Source: IMO-2020-P6
Category: Existence and Construction

Problem:
Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points in the plane such that the distance between any two different points in $\mathcal S$ is at least $1$. Prove there is a line $\ell$ separating $\mathcal S$ such that the distance from any point of $\mathcal S$ to $\ell$ is at least $\Omega\!\bigl(n^{-1/3}\bigr)$. (A line $\ell$ separates a set of points $S$ if some segment joining two points in $\mathcal S$ crosses $\ell$.)

Solution:
Let $\delta$ be the maximum possible distance between two consecutive projections of the points of $\mathcal S$ onto any line $m$. (This maximum is achieved by a standard compactness argument.) We will prove that $\delta \ge \Omega(n^{-1/3})$. This solves the problem, because we can choose a line $m$ achieving this maximum gap $\delta$ and place a line $\ell$ perpendicular to $m$ in the exact middle of the gap. The line $\ell$ separates $\mathcal S$, and the distance from any point of $\mathcal S$ to $\ell$ is at least $\frac{\delta}{2} = \Omega(n^{-1/3})$.

Let $A$ and $B$ be two points in $\mathcal S$ that are farthest apart. Let $R = AB$; since all pairwise distances between points in $\mathcal S$ are at least $1$, we have $R \ge 1$. All points of $\mathcal S$ lie in the intersection of the disks centered at $A$ and $B$ with radius $R$. Consider the circle $\odot(B)$ centered at $B$ with radius $R$. Let $\overline{XY}$ be the chord of $\odot(B)$ such that $\overline{XY} \perp \overline{AB}$ and the distance from $A$ to $\overline{XY}$ is exactly $\frac12$. (Such a chord exists since $R \ge 1 > \frac12$.) Let $\mathcal T$ denote the smaller region bounded by $\odot(B)$ and the chord $\overline{XY}$ (this is the cap containing $A$). 

First, note that $R = AB \le (n-1)\delta < n\delta$, since the $n$ projections of the points in $\mathcal S$ onto the line $AB$ are spaced at most $\delta$ apart. The distance from $B$ to the chord $\overline{XY}$ is $R - \frac12$. The Pythagorean theorem gives the length of the chord as
\[
XY = 2\sqrt{R^{2} - \left(R-\frac12\right)^{2}} = 2\sqrt{R - \frac14} < 2\sqrt{R} < 2\sqrt{n\delta}.
\]

By definition, the region $\mathcal T$ corresponds to an interval of length $\frac12$ along the line $AB$, starting from $A$. Since all points of $\mathcal S$ lie in the intersection of the disks $D(A, R)$ and $D(B, R)$, their projections onto the line $AB$ all lie in the segment $AB$. Let these projections be $x_1 \le x_2 \le \dots \le x_n$, where $x_1 = 0$ corresponds to $A$ and $x_n = R$ corresponds to $B$. The points in $\mathcal T$ are exactly those whose projections satisfy $x_i \le \frac12$. Let $k = |\mathcal T|$ be the number of such points. Then $x_k \le \frac12 < x_{k+1}$. Because the maximum gap between consecutive projections is at most $\delta$, we have
\[
\frac12 < x_{k+1} \le x_1 + k\delta = k\delta.
\]
Thus, the number of points in $\mathcal T$ satisfies
\[
\lvert\mathcal T\rvert = k > \frac{1}{2\delta}.
\]

Next, consider the projections of the points in $\mathcal T$ onto the line $XY$. Any two points $P, Q \in \mathcal T$ are at distance at least $1$. The distance between their projections onto the line $AB$ is at most the width of $\mathcal T$ in the direction of $AB$, which is $\frac12$. Therefore, the distance between their projections onto the line $XY$ is at least
\[
\sqrt{PQ^2 - \left(\frac12\right)^2} \ge \sqrt{1^2 - \frac14} = \frac{\sqrt3}{2}.
\]
The region $\mathcal T$ is bounded by the chord $\overline{XY}$ and the arc of $\odot(B)$. Thus, the projections of all points in $\mathcal T$ onto the line $XY$ lie on the segment $\overline{XY}$. The distance between the extreme projections among these $|\mathcal T|$ points is at most the length of $\overline{XY}$. On the other hand, since the projections are spaced at least $\frac{\sqrt3}{2}$ apart, this distance is at least $\frac{\sqrt3}{2}(\lvert\mathcal T\rvert - 1)$. This implies
\[
XY \ge \frac{\sqrt3}{2}\left(\lvert\mathcal T\rvert - 1\right).
\]

Substituting the lower bound for $|\mathcal T|$, we obtain
\[
XY > \frac{\sqrt3}{2}\left(\frac{1}{2\delta} - 1\right).
\]
Combining this with the upper bound $XY < 2\sqrt{n\delta}$, we have
\[
2\sqrt{n\delta} > \frac{\sqrt3}{4\delta} - \frac{\sqrt3}{2}.
\]
If $\delta \ge \frac{1}{4}$, then $\delta \ge \frac{1}{4} n^{-1/3}$ trivially holds since $n \ge 2$. Otherwise, if $\delta < \frac{1}{4}$, we have $\frac{1}{2\delta} > 2$, so $\frac{\sqrt3}{4\delta} - \frac{\sqrt3}{2} > \frac{\sqrt3}{8\delta}$. This gives
\[
2\sqrt{n\delta} > \frac{\sqrt3}{8\delta} \implies \delta^{3/2} > \frac{\sqrt3}{16\sqrt{n}} \implies \delta > \left(\frac{3}{256}\right)^{1/3} n^{-1/3}.
\]
In either case, we have $\delta \ge \Omega(n^{-1/3})$. This completes the proof.


Grading guidelines:
(Partial) 1. Reformulated the task via projections: defined $\delta$ as the maximum gap between consecutive projections of the $n$ points onto some line, and explained how a gap of size $\delta$ yields a separating line at distance $\gtrsim \delta$ (e.g. by taking a line parallel to the projection line placed in such a gap). 2. Introduced the diameter endpoints $A,B$ (farthest pair) and deduced a global bound $R=AB \le (n-1)\delta$ (or $R < n\delta$) from spacing of projections onto $AB$. 3. Constructed the chord $XY$ of the circle centered at $B$ with $XY \perp AB$ and $\operatorname{dist}(A,XY)=1/2$ (or an equivalent thin cap region $T$), and correctly computed/estimated $XY = 2\sqrt{R^2-(R-1/2)^2}=2\sqrt{R-1/4}$ to relate $XY$ to $R$ and hence to $\delta$. (Almost) 1. Completed the cap-counting strategy (using the thin region $T$ cut off by $XY$) and derived $\delta \ge c n^{-1/3}$, but left a small gap in justifying one of the key spacing/count estimates inside $T$ (e.g. why projections onto $XY$ are at least $\sqrt{3}/2$ apart, or why $|T| \gtrsim (1/2)\delta^{-1}$). 2. Obtained the final inequality $\delta \ge \Omega(n^{-1/3})$ with correct dependence on $n$, but made a localized computational slip in combining bounds (e.g. lost a constant/power when passing from $XY < 2\sqrt{n\delta}$ and $XY > c(\delta^{-1}-1)$ to $\delta \ge c'n^{-1/3}$), while the main argument is otherwise complete. 3. Solution is almost complete, but made minor mistakes which are not negligible.