Succinct Proof: All Pentagons Are Star Shaped

Question

Question: What is a succinct proof that all pentagons are star shaped?

In case the term star shaped (or star convex) is unfamiliar or forgotten:

Definition Reminder: A subset $X$ of $\mathbb{R}^n$ is star shaped if there exists an $x \in X$ such that the line segment from $x$ to any point in $X$ is contained in $X$.

This topic has arisen in the past in my class discussions around interior angle sums; specifically, for star shaped polygons in $\mathbb{R}^2$ we can find their sum of interior angles as follows:

By assumption, there is an interior point $x$ that can be connected to each of the $n$ vertices. Drawing in these line segments, we construct $n$ triangles; summing across all of their interior angles gives a total of $180n^\circ$, but this over-counts the angle sum for the polygon by the $360^\circ$ around $x$. Therefore, the sum of interior angles is $(180n - 360)^\circ = 180(n-2)^\circ$.

This formula for the sum of interior angles holds more generally (often shown by triangulating polygons) but the proof strategy above already fails for some (obviously concave) hexagons.

For example:

The above depicted polygon is not star shaped.

Moreover, it is a fact that any polygon with five or fewer sides is star shaped. And so I re-paste:

Question: What is a succinct proof that all pentagons are star shaped?

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Edit: Since it has come up as a counterexample (of sorts) for each of the first two responses, here is an example of a concave pentagon that may be worth examining in thinking through a proof.

Answer 1

• Assuming(?!) that every polygon has at least one "convex" vertex, let $P$ such a vertex. $P$ "sees" its two neighbors. If $P$ also "sees" a third vertex, $Q$, then connect them with a segment; otherwise, connect $P$'s neighbors with a segment, and rename those neighbors $P$ and $Q$. Either way, diagonal $\overline{PQ}$ separates the interior of the polygon, and it serves as a side of each piece: a triangle and a quadrilateral.

• Repeating the above procedure on the quadrilateral, we conclude that one of its diagonals separates its interior (this time, into two triangles). Necessarily, one of the endpoints of that diagonal is either $P$ or $Q$. We'll say that it's $P$.

• Therefore, $P$ is a vertex of the three sub-triangles, and it can "see" all points in each of those triangles (because triangles are cool like that), which means that it can "see" all points in the original pentagon. $\square$

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Note. Of course, you should never assume. A triangle with edges in one hemisphere of a balloon, but an "interior" that contains the other hemisphere, has no "convex" vertices at all. So, this solution is flawed from its very first word. Overcoming the flaw is left as an exercise to the reader.

Answer 2

(A perhaps more common term than star convex is star-shaped.)

I assume you view the pentagon as a closed set, including its boundary, so that the line of sight from $x$ can touch the boundary and still be "contained."

Here is a proof, maybe still not succinct enough for your purposes. Define a reflex vertex as one at which the internal angle exceeds $180^\circ$. The others are convex vertices.

(1) Any polygon must have at least three convex vertices. Let me take this as given.

(2) Therefore a pentagon $P$ can have at most two reflex vertices.

(3) If $P$ has no reflex vertices, $P$ is convex and easily star-shaped.

(4) If $P$ has one reflex vertex $v$, then $P$ is visible from $v$ and so star-shaped.

(5) If $P$ has two reflex vertices, they are either adjacent or not. If adjacent, then $P$ is visible from the middle convex vertex $v$ and so star-shaped.

(6) If $P$ has two non-adjacent reflex vertices, then let $a,b,c$ be three consecutive vertices, with $a,c$ reflex and $b$ between them convex. Extend the edges $ba$ and $bc$ until they hit the opposite side. Then any point $x$ on that edge between the extensions can see all of $P$:

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Answer 3

A polygon must have at least three convex vertices (since each convex vertex turns by less than $\pi$ and in total the vertices must turn by $2\pi$). Thus there is at least one vertex that is between two convex vertices. This vertex sees all four other vertices, and hence the entire pentagon.

Answer 4

This is several years late, but I think I have a nicer proof. I thought I would post it.

Let's believe we can triangulate the 5-gon into three triangles. Note that these three triangles all must have a vertex $v$ in common: To see this, remove one of the triangles $T$, s.t. we have a 4-gon left. Then whichever way we triangulate the 4-gon every edge will contain a vertex from both of its triangles. In particular, the joint edge of $T$ and the 4-gon will have such a vertex $v$.

Then $v$ sees all three triangles and hence the entire 5-gon.