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April 2

Angle condition for equilateral polygons

By equiangular polygon and its source 5, there is a polynomial condition for a sequence of side lengths $a_{1},\dots ,a_{n}$ to be consistent with an equiangular polygon: namely that $a_{1}+a_{2}x+a_{3}x^{2}+\cdots +a_{n}x^{n-1}$ be divisible by $x^{2}-2x\cos(2\pi /n)+1.$ Is there an analogous condition for a sequence of angles to be consistent with an equilateral polygon? Loraof (talk) 17:06, 2 April 2016 (UTC)[reply]

Take n=5; it's easy to generalize from this. If the angles are α₁, α₂, α₃, α₄, α₅ then first α₁+α₂+α₃+α₄+α₅ = 3π by properties of pentagons in general. Eliminating α₅, there are two other conditions: 1 - cosα₁+cos(α₁+α₂)-cos(α₁+α₂+α₃)+cos(α₁+α₂+α₃+α₄)=0 and sinα₁-sin(α₁+α₂)+sin(α₁+α₂+α₃)-sin(α₁+α₂+α₃+α₄)=0. To prove this, let β₁, β₂, β₃, β₄, β₅ be the corresponding exterior angles. Take the first side to be from 0 to 1 in the complex plane; from which the remaining vertices must be 1+e^iβ₁, 1+e^iβ₁+e^{i(β₁+β₂)}, etc. The 5th vertex will be at the start again iff 1+e^iβ₁+e^{i(β₁+β₂)}+e^{i(β₁+β₂+β₃)}+e^{i(β₁+β₂+β₃+β₄)}=0. Note, this assumes that there isn't any self intersection or similar weirdness going on, otherwise the first condition isn't always true. --RDBury (talk) 23:06, 2 April 2016 (UTC)[reply]

This can be simplified it to

1+e^{i\alpha _{1}}(1+e^{i\alpha _{2}}(1+e^{i\alpha _{3}}(1+e^{i\alpha _{4}}(1+e^{i\alpha _{5}}))))=0

complex numbers can be useful for following the sides of a polygon around. Dmcq (talk) 11:44, 3 April 2016 (UTC)[reply]

Only either use minuses or the exterior angles β. Note also that the same idea proves the original statement about equiangular polygons.--RDBury (talk) 15:55, 3 April 2016 (UTC)[reply]

Thanks! Loraof (talk) 16:22, 3 April 2016 (UTC) Is there by any chance a source I could use in order to put this into the equilateral polygon article? Loraof (talk) 16:24, 3 April 2016 (UTC)[reply]