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Difference between revisions of "2019 AMC 12B Problems/Problem 9"

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==Solution==
 
==Solution==
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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>.
 
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==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.  
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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>.  
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- Robin's solution
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:00, 18 February 2019

Problem

For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$?

$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$

Solution

Note that $\log_2{x} + \log_4{x} > 3$, $\log_2{x} + 3 > \log_4{x}$, and $\log_4{x} + 3 > \log_2{x}$. 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 $4$. $4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow x^2 \cdot x > 64$. Thus, $x > 4$.

Doing the same for the second nets us: $4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$.

Thus, x is an integer strictly between $64$ and $4$: $64 - 4 - 1 = 59$.

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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