TITLE: Covering a polygon with an odd number of sides
QUESTION [13 upvotes]: I have the following elementary problem/question that I do not know how to tackle. It comes with a "math-olympiad-flavor" but I suspect it may be much more difficult than an high-school olympiads problem.
Let $\mathscr{P}$ be a simple polygon (not necessarily convex) in the plane, and such that the number of sides of $\mathscr{P}$ is an odd number $2k+1$.
For each vertex $v$ of $P$ there is an intuitive notion of opposite side (because the number of sides is odd), just label the sides counterclockwise at $v$ from $1$ to $2k+1$, and it would be the one with the label $k+1$.
Now suppose that for each vertex and its corresponding opposite side I consider the triangle the determine in the plane. Let us call $\Delta_v$ this triangle and call it a central triangle.
Is the following:
$$ \mathscr{P} \subseteq \bigcup_{v\in \mathscr{P}} \Delta_v$$
always true? In other words, if one paints all central triangles is it true that the whole polygon is then painted?

REPLY [5 votes]: The statement is true. In fact, we can generalize it a little bit beyond the simple polygons; we need only the weaker condition that its winding number is $1$ (intuitively, that if you "walk around" the polygon you will only make a turn of $360^\circ$, not any other multiple thereof - its angles add up to the right amount). We can also bend the edges however we like.
To see that it doesn't necessarily hold for polygons with higher winding number, consider the case of a 5-pointed star (darker regions indicating those which are swept out twice):
                            
To answer the question, we move from the discrete to the continuous and solve a more generalized problem.
Claim: Let $C$ be a closed curve given by a map $f:[0,1]\to \mathbb{R}^2$ such that $f(0)=f(1)$, and for notational convenience let $f(x)$ denote $f(x\pmod{1})$ in the event that $x\not\in[0,1]$. We wish to show that the union of the line segments connecting $f(x)$ and $f(x+0.5)$ for every $x\in[0,1]$ contains every point around which $C$ has winding number $1$. (Intuitively, this just means that if you stood at that point and turned to watch the path of $C$ from $f(0)$ to $f(1)$, you would make one turn counterclockwise.)

First, let's see that the claim implies our proposition. Given a polygon $\mathscr{P}$ as in the problem with $2k+1$ vertices $P_1,\ldots,P_{2k+1}$, we partition $[0,1]$ into $4k+2$ equal intervals $I_1,I_2,\ldots,I_{4k+2}$, and define $f$ so that for $x\in I_{2j}$, $f(x)$ is constant at the point $P_j$. For $x\in I_{2j+1}$, $f(x)$ traverses the edge from $P_j$ to $P_{j+1}$ (let's say linearly, but any continuous parameterization works). Now, observe that $f$ is indeed a continuous function with $f(0)=f(1)=P_{2k+1}$, and that the curve $C$ is our polygon. Note also that if you are standing in the interior of a simple polygon, it will make a net revolution of $2\pi$ radians around you, which we can choose to be counterclockwise by reversing the order of the vertices (or flipping the signs on this proof - the sign of the winding number doesn't matter, I'm just taking it to be positive for convenience).
Finally, note that as $x$ ranges over the interval $I_{2j}$, the locus of all these line segments is exactly the triangle formed by vertex $P_j$ and its "opposite side" $\overline{P_{j+k}P_{j+k+1}}$ (where the indices are to be interpreted modulo $2k+1$). (When $x$ is in an odd-indexed interval,$x+0.5$ is in an even-indexed one, so these triangles cover every line segment involved.)
So, for the given parameterization, this just says that every point in the interior of the polygon $\mathscr{P}$ is covered by one of these triangles.
Having established that the claim will yield what we want, we move on to the proof.

Proof of claim: Let $q$ be a point not on $C$. Then we can associate each point $f(t)\in C$ with an angle $\theta(t)$ of the vector from $q$ to $f(t)$, parametrized in the usual way relative to the positive $x$-axis (but where we only care about the value modulo $2\pi$).
By some basic covering space theory (I think the Wikipedia article on the winding number sketches this out in more formality), we can uniquely choose the multiple-of-$2\pi$ offsets of $\theta(t)$ so that $\theta(t)$ is continuous and $\theta(0)=0$. $q$ has winding number $1$ iff $\theta(1)=2\pi$. As an example: if $q$ is the origin and $f(t)=(\cos(2\pi t), \sin(2\pi t))$, then $\theta(t)=2\pi t$.
Now, observe that if $\theta(s)-\theta(t)\equiv\pi\equiv-\pi\pmod{2\pi}$, the line segment from $f(s)$ to $f(t)$ contains $q$.
Now, consider $g(t)=\theta(t+0.5)-\theta(t)$. $g$ is a continuous function on $[0,0.5]$, and $g(0)+g(0.5)=\theta(1)-\theta(0)=2\pi$ by assumption on $q$, so by the Intermediate Value Theorem $g$ takes on the value $\pi$ at some point. But this is exactly the condition we need for $q$ to be in our locus of line segments, completing the proof.
(Thanks to Calvin Lin for the "beam" framing of the problem, and thinking about things in terms of a rotating line segment instead of a sum of triangles.)