Difference between revisions of "2019 AMC 12B Problems/Problem 9"
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− | + | Note <math>x=4</math> is a lower bound for <math>x</math>, corresponding to a triangle with side lengths <math>(2,1,3)</math>. If <math>x\leq4</math>, <math>log_2x+log_4x\leq3</math>, violating the triangle inequality. | |
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+ | Note also that <math>x=64</math> is an upper bound for <math>x</math>, corresponding to a triangle with side lengths <math>(6,3,3)</math>. If <math>x\geq64</math>, <math>log_4x+3\leq log_2x</math>, again violating the triangle inequality. | ||
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+ | It is easy to verify all <math>4<x<64</math> satisfy <math>log_2x+log_4x>3</math> and <math>log_4x+3>log_2x</math> (the third inequality is satisfied trivially). The number of integers strictly between <math>4</math> and <math>64</math> is <math>64 - 4 - 1 = 59</math>. | ||
-DrJoyo | -DrJoyo |
Revision as of 14:57, 14 February 2019
Contents
[hide]Problem
For how many integral values of can a triangle of positive area be formed having side lengths ?
Solution
Note is a lower bound for , corresponding to a triangle with side lengths . If , , violating the triangle inequality.
Note also that is an upper bound for , corresponding to a triangle with side lengths . If , , again violating the triangle inequality.
It is easy to verify all satisfy and (the third inequality is satisfied trivially). The number of integers strictly between and is .
-DrJoyo
Solution 2
Note that , , and . The second one is redundant, as it's less restrictive in all cases than the last.
Let's raise the first to the power of . . Thus, .
Doing the same for the second nets us: .
Thus, x is an integer strictly between and : .
- Robin's solution
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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