Difference between revisions of "1991 AIME Problems/Problem 15"

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where <math>a_1,a_2,\ldots,a_n^{}</math> are positive real numbers whose sum is 17. There is a unique positive integer <math>n^{}_{}</math> for which <math>S_n^{}</math> is also an integer. Find this <math>n^{}_{}</math>.
 
where <math>a_1,a_2,\ldots,a_n^{}</math> are positive real numbers whose sum is 17. There is a unique positive integer <math>n^{}_{}</math> for which <math>S_n^{}</math> is also an integer. Find this <math>n^{}_{}</math>.
  
== Solution ==
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__TOC__
We start by recalling the following simple inequality: Let <math>a_{}^{}</math> and <math>b_{}^{}</math> denote two 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> (which can be easily found from the trivial fact that <math>(a_{}^{}-b)^{2}\geq0</math>). Applying this inequality to the given sum, one has
 
  
<math>
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== Solution 1 (Geometric Interpretation)==
\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}}\, ,
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Consider <math>n</math> right triangles joined at their vertices, with bases <math>a_1,a_2,\ldots,a_n</math> and heights <math>1,3,\ldots, 2n - 1</math>. The sum of their hypotenuses is the value of <math>S_n</math>. The minimum value of <math>S_n</math>, 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
</math>
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<cmath>
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S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.
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</cmath>
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Since the sum of the first <math>n</math> odd integers is <math>n^2</math> and the sum of <math>a_1,a_2,\ldots,a_n</math> is 17, we get
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<cmath>
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S_n \ge \sqrt {17^2 + n^4}.
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</cmath>
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If this is an integer, we can write <math>17^2 + n^4 = m^2</math>, for an integer <math>m</math>. Thus, <math>(m - n^2)(m + n^2) = 289\cdot 1 = 17\cdot 17 = 1\cdot 289.</math> The only possible value, then, for <math>m</math> is <math>145</math>, in which case <math>n^2 = 144</math>, and <math>n = \boxed {012}</math>.
  
where we have used the well-known fact that <math>\sum_{k=1}^{n}(2k-1)=n^{2}</math>, and we have defined <math>t=\sum_{k=1}^{n}a_{k}</math>. Therefore, <math>S_{n}\geq(n^{2}+t)/\sqrt{2}</math>. Now, in the present case, <math>t_{}^{}=17</math>, and so it is clear that for <math>n_{}^{}=1</math>, <math>S_{1}^{}</math> does not exist as the sum equals the noninteger value <math>\sqrt{1+17^{2}}>17</math>. This means that <math>n_{}^{}>1</math>. Notice that for <math>n_{}^{}\geq3</math>, <math>S_{n}^{}\geq(n^{2}+t)/\sqrt{2}\geq(3^{2}+17)/\sqrt{2}>18</math>. Is it possible for <math>n_{}^{}=2</math> to have <math>S_{2}^{}=18</math>? The answer is yes. One just needs to solve the quadratic equation for <math>a_{1}^{}</math> obtained from <math>\sqrt{1^{2}+a_{1}^{2}}+\sqrt{3^{2}+(17-a_{1}^{})^{2}}=18</math>, and verify that at least one of the solutions is in the open interval <math>(0_{}^{},17)</math>, which is indeed the case.
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== Solution 2 ==
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The inequality
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<cmath>
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S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.
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</cmath>
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is a direct result of the [[Minkowski Inequality]]. Continue as above.
  
In summary, the answer is <math>n_{}^{}=2</math>.
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== Solution 3 ==
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Let <math>a_{i} = (2i - 1) \tan{\theta_{i}}</math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i} < \frac {\pi}{2}</math>. We then have that
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<cmath>
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S_{n} = \sum_{k = 1}^{n}\sqrt {(2k - 1)^{2} + a_{k}^{2}} = \sum_{k = 1}^{n}(2k - 1) \sec{\theta_{k}}
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</cmath>
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Note that that <math>S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})</math>.
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Note that for any angle <math>\theta</math>, it is true that <math>\sec{\theta} + \tan{\theta}</math> and <math>\sec{\theta} - \tan{\theta}</math> are reciprocals. We thus have that <math>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}}}</math>. By the AM-HM inequality on these <math>n^{2}</math> values, we have that:
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<cmath>
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\frac {S_{n} + 17}{n^{2}}\ge \frac {n^{2}}{S_{n} - 17}\rightarrow S_{n}^{2}\ge 289 + n^{4}
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</cmath>
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This is thus the minimum value, with equality when all the tangents are equal. The only value for which <math>\sqrt {289 + n^{4}}</math> is an integer is <math>n = 12</math> (see above solutions for details).
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=14|after=Last question}}
 
{{AIME box|year=1991|num-b=14|after=Last question}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 01:43, 25 September 2020

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 1 (Geometric Interpretation)

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 an 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 (ProblemsAnswer KeyResources)
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