The General Case
Using the same calculation as in the previous section, we see that any convex polygon with n > 3, the area of the original polygon is given by twice the area of the midpoint polygon minus onefourth the area of the star polygon obtained by skipping one vertex as we go around. In the case where n=6, we obtain two different triangles forming a "star of David". In general, for n even we obtain two polygons and the sum of the areas of the two will give the area of the star with the overlap counted twice. In the case of n odd, there will be a central ngon counted twice, as in the case of the pentagon. Note that if n = 4, the two polygons that we get by taking every second vertex will be bigons, each with zero area so in this case there is no "error term".
Another virtue of working with vectors is that the proof given above works as well for polygons that are "starshaped" with respect to the origin, i.e. if the segment from the origin to any point on the polygon lies inside the polygon. In that case, all the expressions that appear in the summation are positive. One important case of a starshaped polygon is a nonconvex quadrilateral. One of the diagonals must lie completely within the figure and any point on that segment can be chosen as the origin, so that the quadrilateral is starshaped from that point. Note that in this case, the midpoint polygon is still a parallelogram, but this time it lies partially outside the figure. We can translate the portion lying outside the triangle so that it lies inside and shows, as in the case of the convex quadrilateral, that the midpoint parallelogram has area onehalf the original area. At this point, we can note that the formula above works just as well if the origin is not situated in the center of the figure, although in this case, some of the expressions (1/2)(Xi x Xi+1) may be negative. Triangles that are traversed in a counterclockwise sense appear with positive area and those traversed in a clockwise sense appear with negative area. Note that if the polygon is a convex polygon traversed in a clockwise sense, then the midpoint polygon is also traversed in a clockwise sense. The areas in this case are negative, and the same relationships exist as in the cases previously considered. With this interpretation, of oriented areas, we can consider polygons that intersect themselves. For example, a quadrilateral with one crossing point forms a "figureeight" of "hourglass" shape, with one triangular portion traversed in a counterclockwise sense and the other traversed in a clockwise sense. The area will then be the difference between the area of the counterclockwise triangle and the area of the clockwise triangle. In this case, both of the "diagonals" of the selfintersecting quadrilateral lie outside the figure, i.e. on the boundary of the convex quadrilateral determined by the four vertices. The remarkable thing is that the same proofs work. If we choose a "diagonal", then the area of that figure can be thought of as the difference of the areas of the two triangles based on that diagonal, so one region outside the figure is covered once positively and once negatively for a total contribution of zero. The vector proof automatically shows that the oriented area of the midpoint quadrilateral, still a parallelogram, is onehalf the oriented area of the original quadrilateral. Note that for a symmetrical hourglass with zero area, the midpoint parallelogram becomes a doublycovered line segment, also with zero area. We can now go back to consider a geometric interpretation of the error term in the case of a convex pentagon. If we draw the midpoint pentagon and the star pentagon, together they divide the interior of the midpoint pentagon into a central pentagon, five parallelograms that are onehalf the area of five triangles, and five trapezoids that are threefourths the area of the triangular points of the star. If we take twice the area of the midpoint pentagon, we get the sum of the areas of the outer triangles, (3/2) times the area of the pointed triangles, and twice the inner area. Thus twice the area of the midpoint pentagon is the area of the original pentagon plus one half the areas of the pointed triangles plus the area of the central pentagon. These last two summands represent onehalf the area of the star pentagon, as expected.
