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A cyclic polygon is a polygon with vertices upon which a circle $C_0$ can be circumscribed. (All vertices lie on circle $C_0$).

We are given the lengths of the cyclic polygon $\{L_1, L_2,..., L_n\}$. We need to find the coordinates of the vertices $\{(x_1, y_1), (x_2, y_2),...(x_n, y_n)\}$ such that the center of the circle $C_0$ is $(0, 0)$.

Example:

If lengths are $\{2,2,2,2\}$ then one of the possible locations of vertices is

$$ \sqrt2 \, (( -1,1), (-1,1),(1,1),(1,-1)) $$

EDIT1:

Is there Ruler & Compass method to determine its radius and locate the center ?

Narasimham
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maverick
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1 Answers1

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Find the radius $R$ numerically from

$$ t_1=\sin^{-1} \frac{L_1}{2 R } \,,\quad t_2= \sin^{-1 } \frac{L_2}{2 R }\,,\,\,\dots\,\,, t_n=\sin^{-1 } \frac{L_n}{2 R } $$

$$ \pi = t_1+t_2+\dots +t_n$$

Find angle subtended by each sector at center $(t_1,t_2,..).$ Upto rigid rotations/translations of the entire polygon in x-y plane

$$ x_0 = R;\, y_0= 0\,,$$

$$ x_1= R \cos 2 t_1;y_1= R \sin 2 t_1,$$

$$ x_2= R \cos (2 t_1+ 2 t_2) ,\,y2= R \cos (2 t_1+ 2 t_2) \dots$$

Narasimham
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  • can you please explain how can we find the R exactly ? – prime Sep 17 '16 at 14:05
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    You mean by means of an example? Please supply 4 lengths of cyclic quadrilateral or five for pentagon. For the former an example using Mathematica: { L1 = 3, L2 = 4, L3 = 5, L4 = 6.} NSolve[ ArcSin[L1/d] + ArcSin[L2/d] + ArcSin[L3/d] + ArcSin[L4/d] == Pi]; d = 2 R = 6.57457 – Narasimham Sep 17 '16 at 21:12
  • Isn't there an easy way to solve it, like using a pen and a paper – prime Sep 18 '16 at 03:54
  • Guess none. You tagged numerical methods and now seem going geometrical. Edit ok? Else shall change back. ..in other words you ask to solve $ \sin , [\sin^{-1} \frac{L_1}{2 R } + \sin^{-1 } \frac{L_2}{2 R }+....+\sin^{-1 } \frac{L_n}{2 R } ]= 0 $ – Narasimham Sep 18 '16 at 08:25