Difference between revisions of "2006 AIME A Problems/Problem 2"
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== Solution == | == Solution == | ||
| + | By the [[Triangle Inequality]]: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | Also: | ||
| + | |||
| + | <math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | ||
| + | |||
| + | <math>\log_{10} 12\cdot75 > \log_{10} n </math> | ||
| + | |||
| + | <math> n < 900 </math> | ||
| + | |||
| + | Combining these two inequalities: | ||
| + | |||
| + | <math> 6.25 < n < 900 </math> | ||
| + | |||
| + | 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>892</math> | ||
== See also == | == See also == | ||
*[[2006 AIME II Problems]] | *[[2006 AIME II Problems]] | ||
Revision as of 11:39, 14 July 2006
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 
