Sum of Interior and Exterior Angles
Date: 02/14/2003 at 20:11:26 From: Student E Subject: Geometry theorems Is there a theorem for concave polygons about the sum of the interior and the sum of the exterior angles? I know that for convex polygons the sum of the interior angles is (n-2)180 and the sum of the exterior angles is 360. However, my geometry class believes that we might have proven that for concave polygons the sum of the interior angles is still (n-2)180 but the exterior angles' sum is 540. Do you know if this can be found in any textbook or if it has been proven before?
Date: 02/14/2003 at 22:51:16 From: Doctor Peterson Subject: Re: Geometry theorems Hi, E. The theorem you refer to for convex polygons applies to concave polygons as well, IF you take the direction of the exterior angles into account. If you don't, then you may get a variety of sums, depending on how many angles turn in each direction. Take this example, a concave hexagon with two "dents": +----------+ \60 60/ \ / 120+ +120 / \ /60 60\ +----------+ The sum of the interior angles is 60+60+240+60+60+240 = 720 = (6-2)180 as expected. If you walk around the perimeter counterclockwise, you will turn counterclockwise (making positive exterior angles) at all the acute angles, but clockwise (making a negative exterior angle) at the two obtuse angles, where you turn outward from the interior. The sum of the signed exterior angles is therefore 120+120-60+120+120-60 = 360 This is always 360 because in going around a polygon you always make one complete turn, and the sum of the exterior angles in this sense tells how far you have turned. It is this sum that can be used to prove the sum of the interior angles, since each interior angle is the supplement of its exterior angle: sum of A = sum of (180-B) = 180n - sum of B = 180n - 360 = (n-2)180 where A is each interior angle, B is each exterior angle, and n is the number of angles. In our example, the supplement of interior angle A=240 is B=180-240=-60. That's not how we normally think of supplements, but it's the only way that makes sense for reflex angles like this. Now, if we sum the absolute values of the exterior angles, not treating them as signed angles, we get a different answer: 120+120+60+120+120+60 = 600 You might like to play with this and see whether you can get any sum at all by changing the number and size of the negative angles, or if there is some restriction on possible sums. Did you have an actual proof that the sum is 540 degrees for some class of concave polygons, or did you just try it out for some examples and find that to be true for those you tried? If you have a proof, I'd like to see it, so we can figure out when it applies (since it obviously is not true for all concave polygons). - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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