Difference between revisions of "1991 IMO Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
The argument above would not work for <math>a=1</math>, since <math>{\textstyle\sum \frac{1}{k^a}}</math> only converges for <math>a>1</math>. But Osmun Nal argues [https://www.youtube.com/watch?v=3v3CMQS_5Cc in this video] that <math>x_k=k\sqrt{2}-\lfloor k\sqrt{2}\rfloor</math> satisfies the stronger inequality <math>|x_i-x_j|\cdot |i-j|\geq 1</math> for all distinct <math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously. | The argument above would not work for <math>a=1</math>, since <math>{\textstyle\sum \frac{1}{k^a}}</math> only converges for <math>a>1</math>. But Osmun Nal argues [https://www.youtube.com/watch?v=3v3CMQS_5Cc in this video] that <math>x_k=k\sqrt{2}-\lfloor k\sqrt{2}\rfloor</math> satisfies the stronger inequality <math>|x_i-x_j|\cdot |i-j|\geq 1</math> for all distinct <math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously. | ||
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+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1991|num-b=5|after=Last Question}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Revision as of 23:38, 16 November 2023
Contents
[hide]Problem
An infinite sequence of real numbers is said to be bounded if there is a constant such that for every .
Given any real number , construct a bounded infinite sequence such that for every pair of distinct nonnegative integers .
Solution 1
Since , the series is convergent; let be the sum of this convergent series. Let be the interval (or any bounded subset of measure ).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose such that holds for all distinct . The only new cases are when one number (WLOG ) is equal to , so we must guarantee that for all .
Let be the interval , of length . The points that are valid choices for are precisely the points of , so we must show that this set is nonempty. The total length is at most the sum of the lengths . This is .
Therefore the total measure of is , so has positive measure and thus is nonempty. Choosing any and continuing by induction constructs the desired sequence.
Solution 2
The argument above would not work for , since only converges for . But Osmun Nal argues in this video that satisfies the stronger inequality for all distinct ; in other words, this sequence simultaneously solves the problem for all simultaneously.
See Also
1991 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |