt(17)))))

Daunting I know.  But that's basically my point.  I see that such an
expression must exist since 17 is a Fermat number, but I don't see how
to derive it.  If I wanted to try to figure out how to construct a
257- or 65537-gon, I'd be at a loss since without Gauss' formula, I
wouldn't even be able to construct a heptadecagon.  Going from a
trigonometric function to a rational expression in itself seems
impossible.

I've tried using power series expansions of trigonometric functions
evaluated at specific points to try to derive rational expressions to
known-to-be-rational ordinates, but haven't been able to generate
anything finite.

```

```

Date: 12/29/2009 at 05:19:27
From: Doctor Jacques
Subject: Re: Construction of a regular Heptadecagon

Hi David,

I'm afraid I cannot give you the full derivation here (there is a
lot of algebra involved), but I will try to give you an idea of the
concepts involved.

The best way to solve this problem would be to use Galois theory; if
you are familiar with it, you will find a link at the end of this

Actually, Galois theory was not yet discovered when Gauss produced
his result; however, I believe that he used implicitly some of the
symmetry concepts of the theory.

The key ingredient is a result of Lagrange.  Assume that f(x) is a
polynomial with rational coefficients of degree n, with distinct
(possibly complex) roots x_1, ..., x_n.  Let y = g(x_1,...,x_n) be a
polynomial expression (with rational coefficients) in the roots of f
(not all roots need appear explicitly).  There may be up to n! ways
of permuting the x_i in g, and each of these permutations may
produce a value of y.  Lagrange's result says that, if y can take k
distinct values for all those permutations, then y is a root of a
polynomial of degree k with rational coefficients (more precisely,
whose coefficients are rational functions of the coefficients of the
original polynomial).  Actually, there is a little more to it...

For example, consider the quadratic equation f(x) = x^2 - 2x - 1,
whose roots are x_1 = 1 + sqrt(2) and x_2 = 1 - sqrt(2).  The
expression y = x_1 + x_2 takes the same value when we interchange
x_1 and x_2; it is therefore a root of a rational equation of degree
1, which means that it is a rational number (2 in this case).

On the other hand, the expression z = x_1 - x_2 can take two
distinct values when we permute x_1 and x_2; it is therefore a root
of a polynomial of degree 2 (x^2 - 8).  Another example of such an
expression is simply x_1: this expression can take the values x_1
and x_2 when we permute the roots, and it is a root of an equation
of degree 2 (the original equation).

Lagrange used this method to give another derivation of the formulas
for the solution of the cubic and quartic equations (the formulas
were known before, but Lagrange's method does not involve the kind
of "black magic" used in the previous methods).

How can we use those ideas for solving the problem at hand?

From an algebraic point of view, constructing the regular 17-sided
polygon in a unit circle amounts to solving the equation:

z^17 - 1 = 0                  [1]

in the complex plane, by solving a set of quadratic equations (since
these correspond to the ruler and compass constructions).  This
equation has an obvious root z = 1; if we divide it by (z - 1), we
obtain an equation of degree 16:

z^16 + z^15 + ... + z + 1 = 0  [2]

The roots of this equation (called "roots of unity") are
exp(2*m*pi*i/17), where m is any integer not divisible by 17.  If z
is any of the roots, all the roots are given by z^k, where k is any
integer not divisible by 17; as z^17 = 1, the exponent is defined
modulo 17.

Note also that, as |z| = 1, 1/z^k = z^(17-k) is the complex
conjugate of z^k; this implies that z^k + 1/z^k is real; actually,
if z^k = exp(2*m*pi*i/17), we have:

z^k + 1/z^k = 2*cos(2*m*pi/17)     [3]

and this is sufficient to construct the polygon.

Consider now the expressions:

u_0 = z + z^9 + z^13 + z^15 + z^16 + z^8 + z^4 + z^2
u_4 = z^3 + z^10 + z^5 + z^11 + z^14 + z^7 + z^12 + z^6

(The reason for the subscripts in u_0 and u_4 will become apparent
later).

These are polynomial expressions in one of the roots z. If we permute
the roots of [2], only the image of z matters, since it is the only
root that actually appears in u_0 and u_4. Now, replacing z by
another root means replacing z by z^k (and reducing the exponents
modulo 17).  If you try a few cases, you will see that such a
substitution either leaves u_0 and u_4 invariant (for example z ->
z^9), or exchanges them (for example, z -> z^3).  By the remark
above, this implies that u_0 and u_4 are roots of a quadratic
equation.

We can easily find that equation.  Note first that u_0 + u_4 is the
sum of all the powers of z; by [2], this is equal to -1.  With some
patience, you can compute the 64 terms of the expansion of u_0 * u_4
(still using the fact that z^17 = 1), and you will find that this
product equals -4.  The fact that these expressions are rational is
to be expected: because a permutation of the roots of [2] at most
interchanges u_0 and u_4, both expressions remain the same under
such a permutation, and, as we have seen, this implies that they are
rational.  To summarize, we have:

u_0 + u_4 = 1
u_0 * u_4 = -4

and, as we know the sum and product of the u_i, we can find them by

u^2 - u - 4 = 0               [4]

(the fact that the discriminant of that equation is equal to 17 is
no accident).

We can now repeat the process.  We separate each of u_0 and u_4 into
two sets:

v_0 = z + z^13 + z^16 + z^4
v_2 = z^9 + z^15 + z^8 + z^2
v_4 = z^3 + z^5 + z^14 + z^12
v_6 = z^10 + z^11 + z^7 + z^6

such that u_0 = v_0 + v_2 and u_4 = v_4 + v_6.  We now look at
permutations of the roots that leave each of u_0 and u_4 invariant,
i.e., substitutions of the form z -> z^(9k).  If we look at v_0 and
v_2, we can verify that such a permutation either leaves both of
them invariant, or exchanges them (the same is true for v_4 and
v_6).  Using the same kind of argument (and much work), we obtain:

v_0 + v_ 2 = u_0
v_0 * v_2 = -1

and this shows that v_0 and v_2 are the roots of the quadratic:

v^2 - u_0 * v - 1 = 0

Now, there is no algebraic way to distinguish u_0 from u_4: they
are defined by the same equation (this is really the basic idea of
Galois theory).  Therefore, if we replace u_0 by u_4 in the above
equation, the resulting equation will produce v_4 and v_6.  We can
write the single equation:

v^2 - uv - 1 = 0                  [5]

where u stands for either root of [4].  For each root of [4], [5]
will give two values of v, for a total of 4.

As you might suspect, the next step is to separate each of the v_i
in two parts:

w_0 = z + z^16
w_1 = z^13 + z^4
w_2 = z^9 + z^8
w_3 = z^15 + z^2
w_4 = z^3 + z^14
w_5 = z^5 + z^12
w_6 = z^10 + z^7
w_7 = z^11 + z^6

Note that each of the w_i is the sum of two complex conjugate
numbers, and is therefore real.  As the v_i and u_i are sums of these
values, they are also real, which is really helpful when you attempt
the actual geometric construction.

In this case, however, we are faced with a problem.  If we attempt
the construction using w_0 and w_1, for example, we obtain:

w_0 + w_1 = v_0
w_0 * w_1 = v_4

The reason why this is a problem is that the resulting equation
would contain v_0 and v_4; as these correspond to distinct values of
u, we cannot relate them together.  We can describe the procedure as
a tree:

|
---------------------------
u0                         u4
|                          |
-----------                -----------
v0         v2              v4         v6
|          |               |          |
-------    -------         -------    -------
w0     w1  w2     w3       w4     w5  w6     w7

In the construction of any root, we are allowed to use only
"ancestors" of that root; for example, to compute w_0, we may use
v_0 and u_0.  We can also use "uncles", for example, to compute w0,
we could use v_2, because we have v_2 = u_0 - v_0, and both u_0 and
v_0 are ancestors of w_0.  However, we cannot use "cousins": we
cannot use v_4 in the computation of w_0, because v_4 corresponds to
a different choice of u in [4], i.e., a different equation, and we
cannot isolate one specific root of that equation.

However, with (heavy) algebraic manipulations (which did not deter
Gauss), it is possible to express v_4 in terms of v_0, giving:

v_4 = -(v_0^3 -6v_0 + 3)/2

and this allows us to solve the final equation:

w^2 - vw - (v^3 -6v + 3)/2 = 0       [6]

where v is either root of [5].

To summarize, we solve three quadratic equations ([4], [5], and [6])
with real roots, which can be done using ruler and compass.  Each
equation uses one root of a previous equation.  As each equation has
two roots, we get a total of 8 solutions for w, corresponding to the
values of 2*cos(2k*pi/17).

In practice, we can transform the u, v, and w (for example, by
adding constants) to make the geometric construction look "nicer".

As I said before, the proper way to formulate this would be in the
context of Galois theory.  If you are familiar with that, you may
want to read a paper I write on the subject some time ago:

Construction of Regular Polygons
http://home.scarlet.be/j-willekens/store/p17.pdf

The paper also contains the coefficients of the equations to be
solved for a 257-sided polygon; as you will see, it is hopeless to
try to find those coefficients by hand, or to attempt an actual
geometric construction (although it is theoretically possible).

Please feel free to write back if you want to discuss this further.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry
College Modern Algebra

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