Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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Two non-zero [[real number]]s, <math>a</math> and <math>b,</math> satisfy <math>ab = a - b</math>. Which of the following is a possible value of <math>\frac {a}{b} + \frac {b}{a} - ab</math>? | Two non-zero [[real number]]s, <math>a</math> and <math>b,</math> satisfy <math>ab = a - b</math>. Which of the following is a possible value of <math>\frac {a}{b} + \frac {b}{a} - ab</math>? | ||
− | <math>\ | + | <math>\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac { -1 }{2} \qquad \textbf{(C)} \ \frac {1}{3} \qquad \textbf{(D)} \ \frac {1}{2} \qquad \textbf{(E)} \ 2</math> |
− | ==Solution== | + | ==Solution 1== |
− | <math>\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} = \frac{2(a-b}{a-b} =2 \Rightarrow \text{ | + | <math>\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} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}</math>. |
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 and simplifying gives the answer of <math>2.</math> | 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 and simplifying gives the answer of <math>2.</math> | ||
==Solution 2== | ==Solution 2== | ||
− | This simplifies to <math>ab+b-a=0 \Rightarrow (a+1)(b-1) = -1</math>. The two integer solutions to this are <math>(-2,2)</math> and <math>(0,0)</math>. The problem states than <math>a</math> and <math>b</math> are non-zero, so we consider the case of <math>(-2,2)</math>. So, we end up with <math>\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = \boxed{2}~ \ | + | This simplifies to <math>ab+b-a=0 \Rightarrow (a+1)(b-1) = -1</math>. The two integer solutions to this are <math>(-2,2)</math> and <math>(0,0)</math>. The problem states than <math>a</math> and <math>b</math> are non-zero, so we consider the case of <math>(-2,2)</math>. So, we end up with <math>\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = 2 \Rightarrow \boxed{\text{E}}</math> |
+ | |||
+ | ==Solution 3== | ||
+ | Just realize that two such numbers are <math>a = 1</math> and <math>b = \frac{1}{2}</math>. You can see this by plugging in <math>a = 1</math> and then solving for b. With this, you can solve and get <math>2 \Rightarrow\boxed{\text{E}}</math> | ||
+ | |||
+ | ==Solution 4 == | ||
+ | Set <math>a</math> to some nonzero number. In this case, I'll set it to <math>4</math>. | ||
+ | |||
+ | Then solve for <math>b</math>. In this case, <math>b=0.8</math>. | ||
+ | |||
+ | Now just simply evaluate. In this case it's 2. So since 2 is a possible value of the original expression, select <math>\boxed{E}</math>. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | ==Solution 5== | ||
+ | Notice that <math>a=\frac{a}{b}-1</math> and <math>b=1-\frac{b}{a}</math>. Then, <math>\frac{a}{b}=1+a</math> and <math>\frac{b}{a}=1-b</math>. | ||
+ | |||
+ | <math>\frac{a}{b}+\frac{b}{a}-ab=(1+a)+(1-b)-(a-b)=2</math>. The answer is <math>\boxed{E}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=7-RloNHTnXM | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ZWqHxc0i7ro?t=6 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/Q_th4G-xGLo?si=4VwtJirZjREyyQuO | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
+ | |||
==See also== | ==See also== |
Latest revision as of 00:54, 15 July 2024
- The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.
Contents
Problem
Two non-zero real numbers, and
satisfy
. Which of the following is a possible value of
?
Solution 1
.
Another way is to solve the equation for giving
then substituting this into the expression and simplifying gives the answer of
Solution 2
This simplifies to . The two integer solutions to this are
and
. The problem states than
and
are non-zero, so we consider the case of
. So, we end up with
Solution 3
Just realize that two such numbers are and
. You can see this by plugging in
and then solving for b. With this, you can solve and get
Solution 4
Set to some nonzero number. In this case, I'll set it to
.
Then solve for . In this case,
.
Now just simply evaluate. In this case it's 2. So since 2 is a possible value of the original expression, select .
~hastapasta
Solution 5
Notice that and
. Then,
and
.
. The answer is
.
Video Solution
https://www.youtube.com/watch?v=7-RloNHTnXM
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=6
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s
Video Solution by Daily Dose of Math
https://youtu.be/Q_th4G-xGLo?si=4VwtJirZjREyyQuO
~Thesmartgreekmathdude
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.