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

(Solution 1)
Line 15: Line 15:
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
\left(x+ \frac{1}{y}\right) \left(y+ \frac{1}{z}\right) \left(z+ \frac{1}{x}\right) &= xyz + x + y+z+\frac{1}{x}+\frac 1y + \frac 1z + \frac 1{xyz}\
+
\left(x+ \frac{1}{y}\right) \left(y+ \frac{1}{z}\right) \left(z+ \frac{1}{x}\right) &= xyz+x+y+z+\frac 1x +\frac 1y + \frac 1z + \frac 1{xyz}\
(4)(1)\left(\frac 73\right) &= (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\
+
(4)(1)\left(\frac 73\right) &= 4 + 1 + \frac 73 + \frac{(xyz)^2 + 1}{xyz}\
 
6xyz &= 3(xyz)^2 + 3\
 
6xyz &= 3(xyz)^2 + 3\
 
0 &= 3[(xyz)-1]^2
 
0 &= 3[(xyz)-1]^2
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
Thus <math>xyz = 1\ \mathrm{B}</math>
+
Thus <math>xyz = 1 \Rightarrow B</math>
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 23:22, 16 May 2009

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

Multiplying all three expressions together,

\begin{align*} \left(x+ \frac{1}{y}\right) \left(y+ \frac{1}{z}\right) \left(z+ \frac{1}{x}\right) &= xyz+x+y+z+\frac 1x +\frac 1y + \frac 1z + \frac 1{xyz}\\ (4)(1)\left(\frac 73\right) &= 4 + 1 + \frac 73 + \frac{(xyz)^2 + 1}{xyz}\\ 6xyz &= 3(xyz)^2 + 3\\ 0 &= 3[(xyz)-1]^2 \end{align*}

Thus $xyz = 1 \Rightarrow B$

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 $1$.

See also

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