Difference between revisions of "2012 AIME I Problems/Problem 9"
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From this, we see that <math>8yz</math> is the geometric mean of <math>4y^2</math> and <math>16z^2</math>. So, for constant <math>C\ne 0</math>: | From this, we see that <math>8yz</math> is the geometric mean of <math>4y^2</math> and <math>16z^2</math>. So, for constant <math>C\ne 0</math>: | ||
<cmath>\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C</cmath> | <cmath>\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C</cmath> | ||
− | Since <math>\log 4y^2,\log 8yz,\log 16z^2</math> are in | + | Since <math>\log 4y^2,\log 8yz,\log 16z^2</math> are in a geometric progression, so is <math>\log x,\log 2x^4,\log 2x</math>. |
Therefore, <math>2x^4</math> is the geometric mean of <math>x</math> and <math>2x</math> | Therefore, <math>2x^4</math> is the geometric mean of <math>x</math> and <math>2x</math> |
Revision as of 00:26, 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.