Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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If <math>x,y,</math> and <math>z</math> are positive numbers satisfying | If <math>x,y,</math> and <math>z</math> are positive numbers satisfying | ||
| − | <cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = 7 | + | <cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}</cmath> |
Then what is the value of <math>xyz</math> ? | Then what is the value of <math>xyz</math> ? | ||
Revision as of 23:11, 18 September 2020
Problem
If 
and 
are positive numbers satisfying
![\[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\]](http://latex.artofproblemsolving.com/e/4/5/e45acc3f1463f663f3fa0b99ee25e08a65517ac7.png)
Then what is the value of 
?
Solution
Solution 1
We multiply all given expressions to get:
![\[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\]](http://latex.artofproblemsolving.com/3/9/9/39946093baf76ac36c45716580f910fe4040b79a.png)
Adding all the given expressions gives that
![\[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\]](http://latex.artofproblemsolving.com/b/8/b/b8b3ee76ca8fe7a060cceabcc53fac838de13cc6.png)
We subtract 
from 
to get that 
. Hence, by inspection, 
.
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
~AopsUser101
Solution 2
We have a system of three equations and three variables, so we can apply repeated substitution.
![\[4 = x + \frac{1}{y} = x + \frac{1}{1 - \frac{1}{z}} = x + \frac{1}{1-\frac{1}{7/3-1/x}} = x + \frac{7x-3}{4x-3}\]](http://latex.artofproblemsolving.com/9/f/0/9f0d43b0925f112e173739d093abd90c9a6ecf41.png)
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. 
