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Difference between revisions of "2002 AMC 12A Problems/Problem 13"

(New page: == Problem == Two different positive numbers <math>a</math> and <math>b</math> each differ from their reciprocals by <math>1</math>. What is <math>a+b</math>? <math> \text{(A) }1 \qquad ...)
 
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There are in total four such values of <math>x</math>, namely <math>\frac{\pm 1 \pm \sqrt 5}2</math>.  
 
There are in total four such values of <math>x</math>, namely <math>\frac{\pm 1 \pm \sqrt 5}2</math>.  
  
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Out of these, two are positive: <math>\frac{-1+\sqrt 5}2</math> and <math>\frac{1+\sqrt 5}2</math>. We can easily check that both of them indeed have the required property, and their sum is <math>\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{\sqrt 5}</math>.
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Out of these, two are positive: <math>\frac{-1+\sqrt 5}2</math> and <math>\frac{1+\sqrt 5}2</math>. We can easily check that both of them indeed have the required property, and their sum is <math>\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}</math>.
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== Video Solution ==
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https://youtu.be/aprzdDl_KWw
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2002|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2002|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 20:43, 25 August 2022

Problem

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$. What is $a+b$?

$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }\sqrt 5 \qquad \text{(D) }\sqrt 6 \qquad \text{(E) }3$

Solution

Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$.

Hence it is either a solution to $x - \frac 1x = 1$, or to $\frac 1x - x = 1$. Then it must be a solution either to $x^2 - x - 1 = 0$, or to $x^2 + x - 1 = 0$.

There are in total four such values of $x$, namely $\frac{\pm 1 \pm \sqrt 5}2$.

Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$. We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$.

Video Solution

https://youtu.be/aprzdDl_KWw

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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