Difference between revisions of "2006 AIME II Problems/Problem 2"

 
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#REDIRECT [[2006 AIME A 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>.
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== Solution ==
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By the [[Triangle Inequality]]:
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<div style="text-align:center;">
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<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
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<math>\log_{10} 12n > \log_{10} 75 </math>
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<math> 12n > 75 </math>
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<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
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</div>
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Also:
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<div style="text-align:center;">
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<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
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<math>\log_{10} 12\cdot75 > \log_{10} n </math>
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<math> n < 900 </math>
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</div>
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Combining these two inequalities:
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<cmath> 6.25 < n < 900 </cmath>
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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>.
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== See also ==
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{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 18:01, 25 September 2007

Problem

The lengths of the sides of a triangle with positive area are $\log_{10} 12$, $\log_{10} 75$, and $\log_{10} n$, where $n$ is a positive integer. Find the number of possible values for $n$.

Solution

By the Triangle Inequality:

$\log_{10} 12 + \log_{10} n > \log_{10} 75$

$\log_{10} 12n > \log_{10} 75$

$12n > 75$

$n > \frac{75}{12} = \frac{25}{4} = 6.25$

Also:

$\log_{10} 12 + \log_{10} 75 > \log_{10} n$

$\log_{10} 12\cdot75 > \log_{10} n$

$n < 900$

Combining these two inequalities:

\[6.25 < n < 900\]

The number of possible integer values for $n$ is the number of integers over the interval $(6.25 , 900)$, which is $893$.

See also

2006 AIME I (ProblemsAnswer KeyResources)
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
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