A convex polygon has $ n$ sides. A ball is present at every vertex. Two mathematician $A$ and $B$ try to make partition of this polygon into disjoin regions . A region must be in the shape of a convex polygon with its vertices at balls. A and B try to enumerate the various possibilities for mathematician A is asked to present various ways in which the territory can be broken up into an odd number of regions, and B is asked to do the same for an even number of regions.
(a) Is it possible for any positive integer $n$ that the two mathematician come up with an equal number of ways to split the convex polygon ?
(b) Let the number of ways mathematician $A$ comes up with be $S_A$, with $S_B$ being defined similarly. Find $S_A − S_B$in terms of n.
What I have tried:
(a) This is possible for only $n=5$ .
(b) I tried with $n=6$ and get $S_A=21$ and $S_B =23$
How to find General results ??
I will assume that the trivial partition (e.g. partitioning a pentagon into a single pentagon, itself) is a valid partition. In this case, the number of even partitions and odd partitions always differ by exactly one. When $n$ is even, there is one more even partition, and when $n$ is odd, there is one more odd partition.
To see this, it suffices to find a matching which pairs up every even partition with an odd one, with one partition left over. Here is a sketch of the one I came up with.
Take the $n$-gon, and color one edge red. Call the endpoints of the red edge $a$ and $b$, with $a$ occurring clockwise of $b$. There is a unique polygon $P$ in the partition containing this red edge, and two of its vertices are $a$ and $b$. Let $c$ be the vertex of $P$ adjacent to $b$, on the other side of $a$. The basic idea of the bijection is this:
As long as this operation is legal, then it changes the number of polygons by exactly one, which transforms an even bijection into an odd one. Furthermore, this operation is its own inverse; doing it twice gets you back where you started. Therefore, it creates the desired matching.
However, this does not always work. If $\overline{ac}$ is an external edge, then deleting it does not create a valid partition. In this case, we ignore triangle $abc$, leaving an $(n-1)$-gon partitioned into polygons. We then color the edge $bc$ red, and recursively compute the bijection on this smaller polygon.
The only way this entire operation could fail is we keep recursing all the way down until the remaining polygon to be partitioned is just a triangle. There is exactly one way this can happen, and this one way is our exceptional partition.