Prove that there exists a positive constant $c$ such that the following statement is true:

Consider an integer $n>1$, and a set $S$ of $n$ points in the plane such that the distance between any two different points in $S$ is at least $1$. It follows that there is a line $\ell$ separating $S$ such that the distance from any point of $S$ to $\ell$ is at least $c{n}^{-1/3}$.

(A line $\ell$ separates a set of points $S$ if some segment joining two points in $S$ crosses $\ell$.)

Step 1 (Strategy): We show that among all lines in a suitable family, at least one separates $S$ while being far from all points. Consider lines perpendicular to a fixed direction.

Step 2 (Project onto a line): Choose a direction $\theta$ and project all points of $S$ onto the line in direction $\theta$. Let $x_1\le x_2\le \cdots \le x_n$ be the projections.

Step 3 (Gaps between projections): We need a gap $x_{k+1}-x_k\ge 2c{n}^{-1/3}$ for some $k$ with $1\le k\le n-1$ (so the perpendicular line in this gap separates $S$). By the minimum distance condition, the points are spread out.

Step 4 (Counting close pairs): For a random direction $\theta$, the expected number of pairs $(i,j)$ with $|x_i-x_j|<\delta$ can be bounded using the minimum distance condition. Since $|p_i-p_j|\ge 1$ for all pairs, the projections satisfy: at most $O(n/\delta )$ points can have projections within any interval of width $\delta$ (packing argument in strips of width $\delta$).

Step 5 (Conclusion): Choosing $\delta =c{n}^{-1/3}$ for a suitable constant $c$, and averaging over directions $\theta$, one finds a direction and a gap of width $\ge 2\delta$. The perpendicular bisector at this gap gives the desired line $\ell$.

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