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Difference between revisions of "2000 AMC 12 Problems/Problem 11"

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Alternatively, we could test simple values, like <math>(a,b)=\left(1, \frac{1}{2}\right)</math>, which would yield <math>\frac {a}{b} + \frac {b}{a} - ab=2</math>.
 
Alternatively, we could test simple values, like <math>(a,b)=\left(1, \frac{1}{2}\right)</math>, which would yield <math>\frac {a}{b} + \frac {b}{a} - ab=2</math>.
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Another way is to solve the equation for <math>b,</math> giving <math>b = \frac{a}{a+1};</math> then substituting this into the expression yields
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<cmath>\begin{array}{ccl} \dfrac{a}{a/(a+1)} + \dfrac{a/(a+1)}{a} - a \cdot \dfrac{a}{a+1} &=& (a+1) + \dfrac{1}{a+1} - \dfrac{a^2}{a+1} \\&=&\dfrac{(a^2+2a+1)+1-a^2}{a+1} \\&=&\dfrac{2a+2}{a+1} \\ &=& 2.  \end{array} </cmath>
  
 
==See also==
 
==See also==

Revision as of 14:55, 14 June 2013

The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.

Problem

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?

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

Solution

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = 2 \Rightarrow \text{(E)}$.

Alternatively, we could test simple values, like $(a,b)=\left(1, \frac{1}{2}\right)$, which would yield $\frac {a}{b} + \frac {b}{a} - ab=2$.

Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression yields

\[\begin{array}{ccl} \dfrac{a}{a/(a+1)} + \dfrac{a/(a+1)}{a} - a \cdot \dfrac{a}{a+1} &=& (a+1) + \dfrac{1}{a+1} - \dfrac{a^2}{a+1} \\&=&\dfrac{(a^2+2a+1)+1-a^2}{a+1} \\&=&\dfrac{2a+2}{a+1} \\ &=& 2. \end{array}\]

See also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 10 Problems and Solutions
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