Difference between revisions of "2006 AIME II Problems/Problem 2"
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<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | <math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | ||
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− | Also | + | Also, applying the well-known logarithmic property <math>\log_{c} a \cdot \log_{c} b = \log_{c} ab</math>, we have |
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<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | <math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> |
Revision as of 20:45, 30 November 2019
Problem
The lengths of the sides of a triangle with positive area are , , and , where is a positive integer. Find the number of possible values for .
Solution
By the Triangle Inequality:
Also, applying the well-known logarithmic property , we have
Combining these two inequalities:
The number of possible integer values for is the number of integers over the interval , which is .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.