Problem 802

Since the convex hulls of two subsets do not intersect, they lie on opposite sides of some line. Thus, it is necessary to find out in how many ways the given set of points can be divided by a line into two subsets. Let's take a point $O$ in the plane, not lying on any of the lines connecting the given points, and consider the polar correspondence with center $O$. The given points will correspond to $n$ lines, no two of which are parallel and no three of which intersect at one point. As is known (see problem $\underline{60323})$, these lines divide the plane into $1 / 2 n(n+1)+1$ parts, of which $2 n$ are unbounded.

Lemma. Suppose the polars $a, b$ of points $A, B$ divide the plane into four angles. Then the poles of lines intersecting segment $A B$ lie in two vertical angles, and the poles of lines not intersecting segment $A B$ lie in the other two angles.

Proof. Let a line $l$ intersect line $A B$ at point $X$. Then the polar of $X$ passes through the point of intersection of $a$ and $b$. If we rotate $l$ around $X$, then its pole will move along this line, that is, within a pair of vertical angles formed by $a$ and $b$. As point $X$ moves along $A B$, its polar rotates around the point of intersection of $a$ and $b$, transitioning from one pair of vertical angles to another at the moments when $X$ passes through point $A$ or $B$.

It follows from the lemma that two lines divide the given set of points in the same way if and only if their poles either lie in one of the parts into which the plane is divided by the polars of the given points, or lie on opposite sides of all $n$ lines. But the second case is possible if and only if both points lie in unbounded regions. Indeed, if points $P, Q$ lie on opposite sides of all lines, then each of these lines intersects segment $P Q$. Therefore, each of the rays extending this segment lies entirely in one part. Conversely, if point $P$ lies in an unbounded part, then take a ray with its origin at this point, lying entirely in this part and not parallel to any of the $n$ lines. Points of the opposite ray, lying further from $P$ than all points of intersection with the lines, lie on opposite sides of $P$ from these lines.

Thus, the $2 n$ unbounded regions are divided into pairs, each of which corresponds to one way of dividing the given set of points, and each of the other regions corresponds to its own way of dividing. In total, we get $1 / 2 n(n-1)+1$ ways, one of which results in all $n$ points falling into one subset.

## Answer

$1 / 2 n(n-1)$ ways.