Difference between revisions of "1987 IMO Problems/Problem 3"
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== Problem == | == Problem == | ||
− | Let <math> | + | Let <math>x_1 , x_2 , \ldots , x_n </math> be real numbers satisfying <math>x_1^2 + x_2^2 + \cdots + x_n^2 = 1 </math>. Prove that for every integer <math>k \ge 2 </math> there are integers <math>a_1, a_2, \ldots a_n </math>, not all 0, such that <math>| a_i | \le k-1 </math> for all <math>i </math> and |
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== Solution == | == Solution == | ||
− | We first note that by the [[Power Mean Inequality]], <math> \sum_{i=1}^{n} x_i \le \sqrt{n} </math>. Therefore all sums of the form <math> \sum_{i=1}^{n} b_i x_i </math>, where the <math> | + | We first note that by the [[Power Mean Inequality]], <math> \sum_{i=1}^{n} x_i \le \sqrt{n} </math>. Therefore all sums of the form <math> \sum_{i=1}^{n} b_i x_i </math>, where the <math>b_i </math> is a non-negative integer less than <math>k</math>, fall in the interval <math>[ 0 , (k-1)\sqrt{n} ] </math>. We may partition this interval into <math>k^n - 1 </math> subintervals of length <math> \frac{ (k-1)\sqrt{n} }{k^n - 1} </math>. But since there are <math>k^n </math> such sums, by the [[pigeonhole principle]], two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties. |
{{alternate solutions}} | {{alternate solutions}} | ||
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[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Revision as of 19:49, 25 October 2007
Problem
Let be real numbers satisfying . Prove that for every integer there are integers , not all 0, such that for all and
.
Solution
We first note that by the Power Mean Inequality, . Therefore all sums of the form , where the is a non-negative integer less than , fall in the interval . We may partition this interval into subintervals of length . But since there are such sums, by the pigeonhole principle, two must fall into the same subinterval. It is easy to see that their difference will form a sum with the desired properties.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1987 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |