2000 AMC 12 Problems/Problem 20

Problem

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

\[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\]

Then what is the value of $xyz$ ?

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

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, $\boxed{xyz = 1 \rightarrow B}$. \[\] ~AopsUser101

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 $1$.

Video Solution by OmegaLearn

https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s

~ pi_is_3.14

Video Solution

https://youtu.be/ph8o017pw_o

See Also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions

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