$title="The General Case"; $page_name="general"; include "header.php"; ?>
Using the same calculation as in the previous section, we see that for any convex n-gon with n > 2, the area of the midpoint polygon is given by one-half the area of the original polygon plus one-quarter 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" (see figure 9). 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 region 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."
figure_indent("images/star_of_david.jpg", "", "Figure", "9", "For a hexagon the \"star polygon\" is formed by two overlapping triangles. The construction for area ratios still works."); ?>At this point, we can note that the analytic formula 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. In fact, there is no restriction to the application of this formula based on convexity, self-intersection, or orientation; it works for all planar polygons with n > 2.
The general formula in the original discussion can be found here.
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Continue on to the pedagogy section, where you will find the original online discussion and notes on oriented area.
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