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Difference between revisions of "2000 AIME I Problems/Problem 9"

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<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
 
<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
 
\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
 
\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
−
\log_{10}(2000zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
+
\log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
 
\end{eqnarray*}</math></center>
 
\end{eqnarray*}</math></center>
  
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== Solution ==
 
== Solution ==
−
:''Will finish this problem soon. <font style="font-family:Georgia,sans-serif">[[User:Azjps|Azjps]] ([[User talk:Azjps|<font color="green">talk</font>]])</font> 18:12, 31 December 2007 (EST)''
 
 
Since <math>\log ab = \log a + \log b</math>, we can reduce the equations to a more recognizable form:
 
Since <math>\log ab = \log a + \log b</math>, we can reduce the equations to a more recognizable form:
  
−
<cmath>\begin{eqnarray*}- (\log x)(\log y) + \log x + \log y - 1 & = & 3 - 3\log 2 \\
+
<cmath>\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\
−
etc &=& etc
+
-\log y \log z + \log y + \log z - 1 &=& - \log 2\\
 +
-\log x \log z + \log x + \log z - 1 &=& -1\\
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
−
Let <math>x_1, y_1, z_1</math> be <math>\log x, \log y, \log z</math> respectively. Using [[SFFT]], it becomes
+
Let <math>a,b,c</math> be <math>\log x, \log y, \log z</math> respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)
  
−
<cmath>\begin{eqnarray*}(x_1 - 1)(y_1 - 1) &=& 3\log2 - 3\\
+
<cmath>\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\
−
etc &=& etc
+
(b-1)(c-1) &=& \log 2 \\
 +
(a-1)(c-1) &=& 1
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
−
Multiplying the three equations gives  
+
From here, multiplying the three equations gives  
  
−
<cmath>\begin{eqnarray*}(x_1-1)^2(y_1-1)^2(z_1-1)^2 &=& something\\
+
<cmath>\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\
−
(x_1-1)(y_1-1)(z_1-1) &=& \sqrt{something}\end{eqnarray*}</cmath>
+
(a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}</cmath>
  
−
We can now divide each of the previous equations from this equation to get <math>y_1 - 1 = something</math>, so the answer is <math>\boxed{answer}</math>.
+
Dividing the third equation of (*) from this equation, <math>b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1</math>. This gives <math>y_1 = 20, y_2 = 5</math>, and the answer is <math>y_1 + y_2 = \boxed{025}</math>.
 +
 
 +
Alternatively, at (*), notice that the RHS of the first two equations are the same, so <math>a=c</math>. Substituting this into the third equation gives <math>a=c=0,2</math>, which if we solve backwards for <math>y</math> will give us the same answer.  
  
 
== See also ==
 
== See also ==

Revision as of 10:32, 1 January 2008

Problem

The system of equations

$\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\

\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\

\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$. Find $y_{1} + y_{2}$.

Solution

Since $\log ab = \log a + \log b$, we can reduce the equations to a more recognizable form:

\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{eqnarray*}

Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT, the above equations become (*)

\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1 \end{eqnarray*}

From here, multiplying the three equations gives

\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}

Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$. This gives $y_1 = 20, y_2 = 5$, and the answer is $y_1 + y_2 = \boxed{025}$.

Alternatively, at (*), notice that the RHS of the first two equations are the same, so $a=c$. Substituting this into the third equation gives $a=c=0,2$, which if we solve backwards for $y$ will give us the same answer.

See also

2000 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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