2019 AMC 12B Problems/Problem 9

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 $x=4$ is a lower bound for $x$ , corresponding to a triangle with side lengths $(2,1,3)$ . If $x\leq4$ , $\log_2x+\log_4x\leq3$ , violating the triangle inequality.

Note also that $x=64$ is an upper bound for $x$ , corresponding to a triangle with side lengths $(6,3,3)$ . If $x\geq64$ , $\log_4x+3\leq \log_2x$ , again violating the triangle inequality.

It is easy to verify all $4<x<64$ satisfy $\log_2x+\log_4x>3$ and $\log_4x+3>\log_2x$ (the third inequality is satisfied trivially). The number of integers strictly between $4$ and $64$ is $64 - 4 - 1 = 59$ .

Solution 2

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 $4$ : $64 - 4 - 1 = 59$ .

- Robin's solution

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.

Page

Toolbox

Search

2019 AMC 12B Problems/Problem 9

Contents

Problem

Solution

Solution 2

See Also