TITLE: Dividing the edges and diagonals of a polygon among disjoint sub-polygons
QUESTION [9 upvotes]: Let $P$ be a convex $n$-gon ($n$ is odd and $n \geq 5$).
  Determine the smallest $m$ such that all edges and diagonals of $P$ can be covered by the edges 
  of $m$ convex sub-polygons of $P$ which do not share any edges.
($Q$ is a sub-polygon of $P$ if all vertices of $Q$ are vertices of $P$).

This question was proposed at a Chinese contest and maybe it's a known result in graph theory.
For $n=5$, for example, it's not hard to see that $m(5)=3$. If $P:=P_{1}P_{2}P_{3}P_{4}P_{5}$,
$m(5)=3$ can be achieved by 3 convex sub-polygons: 
$P_{1}P_{2}P_{4}P_{5}$; $P_{1}P_{3}P_{4}$; $P_{2}P_{3}P_{5}$.
I also found that $m(7) \leq 6$, but I couldn't go further.
Any suggestion would be appreciated. Thanks in advance.

REPLY [2 votes]: I think the following construction realizes the lower bound in Brian's answer. Let $n=2k+1$ and label the first $2k$ vertices with $\mathbb{Z}_{2k}=\{0,\ldots,2k-1\}$ and the last vertex with $\infty$. First we take all the $(n-1)(n-3)/8=2k(k-1)/4$ 4-gons of the form
\[(i,\,i+j,\,i+k,\,i+k+j)\]
with $i\in\{0,\ldots,2k-1\}$ and $j\in\{1,\ldots,k-1\}$. That covers everything not involving $\infty$ except the "long" diagonals $(i,i+k)$, so we add $(n-1)/2=k$ triangles $(i,i+k,\infty)$ for $i\in\{0,\ldots,k-1\}$.