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Theory

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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

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  • 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 θ=2πk/n, then cos(0θ) + cos(1θ) + cos(2θ) + ... + cos((n-1)θ) = 0. Consequently, if n=2m+1 is odd, then cos(1θ) + cos(2θ) + ... + cos(mθ) = -1/2. This identity can be proven geometrically, or by noting that if S is the sum on the left hand side, then Scos(θ)=S, and since cos(θ)≠1, S=0.

Definitions

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Define θ=2π/17 and then for q=1,...,8 define

αq = cos(qθ) + cos(2qθ) + cos(4qθ) + ... + cos(128qθ)
βq = cos(qθ) + cos(4qθ) + cos(16qθ) + cos(64qθ)
γq = cos(qθ) + cos(16qθ) = cos(qθ) + cos(-qθ) = 2cos(qθ)

These are defined for all integer q, but since the cosine function is periodic and symmetric, we can deduce that αq17+q17-q, and so we only use these eight values of q. Moreover, cos(256qθ)=cos(qθ) and so α2qq and β4qq. In particular,

α1248 = 2cos(qθ) + 2cos(2qθ) + 2cos(4qθ) + 2cos(8qθ) > 0
α3657 = 2cos(3qθ) + 2cos(6qθ) + 2cos(5qθ) + 2cos(7qθ) < 0
β14=2cos(qθ) + 2cos(4qθ)
β28=2cos(2θ) + 2cos(8qθ)
β35=2cos(3qθ) + 2cos(5qθ)
β67=2cos(6qθ) + 2cos(7qθ)

from which it can be seen that γq + γ4q = βq, βq + β2q = αq and αq + α3q = 2cos(qθ) + 2cos(2qθ) + ... + 2cos(8qθ) = -1

Quadratic relations

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Using the identity for cos(A)cos(B) above, it can be verified that

γq2 = γ2q + 2
γqγ4q = β3q
βq2 = ( γq + γ4q ) 2 = 4 + β2q + 2β3q
βqβ2q = -1
αq2 = 6 + αq + 2α3q = 4 - αq

and so αq = ( -1 ± √17 )/2. Consequently

α1248 = ( -1 + √17 )/2 ≈ 1.56
α3657 = ( -1 - √17 )/2 ≈ -2.56

The βq values can be found by verifying that

( β2q - βq )2 = ( βq + βq )2 - 4 βqβq = αq2 + 4 = 8 - αq so that β2q - βq = √( 8 - αq )
Also β2q + βq = αq so that βq = (αq ± √( 8 - αq ) )/2 .

From relative sizes of the cos(kθ) terms, we can deduce

β14=(α1 + √( 8 - α1 ) )/2 = ( -1 + √17 + √(34-2√17))/4 ≈ 2.05
β28=(α1 - √( 8 - α1 ) )/2 = ( -1 + √17 - √(34-2√17))/4 ≈ -0.49
β35=(α3 + √( 8 - α3 ) )/2 = ( -1 - √17 + √(34+2√17))/4 ≈ 0.34
β67=(α3 - √( 8 - α3 ) )/2 = ( -1 - √17 - √(34+2√17))/4 ≈ -2.91

For the γq terms,

( γq - γ4q )2 = ( γq + γ4q )2 - 4γqγ4q = βq2 - 4β3q = 4 + β2q - 2 β3q so that γq - γ4q = √( 4 + β2q - 2 β3q )

and since γq + γ4q = βq it follows that

γq = ( βq ± √( 4 + β2q - 2 β3q ) )/2
cos(qθ) = ( βq ± √( 4 + β2q - 2 β3q ) )/4

In particular

cos(θ) = ( β1 + √( 4 + β2 - 2 β3 ) )/4 =
cos(3θ) = ( β3 + √( 4 + β6 - 2 β9 ) )/4 = ( β3 + √( 4 + β6 - 2 β2 ) )/4 =
cos(5θ) = ( β5 + √( 4 + β10 - 2 β15 ) )/4 = ( β3 + √( 4 + β6 - 2 β2 ) )/4 =

Commentary

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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} ⊂ <16> ⊂ <4> ⊂ <2> ⊂ <6> = Z17*. The α, β and γ values are sums of terms ekθi as k varies over cosets of these subgroups. The algebraic relations between cosets in the chain of subgroups gives the relations between the α, β and γ values, and since the cosets are of index 2 in the next larger coset, the relations are quadratic relations.

Construction of the heptadecagon

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A construction attributed to Richmond (1893) is based on the formulae for cos(3θ) and cos(5θ) given above. Most of the work is done by quadrisecting the angle arctan(4) -- the first bisection gives ratios corresponding to the αq values and the second gives ratios corresponding to the βq values. Finally, these are manipulated via a standard ruler-and-compass square root construction to give the γq values.

Set up the circumcircle and quadrisect arctan(4)

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The identity tan(φ/2) = - cot(φ) + √( 1 + cot2(φ) ) follows from the double-angle formula for tan. Applying this to φ=arctan(4) twice gives

tan(φ/2)=( -1 + √17 )/4 = α1/2 and cot(φ/2)=( 1 + √17 )/4 = -α3/2
tan(φ/4)= -cot(φ/2) + √( 1 + cot2(φ/2) ) = α3/2 + √( 4 + α32 ) / 2 = β3 / 4 = ( -1 - √17 + √(34+2√17))/16

and applying this to ξ = π - φ gives

tan(ξ/2)=-α3/2 and cot(ξ/2)=α1/2
tan(ξ/4)=β1 / 4 = ( -1 + √17 + √(34-2√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