Difference between revisions of "2006 AIME II Problems/Problem 2"
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− | + | == Problem == | |
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+ | The lengths of the sides of a [[triangle]] with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>. | ||
+ | |||
+ | == Solution == | ||
+ | By the [[Triangle Inequality]]: | ||
+ | <div style="text-align:center;"> | ||
+ | <math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math> | ||
+ | |||
+ | <math>\log_{10} 12n > \log_{10} 75 </math> | ||
+ | |||
+ | <math> 12n > 75 </math> | ||
+ | |||
+ | <math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | ||
+ | </div> | ||
+ | Also: | ||
+ | <div style="text-align:center;"> | ||
+ | <math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | ||
+ | |||
+ | <math>\log_{10} 12\cdot75 > \log_{10} n </math> | ||
+ | |||
+ | <math> n < 900 </math> | ||
+ | </div> | ||
+ | Combining these two inequalities: | ||
+ | |||
+ | <cmath> 6.25 < n < 900 </cmath> | ||
+ | |||
+ | The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math>, which is <math>893</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2006|n=I|num-b=1|num-a=3}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 19:01, 25 September 2007
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:
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 I (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 |