How to divide a polygon fairly

Let P be a simple polygon with N vertices, each assigned a weight c (0<i<N) with c€{0 , l } . We define the weight C of P as the added weight of its vertices, i.e., C = c 1 +. . . + c A r Making the assumption that the vertices of P have been sorted along some axis -which can be done in 0(Mog AO time-, we prove that it is possible, in O(N) time, to find two vertices a,b in P, such that the segment ab lies entirely inside the polygon P and partitions it into two polygons, each with a weight not exceeding 2C/3. We also give a list of problems which can be solved efficiently with a divide-and-conquer strategy based on that result