Difference between revisions of "2012 AIME I Problems/Problem 9"
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− | == Problem | + | == Problem == |
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> | ||
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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 an arithmetic progression, so are <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> | ||
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Knowing this, we may substitute <math>\frac{-1}{6}</math> for <math>X</math> in equations <math>(1)</math> and <math>(2)</math>, yielding <math>\frac{-a}{6} = Y + 1</math> and <math>\frac{5a}{6} = Z + 2</math>. Thus, we have that <math>-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7</math>. We are looking for <math>xy^5z = 2^{X+ 5Y + Z}</math>. <math>X = \frac{-1}{6}</math> and <math>5Y + Z = -7</math>, so <math>xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}</math>. The answer is <math>43+6=\boxed{049}</math>. | Knowing this, we may substitute <math>\frac{-1}{6}</math> for <math>X</math> in equations <math>(1)</math> and <math>(2)</math>, yielding <math>\frac{-a}{6} = Y + 1</math> and <math>\frac{5a}{6} = Z + 2</math>. Thus, we have that <math>-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7</math>. We are looking for <math>xy^5z = 2^{X+ 5Y + Z}</math>. <math>X = \frac{-1}{6}</math> and <math>5Y + Z = -7</math>, so <math>xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}</math>. The answer is <math>43+6=\boxed{049}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 4 (Rigorous and easy)== | ||
+ | We know that | ||
+ | <cmath>\frac{\log (4y^2)}{\log (x)} = \frac{\log (16z^2)}{\log (2x)} = \frac{2\log (8yz)}{\log (2x^2)}</cmath> | ||
+ | By the [[Mediant theorem]]. | ||
+ | |||
+ | Substituting into the original equation yields us <math>\frac{2\log (8yz)}{\log (2x^2)} = \frac{\log (8yz)}{\log (2x^4)} \Rightarrow 2\log (2x^4) = \log (2x^2) \Rightarrow x=2^{-1/6}.</math> | ||
+ | For some constant <math>C\not= 0,</math> Let <math>2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = C.</math> Then, we obtain the system of equations | ||
+ | <cmath>y=2^{-13C/12}</cmath> | ||
+ | <cmath>z=2^{-19C/12}</cmath> | ||
+ | <cmath>8yz=2^{C/3}.</cmath> | ||
+ | |||
+ | Adding the first two equations and subtracting the third, we find <math>C=1.</math> Thus, <cmath>xy^5z=2^{-1/6} \cdot 2^{-65/12} \cdot 2^{-19/12}=2^{-43/6} \Rightarrow p+q=\boxed{049}.</cmath> | ||
+ | |||
+ | ~Kscv | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2012aimei/348 | ||
+ | |||
+ | ~ dolphin7 | ||
== See also == | == See also == |
Latest revision as of 21:38, 25 November 2023
Contents
[hide]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 an arithmetic progression, so are .
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 .
Solution 4 (Rigorous and easy)
We know that By the Mediant theorem.
Substituting into the original equation yields us For some constant Let Then, we obtain the system of equations
Adding the first two equations and subtracting the third, we find Thus,
~Kscv
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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