Difference between revisions of "2012 AIME I Problems/Problem 9"
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Let <math>x,</math> <math>y,</math> and <math>z</math> be positive real numbers that satisfy | Let <math>x,</math> <math>y,</math> and <math>z</math> be positive real numbers that satisfy | ||
<cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.</cmath> | <cmath>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.</cmath> |
Latest revision as of 19:21, 24 January 2021
Contents
Problem
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 (that isn't 0, of course), 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 .
Solution 3
Since we are given that , we may assume that , and are all powers of two. We shall thus let , , and . Let . From this we get the system of equations:
Plugging equation into equation yields . Plugging equation into equation and simplifying yields , and substituting for and simplifying yields . But , so , so .
Knowing this, we may substitute for in equations and , yielding and . Thus, we have that . We are looking for . and , so . The answer is .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/348
~ dolphin7
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 |
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