Difference between revisions of "2000 AIME I Problems/Problem 9"
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== Problem == | == Problem == | ||
The system of equations | The system of equations | ||
− | < | + | <cmath>\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}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ | \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ | ||
− | \end{eqnarray*}</ | + | \end{eqnarray*}</cmath> |
has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>. | has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>. |
Revision as of 18:51, 10 March 2015
Problem
The system of equations
has two solutions and . Find .
Solution
Since , we can reduce the equations to a more recognizable form:
Let be respectively. Using SFFT, the above equations become (*)
From here, multiplying the three equations gives
Dividing the third equation of (*) from this equation, . This gives , and the answer is .
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.