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Difference between revisions of "2020 IMO Problems/Problem 6"
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Problem 6. Prove that there exists a positive constant c such that the following statement is true:
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== Problem ==
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Consider an integer n > 1, and a set S of n points in the plane such that the distance between
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Prove that there exists a positive constant <math>c</math> such that the following statement is true:
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any two different points in S is at least 1. It follows that there is a line ℓ separating S such that the
 
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distance from any point of S to ℓ is at least cn^(−1/3)
 
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.
 
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(A line ℓ separates a set of points S if some segment joining two points in S crosses ℓ.)
 
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Note. Weaker results with cn^(−1/3)
 
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replaced by cn^−α may be awarded points depending on the value
 
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of the constant α > 1/3.
 
  
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Proof. For any unit vector <math>v</math>, let <math>a_v=\min_{p\in S} p \cdot v</math> and <math>b_v = \max_{p\in S} p\cdot v</math>. If <math>b_v - a_v\geq n^{2/3}</math> then we can find a line <math>\ell</math> perpendicular to <math>v</math> such that <math>\ell</math> separates <math>S</math>, and any point in <math>S</math> is at least <math>\Omega(n^{2/3}/n) = \Omega(n^{-1/3})</math> away from <math>\ell</math>.
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Consider an integer <math>n > 1</math>, and a set <math>S</math> of n points in the plane such that the distance between any two different points in <math>S</math> is at least <math>1</math>. It follows that there is a line <math>\ell</math> separating <math>S</math> such that the distance from any point of <math>S</math> to <math>\ell</math> is at least <math>cn^{- \frac{1}{3}}</math>.
 +
 
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(A line <math>\ell</math> separates a set of points <math>S</math> if some segment joining two points in <math>S</math> crosses <math>\ell</math>.)
 +
 
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''Note''. Weaker results with <math>cn^{- \frac{1}{3}}</math> replaced by <math>cn^{- \alpha}</math> may be awarded points depending on the value
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of the constant <math>\alpha > \frac{1}{3}</math>.
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== Solution ==
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For any unit vector <math>v</math>, let <math>a_v=\min_{p\in S} p \cdot v</math> and <math>b_v = \max_{p\in S} p\cdot v</math>. If <math>b_v - a_v\geq n^{2/3}</math> then we can find a line <math>\ell</math> perpendicular to <math>v</math> such that <math>\ell</math> separates <math>S</math>, and any point in <math>S</math> is at least <math>\Omega(n^{2/3}/n) = \Omega(n^{-1/3})</math> away from <math>\ell</math>.
  
 
Suppose there is no such direction <math>v</math>, then <math>S</math> is contained in a box with side length <math>n^{2/3}</math> by considering the direction of <math>(1, 0)</math> and <math>(0, 1)</math>, respectively. Hence, <math>S</math> is contained in a disk with radius <math>n^{2/3}</math>. Now suppose that <math>D</math> is the disk with the minimum radius, say <math>r</math>, which contains <math>S</math>. Then, <math>r=O(n^{2/3})</math>. Since the distance between any two points in <math>S</math> is at least <math>1</math>, <math>r=\Omega(\sqrt{n})</math> too.
 
Suppose there is no such direction <math>v</math>, then <math>S</math> is contained in a box with side length <math>n^{2/3}</math> by considering the direction of <math>(1, 0)</math> and <math>(0, 1)</math>, respectively. Hence, <math>S</math> is contained in a disk with radius <math>n^{2/3}</math>. Now suppose that <math>D</math> is the disk with the minimum radius, say <math>r</math>, which contains <math>S</math>. Then, <math>r=O(n^{2/3})</math>. Since the distance between any two points in <math>S</math> is at least <math>1</math>, <math>r=\Omega(\sqrt{n})</math> too.

Revision as of 11:27, 14 May 2021

Problem

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 $cn^{- \frac{1}{3}}$.

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

Note. Weaker results with $cn^{- \frac{1}{3}}$ replaced by $cn^{- \alpha}$ may be awarded points depending on the value of the constant $\alpha > \frac{1}{3}$.

Solution

For any unit vector $v$, let $a_v=\min_{p\in S} p \cdot v$ and $b_v = \max_{p\in S} p\cdot v$. If $b_v - a_v\geq n^{2/3}$ then we can find a line $\ell$ perpendicular to $v$ such that $\ell$ separates $S$, and any point in $S$ is at least $\Omega(n^{2/3}/n) = \Omega(n^{-1/3})$ away from $\ell$.

Suppose there is no such direction $v$, then $S$ is contained in a box with side length $n^{2/3}$ by considering the direction of $(1, 0)$ and $(0, 1)$, respectively. Hence, $S$ is contained in a disk with radius $n^{2/3}$. Now suppose that $D$ is the disk with the minimum radius, say $r$, which contains $S$. Then, $r=O(n^{2/3})$. Since the distance between any two points in $S$ is at least $1$, $r=\Omega(\sqrt{n})$ too.

Let $p$ be any point in $S$ on the boundary of $D$. Let $\ell_1$ be the line tangent to $D$ at $p$, and $\ell_2$ the line obtained by translating $\ell_1$ by distance $1$ towards the inside of $D$. Let $H$ be the region sandwiched by $\ell_1$ and $\ell_2$. It is easy to show that both the area and the perimeter of $H\cap D$ is bounded by $O(\sqrt{r})$ (since $r=\Omega(\sqrt{n})$). Hence, there can only be $O(\sqrt{r})=O(n^{1/3})$ points in $H\cap S$, by that any two points in $S$ are distance $1$ apart. Since the width of $H$ is $1$, there must exist a line $\ell$ parallel to $\ell_1$ such that $\ell$ separates $S$, and any point in $S$ is at least $1/O(n^{1/3}) = \Omega(n^{-1/3})$ away from $\ell$. Q.E.D.

Note. One can also show that $\Omega(n^{-1/3})$ is best possible.

Video solution

https://www.youtube.com/watch?v=dTqwOoSfaAA [video covers all day 2 problems]

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