Difference between revisions of "2019 AMC 12B Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
− | Note that <math>log_2{x} + log_4{x} > 3</math>, <math>log_2{x} + 3 > log_4{x}</math>, and <math>log_4{x} + 3 > log_2{x}</math>. The second one is redundant, as it's less restrictive in all cases than the last. | + | Note that <math>\log_2{x} + \log_4{x} > 3</math>, <math>\log_2{x} + 3 > \log_4{x}</math>, and <math>\log_4{x} + 3 > \log_2{x}</math>. 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 <math>4</math>. <math>4^{log_2{x}} \cdot 4^{log_4{x}} > 64 \Rightarrow x^2 \cdot x > 64</math>. Thus, <math>x > 4</math>. | + | Let's raise the first to the power of <math>4</math>. <math>4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow x^2 \cdot x > 64</math>. Thus, <math>x > 4</math>. |
− | Doing the same for the second nets us: <math>4^{log_4{x}} \cdot 64 > 4^{log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64</math>. | + | Doing the same for the second nets us: <math>4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64</math>. |
Thus, x is an integer strictly between <math>64</math> and <math>4</math>: <math>64 - 4 - 1 = 59</math>. | Thus, x is an integer strictly between <math>64</math> and <math>4</math>: <math>64 - 4 - 1 = 59</math>. |
Revision as of 14:57, 14 February 2019
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.