Difference between revisions of "2000 AMC 12 Problems/Problem 20"

Revision as of 20:19, 4 January 2008

Problem

If $x,y,$ and $z$ are positive numbers satisfying

$x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3$

$\text {(A)}\ 2/3 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 4/3 \qquad \text {(D)}\ 2 \qquad \text {(E)}\ 7/3$

Solution

Solution 1

Multiplying all three expressions together,

$\begin{align*} \left(x+ \frac{1}{y}\right) \left(x+ \frac{1}{y}\right) \left(x+ \frac{1}{y}\right) &= 1 + x + y+z+\frac{1}{x}+\frac 1y + \frac 1z + \frac 1{xyz}\\ (4)(1)\left(\frac 73\right) &= 1 + (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\\ 6xyz &= 3(xyz)^2 + 3\\ 0 &= 3[(xyz)-1]^2 \end{align*}$

Thus $xyz = 1\ \mathrm{B}$

Solution 2

We have a system of three equations and three variables, so we can apply repeated substitution.

$4 = x + \frac{1}{y} = x + \frac{1}{1 - \frac{1}{z}} = x + \frac{1}{1-\frac{1}{7/3-1/x}} = x + \frac{7x-3}{4x-3}$

Multiplying out the denominator and simplification yields $4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0$ , so $x = \frac{3}{2}$ . Substituting leads to $y = \frac{2}{5}, z = \frac{5}{3}$ , and the product of these three variables is $1$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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