Difference between revisions of "1975 IMO Problems/Problem 1"

(Created page with "==Problem== Let <math>x_i, y_i</math> <math>(i=1,2,\cdots,n)</math> be real numbers such that <cmath>x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.</cmath...")
 
 
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==Solution==
 
==Solution==
{{solution}}
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We can rewrite the summation as
 
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<cmath>\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.</cmath>
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Since <math>\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2</math>, the above inequality is equivalent to
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<cmath>\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i.</cmath>
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We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of <math>\sum^n_{i=1}x_iz_i</math>. Obviously, if <math>x_1 = x_2 = \ldots = x_n</math> or <math>y_1 = y_2 = \ldots = y_n</math>, the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of <math>z_i</math>s that are not ordered in the form <math>z_1 \ge z_2 \ge \ldots \ge z_n</math> such that <math>\sum^n_{i=1}x_iz_i</math> is at a maximum out of all possible permutations and is greater than the sum <math>\sum^n_{i=1}x_iy_i</math>. This necessarily means that in the sum <math>\sum^n_{i=1}x_iz_i</math> there exists two terms <math>x_pz_m</math> and <math>x_qz_n</math> such that <math>x_p > x_q</math> and <math>z_m < z_n</math>. Notice that
 +
<cmath>x_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,</cmath>
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which means if we make the terms <math>x_pz_n</math> and <math>x_qz_m</math> instead of the original <math>x_pz_m</math> and <math>x_qz_n</math>, we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality
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<cmath>\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i</cmath>
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is proved, which is equivalent to what we wanted to prove.
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<cmath> </cmath>
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~Imajinary
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1975|before=First Question|num-a=3}}
 
{{IMO box|year=1975|before=First Question|num-a=3}}
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 23:13, 6 November 2020

Problem

Let $x_i, y_i$ $(i=1,2,\cdots,n)$ be real numbers such that \[x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.\] Prove that, if $z_1, z_2,\cdots, z_n$ is any permutation of $y_1, y_2, \cdots, y_n,$ then \[\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.\]

Solution

We can rewrite the summation as \[\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.\] Since $\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2$, the above inequality is equivalent to \[\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i.\] We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of $\sum^n_{i=1}x_iz_i$. Obviously, if $x_1 = x_2 = \ldots = x_n$ or $y_1 = y_2 = \ldots = y_n$, the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of $z_i$s that are not ordered in the form $z_1 \ge z_2 \ge \ldots \ge z_n$ such that $\sum^n_{i=1}x_iz_i$ is at a maximum out of all possible permutations and is greater than the sum $\sum^n_{i=1}x_iy_i$. This necessarily means that in the sum $\sum^n_{i=1}x_iz_i$ there exists two terms $x_pz_m$ and $x_qz_n$ such that $x_p > x_q$ and $z_m < z_n$. Notice that \[x_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,\] which means if we make the terms $x_pz_n$ and $x_qz_m$ instead of the original $x_pz_m$ and $x_qz_n$, we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality \[\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i\] is proved, which is equivalent to what we wanted to prove. \[\] ~Imajinary

See Also

1975 IMO (Problems) • Resources
Preceded by
First Question
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
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