Subfield: Combinatorics


Question: Let $S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $P$ whose vertices are in $S$, let $a(P)$ be the number of vertices of $P$, and let $b(P)$ be the number of points of $S$ which are outside $P$. Prove that for every real number $x$

$$
\sum_{P} x^{a(P)}(1-x)^{b(P)}=1
$$

where the sum is taken over all convex polygons with vertices in $S$.

NB. A line segment, a point and the empty set are considered as convex polygons of 2,1 and 0 vertices, respectively.

Solution:
For each convex polygon $P$ whose vertices are in $S$, let $c(P)$ be the number of points of $S$ which are inside $P$, so that $a(P)+b(P)+c(P)=n$, the total number of points in $S$. Denoting $1-x$ by $y$,

$$
\sum_{P} x^{a(P)} y^{b(P)}=\sum_{P} x^{a(P)} y^{b(P)}(x+y)^{c(P)}=\sum_{P} \sum_{i=0}^{c(P)}\left(\begin{array}{c}
c(P) \\
i
\end{array}\right) x^{a(P)+i} y^{b(P)+c(P)-i}
$$

View this expression as a homogeneous polynomial of degree $n$ in two independent variables $x, y$. In the expanded form, it is the sum of terms $x^{r} y^{n-r}(0 \leq r \leq n)$ multiplied by some nonnegative integer coefficients.

For a fixed $r$, the coefficient of $x^{r} y^{n-r}$ represents the number of ways of choosing a convex polygon $P$ and then choosing some of the points of $S$ inside $P$ so that the number of vertices of $P$ and the number of chosen points inside $P$ jointly add up to $r$.

This corresponds to just choosing an $r$-element subset of $S$. The correspondence is bijective because every set $T$ of points from $S$ splits in exactly one way into the union of two disjoint subsets, of which the first is the set of vertices of a convex polygon - namely, the convex hull of $T$ - and the second consists of some points inside that polygon.

So the coefficient of $x^{r} y^{n-r}$ equals $\left(\begin{array}{l}n \\ r\end{array}\right)$. The desired result follows:

$$
\sum_{P} x^{a(P)} y^{b(P)}=\sum_{r=0}^{n}\left(\begin{array}{l}
n \\
r
\end{array}\right) x^{r} y^{n-r}=(x+y)^{n}=1
$$
Apply induction on the number $n$ of points. The case $n=0$ is trivial. Let $n>0$ and assume the statement for less than $n$ points. Take a set $S$ of $n$ points.

Let $C$ be the set of vertices of the convex hull of $S$, let $m=|C|$.

Let $X \subset C$ be an arbitrary nonempty set. For any convex polygon $P$ with vertices in the set $S \backslash X$, we have $b(P)$ points of $S$ outside $P$. Excluding the points of $X-$ all outside $P$ - the set $S \backslash X$ contains exactly $b(P)-|X|$ of them. Writing $1-x=y$, by the induction hypothesis

$$
\sum_{P \subset S \backslash X} x^{a(P)} y^{b(P)-|X|}=1
$$

(where $P \subset S \backslash X$ means that the vertices of $P$ belong to the set $S \backslash X$ ). Therefore

$$
\sum_{P \subset S \backslash X} x^{a(P)} y^{b(P)}=y^{|X|}
$$

All convex polygons appear at least once, except the convex hull $C$ itself. The convex hull adds $x^{m}$. We can use the inclusion-exclusion principle to compute the sum of the other terms:

$$
\begin{gathered}
\sum_{P \neq C} x^{a(P)} y^{b(P)}=\sum_{k=1}^{m}(-1)^{k-1} \sum_{|X|=k} \sum_{P \subset S \backslash X} x^{a(P)} y^{b(P)}=\sum_{k=1}^{m}(-1)^{k-1} \sum_{|X|=k} y^{k} \\
=\sum_{k=1}^{m}(-1)^{k-1}\left(\begin{array}{c}
m \\
k
\end{array}\right) y^{k}=-\left((1-y)^{m}-1\right)=1-x^{m}
\end{gathered}
$$

and then

$$
\sum_{P} x^{a(P)} y^{b(P)}=\sum_{P=C}+\sum_{P \neq C}=x^{m}+\left(1-x^{m}\right)=1
$$