Sum of some angles in a polygon
Let be a convex polygon which has no two sides which are parallel. For each side consider the vertex of the polygon which is the furthest away from this side. Prove that
IMAR Contest, 2005, Juniors
Proof: Let’s call special segments and let’s find some properties they have.
Claim 1: Each two special segments meet.
Proof: If the two segments have a vertex in common, then they surely meet. Denote the furthest away vertex from the side . If , then the segments and meet.
Suppose that . If is on the same side of as then is further away from than , which is a contradiction.
Therefore and meet, i.e. separates and . The general case is treated the same way.
Claim 2: Suppose is furthest away from and is furthest away from . Then each side on the polygonal path from to has as furthest away point.
Proof: Pick a side which has as further away point . Each special segment must cross the segments and . Because the polygon is convex, this cannot happen unless .
Claim 3: The set of furthest away vertices is made of an odd number of points, and we have
Proof: Denote the set of furthest away points numbered in counterclockwise direction.
Consider the set of sides such that is the furthest away vertex from all these sides. Consider the next furthest away vertex in the counterclockwise direction and its set of furthest away sides, and so on for all furthest away vertices.
It is obvious that whenever and Claim 2 implies that the sides in come after the sides from . Moreover, the endpoints of sets are among .
Suppose the endpoints of are . Then to each of are associated the sets . And because of the second claim the endpoint of is . Therefore . This means that is odd, and because of the construction of the above formula holds.
Why do we have
Consider the polygon . Denote
Obviously we have as the sum of the angles of a triangle. Moreover, we have so summing after we get
and the proof is over.