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

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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
  
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<cmath>x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3</cmath>
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<cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}</cmath>
  
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<math>\text {(A)}\ 2/3 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 4/3 \qquad \text {(D)}\ 2 \qquad \text {(E)}\ 7/3</math>
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Then what is the value of <math>xyz</math> ?
  
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__TOC__
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<math>\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}</math>
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== Solution ==
 
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=== Solution 1 ===
 
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Multiplying all three expressions together,
 
  
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<cmath>\begin{align*}
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== Solution 1 ==
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\left(x+ \frac{1}{y}\right) \left(x+ \frac{1}{y}\right) \left(x+ \frac{1}{y}\right) &= 1 + x + y+z+\frac{1}{x}+\frac 1y + \frac 1z + \frac 1{xyz}\\
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We multiply all given expressions to get:
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(4)(1)\left(\frac 73\right) &= 1 + (4) + (1) + \left(\frac{7}{3}\right) + \frac{(xyz)^2 + 1}{xyz}\\
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<cmath>(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</cmath>
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6xyz &= 3(xyz)^2 + 3\\
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Adding all the given expressions gives that
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0 &= 3[(xyz)-1]^2
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<cmath>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</cmath>
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\end{align*}</cmath>
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We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, by inspection, <math>\boxed{xyz = 1 \rightarrow B}</math>.
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<cmath></cmath>
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~AopsUser101
  
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Thus <math>xyz = 1\ \mathrm{B}</math>
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== Solution 2 ==
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=== Solution 2 ===
 
 
We have a system of three equations and three variables, so we can apply repeated substitution.  
 
We have a system of three equations and three variables, so we can apply repeated substitution.  
  
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Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>.
 
Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>.
  
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== See also ==
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== 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 ==
 
{{AMC12 box|year=2000|num-b=19|num-a=21}}
 
{{AMC12 box|year=2000|num-b=19|num-a=21}}
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:34, 4 November 2022

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 (Problems • Answer Key • Resources)
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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