Difference between revisions of "2019 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need < | + | For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need <cmath>\log_2{x} + \log_4{x} > 3</cmath> <cmath>\log_2{x} + 3 > \log_4{x}</cmath> <cmath>\log_4{x} + 3 > \log_2{x}.</cmath> The second inequality is redundant, as it's always less restrictive than the last inequality. |
Let's raise the first inequality to the power of <math>4</math>. This gives <math>4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64</math>. Thus, <math>x > 4</math>. | Let's raise the first inequality to the power of <math>4</math>. This gives <math>4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64</math>. Thus, <math>x > 4</math>. | ||
− | Doing the same for the | + | Doing the same for the third inequality gives <math>4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64</math> (where we are allowed to divide both sides by <math>x</math> since <math>x</math> must be positive in order for the logarithms given in the problem statement to even have real values). |
+ | |||
+ | Combining our results, <math>x</math> is an integer strictly between <math>4</math> and <math>64</math>, so the number of possible values of <math>x</math> is <math>64 - 4 - 1 = \boxed{\textbf{(B) } 59}</math>. | ||
+ | |||
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+ | ~Minor edits by BakedPotato66 | ||
+ | |||
+ | ==Solution 2 (Somewhat Cheating)== | ||
+ | |||
+ | Using the triangle inequality, you get <math>\log_2{x}+\log_4{x} > 3</math>. Solving for <math>x</math>, you get <math>x > 4</math>. Now we need an upper-bound for <math>x</math> and since we're dealing with bases of <math>2</math> and <math>4</math>, we're looking for answer choices close to a power of <math>2</math> and <math>4</math>. All the answer choices seem to be around <math>64</math>, and plugging that into the inequality <math>3+\log_4{x} > \log_2{x}</math> we see <math>64</math> is the correct number. Now we have <math>64 > x > 4</math> and the number of integers in between is <math>64-4-1 = \boxed{{59 \textbf{(B)}}}</math> --OGBooger | ||
− | |||
==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}} |
Latest revision as of 03:36, 21 July 2024
Problem
For how many integral values of can a triangle of positive area be formed having side lengths
?
Solution
For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need
The second inequality is redundant, as it's always less restrictive than the last inequality.
Let's raise the first inequality to the power of . This gives
. Thus,
.
Doing the same for the third inequality gives (where we are allowed to divide both sides by
since
must be positive in order for the logarithms given in the problem statement to even have real values).
Combining our results, is an integer strictly between
and
, so the number of possible values of
is
.
~Minor edits by BakedPotato66
Solution 2 (Somewhat Cheating)
Using the triangle inequality, you get . Solving for
, you get
. Now we need an upper-bound for
and since we're dealing with bases of
and
, we're looking for answer choices close to a power of
and
. All the answer choices seem to be around
, and plugging that into the inequality
we see
is the correct number. Now we have
and the number of integers in between is
--OGBooger
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.