Difference between revisions of "2020 IMO Problems/Problem 6"
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− | Problem | + | == Problem == |
− | + | Prove that there exists a positive constant <math>c</math> such that the following statement is true: | |
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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 | ||
+ | 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. | ||
− | Let <math>p</math> be any point in <math>S</math> on the boundary of <math>D</math>. Let <math>\ell_1</math> be the line tangent to <math>D</math> at <math>p</math>, and <math>\ell_2</math> the line obtained by translating <math>\ell_1</math> by distance <math>1</math> towards the inside of <math>D</math>. Let <math>H</math> be the region sandwiched by <math>\ell_1</math> and <math>\ell_2</math>. It is easy to show that both the area and the perimeter of <math>H\cap D</math> is bounded by <math>O(\sqrt{r})</math> (since <math>r=\Omega(\sqrt{n})</math>). Hence, there can only be <math>O(\sqrt{r})=O(n^{1/3})</math> points in <math>H\cap S</math>, by that any two points in <math>S</math> are distance <math>1</math> apart. Since the width of <math>H</math> is <math>1</math>, there must exist a line <math>\ell</math> parallel to <math>\ell_1</math> such that <math>\ell</math> separates <math>S</math>, and | + | Let <math>p</math> be any point in <math>S</math> on the boundary of <math>D</math>. Let <math>\ell_1</math> be the line tangent to <math>D</math> at <math>p</math>, and <math>\ell_2</math> the line obtained by translating <math>\ell_1</math> by distance <math>1</math> towards the inside of <math>D</math>. Let <math>H</math> be the region sandwiched by <math>\ell_1</math> and <math>\ell_2</math>. It is easy to show that both the area and the perimeter of <math>H\cap D</math> is bounded by <math>O(\sqrt{r})</math> (since <math>r=\Omega(\sqrt{n})</math>). Hence, there can only be <math>O(\sqrt{r})=O(n^{1/3})</math> points in <math>H\cap S</math>, by that any two points in <math>S</math> are distance <math>1</math> apart. Since the width of <math>H</math> is <math>1</math>, there must exist a line <math>\ell</math> parallel to <math>\ell_1</math> such that <math>\ell</math> separates <math>S</math>, and any point in <math>S</math> is at least <math>1/O(n^{1/3}) = \Omega(n^{-1/3})</math> away from <math>\ell</math>. Q.E.D. |
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+ | Note. One can also show that <math>\Omega(n^{-1/3})</math> is best possible. | ||
+ | ~Shen Kislay kai | ||
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+ | == Video solution == | ||
+ | https://www.youtube.com/watch?v=dTqwOoSfaAA [video covers all day 2 problems] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2020|num-b=5|after=Last Question}} | ||
− | + | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 12:15, 3 September 2024
Contents
[hide]Problem
Prove that there exists a positive constant such that the following statement is true:
Consider an integer , and a set of n points in the plane such that the distance between any two different points in is at least . It follows that there is a line separating such that the distance from any point of to is at least .
(A line separates a set of points if some segment joining two points in crosses .)
Note. Weaker results with replaced by may be awarded points depending on the value of the constant .
Solution
For any unit vector , let and . If then we can find a line perpendicular to such that separates , and any point in is at least away from .
Suppose there is no such direction , then is contained in a box with side length by considering the direction of and , respectively. Hence, is contained in a disk with radius . Now suppose that is the disk with the minimum radius, say , which contains . Then, . Since the distance between any two points in is at least , too.
Let be any point in on the boundary of . Let be the line tangent to at , and the line obtained by translating by distance towards the inside of . Let be the region sandwiched by and . It is easy to show that both the area and the perimeter of is bounded by (since ). Hence, there can only be points in , by that any two points in are distance apart. Since the width of is , there must exist a line parallel to such that separates , and any point in is at least away from . Q.E.D.
Note. One can also show that is best possible. ~Shen Kislay kai
Video solution
https://www.youtube.com/watch?v=dTqwOoSfaAA [video covers all day 2 problems]
See Also
2020 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |