2002 AMC 12A Problems/Problem 13
Contents
[hide]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 |
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