2002 AMC 12A Problems/Problem 13

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 1

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 $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$ .

Solution 2 (speed guess/intuition)

Since the problem is about similarities a number and its reciprocal differing by one, you can guess that the solution has something to do with the golden ratio, $\varphi = \frac{1+\sqrt 5}2$ . Only one of the options has a $\sqrt{5}$ in it, which is $\boxed{(C) \sqrt 5}$ .

Video Solution

https://youtu.be/aprzdDl_KWw

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 13

Contents

Problem

Solution 1

Solution 2 (speed guess/intuition)

Video Solution

See Also