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 positive 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>.
  
 
== 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 17:37, 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 positive real numbers, then $\sqrt{a_{}^{2}+b_{}^{2}}\geq (a+b)/\sqrt{2}$, with equality if and only if $a_{}^{}=b_{}^{}$.

See also

1991 AIME (ProblemsAnswer KeyResources)
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