2019 AMC 12B Problems/Problem 9

Revision as of 14:57, 14 February 2019 by Robinaldabanx (talk | contribs) (Solution 2)

Problem

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

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 $64 - 4 - 1 = 59$.

-DrJoyo

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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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