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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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    $\begingroup$ 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. $\endgroup$
    – Josephine Moeller
    Dec 15, 2012 at 2:09
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    $\begingroup$ @JohnMoeller That definition only specifies two points. I'm asking whether we can strengthen the condition to two vertices. $\endgroup$
    – EuYu
    Dec 15, 2012 at 2:14
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    $\begingroup$ 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. $\endgroup$
    – Clayton
    Dec 15, 2012 at 2:30
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    $\begingroup$ 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. $\endgroup$
    – Joe Z.
    Dec 15, 2012 at 4:03
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    $\begingroup$ @D.R. The existence of $Q$ follows from the Jordan Curve Theorem: it’s the intersection of the interiors of the two curves obtained by walking from $A$ along the segment to $B$ and then around the boundary of the polygon in one of the two directions back to $A$. Since $AB$ is exterior to $P$, there’s a path from $∞$ that crosses $AB$ without touching $P$, and its endpoint lies in the interior of both curves, so $Q$ is nonempty. $\endgroup$
    – Anders Kaseorg
    Oct 24, 2021 at 23:08

2 Answers 2

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Suppose a concave polyhedron exists so that all vertex connections produce lines entirely inside the polyhedron.

Since the object is concave, there are points on two faces with a connecting segment outside of the polyhedron. Now consider the hull of the points for just these two faces. The connecting segment must be inside this hull. Contradiction.

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I will formulate my answer for general convex sets in vector spaces. Then convex closed polytopes (polygons) on the place accrue as a special case.

If $C$ is a convex set on a vector space $X=\mathbb{R}^n$ (One can easily generalise these facts to infinite-dimensional spaces) and $x\in\partial C$ (the boundary of $C$) then, there is a line $\mathcal{H}=\{x| Hx = K\}$ that supports $C$ at $x$, i.e. for all $x\in C$ it is $Hx\leq K$ and for $x\notin C$ it is $Hx>K$ (The relation $\leq$ is meant as the element-wise order of vectors of $X$.) This supporting hyperplane in the case of $\mathbb{R}^2$ is a line, i.e. its angle is $\pi$.

The angle you mentioned is actually the polar angle of the tangent cone $T_C(x)$ wheneven $x$ is a vertex of $C$. If $x$ is not a vertex $T_C(x)$ is a supporting hyper-plane; in general it is a cone that lies completely in a supporting plane of $C$ at $x$, so its angle is less than or equal to $\pi$. This proves your claim in a more general setting.

Back to your question again... you may also use purely Euclidean arguments (In Euclidean geometry by the way, $A_1A_2\ldots A_n$ is convex if it contains all line segments that can be drawn from its vertices. Assume that $\widehat{A_kA_{k+1} A_{k+2}}>180^\circ$. Then $A_kA_{k+2}$ has all its points outside the polygon.

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  • $\begingroup$ I must say that I am much more interested in the Euclidean proof. The last paragraph you provided is one direction of my first claim. Do you have any idea how to prove the converse? If all interior angles are less than $\pi$ then the polygon is convex. $\endgroup$
    – EuYu
    Dec 15, 2012 at 5:26

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