Wikipedia:Reference desk/Archives/Mathematics/2016 April 2

= 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 $$a_1+a_2x+a_3x^2+\cdots + a_nx^{n-1}$$ 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)
 * 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+eiβ1, 1+eiβ1+ei(β1+β2), etc. The 5th vertex will be at the start again iff 1+eiβ1+ei(β1+β2)+ei(β1+β2+β3)+ei(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)
 * 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)
 * 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)

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)