Question

If $x,$ $y,$ and $z$ are positive numbers satisfying $x + \frac{1}{y} = 4, \quad y + \frac{1}{z} = 1, \quad \text{and} \quad z + \frac{1}{x} = \frac{7}{3},$ then what is $xyz?$

$\dfrac{2}{3}$

$1$

$\dfrac{4}{3}$

$2$

$\dfrac{7}{3}$

Solution:

Adding the three equations gives $\left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) + \left(z + \tfrac1x\right) = 4 + 1 + \tfrac73 = \tfrac{22}{3}.$

Multiplying them gives $4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.$

Expanding the product, $\left(x + \tfrac1y\right)\left(y + \tfrac1z\right)\left(z + \tfrac1x\right) = xyz + \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) + \frac{1}{xyz}.$ The middle group is the sum $\tfrac{22}{3},$ so $xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.$

Hence $(xyz - 1)^2 = 0,$ so $xyz = 1.$

Thus, the correct answer is B.

2000 AMC 12 Problem 20

Solution:

Problem 20 in Other Years