2000 AMC 12 Problems/Problem 20
Problem
If 
and 
are positive numbers satisfying
![\[x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3\]](http://latex.artofproblemsolving.com/8/0/9/809268c34d341e924e9b6500bbdc5de487d5445d.png)
Then what is the value of 
?
Solution
Solution 1
Multiplying all three expressions together,
![\begin{align*} \left(x+ \frac{1}{y}\right) \left(y+ \frac{1}{z}\right) \left(z+ \frac{1}{x}\right) &= xyz + x + y+z+\frac{1}{x}+\frac 1y + \frac 1z + \frac 1{xyz}\\ (4)(1)\left(\frac 73\right) &= (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\\ 6xyz &= 3(xyz)^2 + 3\\ 0 &= 3[(xyz)-1]^2 \end{align*}](http://latex.artofproblemsolving.com/8/2/4/8242900c944d695e66210e560bd5f4a2342daf4c.png)
Thus 
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 
.
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 | |
