Let $ \{a_i\}_{i=1}^n$ the $ n$ ordered sides of such a polygon ($ \{a_i\}$ is a permutation of $ 1,2,...,n$). Let R the radius of the circle about which the polygon is circumscribed and $ O$ its center. 1) This situation is impossible if $ n=2\pmod 4$ (and so for example for $ n=1990$) : Let $ A_1,...,A_n$ the n vertices ($ A_{n+1}=A_1$) with $ A_kA_{k+1}=a_k$. Let $ \Delta_i$ the distance between $ A_i$ and the contact (tangent) between the circle and the side $ A_iA_{i+1}$. We have $ \Delta_{k+1}=a_k-\Delta_k$ and so $ \Delta_p=\sum_{k=1}^{p-1}(a_k(-1)^{k-1})+(-1)^{p-1}\Delta_1$ So $ \Delta_1=\Delta_{n+1}=\sum_{k=1}^{n}(a_k(-1)^{k-1})+(-1)^{n}\Delta_1$ So, if $ n=2\pmod 4$, we have $ n$ even and so $ 0=\sum_{k=1}^n(a_k(-1)^{k-1})$ which implies $ \sum_{\text{odd }k\in[1,n]}a_k$ $ =\sum_{\text{even }k\in[1,n]}a_k$ So $ \sum_{k\in[1,n]}a_k$ even, so $ \frac{n(n+1)}{2}$ even, which is wrong when $ n=2\pmod 4$ Q.E.D. 2) This situation is always possible if $ n\geq 4$ and $ n=0\pmod 4$ (and so for example for $ n=1992$) : Choose $ \Delta_1=\frac 12$ and $ a_k=k$ which gives : $ \Delta_{2p}=p-\frac 12$ $ \Delta_{2p+1}=p+\frac 12$ It remains to choose the radius as root of the equation $ \sum_{k=1}^n\arctan(\frac{k}{x})=\pi$ which clearly has root as soon as $ n\geq 3$ Q.E.D.