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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Thus <math>xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}</math> and <math>43+6=\boxed{049}</math>. | Thus <math>xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}</math> and <math>43+6=\boxed{049}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Since we are given that <math>xy^5z = 2^{-p/q}</math>, we may assume that <math>x, y</math>, and <math>z</math> are all powers of two. We shall thus let <math>x = 2^X</math>, <math>y = 2^Y</math>, and <math>z = 2^Z</math>. Let <math>a = \log_{2^X}(2^{Y+1})</math>. From this we get the system of equations: <cmath></cmath> | ||
+ | <math>(1)</math><cmath>a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1</cmath> | ||
+ | <math>(2)</math><cmath>a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2</cmath> | ||
+ | <math>(3)</math><cmath>2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3</cmath> | ||
+ | |||
+ | Plugging equation <math>(1)</math> into equation <math>(2)</math> yields <math>Y + a = Z + 1</math>. Plugging equation <math>(1)</math> into equation <math>(3)</math> and simplifying yields <math>7Y + 2a + 6 = Z + 1</math>, and substituting <math>Y + a</math> for <math>Z + 1</math> and simplifying yields <math>Y + 1 = \frac{-a}{6} </math>. But <math>Y + 1 = aX</math>, so <math>aX = \frac{-a}{6}</math>, so <math>X = \frac{-1}{6}</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 22:38, 25 November 2023
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 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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