Difference between revisions of "2012 AIME I Problems/Problem 9"
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<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> | <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{ | + | Thus <math>xy^5z=2^{-1/6-7}=2^{-43/6}</math> and <math>43+6=\boxed{049}</math>. |
== See also == | == See also == |
Revision as of 00:28, 21 August 2016
Contents
[hide]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 a geometric 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.