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 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 for $n_{}^{}=1$ , $S_{1}^{}$ cannot be a valid solution as the sum equals the noninteger value $\sqrt{1+17^{2}}>17$ . This means that $n_{}^{}>1$ . Notice that if $n_{}^{}\geq3$ , $S_{n}^{}\geq(n^{2}+17)/\sqrt{2}\geq(3^{2}+17)/\sqrt{2}>18$ . Is it possible when $n_{}^{}=2$ to have $S_{2}^{}=18$ ? The answer is yes. One just needs to solve the quadratic equation for $a_{1}^{}$ obtained from $\sqrt{1^{2}+a_{1}^{2}}+\sqrt{3^{2}+(17-a_{1}^{})^{2}}=18$ , and verify that at least one of the solutions is in the open interval $(0_{}^{},17)$ , which is indeed the case.

In summary, the answer is $n_{}^{}=\boxed{002}$ .

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

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