2000 AMC 12 Problem 20

Problem 20 of 25HarderAlgebra

x,

y,

and

z

are positive numbers satisfying

x + \frac{1}{y} = 4,

y + \frac{1}{z} = 1,

and

z + \frac{1}{x} = \frac{7}{3},

then what is

xyz?

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Solution

Adding the three equations gives

\begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered}

Multiplying them gives

4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.

Expanding the product,

\begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned}

The middle group is the sum

\tfrac{22}{3},

xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.

Hence

(xyz - 1)^2 = 0,

xyz = 1.

Thus, the correct answer is B.