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>. | ||
− | 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>. | + | 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>. |
== 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}} |
Revision as of 00:11, 2 July 2013
Problem
Two different positive numbers and each differ from their reciprocals by . What is ?
Solution
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 .
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 |