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Difference between revisions of "1987 IMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
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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.
  
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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.
+
=== Solution 2 ===
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This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find <math>{a_i}</math> where <math>a_i</math> is positive. Let <math>f(a_1, a_2, ..., a_n) = a_1| x_1|+a_2 |x_2|+\cdots+a_n |x_n|\geq0</math>.
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By the [[Cauchy-Schwarz Inequality]],
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<cmath>\begin{split}
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\left(a_1 x_1+a_2 x_2+...+a_n x_n\right)^2
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&\leq \left(a_1^2+a_2^2+...+a_n^2\right)\cdot\left(x_1^2+x_2^2+...+x_n^2\right) \\
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&=a_1^2+a_2^2+...+a_n^2 \\
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&\leq n(k-1)^2
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\end{split}</cmath>
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This implies that <math>\left(a_1 x_1+a_2 x_2+...+a_n x_n\right)^2\leq \sqrt{n} (k-1)</math>, and hence the codomain of <math>f(a_1, a_2, ..., a_n)</math> is <math>\left[0, \sqrt{n} (k-1)\right]</math>. The rest of the proof is similar to Solution 1.
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~[[User:Iraevid13|Iraevid13]]
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 03:36, 28 May 2023

Problem

Let $x_1 , x_2 , \ldots , x_n$ be real numbers satisfying $x_1^2 + x_2^2 + \cdots + x_n^2 = 1$. Prove that for every integer $k \ge 2$ there are integers $a_1, a_2, \ldots a_n$, not all 0, such that $| a_i | \le k-1$ for all $i$ and

$|a_1x_1 + a_2x_2 + \cdots + a_nx_n| \le \frac{ (k-1) \sqrt{n} }{ k^n - 1 }$.

Solution

Solution 1

We first note that by the Power Mean Inequality, $\sum_{i=1}^{n} x_i \le \sqrt{n}$. Therefore all sums of the form $\sum_{i=1}^{n} b_i x_i$, where the $b_i$ is a non-negative integer less than $k$, fall in the interval $[ 0 , (k-1)\sqrt{n} ]$. We may partition this interval into $k^n - 1$ subintervals of length $\frac{ (k-1)\sqrt{n} }{k^n - 1}$. But since there are $k^n$ 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.

Solution 2

This solution is very similar to Solution 1 but uses a slightly different approach for the first part. It suffices to find ${a_i}$ where $a_i$ is positive. Let $f(a_1, a_2, ..., a_n) = a_1| x_1|+a_2 |x_2|+\cdots+a_n |x_n|\geq0$. By the Cauchy-Schwarz Inequality, \[\begin{split} \left(a_1 x_1+a_2 x_2+...+a_n x_n\right)^2 &\leq \left(a_1^2+a_2^2+...+a_n^2\right)\cdot\left(x_1^2+x_2^2+...+x_n^2\right) \\ &=a_1^2+a_2^2+...+a_n^2 \\ &\leq n(k-1)^2 \end{split}\] This implies that $\left(a_1 x_1+a_2 x_2+...+a_n x_n\right)^2\leq \sqrt{n} (k-1)$, and hence the codomain of $f(a_1, a_2, ..., a_n)$ is $\left[0, \sqrt{n} (k-1)\right]$. The rest of the proof is similar to Solution 1.

~Iraevid13

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
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