User:AndrewKepert/17gon

Theory
The most elegant analysis of the constructibility of a regular n-gon requires an understanding of the theory of finite groups. This is outlined in the constructible polygon article. However, it is possible to describe and justify a construction of the heptadecagon based on elementary trigonometry. Within this construction and the associated calculations, there are steps that can be clearly identified as successive quadratic extensions of the field of rational numbers. An equivalent description based on complex roots of unity can also be derived.

Basic identities

 * 2cos(A)cos(B) = cos(A+B) + cos(A-B)
 * For any positive integer n and any integer k not a multiple of n, let &theta;=2&pi;k/n, then cos(0&theta;) + cos(1&theta;) + cos(2&theta;) + ... + cos((n-1)&theta;) = 0.  Consequently, if n=2m+1 is odd, then cos(1&theta;) + cos(2&theta;) + ... + cos(m&theta;) = -1/2.  This identity can be proven geometrically, or by noting that if S is the sum on the left hand side, then Scos(&theta;)=S, and since cos(&theta;)&ne;1, S=0.

Definitions
Define &theta;=2&pi;/17 and then for q=1,...,8 define
 * &alpha;q = cos(q&theta;) + cos(2q&theta;) + cos(4q&theta;) + ... + cos(128q&theta;)
 * &beta;q = cos(q&theta;) + cos(4q&theta;) + cos(16q&theta;) + cos(64q&theta;)
 * &gamma;q = cos(q&theta;) + cos(16q&theta;) = cos(q&theta;) + cos(-q&theta;) = 2cos(q&theta;)

These are defined for all integer q, but since the cosine function is periodic and symmetric, we can deduce that &alpha;q=&alpha;17+q=&alpha;17-q, and so we only use these eight values of q. Moreover, cos(256q&theta;)=cos(q&theta;) and so &alpha;2q=&alpha;q and &beta;4q=&beta;q. In particular,
 * &alpha;1=&alpha;2=&alpha;4=&alpha;8 = 2cos(q&theta;) + 2cos(2q&theta;) + 2cos(4q&theta;) + 2cos(8q&theta;) &gt; 0
 * &alpha;3=&alpha;6=&alpha;5=&alpha;7 = 2cos(3q&theta;) + 2cos(6q&theta;) + 2cos(5q&theta;) + 2cos(7q&theta;) &lt; 0
 * &beta;1=&beta;4=2cos(q&theta;) + 2cos(4q&theta;)
 * &beta;2=&beta;8=2cos(2&theta;) + 2cos(8q&theta;)
 * &beta;3=&beta;5=2cos(3q&theta;) + 2cos(5q&theta;)
 * &beta;6=&beta;7=2cos(6q&theta;) + 2cos(7q&theta;)

from which it can be seen that &gamma;q + &gamma;4q = &beta;q, &beta;q + &beta;2q = &alpha;q and &alpha;q + &alpha;3q = 2cos(q&theta;) + 2cos(2q&theta;) + ... + 2cos(8q&theta;) = -1

Quadratic relations
Using the identity for cos(A)cos(B) above, it can be verified that
 * &gamma;q2 = &gamma;2q + 2
 * &gamma;q&gamma;4q = &beta;3q
 * &beta;q2 = ( &gamma;q + &gamma;4q ) 2 = 4 + &beta;2q + 2&beta;3q
 * &beta;q&beta;2q = -1
 * &alpha;q2 = 6 + &alpha;q + 2&alpha;3q = 4 - &alpha;q

and so &alpha;q = ( -1 &plusmn; &radic;17 )/2. Consequently
 * &alpha;1=&alpha;2=&alpha;4=&alpha;8 = ( -1 + &radic;17 )/2 &asymp; 1.56
 * &alpha;3=&alpha;6=&alpha;5=&alpha;7 = ( -1 - &radic;17 )/2 &asymp; -2.56

The &beta;q values can be found by verifying that
 * ( &beta;2q - &beta;q )2 = ( &beta;q + &beta;q )2 - 4 &beta;q&beta;q = &alpha;q2 + 4 = 8 - &alpha;q so that &beta;2q - &beta;q = &radic;( 8 - &alpha;q )
 * Also &beta;2q + &beta;q = &alpha;q so that &beta;q = (&alpha;q &plusmn; &radic;( 8 - &alpha;q ) )/2.

From relative sizes of the cos(k&theta;) terms, we can deduce
 * &beta;1=&beta;4=(&alpha;1 + &radic;( 8 - &alpha;1 ) )/2 = ( -1 + &radic;17 + &radic;(34-2&radic;17))/4 &asymp; 2.05
 * &beta;2=&beta;8=(&alpha;1 - &radic;( 8 - &alpha;1 ) )/2 = ( -1 + &radic;17 - &radic;(34-2&radic;17))/4 &asymp; -0.49
 * &beta;3=&beta;5=(&alpha;3 + &radic;( 8 - &alpha;3 ) )/2 = ( -1 - &radic;17 + &radic;(34+2&radic;17))/4 &asymp; 0.34
 * &beta;6=&beta;7=(&alpha;3 - &radic;( 8 - &alpha;3 ) )/2 = ( -1 - &radic;17 - &radic;(34+2&radic;17))/4 &asymp; -2.91

For the &gamma;q terms,
 * ( &gamma;q - &gamma;4q )2 = ( &gamma;q + &gamma;4q )2 - 4&gamma;q&gamma;4q = &beta;q2 - 4&beta;3q = 4 + &beta;2q - 2 &beta;3q so that &gamma;q - &gamma;4q = &radic;( 4 + &beta;2q - 2 &beta;3q )

and since &gamma;q + &gamma;4q = &beta;q it follows that
 * &gamma;q = ( &beta;q &plusmn; &radic;( 4 + &beta;2q - 2 &beta;3q ) )/2
 * cos(q&theta;) = ( &beta;q &plusmn; &radic;( 4 + &beta;2q - 2 &beta;3q ) )/4

In particular
 * cos(&theta;) = ( &beta;1 + &radic;( 4 + &beta;2 - 2 &beta;3 ) )/4 = $$\frac{1}{16}\left(-1+\sqrt{17} + \sqrt{34-2\sqrt{17}}+2\sqrt{ 17 + 3\sqrt{17} - \sqrt{34-2\sqrt{17}} - 2 \sqrt{34+2\sqrt{17}} }\right)$$
 * cos(3&theta;) = ( &beta;3 + &radic;( 4 + &beta;6 - 2 &beta;9 ) )/4 = ( &beta;3 + &radic;( 4 + &beta;6 - 2 &beta;2 ) )/4 = $$\frac{1}{16}\left(-1-\sqrt{17} + \sqrt{34+2\sqrt{17}} + 2\sqrt{ 17 - 3\sqrt{17} - \sqrt{34+2\sqrt{17}} + 2\sqrt{34-2\sqrt{17}} }\right)$$
 * cos(5&theta;) = ( &beta;5 + &radic;( 4 + &beta;10 - 2 &beta;15 ) )/4 = ( &beta;3 + &radic;( 4 + &beta;6 - 2 &beta;2 ) )/4 = $$\frac{1}{16}\left(-1-\sqrt{17} + \sqrt{34+2\sqrt{17}} - 2\sqrt{ 17 - 3\sqrt{17} - \sqrt{34+2\sqrt{17}} + 2\sqrt{34-2\sqrt{17}} }\right)$$

Commentary
The reason this works is that 17 is prime and one greater than a power of 2, and so Z17*, the multiplicative group of the finite field is cyclic with 16 elements, and hence is isomorphic to Z16. Consequently Z17* has a chain of subgroups isomorphic to Z8, Z4 and Z2. In this particular case, 6 is a generator of Z17*, and modulo 17, 62=2, 22=4, 42=16 and 162=1, and so the chain of subgroups is {1} &sub; &lt;16&gt; &sub; &lt;4&gt; &sub; &lt;2&gt; &sub; &lt;6&gt; = Z17*. The &alpha;, &beta; and &gamma; values are sums of terms ek&theta;i as k varies over cosets of these subgroups. The algebraic relations between cosets in the chain of subgroups gives the relations between the &alpha;, &beta; and &gamma; values, and since the cosets are of index 2 in the next larger coset, the relations are quadratic relations.

Construction of the heptadecagon
A construction attributed to Richmond (1893) is based on the formulae for cos(3&theta;) and cos(5&theta;) given above. Most of the work is done by quadrisecting the angle arctan(4) -- the first bisection gives ratios corresponding to the &alpha;q values and the second gives ratios corresponding to the &beta;q values. Finally, these are manipulated via a standard ruler-and-compass square root construction to give the &gamma;q values.

Set up the circumcircle and quadrisect arctan(4)
The identity tan(&phi;/2) = - cot(&phi;) + &radic;( 1 + cot2(&phi;) ) follows from the double-angle formula for tan. Applying this to &phi;=arctan(4) twice gives
 * tan(&phi;/2)=( -1 + &radic;17 )/4 = &alpha;1/2 and cot(&phi;/2)=( 1 + &radic;17 )/4 = -&alpha;3/2
 * tan(&phi;/4)= -cot(&phi;/2) + &radic;( 1 + cot2(&phi;/2) ) = &alpha;3/2 + &radic;( 4 + &alpha;32 ) / 2 = &beta;3 / 4 = ( -1 - &radic;17 + &radic;(34+2&radic;17))/16

and applying this to &xi; = &pi; - &phi; gives
 * tan(&xi;/2)=-&alpha;3/2 and cot(&xi;/2)=&alpha;1/2
 * tan(&xi;/4)=&beta;1 / 4 = ( -1 + &radic;17 + &radic;(34-2&radic;17))/16

So the construction below quadrisections the angle arctan(4) internally and externally, resulting in points E and F on the diameter AC of the unit circle such that OE=

still unfinished - need to check algebra to this point, then convert and upload images - I'll get back to this