User:AndrewKepert/17gon

Contents

1 Theory 1.1 Basic identities 1.2 Definitions 1.3 Quadratic relations 1.4 Commentary 2 Construction of the heptadecagon 2.1 Set up the circumcircle and quadrisect arctan(4)

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 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

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

α_q = cos(qθ) + cos(2qθ) + cos(4qθ) + ... + cos(128qθ)

β_q = cos(qθ) + cos(4qθ) + cos(16qθ) + cos(64qθ)

γ_q = 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 α_q=α_17+q=α_17-q, and so we only use these eight values of q. Moreover, cos(256qθ)=cos(qθ) and so α_2q=α_q and β_4q=β_q. In particular,

α₁=α₂=α₄=α₈ = 2cos(qθ) + 2cos(2qθ) + 2cos(4qθ) + 2cos(8qθ) > 0

α₃=α₆=α₅=α₇ = 2cos(3qθ) + 2cos(6qθ) + 2cos(5qθ) + 2cos(7qθ) < 0

β₁=β₄=2cos(qθ) + 2cos(4qθ)

β₂=β₈=2cos(2θ) + 2cos(8qθ)

β₃=β₅=2cos(3qθ) + 2cos(5qθ)

β₆=β₇=2cos(6qθ) + 2cos(7qθ)

from which it can be seen that γ_q + γ_4q = β_q, β_q + β_2q = α_q and α_q + α_3q = 2cos(qθ) + 2cos(2qθ) + ... + 2cos(8qθ) = -1

Quadratic relations

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

γ_q² = γ_2q + 2

γ_qγ_4q = β_3q

β_q² = ( γ_q + γ_4q ) ² = 4 + β_2q + 2β_3q

β_qβ_2q = -1

α_q² = 6 + α_q + 2α_3q = 4 - α_q

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

α₁=α₂=α₄=α₈ = ( -1 + √17 )/2 ≈ 1.56

α₃=α₆=α₅=α₇ = ( -1 - √17 )/2 ≈ -2.56

The β_q values can be found by verifying that

( β_2q - β_q )² = ( β_q + β_q )² - 4 β_qβ_q = α_q² + 4 = 8 - α_q so that β_2q - β_q = √( 8 - α_q )

Also β_2q + β_q = α_q so that β_q = (α_q ± √( 8 - α_q ) )/2 .

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

β₁=β₄=(α₁ + √( 8 - α₁ ) )/2 = ( -1 + √17 + √(34-2√17))/4 ≈ 2.05

β₂=β₈=(α₁ - √( 8 - α₁ ) )/2 = ( -1 + √17 - √(34-2√17))/4 ≈ -0.49

β₃=β₅=(α₃ + √( 8 - α₃ ) )/2 = ( -1 - √17 + √(34+2√17))/4 ≈ 0.34

β₆=β₇=(α₃ - √( 8 - α₃ ) )/2 = ( -1 - √17 - √(34+2√17))/4 ≈ -2.91

For the γ_q terms,

( γ_q - γ_4q )² = ( γ_q + γ_4q )² - 4γ_qγ_4q = β_q² - 4β_3q = 4 + β_2q - 2 β_3q so that γ_q - γ_4q = √( 4 + β_2q - 2 β_3q )

and since γ_q + γ_4q = β_q it follows that

γ_q = ( β_q ± √( 4 + β_2q - 2 β_3q ) )/2

cos(qθ) = ( β_q ± √( 4 + β_2q - 2 β_3q ) )/4

In particular

cos(θ) = ( β₁ + √( 4 + β₂ - 2 β₃ ) )/4 = <math>\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)<math>

cos(3θ) = ( β₃ + √( 4 + β₆ - 2 β₉ ) )/4 = ( β₃ + √( 4 + β₆ - 2 β₂ ) )/4 = <math>\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)<math>

cos(5θ) = ( β₅ + √( 4 + β₁₀ - 2 β₁₅ ) )/4 = ( β₃ + √( 4 + β₆ - 2 β₂ ) )/4 = <math>\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)<math>

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} ⊂ <16> ⊂ <4> ⊂ <2> ⊂ <6> = Z₁₇^*. The α, β and γ values are sums of terms e^kθi 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

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 α_q values and the second gives ratios corresponding to the β_q values. Finally, these are manipulated via a standard ruler-and-compass square root construction to give the γ_q values.

Set up the circumcircle and quadrisect arctan(4)

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