Difference between revisions of "2006 AIME II Problems/Problem 2"
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== Solution == | == Solution == | ||
− | By the [[Triangle Inequality]] | + | By the [[Triangle Inequality]] and applying the well-known logarithmic property <math>\log_{c} a + \log_{c} b = \log_{c} ab</math>, we have that |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math> | <math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math> | ||
Line 14: | Line 14: | ||
<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | <math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math> | ||
</div> | </div> | ||
− | Also | + | |
+ | Also, | ||
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | <math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math> | ||
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<cmath> 6.25 < n < 900 </cmath> | <cmath> 6.25 < n < 900 </cmath> | ||
− | + | Thus <math>n</math> is in the set <math>(6.25 , 900)</math>; the number of positive integer <math>n</math> which satisfies this requirement is <math>\boxed{893}</math>. | |
== See also == | == See also == |
Latest revision as of 13:58, 14 April 2020
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 and applying the well-known logarithmic property , we have that
Also,
Combining these two inequalities:
Thus is in the set ; the number of positive integer which satisfies this requirement 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.