Difference between revisions of "2012 AIME I Problems/Problem 9"
(→Solution) |
(→Solution 2) |
||
Line 28: | Line 28: | ||
<cmath>2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}</cmath> | <cmath>2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}</cmath> | ||
We can plug <math>x</math> in to any of the two equal fractions aforementioned. So, without loss of generality: | We can plug <math>x</math> in to any of the two equal fractions aforementioned. So, without loss of generality: | ||
− | <cmath>\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x)</cmath> | + | <cmath>\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x\right)</cmath> |
<cmath>\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2</cmath> | <cmath>\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2</cmath> | ||
<cmath>\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0</cmath> | <cmath>\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0</cmath> | ||
− | <cmath>\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^ | + | <cmath>\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^{10}z^2=1\implies y^{10}z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}</cmath> |
Thus <math>xy^5z=2^{-1/6-7}=2^{-43/6}</math> and <math>43+6=\boxed{049}</math>. | Thus <math>xy^5z=2^{-1/6-7}=2^{-43/6}</math> and <math>43+6=\boxed{049}</math>. |
Revision as of 21:04, 21 February 2016
Contents
Problem 9
Let
and
be positive real numbers that satisfy
The value of
can be expressed in the form
where
and
are relatively prime positive integers. Find
Solution 1
Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value, so to simplify the problem let us assume without loss of generality that
Then
Solving these equations, we quickly see that
and then
Finally, our desired value is
and thus
Solution 2
Notice that ,
and
.
From this, we see that is the geometric mean of
and
. So, for constant
:
Since
are in an arithmetic progression, so is
.
Therefore, is the geometric mean of
and
We can plug
in to any of the two equal fractions aforementioned. So, without loss of generality:
Thus and
.
See also
2012 AIME I (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.