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

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April 2

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Angle condition for equilateral polygons

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By equiangular polygon and its source 5, there is a polynomial condition for a sequence of side lengths to be consistent with an equiangular polygon: namely that be divisible by 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 α12345 = 3π by properties of pentagons in general. Eliminating α5, there are two other conditions: 1 - cosα1+cos(α12)-cos(α123)+cos(α1234)=0 and sinα1-sin(α12)+sin(α123)-sin(α1234)=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+e1, 1+e1+ei(β12), etc. The 5th vertex will be at the start again iff 1+e1+ei(β12)+ei(β123)+ei(β1234)=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
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]