# Interior angles of a polygon.

I was solving problems from Paul Zeitz's book "The Art and Craft of Problem Solving." There is a problem which states

3.2.11 Fix the proof in Example 2.3.5 on page 45. Show that even a concave polygon has an acute angle that we can use to "snip off' a triangle. Why do we need the extreme principle for this?

Example 2.3.5 he speaks of is to prove that sum of angles in a polygon is 180(n-2) degrees.

I believe he meant an angle less than 180 degrees and not an acute angle since I can think of cases where a concave polygon has no acute angles.

Please give me a hint as to how I can use the extreme principle in order to figure out an angle that is < 180 degrees (instead of acute) among all the internal angles of a polygon. I cannot use the fact that the sum of internal angles of any polygon = 180(n-2), n being number of sides in the polygon. That is because the intention is to prove that the sum is 180(n-2) using the fact that one of the angles is less than 180 degrees so it would end up in a circular logic.

• Hmm. Harder than it looks. I was trying for some of the points on the convex hull but they aren't guaranteed in a snake-like polygon. Certainly acute angles cannot be guaranteed; for any acute angle we can imagine another polygon where the angle is rounded off by using two or three vertices instead of one. – Joffan Jan 9 '15 at 20:51
• To clarify, since I can't edit now, I meant that the points on the convex hull, while their angles will be less than 180 degrees, are not guaranteed to be snippable. – Joffan Jan 9 '15 at 20:58
• @Joffan Thanks a lot! True that it is not guaranteed to be snippable. I just need to prove that an angle exists such that it is less than 180 degrees. Is there a way to prove that? If there is, I would greatly appreciate it if you can please give me a hint. – TryingHardToBecomeAGoodPrSlvr Jan 10 '15 at 0:13