Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | We can multiply all given expressions to get: | |
− | + | <math>(1) xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</math> | |
− | < | + | Adding all the given expressions gives that |
− | + | <math>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</math> | |
− | + | We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, <math>xyz = 1 \rightarrow B</math>. | |
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=== Solution 2 === | === Solution 2 === |
Revision as of 10:59, 6 July 2019
Problem
If and are positive numbers satisfying
Then what is the value of ?
Contents
[hide]Solution
Solution 1
We can multiply all given expressions to get: Adding all the given expressions gives that We subtract from to get that . Hence, .
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 .
Also see
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.