Difference between revisions of "1991 AIME Problems/Problem 15"
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== Solution == | == Solution == | ||
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Interpret the problem geometrically. 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 | Interpret the problem geometrically. 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 | ||
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If this is 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>. | If this is 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>. | ||
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The inequality | The inequality | ||
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is a direct result of the [[Minkowski Inequality]]. Continue as above. | 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 | 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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Revision as of 18:44, 13 June 2018
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
Interpret the problem geometrically. Consider right triangles joined at their vertices, with bases and heights . The sum of their hypotenuses is the value of . The minimum value of , 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 Since the sum of the first odd integers is and the sum of is 17, we get If this is integer, we can write , for an integer . Thus, The only possible value, then, for is , in which case , and .
Solution 2
The inequality is a direct result of the Minkowski Inequality. Continue as above.
Solution 3
Let for and . We then have that Note that that . Note that for any angle , it is true that and are reciprocals. We thus have that . By the AM-HM inequality on these values, we have that: This is thus the minimum value, with equality when all the tangents are equal. The only value for which is an integer is (see above solutions for details).
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.