Difference between revisions of "2002 AMC 12A Problems/Problem 13"
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| − | == Solution == | + | == Solution 1 == |
Each of the numbers <math>a</math> and <math>b</math> is a solution to <math>\left| x - \frac 1x \right| = 1</math>. | Each of the numbers <math>a</math> and <math>b</math> is a solution to <math>\left| x - \frac 1x \right| = 1</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>. | 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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| + | == 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, <math>\varphi = \frac{1+\sqrt 5}2</math>. Only one of the options has a <math>\sqrt{5}</math> in it, which is <math>\boxed{(C) \sqrt 5}</math>. | ||
== Video Solution == | == Video Solution == | ||
Latest revision as of 12:23, 9 November 2024
Problem
Two different positive numbers 
and 
each differ from their reciprocals by 
. What is 
?
Solution 1
Each of the numbers and is a solution to 
.
Hence it is either a solution to 
, or to 
. Then it must be a solution either to 
, or to 
.
There are in total four such values of 
, namely 
.
Out of these, two are positive: 
and 
. We can easily check that both of them indeed have the required property, and their sum is 
.
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, 
. Only one of the options has a 
in it, which is 
.
Video Solution
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
