1991 AIME Problems/Problem 15

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 $a_{}^{}=b_{}^{}$ . 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 $t=\sum_{k=1}^{n}a_{k}$ .

1991 AIME (Problems • Answer Key • Resources)
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1991 AIME Problems/Problem 15

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