2000 AMC 12 Problems/Problem 20

Problem

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

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

Then what is the value of $xyz$ ?

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

Solution

Solution 1

We multiply all given expressions to get: $(1) xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}$ Adding all the given expressions gives that $(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}$ We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$ . Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$ .

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 $1$ .

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.

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