Difference between revisions of "1991 AIME Problems/Problem 15"
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− | 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_{}^{}=3</math>, <math>S_{3}^{}\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>( | + | 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_{}^{}=3</math>, <math>S_{3}^{}\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. |
In summary, the answer is <math>n_{}^{}=2</math>. | In summary, the answer is <math>n_{}^{}=2</math>. |
Revision as of 19:35, 19 April 2007
Problem
For positive integer , define to be the minimum value of the sum where are positive real numbers whose sum is 17. There is a unique positive integer for which is also an integer. Find this .
Solution
We start by recalling the following simple inequality: Let and denote two real numbers, then , with equality if and only if (which can be easily found from the trivial fact that ). Applying this inequality to the given sum, one has
where we have used the well-known fact that , and we have defined . Therefore, . Now, in the present case, , and so it is clear that for , does not exist as the sum equals the noninteger value . This means that . Notice that for , . Is it possible for to have ? The answer is yes. One just needs to solve the quadratic equation for obtained from , and verify that at least one of the solutions is in the open interval , which is indeed the case.
In summary, the answer is .
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |