1991 AIME Problems/Problem 15

Revision as of 19:56, 19 April 2007 by Gabiloncho (talk | contribs) (Solution)

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 $n_{}^{}=1$, $S_{1}^{}$ does not exist 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 $n_{}^{}=2$.

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
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