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 15:55, 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, and
satisfy
. Which of the following is a possible value of
?
Solution
.
Alternatively, we could test simple values, like , which would yield
.
Another way is to solve the equation for giving
then substituting this into the expression yields
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 |