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Consider a closed simple polygon in the plane. It is intuitively obvious that the polygon is convex if and only if all the interior angles measure less than or equal to $\pi$ radians. I have never seen a rigorous proof of this fact and I was wondering if anyone could provide such a proof.

A related question: Given a concave polygon (or more generally a higher dimensional polytope), how can we prove that there will always be two vertices of the polygon which cannot be joined by a line lying entirely inside the polygon?

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    The answer to your related question is by definition. Concave = not convex, and a polytope is convex iff every line segment between any two points in the polytope lies entirely inside the polyhedron. Invert that condition (i.e., "there exists a line segment that exits and reenters the polytope"), and you necessarily (and sufficiently) have a concave polytope.2012-12-15
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    @JohnMoeller That definition only specifies two _points_. I'm asking whether we can strengthen the condition to two _vertices_.2012-12-15
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    Would you please provide us with your definition of a polygon?2012-12-15
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    I think the first question might follow from choosing one of the vertices that have the interior angle greater than $\pi$, triangulating the polygon and checking the value against what is known. That is, for a convex polygon, we know the interior angle has to equal $(n-2)\cdot\frac{\pi}{2}$ from triangulation. I haven't put a ton of thought into it, but this is what I'm thinking.2012-12-15
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    @OlivierBégassat A closed plane figure with straight line boundaries. Feel free to use any reasonable definition of polygon you wish though. A convex polygon is a polygon for which the interior is a convex set.2012-12-15
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    @EuYu Changing the condition to two vertices actually *weakens* it, not strengthens it. Vertices are points too; therefore it is satisfied.2012-12-15
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    @JoeZeng Indeed it does, thank you.2012-12-15
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    @Clayton I'm not too sure what you mean. The interior angles of any simple polygon sums to $(n-2)\frac{\pi}{2}$ regardless of concave or convex.2012-12-15
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    @EuYu: I'm not entirely sure of it. I haven't had a chance to think about it. When I have a mental block (studying for prelims), I'll come back here to think about your problem.2012-12-15
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    I think your question may be equivalent to the following: Given any two rays that meet at a point, the line segment that connects any two points on the ray will always be on the side of the two rays with a smaller angle. Then a polygon is simply the intersection of a bunch of these two-ray configurations.2012-12-15
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    Note that any proof will have to inherently use the simplicity of the polygon, because the theorem isn't true for self-intersecting polygons - imagine taking a heart-shaped (cardiod) loop that wraps twice around the origin and polygonalizing it.2012-12-29

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