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1991 AIME Problems/Problem 15

Revision as of 18:44, 13 June 2018 by Mathlete2017 (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

Interpret the problem geometrically. Consider $n$ right triangles joined at their vertices, with bases $a_1,a_2,\ldots,a_n$ and heights $1,3,\ldots, 2n - 1$. The sum of their hypotenuses is the value of $S_n$. The minimum value of $S_n$, then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of the last right triangle, so \[S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.\] Since the sum of the first $n$ odd integers is $n^2$ and the sum of $a_1,a_2,\ldots,a_n$ is 17, we get \[S_n \ge \sqrt {17^2 + n^4}.\] If this is integer, we can write $17^2 + n^4 = m^2$, for an integer $m$. Thus, $(m - n^2)(m + n^2) = 289\cdot 1 = 17\cdot 17 = 1\cdot 289.$ The only possible value, then, for $m$ is $145$, in which case $n^2 = 144$, and $n = \boxed {012}$.

Solution 2

The inequality \[S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.\] is a direct result of the Minkowski Inequality. Continue as above.

Solution 3

Let $a_{i} = (2i - 1) \tan{\theta_{i}}$ for $1 \le i \le n$ and $0 \le \theta_{i} < \frac {\pi}{2}$. We then have that \[S_{n} = \sum_{k = 1}^{n}\sqrt {(2k - 1)^{2} + a_{k}^{2}} = \sum_{k = 1}^{n}(2k - 1) \sec{\theta_{k}}\] Note that that $S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})$. Note that for any angle $\theta$, it is true that $\sec{\theta} + \tan{\theta}$ and $\sec{\theta} - \tan{\theta}$ are reciprocals. We thus have that $S_{n} - 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} - \tan{\theta_{k}}) = \sum_{k = 1}^{n}\frac {2k - 1}{\sec{\theta_{k}} + \tan{\theta_{k}}}$. By the AM-HM inequality on these $n^{2}$ values, we have that: \[\frac {S_{n} + 17}{n^{2}}\ge \frac {n^{2}}{S_{n} - 17}\rightarrow S_{n}^{2}\ge 289 + n^{4}\] This is thus the minimum value, with equality when all the tangents are equal. The only value for which $\sqrt {289 + n^{4}}$ is an integer is $n = 12$ (see above solutions for details).

See also

1991 AIME (Problems • Answer Key • Resources)
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