Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \left(x+ \frac{1}{y}\right) \left( | + | \left(x+ \frac{1}{y}\right) \left(y+ \frac{1}{z}\right) \left(z+ \frac{1}{x}\right) &= 1 + x + y+z+\frac{1}{x}+\frac 1y + \frac 1z + \frac 1{xyz}\ |
(4)(1)\left(\frac 73\right) &= 1 + (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\ | (4)(1)\left(\frac 73\right) &= 1 + (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\ | ||
6xyz &= 3(xyz)^2 + 3\ | 6xyz &= 3(xyz)^2 + 3\ |
Revision as of 23:35, 9 February 2009
Problem
If and are positive numbers satisfying
Then what is the value of ?
Contents
[hide]Solution
Solution 1
Multiplying all three expressions together,
Thus
Solution 2
We have a system of three equations and three variables, so we can apply repeated substitution.
Multiplying out the denominator and simplification yields , so . Substituting leads to , and the product of these three variables is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |