Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME I Problems/Problem 9"
(→‎Solution)
Line 29: Line 29:
  
 
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>.
 
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 13:21, 19 July 2014

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 $y_1 + y_2 = \boxed{025}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&oldid=62544"