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

m (See also)
(Solution 1)
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== Solution ==  
 
== Solution ==  
 
=== Solution 1 ===
 
=== Solution 1 ===
Multiplying all three expressions together,
+
We can multiply all given expressions to get:
 
+
<math>(1) xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</math>
<cmath>\begin{align*}
+
Adding all the given expressions gives that
\left( x + \frac 1y \right) \left( y + \frac 1z \right) \left( z + \frac 1x \right) &= xyz + x + y + z + \frac 1x + \frac 1y + \frac 1z + \frac 1{xyz}\
+
<math>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</math>
(4)(1)\left(\frac 73\right) &= 4 + 1 + \frac 73 + xyz + \frac 1{xyz}\
+
We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, <math>xyz = 1 \rightarrow B</math>.
2 &= xyz + \frac 1{xyz}\
 
0 &= (xyz - 1)^2
 
\end{align*}</cmath>
 
 
 
Thus <math>xyz = 1 \Rightarrow B</math>
 
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 10:59, 6 July 2019

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 can 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, $xyz = 1 \rightarrow B$.

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 (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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