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

(Solution 1)
(Problem)
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If <math>x,y,</math> and <math>z</math> are positive numbers satisfying
 
If <math>x,y,</math> and <math>z</math> are positive numbers satisfying
  
<cmath>x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3</cmath>
+
<cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = 7/3</cmath>
  
 
Then what is the value of <math>xyz</math> ?
 
Then what is the value of <math>xyz</math> ?

Revision as of 00:11, 19 September 2020

Problem

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

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

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

Also see

2000 AMC 12 (ProblemsAnswer KeyResources)
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

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