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Difference between revisions of "1991 AIME Problems/Problem 15"

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== Solution ==
 
== Solution ==
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We start by recalling the following simple inequality: Let <math>a_{}^{}</math> and <math>b_{}^{}</math> denote two real numbers, then <math>\sqrt{a_{}^{2}+b_{}^{2}}\geq (a+b)/\sqrt{2}</math>, with equality if and only if <math>a_{}^{}=b_{}^{}</math>. Applying this inequality to the given sum, one has
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We start by recalling the following simple inequality: Let <math>a_{}^{}</math> and <math>b_{}^{}</math> denote two real numbers, then <math>\sqrt{a_{}^{2}+b_{}^{2}}\geq (a+b)/\sqrt{2}</math>, with equality if and only if <math>a_{}^{}=b_{}^{}</math> (which can be easily found from the trivial fact that <math>(a-b)^{2}\geq0</math>). Applying this inequality to the given sum, one has
  
 
<math>
 
<math>
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</math>
 
</math>
  
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where we have used the well-known fact that <math>\sum_{k=1}^{n}(2k-1)=n^{2}</math>, and we have defined <math>t=\sum_{k=1}^{n}a_{k}</math>. Therefore, <math>S_{n}\geq(n^{2}+t)/\sqrt{2}</math>.
+
where we have used the well-known fact that <math>\sum_{k=1}^{n}(2k-1)=n^{2}</math>, and we have defined <math>t=\sum_{k=1}^{n}a_{k}</math>. Therefore, <math>S_{n}\geq(n^{2}+t)/\sqrt{2}</math>. Now, in the present case, <math>t_{}^{}=17</math>, and so it is clear that <math>S_{n}=\sqrt{1+17^{2}}>17</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=14|after=Last question}}
 
{{AIME box|year=1991|num-b=14|after=Last question}}

Revision as of 17:59, 19 April 2007

Problem

For positive integer $n_{}^{}$, define $S_n^{}$ to be the minimum value of the sum $\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},$ where $a_1,a_2,\ldots,a_n^{}$ are positive real numbers whose sum is 17. There is a unique positive integer $n^{}_{}$ for which $S_n^{}$ is also an integer. Find this $n^{}_{}$.

Solution

We start by recalling the following simple inequality: Let $a_{}^{}$ and $b_{}^{}$ denote two real numbers, then $\sqrt{a_{}^{2}+b_{}^{2}}\geq (a+b)/\sqrt{2}$, with equality if and only if $a_{}^{}=b_{}^{}$ (which can be easily found from the trivial fact that $(a-b)^{2}\geq0$). Applying this inequality to the given sum, one has

$\sum_{k=1}^{n}\sqrt{(2k-1)^{2}+a_{k}^{2}}\geq \frac{1}{\sqrt{2}}\sum_{k=1}^{n}[(2k-1)+a_{k}]=\frac{n^{2}+t}{\sqrt{2}}\, ,$

where we have used the well-known fact that $\sum_{k=1}^{n}(2k-1)=n^{2}$, and we have defined $t=\sum_{k=1}^{n}a_{k}$. Therefore, $S_{n}\geq(n^{2}+t)/\sqrt{2}$. Now, in the present case, $t_{}^{}=17$, and so it is clear that $S_{n}=\sqrt{1+17^{2}}>17$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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