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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.

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  • $\begingroup$ 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. $\endgroup$
    – Joffan
    Jan 9, 2015 at 20:51
  • $\begingroup$ 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. $\endgroup$
    – Joffan
    Jan 9, 2015 at 20:58
  • $\begingroup$ @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. $\endgroup$
    – TryingHardToBecomeAGoodPrSlvr
    Jan 10, 2015 at 0:13

1 Answer 1

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The convex hull represents the extreme outermost vertices of the polygon. Any vertex on the hull is guaranteed to have an interior angle of no more than 180 degrees, and since the hull is a closed curve not a straight line, some angles at the hull vertices must be less than 180 degrees.

As a simpler application, assuming the points have coordinates, consider the the greatest X-coordinate value at any vertex.

Assuming there is only one point with this X coordinate, the interior angle at this point must be less than 180 degrees to allow the adjacent vertices to have lesser X coordinates.

Assuming there is more than one point with the greatest X coordinate value, take the point amongst those with the greatest Y coordinate. At most one of the adjacent vertices can have the same X coordinate and the other one must have a lesser value. This implies an interior angle of less than 180 degrees once again.

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  • $\begingroup$ @Joffman Thank you very much once again. Based on the argument you have given, I am assuming that this argument holds whether it is a concave or a convex polygon. It is very obvious when the polygon is convex but your argument guarantees the existence of an interior angle less than 180 degrees for a concave polygon as well. Thanks a lot once again. $\endgroup$
    – TryingHardToBecomeAGoodPrSlvr
    Jan 10, 2015 at 4:44
  • $\begingroup$ @Joffman Sorry I did not do that. I am new to Skackexchange so I was not aware of how things work here. Thanks a lot once again for answering my query. :) $\endgroup$
    – TryingHardToBecomeAGoodPrSlvr
    Jan 10, 2015 at 17:25

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