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Difference between revisions of "1990 USAMO Problems/Problem 2"

(New page: ==Problem== A sequence of functions <math>\, \{f_n(x) \} \,</math> is defined recursively as follows: <math> f_1(x) = \sqrt {x^2 + 48}, \quad \mbox{and} \\ f_{n + 1}(x) = \sqrt {x^2 + 6f...)
 
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==Solution==
 
==Solution==
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The solution when n=1 is x=4, and when we plug that into n=2, we get what we got when n=1. Therefore, 4 is a solution for all n. Now we prove that it is the only solution.
 
  
 
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Revision as of 10:47, 18 October 2007

Problem

A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows:

$f_1(x) = \sqrt {x^2 + 48}, \quad \mbox{and} \\ f_{n + 1}(x) = \sqrt {x^2 + 6f_n(x)} \quad \mbox{for } n \geq 1.$

(Recall that $\sqrt {\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$, find all real solutions of the equation $\, f_n(x) = 2x \,$.


Solution

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See Also

1990 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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