Interior Angles of a PolygonDate: 10/21/96 at 12:42:8 From: Wendy G. Rhodes Subject: Concave and convex polygons In geometry there is a theorem that states the following: The sum of the measures of the interior angles of any convex polygon with n sides is (n-2)180 degrees. Does this theorem apply to concave polygons? If not, why? Date: 10/31/96 at 17:48:37 From: Doctor Daniel Subject: Re: Concave and convex polygons Hi Wendy, I've made a picture that should help you out with this: Please refer to this diagram while you're reading my answer. Yes, this theorem does apply to polygons that are non-convex. In order to prove it, I'm actually going to prove something quite similar, but different. This is that any polygon with n vertices (a polygon with n vertices of course has n sides) can be divided into a collection of n-2 triangles which, when put together, fill the entire polygon. (See the figure for help.) All of these triangles have as their corners points which are corners of the original polygon as well. Let's assume that we know that that's true. Then, since each interior angle of the polygon is the sum of angles of the triangles, the total sum of all interior angles of the polygon is the same as the total sum of all interior angles of the triangles. But we know that the sum of angles of a triangle is 180 degrees. So the sum of all angles of the polygon is equal to the number of triangles times the sum of the angles in each, which is equal to (n-2)180. Now, we prove that each polygon can be triangulated by induction. First, we show that each non-triangle polygon with vertices A,B,C,...,X has an "interior diagonal." That's a line segment from one vertex to another which never crosses the boundary of the polygon. (In the picture, BF is an interior diagonal, while BD would not be.) Take any convex angle in the polygon (there must be one, since, for example, there's a rightmost point B in the polygon; since both points (A, C) it's next to are to its left, the angle ABC is convex). Then one of two things is true. Either AC is enclosed by the polygon, at which point AC is our diagonal, or it's not. Suppose AC crosses the boundary of the polygon. (This is actually the case in the figure.) Then there must be a vertex of the polygon inside the triangle ABC. In fact, if we sweep lines parallel to AC from B to AC, there will be a first-crossed vertex inside the triangle (it's F in our example). However, since it's the _first_ crossed vertex, that means that the triangle from B to the intersection of our line with AB and BC will be entirely contained inside the polygon. (In the picture, this is the triangle x1, x2, B.) So the line from B to this vertex is entirely inside the triangle, and thus inside the polygon: it's an internal diagonal. So any non-triangle polygon has an internal diagonal. Now we prove our triangulation theorem: Any polygon of n vertices can be triangulated. Base case: n = 3 is trivial. Induction case: Given our n-gon, P, the internal diagonal AB turns the polygon into two polygons, Q and R; if Q has x vertices, then R has n-x+2 vertices, since we reuse A and B. But our hypothesis says that the Q can be triangulated into x-2 triangles and R into n-x triangles. Since P is the disjoint union of Q and R (except for boundaries), P will be the union of all the triangles of Q and R, which don't intersect. So there will be n-2 triangles, all together, and the induction holds. So any polygon P of n vertices can be triangulated into n-2 triangles, and the formula works as desired for the sum of angles. Thanks for your question! It's a very interesting subject. -Doctor Daniel, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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