content_image/324091.jpg
<img src='content_image/324091.jpg'>

the first option; this gives us a unique solution for θ$_{1}$ and θ$_{2}$ . 

This procedure works whenever g$_{r}$ and g$_{s}$ exist which is certainly true whenever the density disk does not touch a vertex of Ω either at its center or on its boundary because if we avoid these cases, then there is only one graph-like orientation to deal with and ∂ Ω is C ∞ for all the points that enter into the computation. In fact, with a moment’s thought, we can make a stronger statement than this: g$_{r}$ always exists and g$_{s}$ exists as long as the center of the density disk is not a vertex of the polygon. 

We can identify the s values at which g$_{s}$ ( s, r ) does not exist to obtain the arc length positions of the vertices (and therefore obtain side lengths). For a given s corresponding to a vertex, we can find g$_{r}$ and the one-sided derivatives g$_{s}$$_{−}$ and g$_{s}$$_{+}$ . These correspond to the graph-like orientations required by the polygon sides adjacent to the current vertex. 

Referring to Figure 4.11, the one-sided derivatives along with the argument at the beginning of the proof yield the angles θ$_{1}$ , θ$_{2}$ , φ$_{1}$ , and φ$_{2}$ . Thus we can calculate ψ = θ$_{1}$ − φ$_{1}$ which means that the polygon vertex at s has angle π − ψ . 

Doing this for all s corresponding to vertices, we can determine all of the angles of 

