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

Latest revision as of 15:58, 10 November 2023

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).

Solution 4 (Vector)

Observe the sum and apply it to the Cartesian plane, we can see that it is basically calculating distance between the origin and two points. Therefore we can assume every $\sqrt{(2k-1)^2+a_k^2}$ to be a vector from the origin to point $(2k-1, a_k)$ . Now we can do sum inside the vector so we get $((1+3+5+\ldots+2k-1, a_1+a_2+a_3+\ldots+a_k)$ , with $a_1+\ldots+a_k$ to be $17$ and $1+3+\ldots+2k-1$ to be $k^2$ . Then we calculate the length of the vector to be $\sqrt{k^4+289}$ . Since the sum needs to be an integer, we assume $k^4+289$ equals to $s^2$ . Applying difference between squares, we get that $k^2-s=289$ and $k^2+s=-1$ . Therfore $2k^2$ is $288$ and $k=n=12$ to be the final answer.

JINZHENQIAN

@@ Line 4: / Line 4: @@
 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 ==
+__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}\geq 0</math>). Applying this inequality to the given sum, one has
-<math>
+== 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}}\, ,
+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>
+<cmath>
+S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.
+</cmath>
+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
+<cmath>
+S_n \ge \sqrt {17^2 + n^4}.
+</cmath>
+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 <math>S_{n}=\sqrt{1+17^{2}}>17</math>.
+== Solution 2 ==
+The inequality
+<cmath>
+S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.
+</cmath>
+is a direct result of the [[Minkowski Inequality]]. Continue as above.
+== Solution 3 ==
+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
+<cmath>
+S_{n} = \sum_{k = 1}^{n}\sqrt {(2k - 1)^{2} + a_{k}^{2}} = \sum_{k = 1}^{n}(2k - 1) \sec{\theta_{k}}
+</cmath>
+Note that that <math>S_{n} + 17 = \sum_{k = 1}^{n}(2k - 1)(\sec{\theta_{k}} + \tan{\theta_{k}})</math>.
+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:
+<cmath>
+\frac {S_{n} + 17}{n^{2}}\ge \frac {n^{2}}{S_{n} - 17}\rightarrow S_{n}^{2}\ge 289 + n^{4}
+</cmath>
+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).
+== Solution 4 (Vector) ==
+Observe the sum and apply it to the Cartesian plane, we can see that it is basically calculating distance between the origin and two points. Therefore we can assume every <math>\sqrt{(2k-1)^2+a_k^2}</math> to be a vector from the origin to point <math>(2k-1, a_k)</math>.
+Now we can do sum inside the vector so we get <math>((1+3+5+\ldots+2k-1,   a_1+a_2+a_3+\ldots+a_k)</math>, with <math>a_1+\ldots+a_k</math> to be <math>17</math> and <math>1+3+\ldots+2k-1</math> to be <math>k^2</math>. Then we calculate the length of the vector to be <math>\sqrt{k^4+289}</math>. Since the sum needs to be an integer, we assume <math>k^4+289</math> equals to <math>s^2</math>. Applying difference between squares, we get that <math>k^2-s=289</math> and <math>k^2+s=-1</math>. Therfore <math>2k^2</math> is <math>288</math> and <math>k=n=12</math> to be the final answer.
+----JINZHENQIAN
 == See also ==
 {{AIME box|year=1991|num-b=14|after=Last question}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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