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

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== Solution 4 (Vector) ==
 
== Solution 4 (Vector) ==
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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). Now we can do sum inside the vector so we get </math>((1+3+5+\1dots+2k-1  a_1+a_2+a_3+\1dots+a_k), with <math>a_1+\1dots+a_k</math> to be <math>17</math> and <math>1+3+\1dots+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.
+
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 ==
 
== See also ==

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

See also

1991 AIME (Problems • Answer Key • Resources)
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